2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Students can Download 2nd PUC Electronics Chapter 11 Microcontrollers Questions and Answers, Notes Pdf, 2nd PUC Electronics Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

2nd PUC Electronics Microcontrollers One Mark Questions and Answers

Question 1.
What is a microcontroller?
Answer:
Microcontroller is a small computer on a single IC consisting of a relatively simple CPU combined with support functions.

Question 2.
What is an Accumulator?
Answer:
Accumulator is a device which stores a number and which on receipt of another number, adds the two and stores the sum. It can also sense the signal, clear and complement etc.

Question 3.
What is a Program Counter?
Answer:
Program counter is a circuit, which holds the address of a byte in memory. It also specifies the address of the next instruction to be fetched and executed.

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 4.
What is a Stack pointer?
Answer:
Stack is the area of the memoiy (RAM) that is used in some instructions to store and retrieve data quickly. Stack pointer is used to hold an internal RAM address. When a program jumps to some branch the address of the next instruction is stored in Stack.

Question 5.
What is an interrupt circuit?
Answer:
It is an electronic circuit, which breaks in the normal flow of system or routine and flow can be resumed from that point at a later time.

Question 6.
What is the need of clock circuit?
Answer:
Clock circuit generates the internal clock pulses and all the internal operations are synchronised with this pulse.

Question 7.
How much is the ROM of microcontroller 8051?
Answer:
Microcontroller 8051 has on chip ROM of 4k bytes.

Question 8.
How much is the RAM of microcontroller 8051?
Answer:
Microcontroller 8051 has 128 bytes of RAM.

Question 9.
How many serial ports are there in microcontroller 8051?
Answer:
Microcontroller 8051 has one serial port for communication.

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 10.
How many interrupt sources are there in microcontroller 8051?
Answer:
Microcontroller 8051 has six(6) interrupt sources.

Question 11.
What is an addressing mode?
Answer:
Addressing modes specify the way in which the operands are accessed by the instruction.

Question 12.
What is an Operational code (op code)?
Answer:
Operational code (op code) is the short form of expressing the instruction. It consists of mnemonic and operands.

Question 13.
What is the meaning of MOV A, R0?
Answer:
The content of R0 is moved (copied) to register A (accumulator).

Question 14.
What is the meaning of MOV R5, A?
Answer:
Copy the contents of register A(Accomulator) to Register 5.

Question 15.
What is the purpose of register addressing mode?
Answer:
In register addressing mode data is moved between registers A, DPTR, Carry bit and registers R0-R7 are used as part of opcode mnemonics as source or destination.

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 16.
What is the use of direct addressing mode?
Answer:
In direct addressing mode 128 bytes of internal RAM and SFRs may be addressed directly using single byte address assigned to each RAM location and each SFR.

Question 17.
What is the direct address of working register R5 in register bank 2?
Answer:
The direct address of register R5 in register bank 2 is ID.

Question 18.
Write the direct address of the Accumulator.
Answer:
The direct address of accumulator register is OEO.

Question 19.
Write the mnemonic for the operation “Save content of A in RAM location 45H”.
Answer:
The mnemonic for saving content of register A in location 45H is MOV 45H, A.

Question 20.
What is meaning of MOV A, @R1?
Answer:
The meaning of MOV A, @R1 is to copy the contents of RAM location whose address is stored in R1 into accumulator A.

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 21.
How do you write the mnemonic for the operation “copy the contents of RAM location whose address is stored in R0 into accumulator A” in an indirect addressing mode?
Answer:
The mnemonic for copy the contents of RAM location whose address is stored in R0 into accumulator A is MOV A, @ R0.

Question 22.
Which sign is used as the mnemonic for immediate data?
Answer:
The mnemonic for immediate data is hash (#).

Question 23.
What is meaning of MOV A, #22H?
Answer:
The meaning of MOV A, #22H is to load value 22H into accumulator A.

Question 24.
Write the mnemonic for the operation “load the decimal value 66 into R3”.
Answer:
The mnemonic for loading decimal value 66 into R3 is MOV R3, #66.

Question 25.
What is meaning of MOV DPTR, #1246H?
Answer:
The meaning of MOV DPTR, #1246H is to load 16 bit number (1246H) immediately into register DPTR.

Question 26.
What does ‘jump’ instruction do?
Answer:
The jump instruction permanently changes the contents of program counter either conditionally or unconditionally.

Question 27.
What does ‘call’ instruction do?
Answer:
The call instruction temporarily changes the contents of program counter to allow another part of the program to run either conditionally or unconditionally.

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 28.
What is the description of the mnemonic JNZ and ACALL?
Answer:
JNZ is conditional jump instruction, here contents of PC changes only when the result of operation is non zero. ACALL allows to specify 11- bit address in the instruction and calling subroutine within 2k program memory block, starting from the address of the next instruction is possible.

Question 29.
Why is ROM called non volatile?
Answer:
ROM is called non volatile because it does not depend on electrical power to store the numbers.

Question 30.
Why is RAM called volatile?
Answer:
RAM is called volatile because it depends on electrical power to store the numbers, when power is OFF it loses its information.

Question 31.
What is a program?
Answer:
Computer program is a sequence of operations that the computer is to perform.

Question 32.
Expand EEPROM.
Answer:
Electrical Erasable Programmable Read Only Memory.

Question 33.
Expand SRAM.
Answer:
Static Random Access Memory.

Question 34.
Expand RAM.
Answer:
Random Access Memory or Read write memory.

Question 35.
What is data address?
Answer:
Address in memory that is used by the CPU to read or write is called data address.

Question 36.
What is a machine language?
Answer:
Machine language is one in which information is available only in binary form.
2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 37.
Why is assembly language programming called low level programming?
Answer:
Assembly language is low level programming because, it is machine dependant.

Question 38.
How many bits of binary data can a register A hold?
Answer:
Register A can hold 8 bits of binary data.

Question 39.
How many bits of binary data can a register R hold temporarily?
Answer:
Register R can hold 8 bit of binary data temporarily.

Question 40.
How many bits of address can a register PC hold?
Answer:
Register PC can hold 16- bit of address number.

Question 41.
How much “PAL register PC” can hold code address?
Answer:
Practice Assembly Language (PAL) Register can hold code address from 0000H to FFFFH (64K).

Question 42.
What are directives?
Answer:
Assembly language instructions like ADD and MOV are statements called directives.

Question 43.
What does directive do?
Answer:
Instructions tell the computer what to do, while directives give directions to the assembler.

Question 44.
What is a PIC microcontroller?
Answer:
PIC microcontrollers are a family of microcontroller chips produced by microchip technology.

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 45.
Expand PIC.
Answer:
PIC is Peripheral Interface Controller.

Question 46.
How many accumulators are there in PIC microcontroller?
Answer:
PIC microcontrollers has one accumulator.

Question 47.
Mention the maximum operating frequency of PIC16F877.
Answer:
Maximum operating frequency of PIC16F877 is 20MHz.

Question 48.
How many times data can be written in PIC16F877?
Answer:
In PIC16F877 microcontrollers 8k ROM memory in flash technology upto 105 times chip can be reprogrammed.

Question 49.
What is the working voltage range of PIC16F877?
Answer:
PIC microcontrollers has working voltage range of 2.0V to 5.5V.

Question 50.
How much is the EEPROM memory?
Answer:
EEPROM memory has 128 bytes.

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 51.
What is the largest hex value that can be moved into an 8 bit register?
Answer:
FFH is the largest value that can be moved into an 8 bit register.

This Hex Calculator is used to perform addition, subtraction, multiplication and division on two hexadecimal numbers.

2nd PUC Electronics Microcontrollers Two Marks Questions and Answers

Question 1.
Mention two differences between microprocessor and microcontroller.
Answer:

Microprocessor Microcontroller
Most MPs have very less bit handling instructions MCs have many bit handling instructions
MPs are intended for general purpose applications MCs are for a specific application only

Question 2.
What is an addressing mode? Why is it necessary?
Answer:
The CPU can store data may be in registers, memory or in an external source. The ways by which these data source address are specified are called addressing modes. These are necessary because, the data can be accessed using any one of the modes.

Question 3.
Mention the different opcodes used in 8051.
Answer:
MOV dest, sour; PUSH sour; POP dest; XCH dest, sour; ADD A, Rn etc are some of the opcodes used in 8051.

Question 4.
What are the different parts of memory of 8051?
Ans:
The memory is divided into Internal RAM, Intrenal SFR, External RAM, internal and External ROM.

Question 5.
What is the main feature of MOV opcode? Give an example.
Answer:
MOV opcode transfers data within the 8051 memory. MOV A, Rl; MOV R0,22H; MOV A, #54H; MOV R2,#40.

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 6.
What is direct addressing? Explain.
Answer:
In direct addressing mode all 128 bytes of internal RAM and SFRs may be addressed directly using single-byte address assigned to each RAM and SFR registers. MOV R0, 40H save the content of location 40H in R0 register. MOV B, Rl means copy the content of Rl to B.

Question 7.
What is an assembly language?
Answer:
Assembly language fills the gap between low level machine language and high level language. It is written in instruction mnemonics.

Question 8.
Write ALP to represent “load 15H into R2”
Answer:
MOV R2,#15H

Question 9.
Write the summary of unsigned multipliers
Answer:
Multiplication operation use registers A and B as both source and destination addresses for the operation. MUL AB multiplies contents of A and B.

Multiplication Operand 1 Operand 2 Result
Byte x Byte A B A = lower byte
B = High byte

Question 10.
Write the summary of division of unsigned numbers.
Answer:
In the division of unsigned numbers 8051 supports byte over byte only. DIV AB, after division quotient is in A and remainder in B

Division Numerator Denominator Quotient Remainder
Byte/Byte A B A B

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 11.
Mention any two core features of PIC microcontroller.
Answer:
Instruction set simplicity and programmable timer options.

Question 12.
Mention the main features of PIC microcontroller.
Answer:
Operating frequency-0-20MHz, Power supply voltage 2.0 to 5.5V, power saving sleep mode, ADC, Watch dog timer, US ART module etc.

Question 13.
Mention the power supply and power consumption of PIC16F877.
Answer:
Power suppy 2.0V to 5.5V and power consumption is 220mA (2.0V, 4MHz), llmA(2.0V. 32kHz), 50nA standby mode.

Question 14.
Briefly explain Data transfer instruction.
Answer:
In data transfer instructions, data is moved(copied) from the source to destination. Most ofthe instructions in this group does not affect the PSW flags. MOV, PUSH, POP and XCH are the commonly used instructions in this group.

Question 15.
Briefly explain Arithmetic instruction.
Answer:
Various arithmetic operations are carried out in this group of instructions. ADD, SUB, INC, DEC, MUL, DIV are the basic operations used.

Question 16.
Briefly explain Logical instructions.
Answer:
Logical operations like AND, OR, XOR, complement, rotate are performed by these instructions. One of the operand is always register A.

Question 17.
Briefly explain programming instructions.
Answer:
These are also called branching instructions. Conditional and unconditional branching instructions are used. JMP, AJMP, LJMP, CALL, ACALL are some examples.

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 18.
What is the function of register A?
Answer:
The register A is used to store the data temporarily and also to hold the result of arithmetic and logical instructions.

Question 19.
What is the function of register R?
Answer:
Register R stores numbers temporarily.

Question 20.
What is the function of register PC?
Answer:
It is a 16 bit register. Program instruction bytes are fetched from locations in memory that are addressed by PC. The PC is automatically incremented after every instruction byte is fetched and may also be altered by certain instructions.

Question 21.
Briefly explain about structure of Assembly Language.
Answer:
Assembly language is a series of lines of assembly language instructions consisting of mnemonics, one or two operands. The four fields of assembly language are: label; mnemonic; operand; comment.

2nd PUC Electronics Microcontrollers Three Marks Questions and Answers

Question 1.
Mention the reason to write computer instructions in assembly language.
Answer:
a. To speed computer operation. Programs written in assembly language can be stored compactly, and less time is spent fetching the code. High level languages are converted to code by utility programs named compilers. Because of general nature of high level languages, the compilers often produce excess or over head-code.

b. To reduce the size of the program. Assembly language requires no extra overhead code because the assembly language programmer is aware of extra needs of the program for any given situation.

c. To write programs for special situations. Often, particularly when dealing with machine control, no standard programs exist. Robot arms and antilock brakes, for instance, have no standard drivers. It is generally more efficient to write nonstandard driver programs in assembly code, also, when speed of response is critical, assembly-coded programs execute rapidly because of the exact fit of program code to task requirements.

d. To make economical. Small computer systems such as those that are embedded inside other machines, are often produced in large numbers. Reducing code size also reduces the cost of associated ROM chips.

e. To better understand how computer operates. In order to fully understand what is going on “under the hood” of the CPU, you should learn to program the CPU in assembly language.

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 2.
Mention the different programs required ALPs converted into code memory bit charges,
Answer:
The different programs required ALPs converted into code memory bit charges are:
a. An operating system program, which controls the operation of the personal computer used for the entire programming process. DOS is the operating system program used by many PCs. Windows is another.

b. A word processing, often called a text editor program. Programs in assembly language mnemonics are written using the text editor and stored on a disk as files that, normally, end in the extension asm.The asm files are intended for the use of another program called the assembler programs. Any text editor that can produce an ASCII (text) file is suitable for writing assembly language, asm files.

c. An assembler program, which takes the. asm assembly language program file and converts it to a machine code, obj file. The assembler converts the ASCII mnemonic text file into an. obj file that contains machine code instructions to the CPU, in binary form.

d. A testing program, which lets you run and test your program under controlled conditions. Testing your program is the most important step of the programming process. To be able to test program one must be able to execute each instruction and see the results. Utility programs that allow the user to test programs are called debuggers or simulators.

Question 3.
Briefly explain the different bits of binary memories of different registers.
Answer:
Figure shows the different bits of binary memories of different registers.
2nd PUC Electronics Question Bank Chapter 11 Microcontrollers 1

Question 4.
What are the different fields present in ALP (PAL – Practice Assembly language)?
Answer:
PAL contains three internal registers:
Register A: Performs all operations in the CPU Register B: Stores numbers temporarily
Register PC: Holds the address of the next instruction to be executed in code memory. Register A and R can hold 8 bits of binary data and register PC can hold a 16-bit address number. PAL is called an 8-bit computer because the working registers A and R can hold a 1-byte number. PAL register PC limits the number of code byte addresses to 64k because it can hold code address from address 0000H to address FFFFH.

Question 5.
Write ALP to represent load 15H into R2
Answer:
MOV R2, #15h

Question 6.
Write the description of ALP ADDA, R2
Answer:
The content of register R2 is added with the content of accumulator and the the sum is stored in the accumulator.

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 7.
Mention the different types of addressing modes.
Answer:
There are four addressing modes in 8051

  • Immediate addressing mode
  • Register addressing mode
  • Direct addressing mode
  • Indirect addressing mode

Question 8.
Briefly explain data transfer instructions.
Answer:
These instructions move the content of one register to another one. Data can also be transferred to stack with the help of PUSH and POP instructions.
2nd PUC Electronics Question Bank Chapter 11 Microcontrollers 2

Question 9.
Briefly explain arithmetic instructions.
Answer:
These instructions perform several basic arithmetic operations. After execution, the result is stored in the first operand. 8 bit addition, subtraction, multiplication, increment-decrement instructions can be performed in 8051.

Example:
ADDA, Rn
ADDA, #8 bit data
ADDC A, Direct
ADDC A, Rn
ADDC A, @Ri
ADDC A, #8 bit data
SUBB A, Rn
SUBB A, Direct
SUBB A, @Ri
SUBB A, #8 bit data
INC A
INC Rn
INC Direct
INC @Ri
DEC A
DEC Rn
DEC Direct
DEC @Ri
INC DPTR
MUL A B
DIVA B
DA A

Question 10.
Briefly explain logical instructions.
Answer:
These instructions perform logical operations between two register contents on a bit by bit’ basis. After execution, the result is stored in the first operand.
Example:
ANL A, Rn
ANL A, Direct
ANL A, @Ri
ANL A, #8 bit data
ANL Direct, A
ANL Direct, #8 bit data
ORLA, Rn
QRL A, Direct
ORLA, @Ri
ORL A, #8 bit data
ORL Direct, A
ORL Direct, #8 bit data
XRL A, Rn
XRL A, Direct
XRL A, @Ri
XRL A, #8 bit data
XRL Direct, A
XRL Direct, #8 bit data
CLR A
CPLA
RLA
RLC A
RR A
RRC A
SWAP A

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 11.
Briefly explain programming instructions.
Answer:
Assembly language programming consists of writing programs using instr uction mnemonics that are specified by the manufacturer. An instruction consists of a label, an instruction mnemonic, operands and operational comments. The instruction mnemonic specifies the action to be taken by the CPU and the operands specify the addresses of data used in the action. Labels are the names given to the address numbers in program code memory.

Question 12.
Mention three key features,
i. Max operating frequency
ii. Flash programming memory
iii. No. of pins of 16F877.
Answer:
i. Operating speed:

  • DC – 20 MHz oscillator/clock input
  • DC – 200 ns instruction cycle

ii. Flash memory size:

  • 100,000 write Flash endurance
  • 1,000,000 write EEPROM endurance
  • Flash/Data EEPROM retention: > 40 years

iii. No. of pins of 16F877: 28 pins

Question 13.
Mention the value of ROM, RAM and EEPROM of PIC16F877.
Answer:

  • 100,000 write Flash endurance
  • 1,000,000 write EEPROM endurance
  • Flash/Data EEPROM retention: > 40 years

2nd PUC Electronics Microcontrollers Five Marks Questions and Answers

Question 1.
Compare a microprocessor with a microcontroller
Answer:

Microprocessor Microcontroller
μP have many operational codes for moving data and performing operations, but it has only few bit-handling instructions. μC have one or two operational codes for moving data from and performing operations, but it has many bit- handling instructions.
μP is concerned with rapid movement of code and data from external addresses to the chip. μC is concerned with rapid movement of bits within the chip.
μP need external digital parts (Hardware) to perform its tasks as computer. μC can function as computer without addition of external digital parts.
μP are used in general purpose computing devices, supports wide range of applications with general hardware. μC are used in embedded systems and supports only dedicated applications with customized hardware.

Question 2.
Mention the features of 8051.
Answer:
1. Eight-bit CPU with registers A (Accumulator) and B.
2. Sixteen-bit program counter (PC) and data pointer(DPTR)
3. Eight-bit program status word (PSW)
4. Eight-bit stack pointer(SP)
5. Internal ROM or EPROM (8751) of 4k bytes
6. Internal RAM of 128 bytes :

  • Four register banks, each containing eight registers
  • Sixteen bytes, which may be addressed at the bit level
  • Eight bytes of general purpose registers

7. Four 8-bit I/O ports (P0-P3)
8. Two 16-bit timer/counter TO and T1
9. Full duplex serial data receiver/ transmitter : SBUF
10. Control register: TCON, TMOD,SCON,PCON,IP and IE
11. Two external and two internal interrupt sources
12. Oscillator and clock circuits

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 3.
Draw the pin diagram of 8051
Answer:
2nd PUC Electronics Question Bank Chapter 11 Microcontrollers 3

Question 4.
Briefly explain the core features of pic microcontroller(PIC 16F887)
Answer:
2nd PUC Electronics Question Bank Chapter 11 Microcontrollers 4

Question 5.
Briefly explain the steps used in creating a program.
Answer:
The steps to create a program are:

  • Program editor is used to type in the program. The editor must be able to produce ASCII file for many assembler and will have the extension “.asm” or “src”.
  • Assembler on receiving the file in “.asm” form converts the instructions into steps to creat a program machine code. Assembler will produce an object file “.obj” and list file “.1st”.
  • Linker program links one or more object files and produce an absolute object file “.abs”.
  • The “.abs” file is fed to object to hex converter(OH) and create an extension “.hex” and is ready to bum ROM.

Question 6.
Briefly explain signed and unsigned subprogram.
Answer:
Subtraction of unsigned numbers: In 8051 SUBB is used to subtract two numbers by using unsigned representation by setting the bit CY=0 or CY=1. Whereas, in signed representation the sum or difference should not be greater than or less than -128 or 127.

Question 7.
Explain the core features of PIC16F887 microcontroller.
Answer:
24/35 I/O pins with individual direction control:

  • High current source/sink for direct LED drive
  • Interrupt-on-Change pin
  • Individually programmable weak pull-ups
  • Ultra Low-Power Wake-up (ULPWU)

Analog Comparator module with:

  • Two analog comparators
  • Programmable on-chip voltage reference (CVREF) module (% of VDD)
  • Fixed voltage reference (0.6V)
  • Comparator inputs and outputs externally accessible
  • SR Latch mode
  • External Timer 1 Gate (count enable)

A/D Converter:

  • 10-bit resolution and 11/14 channels
  • Timer O: 8-bit timer/counter with 8-bit programmable prescaler

Enhanced Timer 1:

  • 16-bit timer/counter with prescaler
  • External Gate Input mode
  • Dedicated low-power 32 kHz oscillator

Timer2: 8-bit timer/counter with 8-bit period register, prescaler and postscaler

Enhanced Capture, Compare, PWM+ module:

  • 16-bit Capture, max. resolution 12.5 ns
  • Compare, max. resolution 200 ns
  • 10-bit PWM with 1, 2 or 4 output channels, programmable “dead time”, max. frequency 20 kHz
  • PWM output steering control

Capture, Compare, PWM module:

  • 16-bit Capture, max. resolution 12.5 ns
  • 16-bit Compare, max. resolution 200 ns
  • 10-bit PWM, max. frequency 20 kHz

Enhanced USART module:

  • Supports RS-485, RS-232, and LIN 2.0
  • Auto-Baud Detect
  • Auto-Wake-Up on Start bit
  • In-Circuit Serial Programming TM (ICSPTM) via two pins
  • Master Synchronous Serial Port (MSSP) module – supporting 3-wire SPI (all 4 modes) and 12C™ Master and Slave Modes with I2C address mask

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 8.
Briefly explain different addressing modes in the 8051.
Answer:
There are four addressing modes in 8051.

  • Immediate addressing mode: the data source is available immediately as a part of instruction itself. Example: MOV A, #8bit data
  • Register addressing mode: the source and destination register names are parts of the opcode in instruction. Example: MOV A, R0
  • Direct addressing mode: the source and/or destination may be the internal RAM locations. Example: MOV 90h, #50h
  • Indirect addressing mode: the source or destination address may be indicated the content of index register. Example: MOV @Rl,#40h

Problems with Solutions

Question 1.
Write the instructions to move value 34H into register A and value 3FH into register B, then add them together.
Answer:
MOV A, #34h
MOV 0F0, #3Fh
ADD A, 0F0
LCALL 0003h

Question 2.
Write the instructions to add the values 16H and CDH, place the result in register R2.
Answer:
MOV A, #16h
MOV R2, #CDh
ADD A, R2
MOV R2,A
LCALL 0003 h

2nd PUC Electronics Question Bank Chapter 11 Microcontrollers

Question 3.
Add 25H and 34H and put the result in register A.
Answer:
MOV A, #25h
ADDA, #34h
LCALL 0003

Question 4.
Write a program to add two 8-bit numbers and store it in R6. The numbers are 01EH and 01CH.
Answer:
MOV A, #lEh
ADD A, #lCh
MOV R6, A
LCALL 0003h

Question 5.
Subtract 21H from 30H and write the program and solve.
Answer:
MOV A, #30h
SUBB A, #21h
LCALL 0003h

Karnataka SSLC Maths Model Question Paper 6 with Answers (Old Pattern)

Students can Download Karnataka SSLC Maths Model Question Paper 6 with Answers (Old Pattern), Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus SSLC Maths Model Question Paper 1 (Old Pattern)

Time: 3 Hours
Max Marks: 80

I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )

Four alternatives are given to each question. Choose an appropriate answer. Write it along with its alphabet.

(8 × 1 = 8)
Question 1.
The distance between two points p(x1, y1) and q (x2, y2) is given by
Karnataka SSLC Maths Model Question Paper 1 Q1

Question 2.
The degree of polynomial p(x) = x2 – 3x + 4x3 – 6 is
(A) 2
(B) 1
(C) 3
(D) 6

The Polynomial Root Calculator can find multiple roots to polynomial equations with just one click. To open the tool, click on it.

Question 3.
Which one of the following cannot be the probability of an event?
(A) \(\frac { 2 }{ 3 }\)
(B) -1.5
(C) 15%
(D) 0.7

Karnataka SSLC Maths Model Question Paper 6 with Answers

Question 4.
The curved surface area of frustum of a cone is given by
(A) Π (r1 + r2) l
(B) Π (r1 + r2) h
(C) Π (r1 – r2) l
(D) Π (r1 – r2) h

Question 5.
The solutions for the equations x + y = 10 and x – y = 2 are
(A) x = 6, y = 4
(B) x = 4, y = 6
(C) x = 7, y = 3
(D) x = 8, y = 2

Question 6.
In the adjoining figure, TP and TQ are the tangents to the circle with centre O. The measure of ∠PTQ is
Karnataka SSLC Maths Model Question Paper 1 Q6
(A) 90°
(C) 70°
(B) 110°
(D) 40°

Question 7.
The coordinates of origin are
(A) (1, 1)
(B) (2, 2)
(C) (0, 0)
(D) (3, 3)

Question 8.
If the discriminant of quadratic equation b2 – 4ac = 0 then the roots are
(A) Real and distinct
(B) Roots are equal
(C) No Real Roots
(D) Roots are unequal and irrational

(6 × 1 = 6)
Question 9.
State “Basic proportionality theorem”.

Question 10.
Identify the tangent to the circle in the adjoining figure and write its name.
Karnataka SSLC Maths Model Question Paper 1 Q10

Question 11.
State Euclid’s division lemma.

Question 12.
Find the number of zeroes of a polynomial p(x) from the graph given
Karnataka SSLC Maths Model Question Paper 1 Q12

Question 13.
Find the distance of the point p(3, 4) from the origin.

Karnataka SSLC Maths Model Question Paper 6 with Answers

Question 14.
Express 140 as a product of prime factors.

(16 × 2 = 32)
Question 15.
How many two – digit numbers are divisible by 3?

Question 16.
∆ABC ~ ∆DEF , Area of ∆ABC = 64cm2 and area of ∆DEF = 121cm2. If EF = 15.4 cm, Find BC.

Question 17.
Solve for x and y : 2x + y = 6 and 2x – y = 2.

Question 18.
Five years ago, Gouri was thrice as old as Ganesh. Ten years later Gouri will be twice as old as Ganesh. How old are Gouri and Ganesh?

Question 19.
Find the area of the shaded region in the figure, where ABCD is a square of side 14 cm.
Karnataka SSLC Maths Model Question Paper 1 Q19

Question 20.
Construct a pair of tangents to a circle of radius 5 cm Which are inclined to each other at an angle of 60°.

Question 21.
Find the value of k, if the points A (2, 3), B(4, k) and C (6, -3) are collinear.

Question 22.
Prove 3 + √5 is irrational.

Question 23.
Find the zeroes of polynomial p(x) = 6x2 – 3 – 7x.

Question 24.
Find the quadratic polynomial whose sum and product of zeroes are \(\frac { 1 }{ 4 }\) and -1 respectively.

Question 25.
Solve the equation 3x2 – 5x + 2 = 0 by using the formula.

Question 26.
Evaluate : 2 tan2 45° + cos2 30° = sin2 60°

Question 27.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°, Find the height of the tower.

Question 28.
As observed from the top of a 100 m high lighthouse from the sea level the angle of depression of two ships are 30° and 45°. If one of the ships is exactly behind the other on the same side of the lighthouse, find the distance between the two ships (√3 = 1.73)

Question 29.
A die is thrown once. Find the probability of getting a number between 2 and 6.

Question 30.
The volume of a cube is 64 cm2. Find the total surface area of the cube.

(6 × 3 = 18)
Question 31.
Prove that “The tangent at any point of a circle is perpendicular to the radius through the point of contact”.
OR
Prove that “The lengths of tangents drawn from an external point to a circle are equal”.

Question 32.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it, whose sides are \(\frac { 2 }{ 3 }\) of the corresponding sides of the first triangle.

Question 33.
A two digit number is four times the sum of the digits. It is also equal to 3 times the product of digits. Find the number.
OR
The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is \(\frac { 29 }{ 20 }\) Find the original fraction.

Question 34.
If 4tanθ = 3, Evaluate \(\frac { 4sin\theta -cos\theta +1 }{ 4sin\theta +cos\theta -1 }\)
OR
If tan 2A = cot (A – 18°) where 2A is an acute angle. Find the value of A.

Question 35.
Calculate the median for the following data.

Class interval Frequency (F)
0-20 6
20-40 8
40-60 10
60-80 12
80-100 6
100-120 5
120-140 3
n = 50

OR
Calculate the mode for the following frequency distribution table

Class Interval Frequency (F)
5-15 6
15-25 11
25-35 21
35-45 23
45-55 14
55-65 5
n = 80

Question 36.
Construct ‘ogive’ for the following construction.

C31 0-3 3-6 6-9 9-12 12-15
F 9 3 5 3 1

(4 × 4 = 16)
Question 37.
The sum of four consecutive terms which are in an arithmetic progression is 32 and the ratio of the product of the first and the last term in the product of two middle terms is 7.15. Find the number.
OR
In an arithmetic progression of 50 terms, the sum of the first ten terms is 210 and the sum of the last fifteen terms is 2565. Find the arithmetic progression.

Question 38.
Prove that “In a right-angled triangle the square on the hypotenuse is equal to the sum of the square on the other two-siders”.

Question 39.
Solve the equations graphically.
2x – y = 2; 4x – y = 4

Question 40.
A wooden article was made by scooping out a hemisphere from one end of a cylinder and a cone from other ends as shown in the figure. If the height of the cylinder is 40 cm, a radius is 7 cm and height of the cone is 24 cm, find the volume of the wooden article.
Karnataka SSLC Maths Model Question Paper 1 Q40

Solutions

Solution 1.
(B) or (D)
Karnataka SSLC Maths Model Question Paper 1 S1

Solution 2.
(C) 3

Solution 3.
(B) -1, 5

Solution 4.
(A) Π (r1 + r2) l

Solution 5.
(A) x = 6, y = 4

Solution 6.
(C) 70°

Solution 7.
(C) (0, 0)

Solution 8.
(B) Roots are equal

Solution 9.
If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.

Solution 10.
RS

Karnataka SSLC Maths Model Question Paper 6 with Answers

Solution 11.
Let a and b be any two positive integers.
Then there exist two unique whole nos. q and r such that a = bq + r, (0 ≤ r < b)

Solution 12.
4

Solution 13.
Karnataka SSLC Maths Model Question Paper 1 S13

Solution 14.
Karnataka SSLC Maths Model Question Paper 1 S14

Solution 15.
Nos. divisible by 3 = 12,15,18, 21, …….., 99
a = 12, d = 3, T = 99, n = ?
Tn = a + (n – 1)d
⇒ 99 = 12 + (n – 1) 3
⇒ 99 – 12 = 3n – 3
⇒ 87 + 3 = 3n
⇒ 90 = 3n
⇒ n = 30
30 two digits are divisible by 3.

Solution 16.
Karnataka SSLC Maths Model Question Paper 1 S16
Karnataka SSLC Maths Model Question Paper 1 S16.1

Solution 17.
Karnataka SSLC Maths Model Question Paper 1 S17
Substitute the value of x in Eqn 1
2x + y = 6
⇒ 2(2) + y = 6
⇒ 4 + y = 6
⇒ y = 6 – 4 = 2
∴ x = 2, y = 2

Solution 18.
Let the age of Gauri be ‘x’ years.
Let the age of Ganesh be ‘y’ years.
5 years ago
(x – 5) = 3(y – 5)
⇒ x – 5 = 3y – 15
⇒ x – 3y = -15 + 5
⇒ x – 3y = -10 …..(1)
10 years later
(x + 10) = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x – 2y = 20 – 10
⇒ x – 2y = 10 ……(2)
Karnataka SSLC Maths Model Question Paper 1 S18
Substitute the value of y in equation 1
x – 3y = -10
⇒ x – 3(20) = -10
⇒ x – 60 = -10
⇒ x = -10 + 60
⇒ x = 50
Gouri’s Age = 50 yrs, Ganesha’s Age = 20 yrs.

Karnataka SSLC Maths Model Question Paper 6 with Answers

Solution 19.
Karnataka SSLC Maths Model Question Paper 1 S19

Solution 20.
Karnataka SSLC Maths Model Question Paper 1 S20

Solution 21.
Karnataka SSLC Maths Model Question Paper 1 S21

Solution 22.
Let us assume that (3 + √5) is rational
Karnataka SSLC Maths Model Question Paper 1 S22
But √5 is irrational
It contradicts the fact that √5 is irrational
our assumption is wrong.
3 + √5 is irrational.

Solution 23.
p(x) = 6x2 – 7x – 3
By splitting the middle term we get
6x2 – 9x + 2x – 3
3x(2x – 3) + 1 (2x – 3)
(2x – 3)(3x + 1)
for zero’s of p(x) put 2x – 3 = 0 & 3x + 1 = 0
1. 2x – 3=0
2x = 3
x = \(\frac { 3 }{ 2 }\)
2. 3x + 1 = 0
3x = -1
x = \(\frac { -1 }{ 3 }\)
Zeros of p(x) are \(\frac { 3 }{ 2 }\) and \(\frac { -1 }{ 3 }\)

Question 24.
The quadratic polynomial is given by
Karnataka SSLC Maths Model Question Paper 1 S24

Question 25.
3x2 – 5x + 2 = 0
Compare it with the standard form
Karnataka SSLC Maths Model Question Paper 1 S25

Solution 26.
Karnataka SSLC Maths Model Question Paper 1 S26

Solution 27.
Karnataka SSLC Maths Model Question Paper 1 S27

Karnataka SSLC Maths Model Question Paper 6 with Answers

Solution 28.
Karnataka SSLC Maths Model Question Paper 1 S28
Karnataka SSLC Maths Model Question Paper 1 S28.1

Solution 29.
No. of all possible outcome = {1, 2, 3, 4, 5, 6}
n(S) = 6
Let E be the event of getting a no. between 2 and 6.
No. of favourable outcomes n(E) = {3, 4, 5} = 3
P(E) = \(\frac { n(E) }{ n(S) }\) = \(\frac { 3 }{ 6 }\)

Solution 30.
Volume of the cube = l3
64 = l3
l = 4
T.S.A. of the cube = 6l2 = 6 × 42 = 6 × 16 = 96 Sq.cm.

Solution 31.
Theorem: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
Karnataka SSLC Maths Model Question Paper 1 S31
Proof: We are given a circle with centre O a tangent XY to the circle at a point R We need to prove that OP is perpendicular to XY.
Take a point Q on XY other the P and join OQ see in below fig.
The point Q must lie outside the circle. (Why? Note that if Q lies inside the circle XY will become a secant and not a tangent to the circle). Therefore OQ is longer than the radius OP of the circle. That is,
OQ > OR
Since this happens for every point on the line XY except the point R OP is the shortest of all the distances of the point O to the points of XY. So OP is perpendicular to XY. (as shown in theorem).
OR
Theorem: The lengths of tangents drawn from an external point to a circle are equal.
Karnataka SSLC Maths Model Question Paper 1 S31.1
Proof: We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P (see in fig.). We are required to prove that PQ = PR,
For this, we join OR OQ and OR. Then ∠OQP and ∠ORP are right angles because these are angles between the radii and tangents, and according to the theorem, they are right angles.
Now in right triangles OQP and ORP
OQ = OR (Radii of the same circle)
OP = OP (Common)
Therefore, ΔOQP = ΔORP (RHS)
This gives PQ = PR

Solution 32.
Karnataka SSLC Maths Model Question Paper 1 S32

Steps of construction:

  1. Draw BC = 6 cm.
  2. With B as the centre & radius 4 cms. draw an arc.
  3. With C as centre and radius, 5 cm cut the first arc at A
  4. Join AB and AC to get ΔABC
  5. Draw any ray BX below BC making an acute angle with BC.
  6. Locate 3 points B1, B2, B3 as BX such that BB1 = B1B2 = B2B3
  7. Join B3C
  8. Draw a line through B2 parallel to B3C intersecting BC at C’
  9. Draw a line through C’ parallel to AC to intersect BA at A’
  10. Now A’BC’ is the required Δ each of whose sides is \(\frac { 2 }{ 3 }\) of the corresponding sides of ΔABC

Solution 33.
Set the no. be xy.
10x + y = 4(x + y)
10x + y = 4x + 4y
10x – 4x = 4y – y
2x = y
10x + y = 3xy
10x + 2x = 3x(2x)
12x = 6×2
12 = 6x
x = 2
y = 2 × 2 = 4
No. = 24
OR
Karnataka SSLC Maths Model Question Paper 1 S33
Karnataka SSLC Maths Model Question Paper 1 S33.1

Solution 34.
Karnataka SSLC Maths Model Question Paper 1 S34
Karnataka SSLC Maths Model Question Paper 1 S34.1

Solution 35.
Karnataka SSLC Maths Model Question Paper 1 S35
Karnataka SSLC Maths Model Question Paper 1 S35.1

Solution 36.
Karnataka SSLC Maths Model Question Paper 1 S36
Karnataka SSLC Maths Model Question Paper 1 S36.1

Solution 37.
Karnataka SSLC Maths Model Question Paper 1 S37
Karnataka SSLC Maths Model Question Paper 1 S37.1
Karnataka SSLC Maths Model Question Paper 1 S37.2

Solution 38.
Karnataka SSLC Maths Model Question Paper 1 S38
Karnataka SSLC Maths Model Question Paper 1 S38.1

Karnataka SSLC Maths Model Question Paper 6 with Answers

Solution 39.
Karnataka SSLC Maths Model Question Paper 1 S39
Karnataka SSLC Maths Model Question Paper 1 S39.1

Solution 40.
Karnataka SSLC Maths Model Question Paper 1 S40

Karnataka State Syllabus SSLC Maths Model Question Paper 2 (Old Pattern)

Four alternatives are given to each question. Choose an appropriate answer. Write it along with its alphabet.

(8 × 1 = 8)
Question 1.
The co-ordinates of the origin are
(a) (0, 0)
b) (1, 0)
(c) (0, 1)
(d) (1, 1)

Question 2.
If p(x) = x3 – 1 then p(-1) = ……..
(a) 2
(b) -2
(c) 1
(d) 0

Question 3.
The total number of all possible events when a dice is thrown once is.
(a) 4
(b) 5
(c) 6
(d) 8

Question 4.
The area of a square is given by
(a) l × b
(b) l × b × h
(c) πr2
(d) l2

Question 5.
If y = 2x – 1 and x + y = 5, then the value of x is
(a) 2
(b) -2
(c) 1
(d) -1

Question 6.
In the adjoining fig AB and AC are tangents from A. If ∠BAC = 80° then BQC
Karnataka SSLC Maths Model Question Paper 2 Q6
(a) 110°
(b) 100°
(c) 80°
(d) 60°

Question 7.
The graph of an equation cuts x-axis at two points. Then the no. of solutions of that equation is
(a) 0
(b) 1
(c) 2
(d) 3

Question 8.
If the discriminant of a quadratic equation b2 – 4ac > 0 then the roots are
(a) equal
(b) imaginary
(c) unequal & irrational
(d) Real & District

(1 × 6 = 6)
Question 9.
State the converse of “Basic Proportionality theorem”

Question 10.
How are the radius and diameter of a circle related?

Question 11.
Find the HCF of 24 and 36 by factorisation.

Question 12.
If p(x) = x2 – x – 2 find p(o)

Question 13.
Find the distance between the point p (4, 5) and Q (1, 2)

Question 14.
If a no. is divided by 5. which are the remainders?

(2 × 16 = 32)
Question 15.
How many two digits no. are divisible by 4?

Question 16.
∆ABC ~ ∆DEF, Area of ∆DEF = 25 Sqcm. and Area of ∆ABC = 100Sqcm., If DE = 16 cms find AB.

Question 17.
Solve: 3x + 2y = 8, x + 3y = 5

Question 18.
Five years ago A was 7 times as old as B. 20 yrs. later A will be twice as old as B. What are their present ages?

Question 19.
Find the are of the shaded region in the figure given.
Karnataka SSLC Maths Model Question Paper 2 Q19

Karnataka SSLC Maths Model Question Paper 6 with Answers

Question 20.
Construct a tangent to a circle of radius 4 cms. at a point on the circumference.

Question 21.
If the coordinates of the vertices of a Δ are (1, 2), (3, 4) and (5, 6). find the perimeter of the triangle.

Question 22.
Prove that 2 + √3 ls irrational.

Question 23.
Find the zeros of the polynomial p(x) = 2x2 – 3x – 5.

Question 24.
Find the quadratic polynomial whose sum & product of zeros are \(\frac { 1 }{ 3 }\) and 2 respectively.

Question 25.
Solve by using formula 5x2 – 9x + 3 = 0

Question 26.
Evaluate : tan2 45° + 2 cos 30° – sin 60°

Question 27.
A tower stands vertically on the ground. From a point on the ground. 15 mts. away from the foot of the tower, the angle of elevation of the top of the tower is 60°. Find the height of the tower.

Question 28.
An observer 1.5 mts tall is 28.5 mts. away from a chimney. The angle of elevation by the top of the chimney from her eyes is 45°, What is the height of the chimney?

Question 29.
A dice is thrown once. Find the probability of getting a number lying between 3 & 6.

(6 × 3 = 18)
Question 30.
The T.S.A. of a cube is 96 Sqcms. Find the volume of the cube.

Question 31.
Prove that the tangents drawn to a circle from an external point are equal.
OR
Prove that tangents drawn to a circle from an external point subtend equal angles at the centre.

Question 32.
Construct a Δ of sides 6 cms, 9 cms and 12 cm and then another Δ similar to it, whose sides are \(\frac { 2 }{ 3 }\) of the corresponding sides of the first triangle.

Question 33.
Solve the following: 8x + 5y = 9, 3x + 2y = 4
OR
A fraction becames \(\frac { 1 }{ 3 }\) when 1 is subtracted from the numerator & it becomes \(\frac { 1 }{ 4 }\) when 8 is added to its denominator. Find the fraction.

Question 34.
If 4 tanθ = 3, Evaluate \(\left[ \frac { 5sin\theta -cos\theta +2 }{ 5cos\theta +sin\theta -2 } \right]\)

Question 35.
Calculate the median for the following data.

C.I. f
20-40 4
40-60 5
60-80 6
80-100 7
100-120 8
N=30

OR
Calculate the mode for the following frequency distribution.

C.I. f
15-25 1
25-35 6
35-45 15
45-55 18
55-65 9
N=50

Question 36.
Constant OGIVE for the following distribution

C.I. f
3-6 3
6-9 6
9-12 4
12-15 2
15-18 5
N=20

(4 × 4 = 16)
Question 37.
How many two digit numbers are divisible by 3?
OR
Find the 31st term of an AR whose 11th term is 38 and the 16th term is 73.

Question 38.
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides Prove this.

Karnataka SSLC Maths Model Question Paper 6 with Answers

Question 39.
Solve graphically : 3x – y = 5, 2x + y = 5

Question 40.
A metallic sphere of radius 4.2 cms is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Solutions

Solution 1.
(a) (0, 0)

Solution 2.
(b) -2

Solution 3.
(c) 6

Solution 4.
(d) l2

Solution 5.
(a) 2

Solution 6.
(b) 100°

Solution 7.
(c) 2

Solution 8.
(d) Real & distinct.

Solution 9.
It states that “If a line divides the two sides of a triangle in proportion then that line is parallel to the third side

Solution 10.
diameter = 2 × radius

Solution 11.
Karnataka SSLC Maths Model Question Paper 2 S11
24 = 2 × 2 × 2 × 3
36 = 2 × 2 × 3 × 3
HCF = 2 × 2 × 3 = 12

Solution 12.
p(x) = x2 – x – 2
⇒ p(0) = 02 – 0 – 2
⇒ p(0) = -2

Solution 13.
Distance
Karnataka SSLC Maths Model Question Paper 2 S13

Solution 14.
The Remainders are {0, 1, 2, 3, 4}

Solution 15.
The nos. are 12,16, 20, 24, 28, ……. ,96
a = 12, d = 4, Tn = 96, n = ?
Tn = a + (n – 1) d
⇒ 96 = 12 + (n – 1) 4
⇒ 96 – 12 = 4n – 4
⇒ 84 + 4 = 4n
⇒ 4n = 88
⇒ n = 22
There are 22 two digit nos. divisible by 4.

Solution 16.
Karnataka SSLC Maths Model Question Paper 2 S16
Karnataka SSLC Maths Model Question Paper 2 S16.1

Solution 17.
Karnataka SSLC Maths Model Question Paper 2 S17

Solution 18.
Let the present ages of A = x years, B = y years.
Karnataka SSLC Maths Model Question Paper 2 S18
Karnataka SSLC Maths Model Question Paper 2 S18.1
Age of A = 40 years
Age of B = 10 years

Solution 19.
Karnataka SSLC Maths Model Question Paper 2 S19

Solution 20.
Karnataka SSLC Maths Model Question Paper 2 S20

Solution 21.
Karnataka SSLC Maths Model Question Paper 2 S21

Solution 22.
Karnataka SSLC Maths Model Question Paper 2 S22
Karnataka SSLC Maths Model Question Paper 2 S22.1

Solution 23.
2x2 – 3x – 5 = 0
2x2 – 5x + 2x – 5 = 0
x[2x – 5] + 1[2x – 5] = 0
(2x – 5)(x + 1) = 0
if 2x – 5 = 0 or if x + 1 = o
2x = 5 or x = 1
x = \(\frac { 5 }{ 2 }\)
The zeroes are 1 and \(\frac { 5 }{ 2 }\)

Solution 24.
Karnataka SSLC Maths Model Question Paper 2 S24

Karnataka SSLC Maths Model Question Paper 6 with Answers

Solution 25.
It is in the form ax2 + bx + c = 0
a = 5, b = -a, c = 3
Karnataka SSLC Maths Model Question Paper 2 S25

Solution 26.
Karnataka SSLC Maths Model Question Paper 2 S26

Solution 27.
Karnataka SSLC Maths Model Question Paper 2 S27
AB = Tower
CB = distance of the point C from the foot of the tower.
ACB is a Right Angle Triangle
tan60° = \(\frac { AB }{ CB }\)
√3 = \(\frac { AB }{ 15 }\)
AB = 15√3 mtrs
Height of the tower = 15√3 Mtrs

Solution 28.
Karnataka SSLC Maths Model Question Paper 2 S28
AB = Chimney
CD = Observer
ADE = angle of elevation
ADE is the Right angled ∆
AB = AE + EB = AE + 1.5
DE = CB = 28.5
tan45° = \(\frac { AE }{ DE }\)
1 = \(\frac { AE }{ 28.5 }\)
AE = 28.5
Height of the Chimney = AB = AE + EB = 28.5 + 1.5 = 30 mts.

Solution 29.
No. of all Possible outcomes n(s) = 6 {1, 2, 3, 4, 5, 6}
Let E be the vent of getting a no. between 3 and 6.
n(E) = {4, 5} = 2
p(E) = \(\frac { n(E) }{ n(S) }\) = \(\frac { 2 }{ 6 }\)

Solution 30.
T.S.A. of a cube = 6l2
96 = 6l2
⇒ l2 = 16
⇒ l = √16 = 4
Volume of the cube = l3 = 43 = 64 cc.

Solution 31.
Data: O is the centre of the circle. AB and AC are tangents from A
Karnataka SSLC Maths Model Question Paper 2 S31
To prove :
AB = AC
Construction:
Join AO, OB and OC.
Proof:
In Δles AOB and AOC
OA is common
OBA = OCA = 90° (angle between radius and tangent)
OB = OC (radii)
By RHS Postulate
ΔAOB = ΔAOC
AB = AC
OR
Data: O is the centre of the circle AB and AC are tangents from A. Join OA, OB & OC.
Karnataka SSLC Maths Model Question Paper 2 S31.1
To Prove : BQA = QOA
Proof:
In Δles AOB and AOC
OA is common
∠OBA = ∠OCA = 90° (angle between radius & tangent)
OB = OC (radii)
By RHS postulate
ΔAOB = ΔAOC
BOA = COA

Solution 32.
AXY is the required Δ
Karnataka SSLC Maths Model Question Paper 2 S32

Solution 33.
Karnataka SSLC Maths Model Question Paper 2 S33
Karnataka SSLC Maths Model Question Paper 2 S33.1
Karnataka SSLC Maths Model Question Paper 2 S33.2

Solution 34.
Karnataka SSLC Maths Model Question Paper 2 S34

Solution 35.
Karnataka SSLC Maths Model Question Paper 2 S35
Karnataka SSLC Maths Model Question Paper 2 S35.1

Solution 36.
Karnataka SSLC Maths Model Question Paper 2 S36
Karnataka SSLC Maths Model Question Paper 2 S36.1

Solution 37.
List of two-digit nos.
divisible by 3 = 12, 15, 18, ……. , 99.
This is an AP
a = 12, d = 3, Tn = 99, n = ?
Tn = a + (n – 1 ) d
⇒ 99 = 12 + (n – 1)3
⇒ 99 – 12 = 3n – 3
⇒ 87 + 3 = 3n
⇒ 90 = 3n
⇒ n = 30
30 two digit nos. are divisible by 3.
OR
Karnataka SSLC Maths Model Question Paper 2 S37
Substitute the value of d in equation 1
a + 10d = 38
⇒ a + 10(7) = 38
⇒ a + 70 = 38
⇒ a = 38 – 70
⇒ a = -32
⇒ T31 = a + 30d
⇒ T31 = -32 + 30(7)
⇒ T31 = -32 + 210
⇒ T31 = 178

Karnataka SSLC Maths Model Question Paper 6 with Answers

Solution 38.
data: ΔABC || ΔPQR
To Prove that:
Karnataka SSLC Maths Model Question Paper 2 S38
Karnataka SSLC Maths Model Question Paper 2 S38.1

Solution 39.
Karnataka SSLC Maths Model Question Paper 2 S39
Karnataka SSLC Maths Model Question Paper 2 S39.1
Karnataka SSLC Maths Model Question Paper 2 S39.2
Q = (2, 1) intersecting of two lines.
The two lines intersecting at the point (2, 1), so x = 2, y = 1 is the required solution of the pair of linear equations.

Solution 40.
Karnataka SSLC Maths Model Question Paper 2 S40
Karnataka SSLC Maths Model Question Paper 2 S40.1

Karnataka State Syllabus SSLC Maths Model Question Paper 3 (Old Pattern)

Four alternatives are given to each question. Choose an appropriate answer. Write it along with its alphabet.

(8 x 1 = 8)
Question 1.
The coordinates of the mid-point P of the join of the points A(x1, y1) and B(x2, y2) is
Karnataka SSLC Maths Model Question Paper 3 Q1

Question 2.
The quadratic polynomial among the following is
(a) p(x) = x2 + 4x + 3
(b) q(x) = ax + b
(c) g(x) = 4x – 3
(d) f(x) = x3 + 4x2 – 5x + 2

Question 3.
When a dice is thrown twice n(s) = ………
(a) 6
(b) 12
(c) 24
(d) 36

Question 4.
The volume of a sphere is given by
(a) \(\frac { 2 }{ 3 }\) πr3
(b) \(\frac { 4 }{ 3 }\) πr3
(c) 4πr3
(d) 2πr3

Question 5.
The values of x and y which satisfy the given equations are 2x – 3y = 2, 3x – 2y = 8
(a) x = 2, y = 2
(b) x = 3, y = 1
(c) y = 2 , y = 2
(d) x = 2, y = 4

Question 6.
In the given fig AB and AC are tangents from A and BAC = 80° value of BOA = …….
Karnataka SSLC Maths Model Question Paper 3 Q6
(a) 50°
(b) 60°
(c) 70°
(d) 90°

Question 7.
If A(3, 4) and B(5, 6) are two points, then the coordinates of the mid-point of AB are
(a) (4, 5)
(b) (3, 4)
(c) (5, 6)
(d) (6, 7)

Question 8.
The discriminant of the equation ax2 + bx + c = 0
(a) c2 – 4ab
(b) a2 – 4ab
(c) b2 – 4ac
(d) a2 – 2ac

Karnataka SSLC Maths Model Question Paper 6 with Answers

(1 × 6 = 6)
Question 9.
State Pythagoras Theorem

Question 10.
What is the length of the longest chord in a circle of radius 5 cms?

Question 11.
Find the HCF of 135 and 225.

Question 12.
From the graph, find the number of zeroes of the polynomial p(x).
Karnataka SSLC Maths Model Question Paper 3 Q12

Question 13.
Find the distance between A and B if the coordinates of A and B are (2, 3) and (5, 6) respectively.

Question 14.
Express 226 as a product of Prime facotrs.

(2 x 16 = 32)
Question 15.
How many two digit nos. are divisible by 5?

Question 16.
If a pair of corresponding sides of two triangles are 5 cms and 7 cms then find the ratio of the areas of triangles

Question 17.
Solve:
5x + 4y = 14 → (1)
4x + 2y = 10 → (2)

Question 18.
5 Years ago A was thrice as old as his son 20 years, later A will be twice as old as his son. Find their present ages.

Question 19.
Find the area of the sector OAPB
Karnataka SSLC Maths Model Question Paper 3 Q19

Question 20.
Draw a circle of radius 4 cms. construct a tangent to the circle from a point 10 cms away from the centre.

Question 21.
Find the value of K. If the points A (4, 6), B(8, 2k) and C(12, -6) are collinear.

Question 22.
Prove that 5 + √7 is irrational

Question 23.
Find the zeros of the polynomial p(x) = 6x2 + 7x – 3

Question 24.
Find the quadratic polynomial, whose sum & Product of the zeros are \(\frac { 1 }{ 2 }\) & 2 respectively.

Question 25.
Solve by using formula 2x2 – 5x + 3 = 0

Question 26.
Evaluate : 2 sin2 60° + 3 tan2 45° – cos2 30°

Question 27.
The angle of elevation of the top of a tower from a point on the ground, which is 60 M away from the foot of the tower is 30°. Find the height of the tower.

Question 28.
As observed from the top of a 200M height tower from the sea level the angle of depression of two boats are 30° and 45°, if one of the boats is exactly behind the other on the same side of the tower, find the distance between two boats (√3 = 1.73).

Question 29.
A dice is thrown once. Find the probability of getting a perfect square number.

Question 30.
A copper rod of diameter 1 cm and length 8 cms. is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.

(16 x 2 = 32)
Question 31.
Prove that in two concentric circles the chord of the larger circle, which touches the smaller circle is bisected at the point of contact.
OR
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Question 32.
Construct a triangle of side 6 cm., 9 cm and 1 cm. and then a triangle similar to it whose side, are \(\frac { 2 }{ 3 }\) of the corresponding sides of the first triangle.

Question 33.
Solve the following equations by reducing them to a pair of linear equation.
Karnataka SSLC Maths Model Question Paper 3 Q33
OR
5 pencils and 7 books together cost Rs. 50. whereas 7 pencils & 5 books together cost Rs. 46. Find the cost of 1 pencil and 1 book.

Question 34.
If sin 3A = cos (A – 26°). Where 3A is an acute angle, find the value of A.
OR
Karnataka SSLC Maths Model Question Paper 3 Q34

Question 35.
Calculate the median for the following data.

C.I. f
0 – 20 4
20 – 40 6
40 – 60 8
60 – 80 10
80 – 100 4
100 – 120 5
120 – 140 3
n = 40

OR
Calculate the mode for the following frequency distribution

C.I. f
5 – 15 3
15 – 25 8
25 – 35 17
35 – 45 20
45 – 55 11
55 – 65 1
n = 60

Question 36.
Construct ‘OGIVE’ for the following:

C.I. f
0 – 3 8
3 – 6 2
6 – 9 4
9 – 12 2
12 – 15 1
n = 17

(4 x 4 = 16)
Question 37.
Find the 11th term from the last term (towards the first-term) of the AR 10, 7, 4,…….., -62.
OR
In a flower bed there are 23 rose plants in the first row, 21 in the second, 19 in the third & so on. There are 5 rose-plants in the last row. How many rows are there in the flower bed ?

Question 38.
In a triangle if square as one side is equal to the sum of the squares on the other two sides, then the angle opposite to the first side is a right angle, prove this.

Question 39.
Solve Graphically :
2x – y = 5
4x – y = 13

Question 40.
Ram made a bird bath for his garden in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and its radius is 30 cms. Find the T.S.A. of the bird bath.

Solutions

Solution 1.
(b)
Karnataka SSLC Maths Model Question Paper 3 S1

Solution 2.
(a) p(x) = x2 + 4x + 3

Solution 3.
(d) 36

Solution 4.
(b) \(\frac { 4 }{ 3 }\) πr3

Solution 5.
(c) x = 4, y = 2

Solution 6.
(a) 50°

Solution 7.
(a) (4, 5)

Solution 8.
(c) b2 – 4ac

Solution 9.
“In a right-angled triangle the square on the hypotenuse is equal to the sum of the squares as the other two sides”.

Solution 10.
10 cm.

Solution 11.
Karnataka SSLC Maths Model Question Paper 3 S11
By division lemma
225 = 135 x 1 + 90
135 = 90 x 1 + 45
90 = 45 x 2 + 0
The last divisor is 45
HCF = 45

Solution 12.
4

Solution 13.
Karnataka SSLC Maths Model Question Paper 3 S13

Solution 14.
Karnataka SSLC Maths Model Question Paper 3 S14

Solution 15.
Nos. divisible by 5 are 10, 15, 20, ………, 95
They are in AP
a = 10, d = 5, Tn = 95, n = ?
Tn = a + (n – 1)d
⇒ 95 = 10 + (n – 1)5
⇒ 95 – 10 = 5n – 5
⇒ 85 + 5 = 5n
⇒ 90 = 5n
⇒ n = 18
18 two digit nos. are divisible by 5.

Solution 16.
Let ABC and PQR be the triangles.
Karnataka SSLC Maths Model Question Paper 3 S16

Solution 17.
Karnataka SSLC Maths Model Question Paper 3 S17
Substitute the value of x in eqn. 1.
5x + 4y = 14
⇒ 5(2) + 4y = 14
⇒ 10 + 4y = 14
⇒ 4y = 14 – 10
⇒ 4y = 4
⇒ y = 1
∴ x = 2, y = 1

Solution 18.
Let the age of A = x years
Let the age of son = y years
5 years ago
(x – 5) = 3(y – 5)
x – 5 = 3y – 15
x – 3y = -15 + 5
x – 3y = -10 …..(1)
20 years later
(x + 20) = 2(y + 20)
x + 20 = 2y + 40
x – 2y = 40 – 20
x – 2y = 20 …… (2)
From Eqn (1) & (2)
Karnataka SSLC Maths Model Question Paper 3 S18
Present age of the father = 80 years
Age of the son = 30 years.

Solution 19.
Karnataka SSLC Maths Model Question Paper 3 S19

Solution 20.
Karnataka SSLC Maths Model Question Paper 3 S20

Solution 21.
Karnataka SSLC Maths Model Question Paper 3 S21
Karnataka SSLC Maths Model Question Paper 3 S21.1
Karnataka SSLC Maths Model Question Paper 3 S21.2

Solution 22.
Let us assume that 5 + √7 is rational.
Karnataka SSLC Maths Model Question Paper 3 S22
but √7 is irrational.
If contradicts the fact that √7 is irrational
our assumption is wrong.
(5 + √7) is irrational.

Solution 23.
p(x) = 6x2 + 7x – 3
By splitting the middle term
6x2 + 9x – 2x – 3
⇒ 3x(2x + 3) – 1(2x + 3)
⇒ (2x + 3) (3x – 1)
for zeroes of p(x) put
2x + 3 = 0 and 3x – 1 = 0
x = \(\frac { -3 }{ 2 }\) and x = \(\frac { 1 }{ 3 }\)
zeroes of p(x) are \(\frac { -3 }{ 2 }\) and x = \(\frac { 1 }{ 3 }\)

Solution 24.
Karnataka SSLC Maths Model Question Paper 3 S24
Karnataka SSLC Maths Model Question Paper 3 S24.1

Solution 25.
Compare this with ax2 + bx + c = 0
a = 2, b = -5, c = 3
Karnataka SSLC Maths Model Question Paper 3 S25

Solution 26.
Karnataka SSLC Maths Model Question Paper 3 S26

Karnataka SSLC Maths Model Question Paper 6 with Answers

Solution 27.
Karnataka SSLC Maths Model Question Paper 3 S27
Karnataka SSLC Maths Model Question Paper 3 S27.1

Solution 28.
Karnataka SSLC Maths Model Question Paper 3 S28

Solution 29.
No. of all possible outcomes n(S) = 6
Let ‘E’ be the event of getting a perfect square no.
n(E) = 2 {1, 4}
p(E) = \(\frac { n(E) }{ n(S) }\) = \(\frac { 2 }{ 6 }\)

Solution 30.
Karnataka SSLC Maths Model Question Paper 3 S30

Solution 31.
Karnataka SSLC Maths Model Question Paper 3 S31
Data: Two concentric circles C1 and C2 have the same centre O.
AB is the chord of the larger circle C2.
It touches C1 at P
To prove AP = PB
Construction: Join OP
Proof: AB is the tangent to C1 at P and OP is the radius.
OP ⊥ AB
AB is the chord for C2 & OP ⊥ AB
OP is the bisector of AB, as the perpendicular from the centre bisects the chord.
i.e., AP = BR
OR
Karnataka SSLC Maths Model Question Paper 3 S31.1
Data: ‘O’ is the centre of the circle XY is a tangent at P.
To prove: QP ⊥ XY
Construction: Take a point Q as XY other than P and join OQ.
Proof: The point P lies outside the circle (if it lies inside XY becomes a secant).
OQ is longer than OP (radius)
OQ > OP
Since this happens for every point on XY except the point P
OP is the shortest of all the distances from the point O to the points on XY.
So, OP is the perpendicular to XY.

Solution 32.
Karnataka SSLC Maths Model Question Paper 3 S32
ΔABC is required Δ
ΔA’BC’ is required Δ

Solution 33.
Karnataka SSLC Maths Model Question Paper 3 S33
Karnataka SSLC Maths Model Question Paper 3 S33.1
Karnataka SSLC Maths Model Question Paper 3 S33.2

Solution 34.
sin3A = cos(A – 26°)
⇒ sin 3A = cos (90 – 3A)
⇒ cos (90 – 3A) = cos (A – 26)
⇒ 90 – 3A = A – 26
⇒ 90 + 26 = A + 3A
⇒ 116 = 4A
⇒ A = 29°
OR
Karnataka SSLC Maths Model Question Paper 3 S34

Karnataka SSLC Maths Model Question Paper 6 with Answers

Solution 35.
Karnataka SSLC Maths Model Question Paper 3 S35
Karnataka SSLC Maths Model Question Paper 3 S35.1

Solution 36.
Karnataka SSLC Maths Model Question Paper 3 S36
Karnataka SSLC Maths Model Question Paper 3 S36.1

Solution 37.
If we write the given AP in the reverse order, then a = -62 and d = 3.
Now find the 11th term with these a and d.
so, a11 = -62 + (11 – 1) x 3
a11 = -62 + 30
an = -32
OR
The number of rose plants in 1st, 2nd, 3rd, ……… rows are
23, 21, 19, ………. 5
It forma an AP
Let the number of rows = n
a = 23, d = -2, an = 5
an = a + (n – 1)d
⇒ 5 = 23 + (n – 1)(-2)
⇒ 5 – 23 = -2n + 2
⇒ -18 – 2 = -2n
⇒ -20 = -2n
⇒ n = 10
No. of rows = 10

Solution 38.
Karnataka SSLC Maths Model Question Paper 3 S38
AC2 = AB2 + BC2
we need to prove ∠B = 90°
construct ΔPQR , right angled at Q such that
PQ = AB & QR = BC
Now from ΔPQR , we have
PR2 = PQ2 + QR2 (by pythogora’s theorem)
PR2 = AB2 + BC2 (by construction)
AC2 = AB2 + BC2 (data)
AC = PR
In ΔABC and ΔPQR
AB = PQ (by construction)
BC = QR (by construction)
AC = PR (Proved)
ΔABC = ΔPQR
∠B = ∠Q (CPCT)
∠Q = 90° (by Construction)
∠B = 90°

Solution 39.
Karnataka SSLC Maths Model Question Paper 3 S39
Karnataka SSLC Maths Model Question Paper 3 S39.1

Solution 40.
Karnataka SSLC Maths Model Question Paper 3 S40
Let h be the height of the cylinder and r the common radius of cylinder
hemi-sphere T.S.A. of bird-bath = CSA of cylinder + CSA of the hemisphere.
= 2πrh + 2πr2
= 2πr (h + r)
= 2 x \(\frac { 22 }{ 7}\) x 30 (145 + 30)
= 2 x \(\frac { 22 }{ 7}\) x 30 x 175
= 33000 cm2

Karnataka SSLC Maths Model Question Paper 6 with Answers

Karnataka State Syllabus SSLC Maths Model Question Paper 4 (Old Pattern)

Four alternatives are given to each question. Choose an appropriate answer. Write it along with its alphabet.

(1 × 8 = 8)
Question 1.
The distance between two points A (1, 7) and B (4, 2) is
(a) √34
(b) √43
(c) √68
(d) √86

Question 2.
If p(x) = x2 + x + 1 then the value of p(-1) is
(a) -1
(b) 1
(c) 2
(d) -2

Question 3.
Which is correct for the event E?
(a) P(\(\bar { E }\)) = 1 + P(E)
(b) P(\(\bar { E }\)) = P(E)
(c) P(\(\bar { E }\)) = 1 – P(E)
(d) P(E) = 1 + P(\(\bar { E }\))

Question 4.
The volume of a cylinder is 154 cc. and the radius is 7 cms. then the height is
(a) 2 cms
(b) 3 cms
(c) 4 cms
(d) 1 cm

Question 5.
The value of x and y for the following equations are x + y = 15, x – 7 = 1
(a) (8, 7)
(b) (7, 8)
(c) (10, 5)
(d) (5, 10)

Question 6.
ABCD is a cyclic quadrilateral. A = 70°, then C = ?
(a) 100°
(b) 110°
(c) 120°
(d) 70°

Question 7.
OA is a radius in the circle. If coordinates of A are (2, 3) then OA =…
Karnataka SSLC Maths Model Question Paper 4 Q7
a) √9
b) √4
c) √13
d) √5

Question 8.
If the discriminant of a quadratic equation b2 – 4ac < 0 then the roots are
(a) Real and distinct
(b) equal
(c) not real
(d) unequal&rational

(1 × 6 = 6)
Question 9.
In ΔABC, XY || BC, If AX = 2 cm, AB = 5 cms and AY = 4 cms. then AC = ………
Karnataka SSLC Maths Model Question Paper 4 Q9

Question 10.
Identify the largest chord in the given figure.
Karnataka SSLC Maths Model Question Paper 4 Q10

Question 11.
Find the HCF of 25 and 15

Question 12.
From the graph find the number of zeros of the polynomial p(x)
Karnataka SSLC Maths Model Question Paper 4 Q12

Question 13.
Which are the coordinates of the origin?

Question 14.
Express 200 as a product of prime factors.

(16 × 2 = 32)
Question 15.
How many 2 digit numbers are divisible by 9.

Question 16.
Diagonals of a trapezium ABCD with AB || DC intersect at O. If AB = 2CD. Find the ratio of the areas of triangles AOB and COD.

Question 17.
Solve for x and y
2x + y = 4
3x + 4y = 6

Question 18.
Half the perimeter of a rectangle whose length is 4 m. more than its width is 36 m. Find the dimensions of the garden.

Question 19.
Find the area of the shaded region in the fig given. If each side of the square is 14 cm.
Karnataka SSLC Maths Model Question Paper 4 Q19

Question 20.
Draw a circle of radius 4 cms. Draw two radii in it such that the angle between them is 130°. Construct two tangents at the ends of radii.

Question 21.
The coordinates of the vertices of a triangle are (2, 3), (4, 5) and (6, 9). Find its area.

Question 22.
Prove that (5 + √3) is irrational.

Question 23.
Find the zeros of the polynomial p(x) = 5x2 – 2x – 3

Question 25.
Solve the given equation using formula x2 – 6x – 4 = 0

Question 26.
If A = √2 – 1 show that \(\frac { tanA }{ 1+{ tan }^{ 2 }A } =\frac { \surd 2 }{ 4 }\)

Question 27.
From a point an the ground, the angles of elevation of the bottom and top of a tower fixed at the top of 20 m high building is 45° & 60° respectively. Find the height of the tower.

Question 28.
The angle of elevation of an aeroplane from a point on the ground is 60°. After 15 seconds flight the elevation changes to 30°. If the aeroplane is flying at a height of 1500√3 m, find the speed of the plane in km/hr.

Question 29.
12 defective pens are mixed with 132 good ones one pen is taken out at random from this lot. Find the probability that the pen is taken out is a good one.

(6 × 3 = 18)
Question 30.
Two cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Question 31.
Prove that in two concentric circles the chord of the longer circle, which touches the smaller circle is bisected at the point of contact.
OR
Prove that the tangents drawn to a circle from an external point are equal.

Question 32.
Construct a Δ of sides, 8 cms., 10 cms & 12 cms. and then A similar to it, whose sides are \(\frac { 3 }{ 4 }\) the corresponding sides of the first Δ

Question 33.
90% and 97% acid solutions are mixed to get 21 Its. of 95% pure acid solution. Find the amount of each type of acid to be mixed to form the mixture.
OR
One says “give me a hundred friend I shall then become twice as rich as you”. The other says “If you give me 10,1 shall be 6 times as rich as you”. Find the amount of their capital.

Question 34.
If A, B, C are interior angles of a triangle ABC show that
Karnataka SSLC Maths Model Question Paper 4 Q34

Question 35.
Calculate the median for the following data

C.I. f
65 – 85 4
85 – 105 5
105 – 125 13
125 – 145 20
145 – 165 14
165 – 185 8
185 – 205 4

OR
Calculate the mode for the following data.

C.I. f
1000 – 1500 24
1500 – 2000 40
2000 – 2500 33
2500 – 3000 28
3000 – 3500 30
3500 – 4000 22
4000 – 4500 16
4500 – 5000 07

Question 36.
Draw the ‘less than type OGIVE for the following data.

C.I. f
50 – 60 6
60 – 70 5
70 – 80 9
80 – 90 12
90 – 100 6

(4 × 4 = 16)
Question 37.
Which term of the A.P 45, 41, 37, 33, ………… is the first negative term?
OR
Karnataka SSLC Maths Model Question Paper 4 Q37

Question 38.
Prove that in a Δ, if square as one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Question 39.
Solve Graphically
x + 3y = 6
2x – 3y = 12

Question 40.
The base radius and height of a right circular solid cone are 2 cms and 8 cms respectively. It is melted & recast into spheres of diameter 2 cms. each. Find the number of spheres formed.

Solutions

Solution 1.
(a) √34

Solution 2.
(b) 1

Solution 3.
(c) P(\(\bar { E }\)) = 1 – P(E)

Solution 4.
(d) 1 cm

Solution 5.
(a) (8, 7)

Solution 6.
(b) 110°

Solution 7.
(c) √13

Solution 8.
(c) not real

Solution 9.
Karnataka SSLC Maths Model Question Paper 4 S9

Solution 10.
XY

Solution 11.
25 = 5 × 5
15 = 5 × 3

Solution 12.
2

Solution 13.
(0, 0)

Solution 14.
Karnataka SSLC Maths Model Question Paper 4 S14
200 = 2 × 2 × 2 × 5 × 5 = 23 × 52

Solution 15.
Two digit Nos. are 18, 27, 36, ……. 99
It is in AP
a = 18, d = 9, an = 99, n = ?
an = a + (n – 1) d
⇒ 99 = 18 + (n – 1)9
⇒ 99 – 18 = 9n – 9
⇒ 81 + 9 = 9n
⇒ 90 = 9n
⇒ n = 10
There are 10 two digit nos.

Solution 16.
Area of ∆AOB and ∆COD
AÔB = COD (V.O.A.)
COO = OA (alternate angles)
By AA Similarity criterion
∆AOB ~ ∆DOC
Karnataka SSLC Maths Model Question Paper 4 S16

Solution 17.
Karnataka SSLC Maths Model Question Paper 4 S17
Substitute the value of x in equation 1
2x + y = 4
⇒ 2(2) + y = 4
⇒ 4 + y = 4
⇒ y = 4 – 4 = 0
x = 2, y = 0

Solution 18.
Karnataka SSLC Maths Model Question Paper 4 S18
l + b = 36
⇒ l = 36 – b
l = b + 4
b + 4 = 36 – b
⇒ b + b = 36 – 4
⇒ 2b = 32
⇒ b = 16
l = b + 4
⇒ l = 16 + 4
⇒ l = 20
Length = 20 m
Width = 16 m

Solution 19.
Karnataka SSLC Maths Model Question Paper 4 S19
Karnataka SSLC Maths Model Question Paper 4 S19.1

Solution 20.
PA and PB are tangents to the circle at the ends of radii
Karnataka SSLC Maths Model Question Paper 4 S20

Solution 21.
Karnataka SSLC Maths Model Question Paper 4 S21

Karnataka SSLC Maths Model Question Paper 6 with Answers

Solution 22.
Let us assume that (5 + √3)
Karnataka SSLC Maths Model Question Paper 4 S22
RHS is rational
⇒ √3 is rational
But √3 is irrational
It is contradictory to the fact that √3 is irrational
Hence our assumption is wrong.
(5 + √3) is irrational.

Solution 23.
To get the zeros of the, polynomial
put p(x) = 0
⇒ 5x2 – 2x – 3 = 0
⇒ 5x2 – 5x + 3x – 3 = 0
⇒ 5x(x – 1) + 3(x – 1) = 0
⇒ (x – 1)(5x + 3) = 0
x – 1 = 0 or 5x + 3 = 0
x = 1 or x = \(\frac { -3 }{ 5 }\)
Zeros of the polynomial are 1 and \(\frac { -3 }{ 5 }\)

Solution 24.
Here a = a, b = -5, c = c
Let the zeros be α and β then
Karnataka SSLC Maths Model Question Paper 4 S24

Solution 25.
Karnataka SSLC Maths Model Question Paper 4 S25
Karnataka SSLC Maths Model Question Paper 4 S25.1

Solution 26.
Karnataka SSLC Maths Model Question Paper 4 S26

Solution 27.
Karnataka SSLC Maths Model Question Paper 4 S27
Karnataka SSLC Maths Model Question Paper 4 S27.1

Solution 28.
Karnataka SSLC Maths Model Question Paper 4 S28
Karnataka SSLC Maths Model Question Paper 4 S28.1
Karnataka SSLC Maths Model Question Paper 4 S28.2

Solution 29.
No. of defective pens = 12
No. of good pens = 132
Total no. of pens = 144
No. of all possible outcomes = 144
Let E be the event that the pen taken out is a good one
No. of outcomes favourable to E = 132
Karnataka SSLC Maths Model Question Paper 4 S29

Solution 30.
Karnataka SSLC Maths Model Question Paper 4 S30
Let each edge of the cube be a
Volume of each cube = a3
64 = a3
⇒ a = 4
For the resulting cuboid l = 4 + 4 = 8 cm
breadth = 4 cm
Height = 4 cm.
Surface area of the resulting cuboid = 2(lb + bh + hl)
= 2[(8 × 4) + (4 × 4) + (4 × 8)]
= 2[32 + 16 + 32]
= 2 [80]
= 160 cm2

Solution 31.
C1 and C2 are the concentric circle with centre O A chord AB of the larger circle C touches the smaller circle C2 at P
Karnataka SSLC Maths Model Question Paper 4 S31
To prove: AP = BP
Construction: Join OP
Proof: AB is a tangent to the circle C2 at P and OP is the radius.
OP ⊥ AB
AB is the chord of circle Q & OP ⊥ AB
OP is the bisector of chord AB
AP = PB
OR
Data: O is the centre of the circle. AB and AC are tangents from A
Karnataka SSLC Maths Model Question Paper 4 S31.1
To prove AB = AC
Construction: Join OA, OB and OC
Proof: In ΔAOB and ΔAOC
ABO = ACO = 90° [Angle between radius & tangent]
OB = OC (Radii of the circle)
OA is common
By RHS criterian
ΔAOB = ΔAOC
AB = AC

Solution 32.
A’CC’ is the required triangle
Karnataka SSLC Maths Model Question Paper 4 S32

Solution 33.
Let x Lts of 90% pure acid solution & Y Lts of 97% pure acide solution be mixed,
total volume of the mixture = (x + y) Lts
x + y = 21 ……… (1)
90% of x + 97% of y = 95% of 21
Karnataka SSLC Maths Model Question Paper 4 S33
Substituting the value of y in eqn 1
x + y = 21
⇒ x + 15 = 21
⇒ x = 21 – 15 = 6
⇒ x = 6
Quantity of 90% pure acid solution = 6 Lts.
Quantity of 97% pure acid solution = 15 Lts.
OR
Let the amounts with them be respectively.
x and y
x + 100 = 2(y – 100)
⇒ x + 100 = 2y – 200
⇒ x – 2y = -200 – 100
⇒ x – 2y = -300 ……. (1)
y + 10 = 6(x – 10)
⇒ y + 10 = 6x – 60
⇒ y – 6x = -60 – 10
⇒ -6x + y = -70
⇒ 6x – y = 70 ……… (2)
From Equation 1
x = 2y – 300
Substituting the value of x in equation 2
6x – y = 70
⇒ 6(2y – 300) – y = 70
⇒ 12y – 1800 – y = 70
⇒ 11y = 70 + 1800
⇒ 11y = 1870
⇒ y = 170
Substitute the value Y in (1)
x = 2y – 300
⇒ x = 2(170) – 300
⇒ x = 340 – 300
⇒ x = 40
Their capitals are 40 Rs. & 170 Rs.

Karnataka SSLC Maths Model Question Paper 6 with Answers

Solution 34.
Because A, B, C are the angles of a ∆ABC
A + B + C = 180°
⇒ B + C = (180 – A)
Karnataka SSLC Maths Model Question Paper 4 S34
Karnataka SSLC Maths Model Question Paper 4 S34.1

Solution 35.
Karnataka SSLC Maths Model Question Paper 4 S35
Karnataka SSLC Maths Model Question Paper 4 S35.1
Karnataka SSLC Maths Model Question Paper 4 S35.2

Solution 36.
Karnataka SSLC Maths Model Question Paper 4 S36
Karnataka SSLC Maths Model Question Paper 4 S36.1

Solution 37.
a = 45
d = a2 – a1 = 41 – 45 = -4
Let nth term of the AP be the first negative term then an < 0
a + (n – 1)d < 0
⇒ 45 + (n – 1) (-2) < 0
⇒ 45 – 4n + 4 < 0
⇒ 49 – 4n < 0
⇒ 49 < 4n
⇒ n > \(\frac { 49 }{ 4 }\)
⇒ n > 12\(\frac { 1 }{ 4 }\)
least positive integral value of n = 13
13th term is the first negative term.
OR
Let ‘a’ be the first term and d the common difference
a4 = a + (4 – 1) d = a + 3d
a6 = a + 5d
a7 = a + 6d
a8 = a + 7d
Karnataka SSLC Maths Model Question Paper 4 S37

Solution 38.
Karnataka SSLC Maths Model Question Paper 4 S38
data : ABC is a ∆ in which AC2 = AB2 + BC2
To prove: ∠B = 90°
Construction: Construct ∆PQR right angled at Q such that
PQ = AB & RQ = CB
Proof: PR2 = PQ2 + RQ2 (by Pythagoras theorem)
PR2 = AB2 + BC2 (by construction)
But AC2 = AB2 + BC2 (data)
AC = PR
In ∆ABC and ∆PQR
AB = PQ
BC = QR (by construction)
AC = PR (proved)
∆ABC = ∆PQR
∠B = ∠Q
but ∠Q = 90° (by construction)
∠B = 90°

Solution 39.
Karnataka SSLC Maths Model Question Paper 4 S39
Karnataka SSLC Maths Model Question Paper 4 S39.1

Solution 40.
Karnataka SSLC Maths Model Question Paper 4 S40

Karnataka State Syllabus SSLC Maths Model Question Paper 5 (Old Pattern)

Four alternatives are given to each question. Choose an appropriate answer. Write it along with its alphabet.

(1 × 8 = 8)
Question 1.
The distance between the points (2, 3) and (4, 1) is
(a) 3√2
(b) 2√3
(c) √2
(d) 2√2

Question 2.
The cubic polynomial among the following is
(a) p(x) = x2 – x – 2
(b) q(x) = 2x2 + x – 7
(c) r(x) = x3 + x2 – 1
(d) s(x) = x4 – x3 + x2 – 2

Question 3.
The probability of a certain event is
(a) 0
(b) 1
(c) \(\frac { 1 }{ 2 }\)
(d) \(\frac { 1 }{ 3 }\)

Question 4.
The volume of a sphere of radius 21 cm is
(a) 38808 cm3
(b) 80838 cm3
(c) 83808 cm3
(d) 88380 cm3

Question 5.
If y = x – 2 and x + y = 8, then the values of x and y are respectively,
(a) -5, 3
(b) -3, 5
(c) 3, 5
(d) 5, 3

Question 6.
In the adjoining figure, AB and AC are tangents from A. Then ABC =?
Karnataka SSLC Maths Model Question Paper 5 Q6
(a) 30°
(b) 40°
(c) 50°
(d) 60°

Question 7.
The distance between points (a, b) and (-a, -b) is
(a) \(2\sqrt { { a }^{ 2 }+{ b }^{ 2 } }\)
(b) \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } }\)
(c) 2\(\sqrt { { a }+{ b } }\)
(d) \(\sqrt { { a }+{ b } }\)

Question 8.
The roots of the equation 2x2 – 200 = 0 are
(a) ±20
(b) ±10
(c) 20
(d) 10

(6 × 1 = 6)
Question 9.
State the converse of Basic proportionality theorem.

Question 10.
AB is a chord in a circle of radius 5 cms. OP ⊥ AB. If OP = 4 cms then AB = ……..
Karnataka SSLC Maths Model Question Paper 5 Q10
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 6 cm

Question 11.
The product of HCF and LCM of two numbers is 75. Then the product of the numbers is
(a) 125
(b) 25
(c) 75
(d) 57

Question 12.
The zeroes of the Polynomial p(x) = x2 + 4x + 3 are
(a) -3, -1
(b) 1, 3
(c) -1, 3
(d) -3, 1

Karnataka SSLC Maths Model Question Paper 6 with Answers

Question 13.
If a point P (x, y) divides the line segment joining, A (x1, y1) and B (x2, y2) in the ratio m1 and m2, then the coordinates of P are
Karnataka SSLC Maths Model Question Paper 5 Q13
Karnataka SSLC Maths Model Question Paper 5 Q13.1
Karnataka SSLC Maths Model Question Paper 5 Q13.2
Karnataka SSLC Maths Model Question Paper 5 Q13.3

Question 14.
Express 120 as a product of prime factors.

(16 × 2 = 32)
Question 15.
Write the first term and common difference for the following A.P.
\(\frac { 1 }{ 3 }\) , \(\frac { 5 }{ 3 }\) , \(\frac { 9 }{ 3 }\) , \(\frac { 13 }{ 3 }\) , ……..

Question 16.
In the figure of ∆ABE = ∆ACD show that ∆ADE ~ ∆ABC
Karnataka SSLC Maths Model Question Paper 5 Q16

Question 17.
Solve for x and y
3x + 2y = 3
2x + 3y = 2

Question 18.
If twice the son’s age is added to the age of his father the sum is 90. If twice the father’s age is added to the age of the son the sum is 120. Find their ages.

Question 19.
In the figure given O is the centre of the bigger circle and AC is its diameter. Another circle with BA as the diameter is drawn.
If AC = 54 cms & BC = 10 cms. Find the area of the shaded region.
Karnataka SSLC Maths Model Question Paper 5 Q19

Question 20.
Construct a tangent of length 4 cm. from an external point to a circle of radius 3 cms.

Question 21.
Find the relation between x and y, such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).

Question 22.
Prove that 7 + √5 is irrational.

Question 23.
Find the quadratic polynomial, whose zeroes are (2 + √3) and (2 – √3)

Question 24.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time & product of its zeroes as 2, -7, -14 respectively.

Question 25.
Solve the equation 2x2 – 5x + 3 = 0 using formula.

Question 26.
Evaluate:
Karnataka SSLC Maths Model Question Paper 5 Q26

Question 27.
From the top of a 7 m high building, the angle of elevation of the top of a tower is 60° & the angle of depression of its foot is 45°. Find the height of the tower.

Question 28.
From the top of a lighthouse, the angles of depression of two ships on the opposite sides of it are observed to be 30° and 60°. If the height of the lighthouse is h mts & the line joining the ships passes through the foot of the lighthouse find the distance between the ships.

Question 29.
A lot of 20 bulbs contains 4 defective ones, one bulb is drawn at random from the lot. What is the probability that the bulb is defective?

Question 30.
The sum of the radius of base and height of the solid right circular cylinder is 37 cms. If the total surface area of the solid cylinder is 1628 cm2. Find the volume of the cylinder.

(6 × 3 = 18)
Question 31.
A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle. Prove this.

Question 32.
Draw a line segment of length 4.5 cms and divide it is the ratio 3 : 6 & measure each part.

Question 33.
Pocket money of A and B are in the ratio 6 : 5 & the ratio of their expenditure are in the ratio 4 : 3. If each of them saves Rs. 50 at the end of the month find their pocket money.
OR
The sum of a two-digit number & the no. obtained by reversing the digits is 99. If the no. obtained by reversing the digits is 9 more than the original no. find the no.

Question 34.
If tan2A = cot (A – 18°) where 2A is an acute angle, find A.
OR
Show that tan 48°. tan 23°. tan 42 °. tan 67° = 1

Question 35.
Calculate the median for the following data.

C.I. f
1500 – 2000 14
2000 – 2500 50
2500 – 3000 60
3000 – 3500 86
3500 – 4000 74
4000 – 4500 62
4500 – 5000 48

OR
Calculate the mode for the following data.

C.I. f
150 – 155 12
155 – 160 13
160 – 165 10
165 – 170 8
170 – 175 5
175 – 180 2

Question 36.
Contant OGIVE for the following distribution.

C.I. f
50 – 60 6
60 – 70 5
70 – 80 9
80 – 90 12
90 – 100 6

(4 × 4 = 16)
Question 37.
If the 3rd and 9th terms of an A.P. are 4 and -8 respectively, which term of this AP is zero?
OR
If (2n + 3) is the nth term of an A.P. find (i) first term, (ii) common difference, (iii) 15th term.

Question 38.
Prove that in a right-angled triangle, the square as the hypotenuse is equal to the sum of the squares on the other two sides?

Question 39.
Solve the following in graphically:
3x – y = 3
2x + y = 2

Question 40.
The cost of painting the total outer surface of a closed cylindrical tank at 60 per square cm. is Rs. 237.60. The height of the tank is 6 times the radius of the base. Find the height & radius of the tank.

Solutions

Solution 1.
(d) 2√2

Solution 2.
(c) r(x) = x3 + x2 – 1

Solution 3.
(b) 1

Solution 4.
(a) 38808 cm3

Solution 5.
(d) (5, 3)

Solution 6.
(c) 50°

Solution 7.
(a) \(2\sqrt { { a }^{ 2 }+{ b }^{ 2 } }\)

Solution 8
(b) ±10

Solution 9.
If a line divides the two sides of a triangle proportionally then that line is parallel to the third side.

Solution 10.
(d) 6 cms

Solution 11.
(c) 75

Solution 12.
(a) (-3, -1)

Karnataka SSLC Maths Model Question Paper 6 with Answers

Solution 13.
Karnataka SSLC Maths Model Question Paper 5 S13

Solution 14.
Karnataka SSLC Maths Model Question Paper 5 S14

Solution 15.
Karnataka SSLC Maths Model Question Paper 5 S15

Solution 16.
ΔABE = ΔACD (data)
AB = AC …… (1) (CPCT)
AE = AD (CPCT)
⇒ AD = AE ….. (2) (CPCT)
Dividing (1) & (2)
\(\frac { AB }{ AD }\) = \(\frac { AC }{ AE }\) …..(3)
also DAE = BAC ….. (4) (Common angle)
From (3) & (4)
ΔADE ~ ΔABC (SAS Similarity criterian)

Solution 17.
Karnataka SSLC Maths Model Question Paper 5 S17
Substitute the value of y in equation (1)
3x + 2y = 3
⇒ 3x + 2(0) = 3
⇒ 3x + 0 = 3
⇒ 3x = 3
⇒ x = 1
∴ x = 1, y = 0

Solution 18.
Let the age of father = x years
Let the age of the son = y years
Karnataka SSLC Maths Model Question Paper 5 S18
Substitute the value of y in equation 1
x + 2y = 90
⇒ x + 2(20) = 90
⇒ x + 40 = 90
⇒ x = 90 – 40 = 50
Age of the father = 50 years
Age of the son = 20 years

Solution 19.
Area of the shaded region = Area of the circle with AC as diameter – Area of the circle with AB as the diameter
Area of the shaded region
Karnataka SSLC Maths Model Question Paper 5 S19
Karnataka SSLC Maths Model Question Paper 5 S19.1

Solution 20.
Draw the rough figure & calculate OA using Pythagora’s theorem. Then construct the tangent.
Karnataka SSLC Maths Model Question Paper 5 S20

Solution 21.
Let P → (x, y)
A → (3, 6) & B → (-3, 4)
PA = PB (data)
PA2 = PB2
(3 – x)2 + (6 – y)2 = (-3 – x)2 + (4 – y)2
⇒ 9 + x2 – 6x + 36 + y2 – 12y = 9 + x2 + 6x +16 + y2 – 8y
⇒ -6x + 36 – 12y = 6x + 16 – 8y
⇒ -12x – 4y + 20 = 0 (÷4)
⇒ 3x + y – 5 = 0
This is required relation.

Solution 22.
Let us assume that 7 + √5 is rational
Karnataka SSLC Maths Model Question Paper 5 S22
RHS is rational
⇒ LHS is also rational but √5 is irrational.
It is contradictory to the fact that √5 is irrational.
our assumption is wrong
7 + √5 is irrational

Solution 23.
S = Sum of the zeros = 2 + √3 + 2 – √3 = 4
P = Product of the zeroes = (2 + √3)(2 – √3) = 4 – 3 = 1
The required quadratic polynomial is
k[x2 – Sx + P] where ≠ 0 is real
⇒ k[x2 – 4x + 1]

Solution 24.
Let α, β & γ be the zeroes of the required cubic polynomial
Karnataka SSLC Maths Model Question Paper 5 S24

Solution 25.
2x2 – 5x + 3 = 0
It is in the form ax2 + bx + c = 0
a = 2, b = -5, c = 3
Karnataka SSLC Maths Model Question Paper 5 S25

Solution 26.
Karnataka SSLC Maths Model Question Paper 5 S26

Solution 27.
Karnataka SSLC Maths Model Question Paper 5 S27
In the Rt. angled ΔABD
tan 45° = \(\frac { AB }{ BD }\)
1 = \(\frac { AB }{ BD }\)
BD – AB = 7m
AE = 7m
In the Rt. angled ΔAEC
tan60° = \(\frac { CE }{ AE }\)
√3 = \(\frac { CE }{ 7 }\)
CE = 7√3
Height of the tower = CD
= DE + CE
= AB + CE
= 7 + 7√3
= 7(√3 + 1)

Karnataka SSLC Maths Model Question Paper 6 with Answers

Solution 28.
Karnataka SSLC Maths Model Question Paper 5 S28

Solution 29.
Total no. of bulbs = 20
No. of all possible outcomes = 20
Let E1 be the event that the bulb-drawn at random from the lot is defective
No. of outcomes favourable to E1 is 4 Since there are 4 defective bulbs.
Karnataka SSLC Maths Model Question Paper 5 S29

Solution 30.
r + h = 37
T.S.A. of the cylinder = 2πrh + 2πr2 = 2πr (h + r)
But 2πr (h + r) = 1628
Karnataka SSLC Maths Model Question Paper 5 S30

Solution 31.
Karnataka SSLC Maths Model Question Paper 5 S31
Data: A radius OP of a circle is drawn. Through the circle is drawn. Through the end point P of this radius a line AB is drawn the perpendicular to radius OR
To prove: AB is a tangent to the circle at P
Proof: Let Q be any point different from P on this line.
Now OP ⊥AB (data)
OP is the shortest of all the distances from the point O to the line APB.
OP < OQ ⇒ OQ > Radius OP
Q is an exterior point of the circle i.e. Q lies outside the circle, for all positions of Q different from P
Line AB meets the circle only at the point P Line AB is a tangent to the circle at P

Solution 32.
AC : CB = 3 : 6
AB = 4.5 cms.
AC = 1.5 cms
CB = 3 cms.
Karnataka SSLC Maths Model Question Paper 5 S32

Solution 33.
Let the pocket money of A and B respectively = 6x and 5x
Let their expenditure be 4y and 3y respectively.
Monthly savings of A and B = (6x – 4y) & (5x – 3y)
Karnataka SSLC Maths Model Question Paper 5 S33
The pocket money of A = 6x = 6 × 25 = Rs. 150
The pocket money of B = 6x = 5 × 25 = Rs. 125
OR
Let the two digit no. be yx = 10y + x
as reversing the digits it becomes 10x + y
10y + x + 10x + y = 99
⇒ 11x + 11y = 99
⇒ x + y = 9 …….. (1)
According to the second condition
10x + y = 10y + x + 9
⇒ 9x – 9y = 9
⇒ x – y = 1 ……… (2)
Karnataka SSLC Maths Model Question Paper 5 S33.1
Substitute the value of x in equation (2)
x – y = 1
⇒ 5 – y = 1
⇒ y = 4
The required no. is yx = 45

Solution 34.
tan 2A = cot (A – 18°)
cot(90 – 2A) = cot( A – 18) [∴ tan θ = cot (90 – θ)]
90 – 2A = A – 18
90 + 18 = A + 2A
108 = 3A
A = 36°
OR
LHS = tan 45° tan 23° tan 42° tan 67°
= tan(90 – 42) tan 23° tan 42° tan(90 – 23°)
= cot42° tan 23° tan 42° cot 23°
Karnataka SSLC Maths Model Question Paper 5 S34

Solution 35.
Karnataka SSLC Maths Model Question Paper 5 S35
Karnataka SSLC Maths Model Question Paper 5 S35.1
Karnataka SSLC Maths Model Question Paper 5 S35.2

Solution 36.
Karnataka SSLC Maths Model Question Paper 5 S36
Karnataka SSLC Maths Model Question Paper 5 S36.1

Solution 37.
Let the first term and common difference of the AP be a and d respectively.
T3 = 4 ⇒ a + 2d = 4 …… (1)
T9 = -8 ⇒ a + 8d = -8 ……. (2)
Solving Equation 1 & 2
d = -2
Substitute the value of d in equation 1
a + 2d = 4
⇒ a + 2(-2) = 4
⇒ a – 4 = 4
⇒ a = 8
Let the nth term be zero
Tn = 0
a + (n – 1)d = 0
⇒ 8 + (n – 1)(-2) = 0
⇒ 8 – 2n + 2 = 0
⇒ 10 – 2n = 0
⇒ 10 = 2n
⇒ n = 5
5th term of the AP = zero
OR
a. an = 2n + 3
put n = 1
a1 = 2(1) + 3
⇒ a1 = 5
first term = 5
Put n = 2
b. a2 = 2(2) + 3
⇒ a2 =4 + 3
⇒ a2 = 7
d = a2 – a1
⇒ d = 7 – 5
⇒ d = 2
Common difference = 2
c. Tn = (2n + 3)
Put n = 15
T15 = 2(15) + 3 = 30 + 3 = 33
Fifteenth term = 33

Solution 38.
data: ABC is a ∆ in which B = 90°
To Prove: AC2 = AB2 + BC2
Construction: Draw BD ⊥ AC
Proof: In ∆ABC and ∆ABD
A is common
Karnataka SSLC Maths Model Question Paper 5 S38
Karnataka SSLC Maths Model Question Paper 5 S38.1

Solution 39.
Karnataka SSLC Maths Model Question Paper 5 S39

Karnataka SSLC Maths Model Question Paper 6 with Answers

Solution 40.
Karnataka SSLC Maths Model Question Paper 5 S40
Karnataka SSLC Maths Model Question Paper 5 S40.1

Karnataka SSLC Maths Model Question Papers

1st PUC Physics Question Bank Chapter 4 Motion in a Plane

You can Download Chapter 4 Motion in a Plane Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Physics Question Bank Chapter 4 Motion in a Plane

1st PUC Physics Motion in a Plane TextBook Questions and Answers

Question 1.
State, for each of the following physical quantities, if it is a scalar or a vector: volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.
Answer:
Scalar: Volume, mass, speed, density, number of moles, angular frequency.
Vector: Acceleration, velocity, displacement, angular velocity.

Question 2.
Pick cut the two scalar quantities in the following list:
force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, relative velocity.
Answer:
Scalar quantities: Work, current.

Question 3.
Pick out the only vector quantity in the following list:
Temperature, pressure, impulse, time, power, total, path length, energy, gravitational potential, coefficient of friction, charge.
Answer:
Vector quantity: Impulse.

KSEEB Solutions

question 4.
State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:

  1. adding any two scalars,
  2. adding a scalar to a vector of the same dimensions,
  3. multiplying any vector by any scalar,
  4. multiplying any two scalars,
  5. adding any two vectors,
  6. adding a component of a vector to the same vector.

Answer:

  1. No, scalars must represent same physical quantity.
  2. No, vector can be added only to another vector.
  3. YesYes
  4. The two vectors must represent the same physical quantity.
  5. Adding a component of a vector to same vector is meaningless.

Question 5.
Read each statement below carefully false:

  1. The magnitude of a vector is always a scalar,
  2. each component of a vector is always a scalar,
  3. the total path length is always equal to the magnitude of the displacement vector of a particle,
  4. the average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of the average velocity of the particle over the same interval of time,
  5. Three vectors not lying In a plane can never add up to give a null vector.

Answer:

  1. True. The magnitude of a vector gives its length, which is always a scalar.
  2. False. Each component of a vector is a vector by itself.
  3. False. The total path length is always greater than or equal to the magnitude of the displacement vector of the particle.
  4. True. Since the path length is always greater than or equal to the magnitude of the displacement vector of the particle, the average speed is always greater than or equal to the magnitude of the average velocity.
  5. True. If the vectors are not coplanar, their addition will always yield some nonzero component in at least one direction.

Question 6.
Establish the following vector inequalities geometrically or otherwise:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 1
When does the equality sign above apply?
Answer:
a) Geometrical solution:-
Consider the triangle shown in the figure.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 2
which is true by the triangle inequality (Sum of the lengths of any two sides of a triangle is always greater than the length of the third side.)
Equality holds in the case of a degenerate triangle (collinear vectors).
Hence proved.
Analytical solution:-
We must prove that
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 65 sqaure both sides,
To prove that
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 4
i.e., to prove that
a² + b² + 2ab cos θ ≤ a² + b² + 2ab,
( here \(|\overrightarrow{\mathrm{a}}|\) = a , \(|\overrightarrow{\mathrm{b}}|\) = b)
i.e., to prove that
2ab cos θ ≤ 2ab i.e., TPT cosθ ≤ 1
Since the range of cose is [-1, 1], the above inequality is true. The equality holds when θ = 0° (collinear vectors).
Hence Proved.

b) Geometrical solution:
Consider the triangle shown in the figure.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 5
Which is true by the triangle inequality. (Difference in the lengths of any two sides of a triangle is always lesser than the length of the third side.) The equality holds for a degenerate triangle (collinear vectors).
Proved.
Analytical solution:-
We are to prove that
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 6
Square both sides we must prove that
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 7
i.e., To prove that
a² + b² + 2ab cos θ ≥ a² + b² – 2ab
i.e., to prove that 2ab cos θ ≥ – 2 ab
to prove that cos θ ≥ -1
Since the range of cos θ is [-1, 1], the above inequality is true.
The equality holds when θ = 180° (collinear vectors)
Hence Proved.

c) geometrical solution:-
Consider the triangle shown in the figure.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 8
we must prove that \(|\overrightarrow{\mathrm{c}}| \leq|\overrightarrow{\mathrm{a}}|+|\overrightarrow{\mathrm{b}}|\)
or to prove that \(|\overrightarrow{\mathrm{c}}| \leq|\overrightarrow{\mathrm{a}}|+|\overrightarrow{\mathrm{- b}}|\)
(Since \(|-\vec{b}|=|\vec{b}|\)) which is true by the triangle inequality. (Sum of the lengths of any two sides of a triangle is always greater than the length of the third side.)
The equality holds when the triangle is degenerate (collinear vectors).
Hence proved.
Analytical solution:-
We must prove that \(|\vec{a}-\vec{b}| \leq|\vec{a}|+|\vec{b}|\)
Square both sides.
To prove that \((|\vec{a}-\vec{b}|)^{2} \leq(|\vec{a}|+|\vec{b}| |)^{2}\)
i.e., To prove that a² + b² – 2ab cos θ ≤ a² + b² + 2ab
i.e., to prove that -cos θ ≤ 1
Since the range of cos θ is [-1, 1], the range of -cos θ is also [-1, 1].
∴ The above inequality is true. Equality holds when θ = 180° (collinear vectors).
Hence proved.

d) Geometrical solution:-
Consider the triangle shown in the figure.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 9
i.e., To prove that \(|\vec{c}| \geq|| \vec{a}|-|-\vec{b}||\)
(Since \(|-\vec{b}|=|\vec{b}|\))
Which is true by the triangle inequality. (Difference in the lengths of any two sides of a triangle is always lesser than the length of the third side.)
The equality holds in the case of a degenerate triangle (collinear vectors).
Hence proved.
Analytical solution:-
We must prove that \(|\vec{a}-\vec{b}| \geq|| \vec{a}|-| \vec{b}||\)
Square both sides.
To prove that \((|\vec{a}-\vec{b}|)^{2} \geq(|| \vec{a}|-| \vec{b}||)^{2}\)
i.e., To prove that a² + b² – 2ab cose ≥ a² + b² – 2ab
i.e., to prove that cos θ ≤ 1
Since the range of cos θ is [-1, 1],
∴ The above inequality is true.
Equality holds when θ= 0 (collinear vectors).
Hence proved.

KSEEB Solutions

Question 7.
Given \(\vec{a}+\vec{b}+\vec{c}+\vec{d}=\vec{0}\). which of the
following statements are correct:
(a) \(\vec{a}, \vec{b}, \vec{c}\) and \(\vec{d}\) must each be a null vector,

(b) The magnitude of \((\vec{a}+\vec{c})\) equals the magnitude of \((\vec{b}+\vec{d})\),

(c) The magnitude of \(\vec{a}\) can never be greater than the sum of the magnitudes of \(\vec{b}, \vec{c}\) and \(\vec{d}\)

(d) \(\vec{b} + \vec{c}\) must lie in the plane of \(\vec{a}\) and \(\vec{d}\) if \(\vec{a}\) and \(\vec{d}\) if they are not collinear, and in the line of \(\vec{a}\) and \(\vec{d}\), if they are collinear?
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 10
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 11
c) Correct. By an extension of the triangle inequality (sum of the lengths of any (n-1) sides of an n – sided closed polygon is always greater than the length of the remaining side.)

d) Correct. Given \(\vec{a}+\vec{b}+\vec{c}+\vec{d}=\vec{0}\)
⇒ \((\vec{b}+\vec{c}) = -(\vec{a}+\vec{d})\)
If \(\vec{a}\) and \(\vec{d}\) are not collinear, then \((\vec{a}+\vec{d})\) defines a plane. Clearly \((\vec{b}+\vec{c})\) has to lie in this plane for the given condition to hold good. Now if \(\vec{a}\) and \(\vec{d}\) are collinear, \(\vec{a}\) + \(\vec{d}\) is a line and clearly \((\vec{b}+\vec{c})\) must be collinear with this line for the given condition to hold good.

Question 8.
Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 12
Answer:
The magnitude of the displacement vector for each girl = Shortest distance between points P and Q = Diameter PQ
= 2 × 200 m = 400 m
Since girl B skates along the diameter PQ, for her, the magnitude of the displacement vector is equal to the actual length of path skated.

Question 9.
A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 min, what is the
(a) net displacement
(b) average velocity, and
(c) average speed of the cyclist?
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 13
Answer:
a) Since the cyclist returns to the starting point, his net displacement is zero.
b) The average velocity = \(\frac{\text { Net displacement }}{\text { Time taken }}\) = 0
c) Average speed = \(\frac{\text { length of the path }}{\text { Time taken }}\)
= \(\frac{\mathrm{OP}+\mathrm{arc} \mathrm{PQ}+\mathrm{QO}}{10 \mathrm{min}}\)
= \frac{1 \mathrm{km}+1/{4} \times 2 \pi \times 1 \mathrm{km}+1 \mathrm{km}}{1 / 6 \mathrm{hour}}[/latex] = 21.4 km / hour.

Question 10.
On open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 14
Let the origin be O. The path of the motions is as shown in the figure.
a) After 3 rd turn:-
In triangle ABC, AC = AB cos (∠ACB) + BC cos (∠BAC)
∠ACB = ∠BAC = 30° (Since AB = BC and ∠ABC = 120°).
∴ AC = 500 cos 30° + 500 cos 30°
AC = 866 m
In triangle OAC,
OC = \(\sqrt{\mathrm{OA}^{2}+\mathrm{AC}^{2}}\) (Since ∠OAC – 90°)
= \(\sqrt{500^{2}+866^{2}}\) = 1000 m
∴ Magnitude of the displacement = 1 km
Angle with the original direction
= tan-1 \(\left(\frac{\mathrm{AC}}{\mathrm{OA}}\right)\) = tan-1 \(\left(\frac{866}{500}\right)\) = 60°
Path length = 3 × 500 m = 1500 m Clearly magnitude of displacement < path length b) After sixth turn:- The net displacement here = 0 (because the motionist returns to his starting point) Angle with the original direction = 0° Path length = 6 × 500 m = 3 km Clearly, path length > magnitude of the displacement.

c) After the eighth turn:-
In triangle OAB,
OB = \(\sqrt{\mathrm{OA}^{2}+\mathrm{AB}^{2}+2 . \mathrm{OA} . \mathrm{AB} . \cos \theta}\)
= \(\sqrt{500^{2}+500^{2}+2 \times 500 \times 500 \times \cos 60^{\circ}}\)
= 866 m
Net displacement = 866 m
Angle with the original direction
= \(\tan ^{-1}\left(\frac{A B \sin \theta}{O A+A B \cos \theta}\right)\)
= \(\tan ^{-1}\left(\frac{500 \sin 60^{\circ}}{500+500 \cos 60^{\circ}}\right)\)
= \(\tan ^{-1}\left(1/{\sqrt{3}}\right)\)
= 30°
Path length = 8 × 500 m = 4 km
Clearly, path length > Magnitude of net displacement.

Question 11.
A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cab man takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is

  1. the average speed of the taxi,
  2. the magnitude of average velocity?
  3. Are the two equal?

Answer:
1. Average speed of the taxi
= \(\frac{\text { Path length }}{\text { Time taken }}\)
= \(\frac{23 \mathrm{km}}{28 \mathrm{min}}\) = \(\frac{23 \mathrm{km}}{28 \mathrm{28}_{60} \mathrm{h}}\) = 49.3 km h-1

2. Magnitude of Average Velocity
= \(\frac{\text { Net displacement }}{\text { Time taken }}\)
= \(\frac{10 \mathrm{km}}{28 \mathrm{min}}\) = \(\frac{10 \mathrm{km}}{28 \mathrm{28}_{60} \mathrm{h}}\) = 21.4 km h-1

3. Clearly, average speed is greater than magnitude of average velocity. The two are not equal.

KSEEB Solutions

Question 12.
Rain is falling vertically with a speed of 30 ms-1. A woman rides a bicycle with a speed of 10 ms-1 in the north to south direction. What is the direction in which she should hold her umbrella?
Answer:
The direction of θ is given by
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 15
= 18° with the vertical.

Question 13.
A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?
Answer:
Time taken to cross the river
= \(\frac{\text { Width of the river }}{\text { Speed in this direction }}\)
= \(\frac{1 \mathrm{km}}{4 \mathrm{km} \mathrm{h}^{-1}}\) = \(1 / 4\) h = 15 min
Distance along the river covered in this time
= Speed of the river x Time taken to cross the river
= 3 km h-1 × 15 min
= 0.75 km.

Question 14.
In a harbour, wind Is blowing at the speed of 72 km/h, and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/ h to the north, what is the direction of the flag on the mast of the boat?
Answer: When the boat moves, the flag experiences wind blowing at 51 km h-1 in the south direction. Let the flag be at an angle 0 with the East direction.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 16
∴ The flag blows along the east direction.

Question 15.
The ceiling of a long hall is 25 m high. What Is the maximum horizontal distance, that a ball thrown with a speed of 40 m s-1 can go without hitting the ceiling of the hall?
Answer:
Given hm = 25 m , g = 9.8 ms-2 v0 = 40 ms-1
hm = \(\frac{\left(v_{0} \sin \theta_{0}\right)^{2}}{2 g}\)
25 = \(\frac{\left(40 \sin \theta_{0}\right)^{2}}{2 \times 9.8}\)
i.e. (40 sin θ0)² = 490
⇒ 40 sin θ0 = \(7 \sqrt{10}\)
⇒ sin θ0 = \(\frac{7 \sqrt{10}}{40}\)
⇒ θ0 = 33.6°
Maximum horizontal distance
R = \(\frac{v_{0}^{2}}{g}\) Sin 2 θ
= \(\frac{40^{2}}{9.8}\) sin(2 × 33.6°)
= 150.5 m

Question 16.
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?
Answer:
Given R = \(\frac{v_{0}^{2}}{g}\) sin 2 θ = 100 m
Since R is maximum, θ = 45°
R = \(\frac{v_{0}^{2}}{g}\) = 100 m
Maximum height is achieved when throwing angle is 90°.
hmax = \(\frac{\left(v_{0} \sin 90^{\circ}\right)^{2}}{29}\)
= \(\frac{v_{0}^{2}}{g}\) = \(\frac{1}{2} \times \frac{v_{0}^{2}}{g}\)
1/2 × 100 m = 50 m.
∴ Maximum height hmax = 50m.

Question 17.
A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?
Answer:
Radius of the circle r = 80 cm = 0.8m Angular speed
υ = \(\frac{14 \times 2 \pi r}{t}\)
υ = \(\frac{14 \times 2 \times 22 / 7 \times 0.8}{25}\)
υ = 2.816 ms-1
Acceleration = \(\frac{v^{2}}{r}\) = \(\frac{2.816^{2}}{0.8}\)
= 9.9 ms-2
Direction of acceleration: Towards the centre of the circle and along the radius.

Question 18.
An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Answer:
Radius r = 1.00 km = 1000 m
Speed v = 900 km h-1 = 250 ms-1
Centripetal acceleration
a = \(\frac{v^{2}}{r}=\frac{250^{2}}{1000}\)
a = 62.5 ms-2
Acceleration due to gravity g = 9.8 ms-2.
a = \(\frac{62.5}{9.8}\) = 6.4 g

question 19.
Read each statement below carefully and state, with reasons, if it is true or false:

  1. The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre
  2. The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point
  3. The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector

Answer:

  1. False. The statement is true only if the particle is executing circular motion at a uniform speed (no tangential acceleration).
  2. True. The instantaneous velocity is in the direction tangential to the instantaneous path of the particle.
  3. True. Consider the two points and on the circle as shown in the figure.

1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 17
At both these points, acceleration is towards the centre P. The net acceleration is O (equal magnitude but of opposite directions). Similarly, over one cycle, the net acceleration is a null vector.

Question 20.
The position of a particle is given by
\(r=3.0 t \hat{i}-2.0 t^{2} \hat{j}+4.0 \hat{k} m\)
where t is in seconds and the coefficients have the proper units for r to be in meters.

  1. Find the v and an of the particle?
  2. What is the magnitude and direction of velocity of the particle at t = 2.0 s?

Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 18
= 8.54 ms-1
Direction θ = \(\tan ^{-1}\left(\frac{v_{y}}{v_{x}}\right)\)
= \(\tan ^{-1}\left(\frac{-8}{3}\right)\)
= – 69.44° with the horizontal (x – axis)

KSEEB Solutions

Question 21.
A particle starts from the origin at t = 0 s with a velocity of 10.0 J m/s and moves in the x-y plane with a constant acceleration of (8.0 I + 2.0 J) m s-2.

  1. At what time is the x – coordinate of the particle 16 m? What is the y – coordinate of the particle at that time?
  2. What is the speed of the particle at the time?

Answer:
1. Along x – direction,
vx(0) = 0, ax = 8 ms-22
rx(t) = rx(0) + vx(o) t + 1/2 ax
= 0 + 0 + 1/2 × 8 × t² = 4 t²
Given rx(t1) = 16 m
4 t² = 16m
⇒ t = 2 s
∴ At time t = 2.0 s, x – coordinate of the particle = 16 m
Along y – direction,
vy(o) = 10 ms-1, ay = 2 ms-2
ry(t) = ry(0) + vy(0)t +1/2 ay
0 + 10 t + 1/2 × 2 × t²
= 10 t + t²
ry(2) = 10(2) + (2)² = 24 m
∴ y – coordinate at time t = 2.0 s is 24 m

2. Along x – direction,
Vx(t) = vx (0) + axt
= 0 + 8t = 8t
vx(2) = 16 ms-1
Along y – direction,
vy(t) = vy(0) + ayt
= 10 + 2t
vy (2) = 14 ms-1
Speed at time t a 2.0 s is \(\sqrt{\left(v_{x}(2)\right)^{2}+\left(v_{y}(2)\right)^{2}}\)
= \(\sqrt{16^{2}+14^{2}}\) = 21.26 ms-1

Question 22.
I and J are unit vectors along the x- and y-axis respectively. What is the magnitude and direction of the vectors I+ J, and I – J? What are the components of a vector A = 2I+3J along with the directions of i + j and i – j? [You may use graphical method]
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 19
Question 23.
For any arbitrary motion in space, which of the following relations are true:

  1. vaverage = (1/2) (v (t1 + v (t2)))
  2. vaverage = [r(t2) – r(t1)] /(t2 – t2)
  3. v (t) = v (0) + a t
  4. r (t) = r (0) + v (0) t + (1/2) a t²
  5. aaverage = [v(t2) – v(t1)]/(t2 – t1)

(The ‘average’ stands for an average of the quantity over the time interval t1 to t2)
Answer:

  1. False. Average velocity is not the arithmetic mean of the individual velocities.
  2. True. By definition.
  3. False. This is true only if the acceleration \(\overrightarrow{\mathrm{a}}\) is constant.
  4. False. This is true only if the acceleration \(\overrightarrow{\mathrm{a}}\) is constant.
  5. True. By definition.

Question 24.
Read each statement below carefully and state, with reasons and examples, if it is true or false:
A scalar quantity is one that

  1. is conserved in a process
  2. can never take negative values
  3. must be dimensionless
  4. does not vary from one point to another in space
  5. has the same value for observers with different orientations of axes.

Answer:

  1. False. Mass is a scalar quantity and it is not conserved in a nuclear reaction.
  2. False. Angle is a scalar quantity and it can take negative values.
  3. False. Mass, current, etc. are scalar quantities that have dimensions.
  4. False, Temperature, electric potential, etc. are scalar quantities that vary from point to point in a medium.
  5. True. Change of axis does not change a scalar quantity. For example, mass is a scalar quantity that is independent of the axis chosen.

Question 25.
An aircraft is flying at a height of 3400 m above the ground if the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30°, what is the speed of the aircraft?
Answer:
Let O be the observation point. A and B are the initial and final positions of the aircraft.
The angle subtended θ = 30°
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 20
= \(\frac{\text { Distance coverd }}{\text { Time taken }}=\frac{1960 \mathrm{m}}{10 \mathrm{s}}\) = 196 ms-1.

1st PUC Physics Motion in a Plane Additional Exercises Questions and Answers

Question 26.
A vector has magnitude and direction. Does it have a location in space? Can it vary with time Will two equal vectors a and b at different locations in space necessarily have Identical physical effects? Give examples in support of your answer.
Answer:
Yes, a vector has a defined location in space. It may or may not vary with time. For example, \(3 \hat{i}+4 \hat{j}\) is time independent whereas \(4 t \hat{i}-7 \sqrt{t} \hat{k}\) is time dependent.
Two vectors at different locations need not have the same physical effects. Consider the weight of a person in newtons, ‘k’ newtons of weight have different meanings on earth and on the moon, and so they have different physical effects.

Question 27.
A vector has both magnitude and direction. Does it mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector?
Answer:
A quantity that has magnitude and direction need not necessarily be a vector. Consider finite rotations of a particle about an axis. 2 complete rotations bring back the particle to its initial position but these rotations do not add up as per. Parallelogram law of addition. Hence, rotation is not a vector.

Question 28.
Can you associate vectors with

  1. the length of a wire bent into a loop,
  2. a plane area,
  3. a sphere? Explain.

Answer:

  1. No.
  2. Yes. The area \(\overrightarrow{\mathrm{A}}\) can be defined as \(\overrightarrow{\mathrm{A}}\) = \(\overrightarrow{I} \times \overrightarrow{\mathbf{b}}\) where \(\overrightarrow{i}\) and \(\overrightarrow{b}\) are the length and breadth vbctors.
  3. No.

KSEEB Solutions

Question 29.
A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to the fixed, and neglect air resistance.
Answer:
Let ‘u’ be the muzzle speed.
0=30°
Range R = \(\frac{\mathfrak{u}^{2} \sin 2 \mathfrak{\theta}}{\mathfrak{g}}\)
3000 m = \(\frac{\mathrm{u}^{2}}{\mathrm{g}}\) × sin (2 × 30°)
⇒ \(\frac{\mathrm{u}^{2}}{\mathrm{g}}\) = \(\frac{3000}{\sin 60^{\circ}}\) = 3464 m.
Maximum range possible Rmax = \(\frac{\mathrm{u}^{2}}{\mathrm{g}}\)
= 3.46 km
∴ It is not possible to hit a target 5 km away.

Question 30.
A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an antiaircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m s-1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit?
(Take g = 10 m s-2).
Answer:
Plane’s altitude
h = 1.5 km = 1500 m
Plane’s speed
Vp = 720 km/h = 200 ms-1
Shell’s speed
Vs = 600 ms-1
Let the shell be fired at an angle θ with the vertical.
Horizontal distance covered by the shell in time t = Horizontal distance covered by the plane in time t
⇒ Vs sin θ .t = Vpt
⇒ 600 sin θ = 200
⇒ θ = sin-1(1/3) = 19.5°
Minimum altitude to ensure no hit = Maximum height achieved by the shell
= \(\frac{\mathrm{u}^{2} \cos ^{2} \theta}{2 \mathrm{g}}\)
= \(\frac{600^{2} \cos ^{2}\left(19.5^{\circ}\right)}{2 \times 10}\)
= 1600 m
= 16 km.

Question 31.
A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Answer:
Speed of the cyclist v = 27 km/h
= 7.5 ms-1.
Radius of the turn r = 80 m
Centripetal acceleration ac = \(\frac{v^{2}}{r}=\frac{7.5^{2}}{80}\)
= 0.70 ms -2
Tangential retardation ar = \(\frac{0.5}{1}\) = 0.5 ms-2
Net acceleration a = \(\sqrt{a_{c}^{2}+a_{r}^{2}}\)
= \(\sqrt{0.70^{2}+0.5^{2}}\)
= 0.86 ms-2
Angle θ with the tangent is given by
θ = tan-1 \(\left(\frac{\mathrm{a}_{\mathrm{r}}}{\mathrm{a}_{\mathrm{t}}}\right)\)
θ = tan-1\(\left(\frac{0.70}{0.50}\right)\)
θ = 54.5°.

An online projectile motion calculator allows you to compute the velocity, maximum height, and flight parameters at a given time in a fraction of a second.

Question 32.
(a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 21
(b) Shows that the projection angle θ for a projectile launched from the origin is given by
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 22
where the symbols have their usual meaning.
Answer:
(a) Let the horizontal component of the initial velocity be vox and the vertical component be voy.
During motion, the horizontal component is still vox but the vertical component
= Voy – gt
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 23
(b) For the projectile
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 24

1st PUC Physics Motion in a Plane One Mark Questions and Answers

Question 1.
What is the work done by a body in circular motion?
Answer:
Zero.

Question 2.
What is the trajectory of a projectile?
Answer:
Parabola.

Question 3.
What is velocity of projection?
Answer:
The velocity with which the body is projected is called velocity of projection.

Question 4.
What is angle of projection?
Answer:
The angle between the direction of projection and the horizontal drawn at that point is called the angle of projection.

Question 5.
What is range of projectile?
Answer:
The maximum horizontal distance covered by the projectile is called the range.

Question 6.
What is time of flight?
Answer:
The time taken to describe the range is called the time of flight.

Question 7.
When is the range of a projectile is maximum?
Answer:
When the angle of projection is 45° the range of a projectile is maximum.

Question 8.
Why should an athlete throw the javelin or shot put approximately at an angle of 45°?
Answer:
An athlete must throw shot – put or a javelin approximately at an angle 45°to achieve the maximum range.

Question 9.
What is uniform circular motion?
Answer:
Motion of a body along a circular path with constant speed is called uniform circular motion.

KSEEB Solutions

Question 10.
What is the direction of motion (or velocity) of a body in uniform circular motion?
Answer:
The direction of motion of the body at any instant will be along the tangent to the circular path.

Question 11.
Define angular displacement.
Answer:
The angle through which the radius vector rotates is called angular displacement.

Question 12.
Define angular velocity.
Answer:
The rate of angular displacement is called angular velocity.

Question 13.
Define period of revolution of a body.
Answer:
It is the time taken by the body to complete one revolution.

Question 14.
Define frequency of revolution of a body.
Answer:
Number of revolutions completed by the body in one second is called frequency of revolution.

Question 15.
Define centripetal acceleration.
Answer:
In uniform circular motion, the acceleration produced on the body acts along the radius, towards the centre of the circular path and is known as centripetal acceleration or radial acceleration.

Question 16.
What is the source of centripetal force for the planets revolving round the sun?
Answer:
The gravitational force of attraction between the sun and the planets provides the centripetal force for the revolution of the planets round the sun.

Question 17.
What Is banking of roads?
Answer:
Raising the outer edge of the road to get the required centripetal force for a vehicle is known as banking of roads (or track).

Question 18.
What is the work done by centripetal force?
Answer: Zero.

Question 19.
A body moves along a circle of radius 3m. What is the distance travelled. When it completes one circle?
Answer:
Given: radius, r = 3m.
Distance travelled by. the body when it completes one circle = circumference of the circle
= 2 π r = 2 × 3.4 × 3
= 18.84 m

Question 20.
Give an example for three dimen-sional motion.
Answer:
Flight of Aeroplane or movement of gas molecule in space.

Question 21.
Give an example of centripetal force.
Answer:
Electrostatic force between an electron and nucleus provides centripetal force for the revolution of the electron in its orbit.

Question 22.
Which vector quantity becomes zero at highest point of motion of projectile?
Answer:
Vertical component of velocity becomes zero at the highest point.

Question 23.
State the law of parallelogram of two forces.
Answer:
If two forces acting at a point are represented both in magnitude and direction by the two adjacent sides of a parallelogram then their resultant is represented by the diagonal of the parallelogram drawn from the same point.

Question 24.
What will be the effect on the horizontal range of a projectile when its initial speed is doubled keeping the angle the same?
Answer:
Since R α v0², doubling the initial speed will increase the range by a factor of 2² = 4 times.

KSEEB Solutions

Question 25.
What is the minimum number of forces acting on a object in a plane that can produce a zero resultant force?
Answer:
Two. (The forces should have equal magnitude but opposite direction).

Question 26.
A unit vector Is represented by \(\mathbf{a} \hat{\mathbf{i}}+\mathbf{b} \hat{\mathbf{j}}+\mathbf{c} \hat{\mathbf{k}}\). If the values of a and b are
0.6 and 0.8 respectively, find the value of C.
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 25
⇒ c = 0

1st PUC Physics Motion in a Plane Two Mark Questions and Answers

Question 1.
When is a body said to have two-dimensional motion? Give an example.
Answer:
Motion of a particle in a plane is known as two-dimensional motion. Example:

  • A car moving along a zig-zag path on a road.
  • Motion of the planet around the sun in its orbit.

Question 2.
What is a projectile? Give an example.
Answer:
Anybody was thrown into space, that moves under the effect of gravity is called a projectile.
E.g.:

  • A bullet fired from the gun.
  • A javelin thrown by an athlete.

Question 3.
Obtain the relation between angular velocity and linear velocity.
Answer:
Consider a particle moving in a circle of radius ‘r’ with uniform angular velocity ‘ ω ’ and linear speed ‘V’. Let the particle moves from A to B in time ‘t’ seconds through a small distance ‘s’ on the circumference. Let θ be the angular displacement.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 26
Angular velocity ω = \(\frac{\theta}{\mathfrak{t}}\)
Angular displacement θ = \(\frac{s}{\mathfrak{r}}\)
∴ ω = \(\frac{\mathrm{s}}{\mathrm{tr}}\)
But \(\frac{s}{\mathfrak{t}}\) = v, linear velocity
∴ ω = \(\frac{\mathrm{v}}{\mathrm{r}}\) or
v = r ω
linear velocity = radius × angular velocity.

Question 4.
What is angle of banking? Give the expression for it.
Answer:
The angle through which the outer edge of the roads are raised is called the angle of banking. The angle of banking is given by,
θ = tan-1\(\left(\frac{v^{2}}{r g}\right)\)
where m is the mass, v is the velocity and r is the radius.

Question 5.
A body travels one round in a circle of radius ‘R’- What is the

  • displacement and.
  • distance traveled.

Answer:

  • The displacement of a body = 0
  • The distance travelled = 2nR

Question 6.
Write the expressions for the maximum height reached by a projectile, and explain the terms.
Answer:
The expression for the maximum height reached by the projectile is
\(\mathrm{H}=\frac{u^{2} \sin ^{2} \theta}{2 \mathrm{g}}\)
Where u, is the initial velocity of the projectile, θ is the angle between the horizontal & the initial velocity, g is the acceleration due to gravity.

Question 7.
What is banking of roads? Give an example of it.
Answer:
Vehicles while taking a turn cannot incline themselves to one side to get the required centripetal force. To provide necessary centripetal force without slipping. The outer edge of the road is raised over the inner edge.

Question 8.
Why a cyclist bends inward when to Is crossing a curve in a road?
Answer:
He bends inward to get sufficient centripetal force.

KSEEB Solutions

Question 9.
Two equal forces have their resultant equal to either. What is the inclination between them?
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 27
= A
⇒ 2A² (1 + cos θ) = A²
⇒ 1 + cos θ = 1/2
⇒ cos θ = -1/2
⇒ θ = 120°

Question 10.
A cyclist has to bend a little Inwards from his vertical position while turning. Why?
Answer:
Bending provides a component of the normal reaction force from the ground to provide the cyclist the necessary centripetal force for turning. Hence, bending is necessary.

Question 11.
Suppose you have two forces \(\overrightarrow{\mathrm{F}}\) and \(\overrightarrow{\mathrm{F}}\). How would you combine them in order to have resultant forces of magnitudes

  1. zero
  2. 2 \(\overrightarrow{\mathrm{F}}\), and
  3. \(\overrightarrow{\mathrm{F}}\)?

Answer:

  1. If the forces act in opposite directions, the resultant is zero.
  2. If the forces act in the same direction, resultant \(\overrightarrow{\mathrm{R}}\) = 2 \(\overrightarrow{\mathrm{F}}\).
  3. For \(\overrightarrow{\mathrm{R}}\) = 2 \(\overrightarrow{\mathrm{F}}\)
    \(\sqrt{\mathrm{F}^{2}+\mathrm{F}^{2}+2 \mathrm{F} \cdot \mathrm{F} \cdot \cos \theta}\) = F
    ⇒ 2F² (1 + cos θ) = F²
    ⇒ 1 + cos θ = 1/2
    ⇒ θ = 120°

Question 12.
Prove the following statement: “For elevation which exceeds or falls short of 45° by equal amounts, the range Is equal”.
Answer:
At 45° – θ, the range is
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 28
At 45° + θ, the range is
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 29
⇒ R2 = R1 Proved.

Question 13.
What is the distance traveled by a point during time ‘t’, if it moves in x – y plane according to the relation x = a sin cot and y = a(1 – cos cot ωt)?
Answer:
Distance travelled d(t) =
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 30

Question 14.
A lady walking due east on a road with velocity 10 ms-1 encounters rain falling vertically with a velocity of 30ms-1. At what angle should she hold her umbrella to protect herself from the rain?
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 31
⇒ θ = 18°16′ with the vertical direction.

Question 15.
If the vectors of equal magnitude add to either of them by magnitude, what is the angle between them?
Answer:

  1. If the forces act in opposite directions, the resultant is zero.
  2. If the forces act in the same direction, resultant \(\overrightarrow{\mathrm{R}}\) = 2 \(\overrightarrow{\mathrm{F}}\).
  3. For \(\overrightarrow{\mathrm{R}}\) = 2 \(\overrightarrow{\mathrm{F}}\)
    \(\sqrt{\mathrm{F}^{2}+\mathrm{F}^{2}+2 \mathrm{F} \cdot \mathrm{F} \cdot \cos \theta}\) = F
    ⇒ 2F² (1 + cos θ) = F²
    ⇒ 1 + cos θ = 1/2
    ⇒ θ = 120°

Question 16.
A swimmer can swim with velocity of 10km h-1 w.r.t. the water flowing in a river with velocity of 5 km h-1. In what direction should he swim to reach the
point on the other bank just opposite to his starting point?
Answer:
Let the angle of the swimmer be e w.r.t the horizontal.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 66
⇒ cos θ = 1/2
⇒ θ = 60°

Question 17.
The angle between vector \(\overrightarrow{\mathbf{A}}\) and \overrightarrow{\mathbf{B}} is 60°. What is the ratio of \(\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}}[latex] and [latex]|\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}|[latex]?
Answer:
Required ratio [latex]=\frac{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}}{\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}}\)
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 33
= cot θ
= cot (60°)
= \(1 / \sqrt{3}\)

Question 18.
Two forces P=10 N and Q = 15 N are acting at a point making an angle 30° with each other. What is the cross product of P
and Q?
Answer:
P= 10N, Q = 15N, θ =30°
From the equation P × Q = P.QSine
= 10.15 Sin 30° = 75N.

1st PUC Physics Motion in a Plane Three Mark Questions and Answers

Question 1.
Find a unit vector parallel to the vector \(3 \hat{i}+7 \hat{j}+4 \hat{k}\)
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 34
Unit vector parallel to \(\overrightarrow{\mathrm{a}}=\frac{\overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}}|}\)
\(=\frac{3 \hat{i}+7 \hat{j}+4 \hat{k}}{\sqrt{74}}\)

Question 2.
What is a projectile? Show that the path followed by an oblique projectile Is a parabola.
Answer:
Anybody was thrown into space, that moves under the effect of gravity is called a projectile.
E.g.:
1. A bullet fired from the gun.
2. A javelin thrown by an athlete., Consider a particle thrown up at an angle θ to the horizontal with a velocity u. The velocity of projection can be resolved into two component u cos θ and usinq along OX and OY respectively.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 35
The horizontal component remains constant while the vertical component is affected by gravity.
Let P(x, y) be the position of the projectile after ‘t’ seconds.
Then distaince travelled ‘x’ along OX is given by,
x = horizontal component of the velocity time
x = u cos θ × t ………… (1)
The distance travelled along the vertical direction in same time‘t’ is,
y = (u sin θ) t – 1/2 gt² ……………… (2)
Substituting, t = \(\frac{x}{u \cos \theta}\) from (1) in (2),
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 67
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 36…………………..(3)
where a = tan θ, \(\mathrm{b}=\frac{\mathrm{y}=\mathrm{gax}-\mathrm{bx}^{2}}{2 \mathrm{u}^{2} \cos ^{2} \mathrm{\theta}}\) are constant for given values of θ, u and g.
The equation (3) represents a parabola. Therefore trajectory of a projectile is a parabola.

Question 3.
Establish a relation between linear velocity and angular velocity in a uniform circular motion and explain the direction of the velocity.
Answer:
The distance ‘s’ covered by a body travelling in an arc of radius Y and turning its radial line by ‘θ’ is given by
s = r θ
Differentiating both sides w.r.t. time, we have
\(\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{r} \frac{\mathrm{d} \theta}{\mathrm{dt}}\)
i.e., v = rw
or Linear velocity = radius × angular velocity.
At each point the body moves along the tangent.
The presence of centripetal force
F =\(\frac{m v^{2}}{r}\) makes the body to travel in the circular path.
Thus, the direction of velocity is always along the tangent at any point in the circular path.

Question 4.
Find the magnitude and direction of the resultant of two forces p and Q in terms of their magnitudes and angle Q between them.
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 37
P and Q are two forces acting at a point O; making an angle θ. They are represented by OA and OB. the diagonal OC of the parallelogram OBCA represents the resultant R.
Draw CD perpendicular to extended line of OA. In the right angle triangle ODC.
OC² = OD² + CD²
= (OA + AD)² + CD²
= OA² + 2OA AD + AD² + CD²
= OA² + 2OA. AC cos θ + AC²
R² = P² + 2PQcos θ + Q² v
From triangle ADC,
AC² = AD² + CD² & \(\frac{\mathrm{AD}}{\mathrm{AC}}\) = cosq
Magnitude of the resultant
R = \(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cdot \cos \theta}\)
To find the direction:
If a is the angle made by R with P, then from the right angled triangle ODC
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 38

Question 5.
In long jump, does it matter how high you jump? What factors determine the span of the jump?
Answer:
In order to increase the span of the jump, it is necessary for the horizontal component of the velocity to be more than the vertical component. The height of the jump does not matter.
The factors that determine the span of the jump are:

  1. angle of elevation at the time of jumping.
  2. Speed of the runner at the time of jumping.

Question 6.
Determine a unit vector which Is per-pendicular to both \(\vec{A}=2 \hat{i}+\hat{j}+\hat{k}\) and \(\overrightarrow{\mathbf{B}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\)
Answer:
The unit vector \(\hat{n}\) is given by
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 39

Question 7.
An accelerating train is passing over a high bridge. A stone is dropped from the train at an instant when Its speed is 10 ms-1 and acceleration is 1ms-2. Find the horizontal and vertical components of the velocity and acceleration of the stone one second after it is dropped.Take g = 10 ms-2.
Answer:
Horizontal acceleration = 0 (because there is no force acting on the stone in the horizontal direction to provide any acceleration)
Horizontal velocity = 10 ms-1.
Vertical acceleration = g = 10ms-2.
Vertical velocity = ut + 1/2 gt²
= 0(1) + 1/2 (10)(1)²
= 5 ms-1.

Question 8.
A projectile is projected with velocity ‘u’ making an angle θ with the horizontal direction. Find:

  1. Time of flight
  2. Horizontal range

Answer:
1. Let H be the maximum height reached by the projectile in time t1
For vertical motion,
The initial velocity = usin θ
The final velocity = 0
Acceleration = – g
using, v² = u² + 2as
0 = u² sin2 θ – 2gH
2gH =u²sin2 θ
u² sin² θ
H = \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{g}}\)

2. Let t, be the time taken by the projectile to reach the maximum height H.
For vertical motion,
initial velocity =usin0
Final velocity at the maximum height = 0
Acceleration a = – g
Using the equation v = u + at1
0 = usin θ – gt1
gt1 = usin θ
t1 = \(\frac{\mathrm{usin} \theta}{\mathrm{g}}\)
Let t2 be the time of descent.
But t1 = t2
i.e. time of ascent = time of descent.
Time of flight T = t1 + t2 = 2t1
∴ T = \(\frac{2 u \sin \theta}{g}\)

Question 9.

  1. Show that for two complementary angles of projection of a projectile with the same velocity, the horizontal ranges are equal.
  2. For what angle of projectile is the rangle maximum?
  3. For what angle of projection of a projectile are the horizontal range and height attained by the projectile equal?

Answer:
1. For projection angle θ,
R1 = \(\frac{u^{2} \sin 2 \theta}{g}\)
For projection angle 90° – θ,
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 40
But sin (180° – 2θ) = sin2θ
∴ R2 = \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) = R1

2. Range is maximum when projection angle = 45°.

3. Given R = H
⇒ \(\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}}\) = \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{g}}\)
⇒ 2 sin θ cos θ = \(\frac{\sin ^{2} \theta}{2}\)
⇒ cot θ = 1/4
θ = cot-1(1/4)
⇒ θ = 75.96°.

KSEEB Solutions

Question 10.
What is meant by centripetal acceleration? Derive the formula for centripetal acceleration.
Answer:
In uniform circular motion, the acceleration produced on the body acts along the radius, towards the centre of the circular path. This acceleration is known as centripetal acceleration or radial acceleration.
Let a particle of mass m be moving around the circle of radius r with a uniform speed ‘v’.
Let the particle moves from A to B in a time ‘t’ seconds covering a small angle θ.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 41
At the point A the velocity of the particle is along the tangent AC and therefore there is no component along AO (i.e. along radius).
At the point B the velocity of the particle is along BD. This can be resolved into two components v cos θ and v sin θ. Since θ is small sin θ ≅
θ and cos θ ≅ 1.
∴ change in velocity parallel to AC
= v cos θ – v = v -v = 0. ∵ cos θ ≅ 1.
∴ There is no acceleration along the tangent,
Change in velocity parallel to OA is
= vsin θ – 0 = v θ ( sinθ ≅ θ)
∴ Acceleration along AO = \(\frac{\mathrm{v} \theta}{\mathrm{t}}\)
i.e., centripetal acceleration = \(\frac{\mathrm{v} \theta}{\mathrm{t}}\) = v ω
∴ centripetal force = mass × centripetal acceleration
F=m v ω
F = mw²r = \(\frac{m v^{2}}{r}\) ∵v = ω r

Question 11.
From the top of a tower of 100m height, a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velocity of 25ms-1. Find when and where the two balls will meet. Take g = 9.8 ms-2
Answer:
Let AC = x
Then BC = 100 – x
x = ut + 1/2 g t²
x = 4.9 t² …………… (1)
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 42
BC = ut – 1/2 g t²
= 25 t – 1/2 × 9.8 t²
100 – x = 25 t – 4.9 t² ………………. (2)
Substituting 4.9 t² = x from (1) in (2)
100 – 2 = 25 t – x
⇒ t = 4 seconds
x = 4.9 t²
= 4.9 × 4²
= 78.4 m from the top or 21.6 m from the ground.

1st PUC Physics Motion in a Plane FourFive Marks Questions and Answers

Question 1.
Show that the trajectory of a projectile is a parabola.
Answer:
Consider a particle thrown up at an angle q to the horizontal with a velocity u. The velocity of projection can be resolved into two component u cos θ and u sin θ along OX and OY respectively.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 43
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 44
The horizontal component remains constant while the vertical component is affected by gravity.
Let P(x, y) be the position of the projectile after ‘t’ seconds.
Then distance travelled ‘x’ along OX is given by,
x = horizontal component of the velocity × time
x = ucos θ × t …………………. (1)
The distance travelled along the vertical direction in same time ‘t’ is,
y = (usin θ) t – 1/2gt² ………… (2)
Substituting, t = \(\frac{x}{u \cos \theta}\) from (1) in (2),
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 45
y = ax – bx² …………………… (3) where
a = tan θ, \(\mathrm{b}=\frac{\mathrm{y}=\mathrm{gax}-\mathrm{bx}^{2}}{2 \mathrm{u}^{2} \cos ^{2} \mathrm{\theta}}\), are constants
for given values of θ, u and g.
Equation (3) represents a parabola. Therefore trajectory of a projectile is a parabola.

Question 2.
Derive an expression for the maximum height reached, time of flight and range of a projectile.
Answer:
(i) Let H be the maximum height reached by the projectile in time t1 For vertical motion,
The initial velocity = usin θ
The final velocity = 0
Acceleration = – g
∴ using, v² = u² + 2as
0 = u²sin² θ – 2gH
2gH = u²sin² θ
H = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)

(ii) Let t, be the time taken by the projectile to reach the maximum height H.
For vertical motion,
initial velocity = usin θ
Final velocity at the maximum height = 0
Acceleration a = – g
Using the equation v = u + at1
0 = usin θ – gt1
gt1 = usin θ
t1 = \(\frac{\mathrm{usin} \theta}{\mathrm{g}}\)
Let t2 be the time of descent.
But t1 = t2
i.e. time of ascent = time of descent.
∴ Time of flight T = t1 + t2 = 2t1
∴ T = \(\frac{2 u \sin \theta}{\dot{g}}\)

(iii) Let R be the range of the projectile in a time T. This is covered by the projectile with a constant velocity ucos θ.
Range = horizontal component of
velocity × Time of flight
i.e, R = ucos θ. T
R = ucos θ. \(\frac{2 u \sin \theta}{g}\)
R= \(\frac{u^{2} \sin 2 \theta}{g}\)
∵ 2 sin θ.cos θ = sin2 θ.

Question 3.
Derive an expression for the centripetal force.
Answer:
Let a particle of mass m be moving around the circle of radius r with a uniform speed ‘V’.
Let the particle moves from A to B in a time ‘t’ seconds covering a small angle θ.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 46
At the point A the velocity of the particle is along the tangent AC and therefore there is no component along AO (i.e. along radius).
At the point B the velocity of the particle is along BD. This can be resolved into two components v cos θ and v sin θ. Since θ is small sin θ ≅
θ and cos θ ≅ 1.
∴ change in velocity parallel to AC
= v cos θ – v = v -v = 0. ∵ cos θ ≅ 1.
∴ There is no acceleration along the tangent,
Change in velocity parallel to OA is
= vsin θ – 0 = v θ ( sinθ ≅ θ)
∴ Acceleration along AO = \(\frac{\mathrm{v} \theta}{\mathrm{t}}\)
i.e., centripetal acceleration = \(\frac{\mathrm{v} \theta}{\mathrm{t}}\) = v ω
∴ centripetal force = mass × centripetal acceleration
F=m v ω
F = mw²r = \(\frac{m v^{2}}{r}\) ∵v = ω r.

KSEEB Solutions

Question 4.
Define centripetal force and give example.
Answer:
In uniform circular motion the force on the rotating body acts along the radius towards the centre of the circular path and is known as centripetal force.
Eg:

  1. In the case of a stone rotated round the circle by means of a string, the centripetal force is provided by the tension in the string.
  2. The force of gravitational attraction towards the sun is the centripetal force keeping the planets orbiting round the sun.
  3. In an atom the electrons revolve round the nucleus in the circular orbit the centripetal force is provided by the electrostatic force of attraction between the nucleus and the electrons.

Question 5.
What is the angle of banking for a curve road radius 180m suitable for a maximum speed of 30m? Calculate the maximum speed to be maintained when the angle of banking is
θ = 30° (g = 9.8ms1) (Rural 2005)
Answer:
Radius of the curve road, r =180m, Maximum speed v=30ms-1
Angle of banking 0=? Acceleration due to gravity, g =9.8ms-2
From the equation,
tan θ = \(\frac{v^{2}}{r g}\) = \(\frac{30^{2}}{180 \times 9.8}\) = \(=\frac{900}{1764}\)
= 0.5102
θ = tan-1(0.5102)
= 27° 1°
When the angle of banking θ = 30°,
From the equation tan θ = \(\frac{v^{2}}{r g}\)
tan θ 30° = \(\frac{v^{2}}{180 \times 9.8}\)
v2 = 180 × 9.8 × tan30°
= 1764 × 0.5774 = 1018.5
v = 32ms-1

Question 6.
A bullet fired from rifle attains a maximum height of 50m and crosses a range of 200m. Find the angle of projection.
Answer:
Maximum height attained by the bullet, H=50m,
Horizontal range R=200m,
Angle of projection 0 =?
From the equation H = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\)
50 = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\) ……………….(1)
From the equation R = \(\frac{u^{2} 2 \sin \theta \cos \theta}{g}\)
200 = \(\frac{u^{2} 2 \sin \theta \cos \theta}{g}\) ………………. (2)
Divide equ (1) by equ (2), we have
\(\frac{50}{200}=\frac{\sin \theta}{4 \cos \theta}\)
i.e \(\frac{1}{4}\) = \(\frac{1}{4}\) tan θ Thus tan θ = 1
θ = tan-1(1.0000) = 45°.

Question 7.
A train of mass 10,000 kg moving at 72km ph rounds a curve whose radius of curvature is 200m. What is its acceleration? What is the centripetal force?
Answer:
m = 10,000kg, v = 72 km/h
= \(\frac{72 \times 1000}{60 \times 60}\) = 20m/s
r = 200 m.
Acceleration a = v ω
= \(v\left(\frac{v}{r}\right)\)
= \(\frac{v^{2}}{r}=\frac{(20)^{2}}{200}\) = 2m/s2
Centripetal force, F = \(\frac{m v^{2}}{r}\)
= \(\frac{10000 \times(20)^{2}}{200}\) = 20000 N.

KSEEB Solutions

Question 8.
An object is projected with a velocity of 60ms-1 in a direction making an angle of 60° with the horizontal. Find

  1. the maximum height
  2. the time taken to reach maximum height
  3. the horizontal range

Answer:
u = 60m/s, θ = 60°.
1. Maximum height
H = \(\frac{u^{2} \sin ^{2}(\theta)}{2 g}\) = \(\frac{60^{2} \sin ^{2}(60)}{2 \times 9.8}\) = 137.74 m.

2. Time to reach maximum height
t = \(\frac{u \sin (\theta)}{g}\) = \(\frac{60 \times \sin (60)}{9.8}\) = 5.30s.

3. Range
R = \(\frac{u^{2} \sin (2 \theta)}{g}\) = \(\frac{60^{2} \sin (2 \times 60)}{9.8}\) = 318.12 m

Question 9.
Drive an expression for magnitude and direction of the resultant of two forces acting at a point.
Answer:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 47
P and Q are two forces acting at a point O, making an angle θ. They are represented by OA and OB. the diagonal OC of the parallelogram OBCA represents the resultant R.
Draw CD perpendicular to extended line of OA. In the right angle triangle ODC.
OC² = QD² + CD²
= (OA + AD)² + CD²
= OA² + 2.0A AD + AD² + CD²
= OA² + 20A. AC cosq + AC²
R² = P² + 2PQcos θ + Q²
∵ From triangle ADC,
AC² = AD²+ CD² & \(\frac{A D}{A C}\) = cos θ
Magnitude of the resultant
R = \(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cdot \cos \theta}\)
To find the direction:
If a is the angle made by R with P, then from the right angled triangle ODC
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 68

Question 10.
A projectile shot at an angle of 60° above the horizontal ground strikes a vertical wall 30 m away at a point 15m above the ground. Find the speed with which the projectile was launched and the speed with which it strikes the wall. Take g = 10 ms-2
Answer:
Let the initial velocity of the projectile be u
Horizontal component = ux
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 49
Horizontal distance, d = ux t
i.e., 30 = \(\frac{\mathrm{u}}{2}\) t
⇒ t = \(\frac{60}{\mathrm{u}}\) ………… (1)
Vertical component uy = usin60°
= \(\frac{\mathrm{u} \sqrt{3}}{2}\)
Vertical displacement s = uyt – 1/2 g t²
15= \(\frac{\mathrm{u} \sqrt{3}}{2}\) t – 1/2 × 10 × t²
Substitute t = 60/u from (1),
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 69
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 50
⇒ u = 22.07 ms-1
∴Speed of launch = 22.07 ms-1
At the time of striking the wall,
vx = ux = 22.07 cos 60°
= 11.04 ms-1
vy = uy – gt
= 22.07 sin 60° – 10 × \(\left(\frac{60}{22.07}\right)\)
vy = -8.07 ms-1
Resultant speed v = \(\sqrt{v_{x}^{2}+v_{y}^{2}}\)
= \(\sqrt{11.04^{2}+(-8.07)^{2}}\)
= 13.68 ms-1</sup

Question 11.
Define projectile. Show that the path of projectile is a parabola. Find the angle of projection at which the horizontal range and maximum height of the projectile are equal.
Answer:
Consider a particle thrown up at an angle q to the horizontal with a velocity u. The velocity of projection can be resolved into two component u cos θ and u sin θ along OX and OY respectively.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 51
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 52
The horizontal component remains constant while the vertical component is affected by gravity.
Let P(x, y) be the position of the projectile after ‘t’ seconds.
Then distance travelled ‘x’ along OX is given by,
x = horizontal component of the velocity × time
x = ucos θ × t …………………. (1)
The distance travelled along the vertical direction in same time ‘t’ is,
y = (usin θ) t – 1/2gt² ………… (2)
Substituting, t = \(\frac{x}{u \cos \theta}\) from (1) in (2),
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 53
y = ax – bx² …………………… (3) where
a = tan θ, b = \(\frac{\mathrm{g}}{2 \mathrm{u}^{2} \cos ^{2} \theta}\), are constants
for given values of θ, u and g.
Equation (3) represents a parabola. Therefore trajectory of a projectile is a parabola.

1st PUC Physics Motion in a Plane Numerical Problems Questions and Answers

Question 1.
A ball is thrown Into air with a speed of 62 ms-1 at an angle of 45° with the horizontal. Calculate

  1. the maximum height attained
  2. the horizontal range
  3. the time of flight and
  4. the velocity of the ball after 4 seconds.

Solution:
1. Maximum height, H = \(\frac{u^{2} \sin ^{2} \theta}{2 g}\) Here,
u = 62 ms-1, θ = 45°, g = 9.8 ms-2
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 54
= \(\frac{62 \times 62}{4 \times 9.8}\) = 98.06 m.

2. Horizontal range. R is given by,
R = \(\frac{u^{2} \sin 2 \theta}{g}\)
= \(\frac{62 \times 62 \times \sin 90}{9.8}\)
= \(\frac{62 \times 62}{9.8}\) = 392.2m.

3. The time of flight,
T = \(\frac{2 u \sin \theta}{g}\)
= \(\frac{2 \times 62 \times \sin 45}{9.8}\)
= \(\frac{2 \times 62}{\sqrt{2} \times 9.8}\) = 8.95 s.

4. The horizontal component of the velocity remains constant throughout.
∴ Horizontal component of the velocity after 4 seconds vx = ucos θ
Here u = 62 ms-1 and θ = 45°
∴ vx = 62 cos45° = \(\frac{62}{\sqrt{2}}\) = 44.3 s.
The vertical component of the velocity changes due to the acceleration due to gravity.
We have vy = uy + at,
Here, uy = usin θ; a = -g; t = 4 s;
From vy = usin θ – 4g
∴ vy = 62 sin45 – 9.8 × 4
= \(\frac{62}{\sqrt{2}}\) -39.2 = 4.64 ms-1
Magnitude of the resultant velocity after 4 seconds is,
v = \(\sqrt{v_{x}^{2}+v_{y}^{2}}\)
= \(\sqrt{(44.3)^{2}+(4.64)^{2}}\) = 44.5 ms-1
Direction of the velocity is given by,
tan a = \(\frac{v_{y}}{v_{x}}\) = \(\frac{4.64}{44.3}\) = 0.104
∴ a = tan-1(0.104) = 5°5’

Question 2.
A bullet is fired at an angle of 60° with the vertical with certain velocity hits the ground 3 km away is it possible to hit the target 5 km away by adjusting the angle of projection? If not, what must be the velocity of projection for the same angle of projections?
Solution:
Let ‘u’ be the velocity of projection.
Angle of projection ‘θ’ = 90 – 60° = 30°
Horizontal Range,
R = \(\frac{u^{2} \sin 2 \theta}{9}\) = \(\frac{u^{2} \sin 60}{g}\)
∴ u2 = \(\frac{\mathrm{Rg}}{\sin 60}\) …………… (1)
For a given velocity, the maximum range attained is given by making sin2 θ =1, and
Rmax \(\frac{u^{2}}{g}\).
Substituting for u² from (1), we have
Rmax = \(\frac{\mathrm{R}}{\sin 60}\)
= \(\frac{3}{0.866}\) = 3.46km
Since the maximum range is less than 5 km, it is not possible to hit the target 5km away.
From equation (1), we have,
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 55
= 56582
∴ u = 237.9 ms-1
Thus the bullet will hit the target if it is projected with a velocity of 237.9 ms-1

Question 3.
A stone of mass 0.5 kg is tied to one end of a string and whirled along the horizontal circle of 1.2 m radius, If the period of rotation is 5 s, calculate the tension in the string. If breaking tension is 1.5 N, calculate the maximum speed with which it is rotated and also calculate the corresponding period of rotation.
Solution:
When the mass is rotated along a horizontal circle, the centripetal force is equal to the tension of the string.
Hence, T= mw²r,
Here, m = 0.5 kg,
w = \(\frac{2 \pi}{T}\) = \(\frac{2 \pi}{5}\)
r = 1.2 m.
∴ T = 0.5 × \(\left(\frac{2 \pi}{5}\right)^{2}\) × 1.2
= \(\frac{0.5 \times 4 \pi^{2} \times 1.2}{25}\)
= 0.9475 N
The maximum tension which the string can withstand without breaking is 1.5 N and the corresponding velocity ‘v’ is given by, 1.5 = \(\frac{m v^{2}}{r}\)
∴ v2 = \(\frac{1.5 \times r}{m}\) or
v = \(\sqrt{\frac{1.5 \times 1.2}{0.5}}\)
= 1.89 ms-1
The corresponding angular velocity is given by,
w = \(\frac{\mathrm{v}}{\mathrm{r}}\) = \(\frac{1.89}{1.2}\)
= 1.575 rads-1
∴ The period of rotation,
T = \(\frac{2 \pi}{1.575}\) = 3.99 s.

Question 4.
At what angle should a cyclist lean over from the vertical while negotiating a curve of radius 58m with a speed of 15 km/hour?
Solution:
The angle of banking is given by,
tan θ = \(\frac{v^{2}}{r g}\)
Here, v = \(\frac{15000}{60 \times 60}\) = 4.17 ms-1
∴ tan θ = \(\frac{(4.17)^{2}}{58 \times 9.8}\) = 0.03059
θ = tan-1(0.03059)
= 1°45’.

KSEEB Solutions

Question 5.
A train is moving round a curve of 52m radius and the distance between the rails is 1.2 m by how much should the outer rails be raised above the inner one so that a train running at the rate of 45 kmhr-1 may not skid.
Solution:
The angle of banking is given by,
tan θ = \(\frac{v^{2}}{r g}\)
Here, v = \(\frac{45000}{60 \times 60}\) = 12.5 ms-1 and
r = 52 m
∴ tan θ = \(\frac{(12.5)^{2}}{52 \times 9.8}\) = 0.3066
θ = tan-1(0.3066)
= 17°2’
If ‘h’ is the height of the outer rail above the inner one, then
h = 1.2’sin 17.05°
= 0.352 m.

Question 6.
A golf ball leaves the tee with a velocity of 50ms-1 at an angle of 300 with the horizontal. Find its

  1. time of flight
  2. Range of projectile and
  3. The velocity with which it hits the ground at the end of its flight.

Answer:
u = 50m/s, θ = 30°.
1. Time of flight
T = \(\frac{2 u \sin (\theta)}{g}\) = \(\frac{2 \times 50 \times \sin (30)}{9.8}\) = 5.10 s

2. Range R = \(\frac{u^{2} \sin (2 \theta)}{g}\) = \(\frac{50^{2} \sin \left(2 \times 30^{\circ}\right)}{9.8}\) = 220.92 m

3. Velocity at the end = velocity of projection = 50 m/s

Question 7.
The greatest and the least resultants of two forces acting at a point are 5 N and1N respectively. Find the forces. What is the resultant of these two forces when they act at an angle of 50°?
Solution:
Let P and Q be the two forces. The greatest resultant is
P + Q = 5 …………… (1)
The minimum resultant is,
P – Q = 1…………….. (2)
Adding (1) and (2)
2P = 6
P =3N
∴ Q = 2 N
When they are acting at an angle,
θ =50°
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 56
= 4.55 N
The direction of the resultant is,
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 57
= 0.3575
∴ a = 19.67° = 19° 40′
Therefore the resultant is making an angle of 19°40′ with 3N.

Question 8.
Four forces of magnitudes 2N, 3N, 4N, and 5N are acting on a body at a point are inclined at 30°, 60°, 90°, and 120° respectively with the horizontal. Find their resultant.
Solution:
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 58
Given, P1 =2N, P2 = 3N
P3 = 4N and P4 = 5N
These forces can be resolved into components along the X and Y directions.
Rx= P1 cos θ1 + P2 cos θ2 + P3 cos θ3 + P4 cos θ4
= 2cos 30° + 3cos 60 + 4cos 90 + 5 cos 120
= 2\(\frac{\sqrt{3}}{2}\) + 3 × \(\frac{1}{2}\) + 4 × 0 + 5 × \(\frac{1}{2}\) = 0.732 N

Ry= P1 sin θ1 + P2 sin θ2 + P3 sin θ3 + P4 sin θ4
= 2 sin 30 + 3 sin 60 + 4 sin 90 + 5 cos 120
= 2 × \(\frac{1}{2}\) + 3 × \(\frac{\sqrt{3}}{2}\) + 4 × 1 + 5 × \(\frac{\sqrt{3}}{2}\)
= 1 + 1.5\(\sqrt{3}\) + 4 + 2.5\(\sqrt{3}\)
= 1 + 2.59 + 4 + 4.33
= 11.92 N
∴ Resultant force R = \(\sqrt{\mathrm{R}_{\mathrm{x}}^{2}+\mathrm{R}_{\mathrm{Y}}^{2}}\)
= Vo.7322 +11.92 2
= 11.96 N.

Question 9.
The resultant of two forces acting at 60° is \(\sqrt{13}\) N. When the same forces act at 90° the resultant is \(\sqrt{10}\) N. Find the magnitude of the two forces.
Solution:
Let P and Q be the two forces. When the angle between them is 60°, resultant force is \(\sqrt{13}\) N. Thus
\(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cos 60^{\circ}}\) = \(\sqrt{13}\)
i.e.\(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+\mathrm{PQ}}\) = \(\sqrt{13}\)
P + Q + PQ = 13 ……………(1)
When the angle is 90° resultant force is
\(\sqrt{10}\) N.
\(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}+2 \mathrm{PQ} \cos 90^{\circ}}\) = \(\sqrt{10}\)
or \(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}}\) = \(\sqrt{10}\)
P2+ Q2 =10 ……………….. (2)
Substituting (2) in (1)
10 + P = 13
PQ = 3 ……………………(3)
(P + Q)² = P²+ Q² + 2PQ
= 10 + 2 × 3 = 16
P + Q =4 ………………. (4)
(P – Q)² = P² + Q² – 2PQ
= 10 – 6 = 4
P – Q = 2 ………………. (5)
2P =6; 2Q = 2
P = 3N and Q = 1N.

KSEEB Solutions

Question 10.
The sum of the two forces is 8 kg wt and the magnitude of the resultant which is at right angles to the smaller force is 4 kgwt. Find the two forces.
Solution:
Let P be the smaller force.
P + Q = 8 kg wt.,
a = 90°, R = 4 kg wt.,
tan α = \(\frac{\mathrm{Q} \sin \theta}{\mathrm{P}+\mathrm{Q} \cos \theta}\)
tan 90° = \(\frac{\mathrm{Q} \sin \theta}{\mathrm{P}+\mathrm{Q} \cos \theta}\)
∞ = \(\frac{\mathrm{Q} \sin \theta}{\mathrm{P}+\mathrm{Q} \cos \theta}\)
⇒ P + Q cos θ = 0
cos θ = \(-\frac{P}{Q}\)
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 59
Q² – P²= 16
(Q-P)(Q+P) = 16
(Q-P) × 8 = 16
Q – P = 2
P + Q = 8
∴ 2Q = 10 ⇒ Q = 5 kg wt. and P = 3 kg wt.

Question 11.
The resultant of two forces acting at an angle of 150° is perpendicular to the smaller force. If the smaller force is 12\(\sqrt{3}\) N, find the greater force and its resultant.
Solution:
θ = 150°,
Let P be the smaller force P = 12\(\sqrt{3}\) N
a = 90°, Q = ?, R = ?
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 60

Question 12.
Two equal, non parallel forces acting at a point on a body. The square of the resultant is found to be three times the product of the forces, what is the angle between them.
Answer:
Given, P = Q & R² = 3PQ = 3P²
we have R² = P² + Q² + 2PQ cos θ
i.e. 3P²= P² + P² + 2P² cos θ
i.e. 3P² = 2P² + 2P² cos θ
1P² = 2P²cos θ
\(\frac{1}{2}\) = cose
⇒ θ = cos-1(\(\frac{1}{2}\)) = 60°.

Question 13.
The greatest and least resultant of two forces acting at a point are 17N and 7N respectively. Find the two forces. If these forces act at right angles, find the magnitude of their resultant.
Answer:
For maximum resultant
P + Q=17N ……………….. (1)
For minimum resultant
P – Q = 7N ……………….. (2)
Eq(1) + (2) gives 2P = 24 Or P = 24/2 = 12N.
From (1), Q = 17 – 12 = 5N
When P and Q are in perpendicular directions,
R = \(\sqrt{\mathrm{P}^{2}+\mathrm{Q}^{2}}\)
= \(\sqrt{12^{5}+5^{2}}\) = 13N(∵ cos 90° = 0).

Question 14.
Two equal forces are acting at a point and angle between them in 60°. Calculate magnitude of Resultant of these forces.
Answer:
Q = P and θ = 60°.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 61

1st PUC Physics Motion in a Plane Hard Questions and Answers

Question 1.
The total speed v1 of a projectile at its greatest height is \(\sqrt{\frac{6}{7}}\) of its speed v2 at half its greatest height. Find the angle of projection.
Answer:
Velocity at the highest point = ucos θ = v1
Hmax = \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{g}}\)
Using v = u – 2gh,
At h = \(\frac{\mathrm{H}_{\mathrm{max}}}{2}\), vertical component of v2 is
given by,
v2y2 = u2sin2θ – 2g × \(\frac{\mathrm{H}_{\mathrm{max}}}{2}\)
v2y2 = u2 sin2θ – \(\mathrm{g}\left(\frac{\mathrm{u}^{2} \sin 2 \theta}{2 \mathrm{g}}\right)\)
⇒ v = \(\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2}\)
Horizontal component v2 is
v2x = v1 = ucosθ
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 62
Squaring both sides,
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 70
⇒ 7 cos²θ = 6 cos²θ + 3 sin² θ
⇒ cos²θ = 3 sin²θ
⇒ tan²θ = 1/3
⇒ tan θ = ± \(1 / \sqrt{3}\)
⇒ θ = ± 30°
θ = 30°
(because negative angle is not possible)

Question 2.
Prove that the path of one projectile as seen from another projectile is a straight line.
Answer:
The coordinates of first projectile as seen from the second projectile are
X = x1 – x1
= u1cos θ1 t – u2cos θ2 t
X = (u1cos θ1 – u2cos θ2)t
Y = y1 – y2
= u1sin θ1 t – 1/2 gt² – (u2sin θ2 t – 1/2 gt²)
Y = (u1 sin θ1 – u2sin θ2)t
∴ \(\frac{Y}{X}=\frac{\left(u_{1} \sin \theta_{1}-u_{2} \sin \theta_{2}\right) t}{\left(u_{1} \cos \theta_{1}-u_{2} \cos \theta_{2}\right) t}\)
= a constant, say m
This equation is of the form Y = mX, which is the equation of a straight line. Thus the path of a projectile as seen from another projectile is a straight line.

Question 3.
A particle is projected with some speed at an angle a to the horizontal from the end B of the horizontal base BC of a triangle ABC. It rises to the vertex A and after just grazing it, falls down to reach point C of the base BC. If the base angles of the triangle are β and γ, show that 4 cot α = cot β + cot γ.
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 63
Answer:
From the figure,
cot β = \(\frac{\mathrm{BD}}{\mathrm{h}_{\max }}\)
and cot γ = \(\frac{\mathrm{DC}}{\mathrm{h}_{\max }}\)
cot β + cot γ = \(\frac{\mathrm{BD}+\mathrm{DC}}{\mathrm{h}_{\max }}\)
1st PUC Physics Question Bank Chapter 4 Motion in a Plane img 64
\(\frac{\text { Range }}{h_{\max }}\)
⇒ cot β + cot γ = \(\frac{\left(u^{2} \sin 2 \alpha\right) / g}{\left(u^{2} \sin ^{2} \alpha\right) / 2 g}\)
⇒ cot β + cot γ = 2 × \(\frac{2 \sin \alpha \cos \alpha}{\sin ^{2} \alpha}\)
cot β + cot γ = 4 cot α.

1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation

You can Download Chapter 3 Classification and Tabulation Questions and Answers, Notes, 1st PUC Statistics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation

1st PUC Statistics Classification and Tabulation Two Marks Questions and Answers

Question 1.
What is classification of the data?
Solution:
Classification is the process of arranging the data in to groups or classes according to common

Question 2.
What are the objectives of classification?
Solution:

  • To reduce the size of the data
  • To bring the similarities together.
  • To enable further statistical analysis

KSEEB Solutions

Question 3.
What are the basis/types of classification?
Solution:

  1. Chronological classification
  2. Geographical classification
  3. Qualitative classification
  4. Quantitative classification.

Question 4.
Give the formula of Sturge’s to find the number of classes. Or Give the formula used to determine the number of classes.
Solution:
The numbers of classes are obtained using Sturges’s Rule: k = 1 + 3.22 log N;
N-number of items.

Question 5.
For what purpose is correction factor used in frequency distribution?
Solution:
To get a better continuity between the class interval of a frequency distribution exclusive class intervals are used, so, if the frequency distribution is in inclusive class intervals isconverted into exclusive class intervals using correction factor.

Question 6.
What are the guidelines of classification?
Solution:
following are some guidelines following while classification:

  • The number of classes should generally between 4 and 15.
  • Exclusive classes should be formed for better continuity between the class intervals.
  • The width of the classes should be usually kept constant throughout the distribution.
  • Avoid open-end classes.
  • The classes should be arranged in ascending or descending order.
  • The lower limit of the first class should be either ‘o’ or multiple of 5.

Question 7.
Define the term tabulation
Solution:
Tabulation is a systematic arrangement of the classified data in to columns and rows of a table

Question 8.
Mention the parts of a Table
Solution:
Table No., Title of the table, Headnote/Sub-title, Captions, Stubs, Body of the table, Footnote & Source.

KSEEB Solutions

Question 9.
What are the objects of Tabulation?
Solution:
The object/Purpose of tabulation are:-

  • It simplifies the complex data
  • To facilitate for comparison
  • To give an identity to the data
  • To reveals trend and patterns of the data

Question 10.
How the number of classes using Prof. H.A.Sturge’s?
Solution:
The number of classes are obtained using Sturges’s Rule: k = 1 + 3.22 Log N

Question 11.
What are inclusive & exclusive class intervals?
Solution:
If in a class, both lower and upper limits are included in the same class are inclusive class intervals, eg. 0-9, 10-19, 20-29…
If in a class, the lower limit is included in the same class but the upper limit is included in the next class are exclusive class intervals, eg. 0-10, 10-20, 20-30…

Question 12.
Define frequency distribution
Solution:
A systematic presentation of the values taken by a variable and the corresponding frequencies is called Frequency distribution.

KSEEB Solutions

Question 13.
What are Marginal and Conditional frequency distributions?
Solution:
If in a bivariate frequency distribution, if the distribution of only one variable is considered, the distribution is called marginal frequency distribution.
If in a bivariate frequency distribution, if the distribution of only one variable is formed subject to the condition of the other variable it is called conditional frequency distribution.

Question 14.
What is the tabulation of the data?
Solution:
Tabulation is a process of systematic arrangement of the classified data in rows and columns, in the form of a table.

Question 15.
What are the parts of a table?
Solution:
The parts of a table are: Table No., Title of the table, Headnote/Sub-title, Captions, Stubs, Body of the table, Footnote & Source

Question 16.
What is the purpose of ‘table number’ in tabulation?
Solution:
A number should be given to each and every table, in order to distinguish and also for easy reference.

Question 17.
What are captions and stubs of a table?
Solution:
Captions: Column headings are called captions. They explain what the column represents. Captions are always written in one or two words on the top of each column.
Stubs: Row headings are called Stubs. They explain what the row represents. Stubs are usually written in one or two words at the left extreme sid/of each row.

KSEEB Solutions

Question 18.
What is headnote of table?
Solution:
It is a brief explanatory note or statement given just below the heading of the table put in a bracket. The statistical units of measurements, such as in ‘000s, Rs., Million tones, crores, Kgs., etc., are usually put in bracket.

Question 19.
What is indicated by source note of a table?
Solution:
Below foot note or below the table, source of the data may be mentioned for verification to the reader, regarding publications, organizations, pages, Journals etc,

1st PUC Statistics Classification and Tabulation Five Marks Questions and Answers

Question 1.
Explain chronological classification and geographical classification of data with examples.
Solution:
Temporal/Chronological classification: when the data enumerated over a different period of time, the type of classification is called chronological classification. The above type of classification is called as time series, e.g. Time series of the population, is listed in chronological order starting with the earliest period.
Table showing the population of India
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 1
Geographical classification :
In this type of classification the data are classified on the basis of geographical or locational or area wise differences between various items. Such as cities, districts, states etc. e.g., production of sugar in India may be presented state wise in the following manner:-
Table showing the production of sugar in India
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 2

KSEEB Solutions

Question 2.
Explain qualitative and quantitative classification with examples.
Solution:
Qualitative classification: Classification of the different units on the basis of qualitative characteristics (Called Attributes). Such as sex, literacy, employment etc.
e. g., the members of a club can be classified on the basis of sex wise distribution as follows:-
Table showing the sex distribution of members of a club
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 3
Quantitative classification: classification of the number of units on the basis of quantitative data, such as according to Height, weight, Wages, Age (years), Number of children, etc, Thus the groups of a student may be classified on their heights as follows:-
Table showing the heights of students
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 4

Online midpoint calculator and find the midpoint of aline segment joining two points using the midpoint calcultor in just one click.

Question 3.
Define the following terms :
i. Frequency, class frequency:
Solution:
Frequency refers to the number of times an observation repeated (f). The number of observations corresponding to a particular class is known as the class Frequency Class frequency is a positive integer including zero

ii. Class limits:
Solution:
Class limits- Lowest and the highest values that are taken to define the boundaries of the class are called class limits

iii. Range of the class: It is the difference between highest and lowest value in the data, i.e., Range = H.V. – L.V.

iv. Width of the class: The difference between the upper and lower limits of class called width of the class. It is denoted by c or i.
i/c = Upper limit(UL) – Lower limit(LL)

v. Class mid point
Solution:
The central value of a class called mid value/point; \(\mathrm{m} / \mathrm{x}=\frac{\mathrm{LL}+\mathrm{UL}}{2}\)

vi. Define Frequency Density:
Solution:
The frequency per unit of class interval is the frequency density.
i.e. frequency density = \(\frac{\text { Frequency the class }}{\text { width of the class }}=f / w\)

vii. Relative frequency:
Solution:
Relative frequency.. is the ratio of frequency of class to the total frequency of the distribution
Relative Frequency \(\frac{\mathrm{f}}{\mathrm{N}}\)

viii. Class interval-inclusive, exclusive and open-end classes:
Solution:
If in a class, lower as well as upper limits are included in the same class are called Inclusive class
e. g. 30-39,40-49….
If in a class, the lower limit is included in the same class and upper limit is included in the next class, such a class is called Exclusive class, eg. 30-40, 40-50…
If in a class, the lower and upper limits of the class is not specified are called open end classes” e.g. less than/below, or more than/ above 100

ix. Cumulative frequency- less than and more than cumulative frequency:
Solution:
The added up frequencies are called cumulative frequencies.
The number of observations with values less than upper limit is less than cumulative frequency. (l.c.f) i.e. Frequencies added from the top.
The number of observations with values more than lower class limit is more than cumulative frequency (m.c.f)

x. Correction factor:
Solution:
It is half of the difference between lower limit of a class and upper limit of the preceding class. Thus,
Solution:
Correction facctor (C.F) = \(\frac{\text { Lower limit of a class-Upper limit of the precending class }}{2}\)

xi. A Frequency distribution Discrete, Continuous, Bi-variate, Marginal
Solution:
A systematic presentation of the values taken by a variable and the corresponding frequencies is called frequency is called Frequency distribution.
While framing a frequency distribution, if class intervals are not considered, is called discrete frequency distribution.
1. Example:
The number of families according to number of children .
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 5
While framing a frequency distribution, if class intervals are considered, is called continuous frequency distribution.

2. Example:
The following table showing the weight (kgs.) of persons
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 6
A frequency distribution formed on the basis of two related variables is called bi-variate frequency distribution.
For example, we want to classify data relating to the Height and Weights of a group of individuals, Income and Expenditure of a group of individuals, Ages of Husbands and Wives, Ages of mothers and Number of children, etc .
In a Bivariate frequency distribution, the frequency distribution of only one of the variables is considered, it is marginal frequency distribution.

KSEEB Solutions

Question 4.
Mention/what are the rules/principles of formation of Frequency of distribution?
Solution:

  1. The lower limit of the first class should be either 0 or a multiple of 5
  2. Exclusive classes should be formed for better continuity
  3. The number of classes should be generally between 4 & 15
  4. The width of the classes should be kept constant throughout the distribution
  5. Avoid open-end classes
  6. The classes should be arranged in ascending or descending order.

Question 5.
Mention the rules/principles of the tabulation
Solution:

  1. The size of the table should be according to the size of the paper
  2. The stubs and captions should be arranged logically according to the alphabetical, Chronological, Geographical order and items to be arranged according to the size.
  3. If desired to locate the particular quantity in the body of the table that should be showed by thick colored inks or shaded off.
  4. The table should not be overloaded with number of characteristics, rather can be prepare another table.
  5. The table should be complete in all respects by captions and stubs, titles heading and no cell is left blank if left do not put ‘o’ but give (—) dash marks or write N.A
  6. Miscellaneous column can be provided for the data which do not fit to the data, such as Ratio, percentages.
  7. Ditto ( “ ) marks should not be used, as they may confuse with the no. 11
  8. Footnote may contain about errors, omissions, remarks about the data.
  9. Sources if provided regarding publications, organizations, pages, Journals etc.

Question 6.
The employees of a college can be classified on the basis of sex wise distribution as follows:-
Solution:
Table showing the sex distribution of employess of a college.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 7

Question 7.
The Employees of a college can be classified according to their occupations as :
Solution:
Table-1
The blank table given below represents the number of employees with different occupation in a college

Occupations Number of employees
Teaching staff
clerks
Attenders
Security men
Total

Question 8.
The employees of a commercial Bank can be classified according to their occupations and sex is :
Solution:
Table-2
The blank table given below represents the number of employees with different occupation and their sex in a commercial Bank
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 8

Question 9.
The employees of a college can be classified according to their occupations , sex and their marital status is :
Solution:
Table-3 : The blank table given below represents the number of employees with different occupation, sex and their marital status in a college
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 9

Question 10.
In a survey of 40 families in a certain locality, the number of children per family was recorded and the following data were obtained.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 10
Represent the data in the form of a discrete frequency distribution.
Solution:
Frequency distribution of the number of children.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 11

Question 11.
Prepare a frequency table from the following table regarding the number fatal accidents occurred in a day in Bangalore in June 2010.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 12
Solution:
Frequency distribution of the number of children.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 13
Question 12.
From the following paragraph prepare a discrete frequency table with the number of letters present in the words .
“Success in the examination confers no right to appointment unless government is satisfied, after such enquiry as may be deemed necessary that the candidate is suitable for appointment to the public service”
Solution:
The number of digits in the above statement: Highest digit = 11 and Lowest digit = 2
Frequency distribution of the number of letters in the words present in the statement
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 14

KSEEB Solutions

Question 13.
The following are the marks obtained by 50 college students in a certain test.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 15
Take suitable width of the class interval marks using struge’s rule.
Solution:
Here, N= 50; Range = H.V.-L. V. = 49-12 = 37
The number classes as per Sturge’s rule are obtained as follows:
Number of class intervals (K) = 1 + 3.322 logN= 1 + 3.22 log 50 = 1 + (3.22 × 1.6990) = 6.47=7
classes (Approx.) Size/width of class intervals – e = \(\frac{\text { Range }}{\text { Number of class intervals }}=\frac{37}{7}\)
5.28 = 6 (Approx.)
The size/width of each class is 6 and there are 7 classes. Thus, the required continuous frequency distribution with exclusive class intervals width is prepared as :
Frequency distribution of marks of students
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 16

Question 14.
The following data gives ages of 32 individuals in a locality. Using Sturge’s rule form a frequency table with exclusive type of class intervals.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 17
Solution:
Here N= 32; Range = H .V.- L.V. = 59- 01 = 58
The number classes as per Sturge’s rule are:
Number of class intervals (K) =1 + 3.22 logN = 1 + 3.22 log (32) = 1+(3.22 × 1.5051) = 5.85=6
classes (Approx.) Size/width of class intervals – e = \(\frac{\text { Range }}{\text { Number of class intervals }}=\frac{58}{6}\) = 10
(approx). Teh size / width of each class is 10 and there are 6 classes.
Frequency distribution of marks of students
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 118

KSEEB Solutions

Question 15.
The following are the marks obtained by 50 students in statistics; prepare a frequency table with class intervals of 10 marks.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 19
Solution:
Range: H.V-L.V = 93 – 23 = 70; take width as 10 marks, and then the number of classes will be: 70/10=7.
Frequency distribution of marks of students in statistics test
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 20

Question 16.
Prepare a bivariate frequency distribution of the marks in Accountancy & Statistics:
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 21
Solution:
Here the both variables are discrete in nature no need to prepare class interval.
Bi-variate frequency table showing Ages (years) of Mothers and Number of children
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 22

Question 17.
Below are the ages of husbands and wives prepare a bivariate frequency distribution with suitable width :
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 23
Here both are continuous variables form the class intervals as below :
Solution:
Ages of Husbands: Highest age = 47, Lowest age = 25, Difference = 22/(i)5width =5. classes. Ages of wife: Highest age = 47, Lowest age = 21, Difference = 26/(i)5width = 6 (Approx.) classes.
Let X and Y be the ages of Husbands and ages of Wives.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 24

KSEEB Solutions

Question 18.
Below are given the marks obtained by a batch of 20 students in mathematics and statistics:
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 25
Solution:
Marks in Mathematics: Highest Marks = 72, Lowest marks = 25, Difference = 47/(i)l Owidth 5 = 5 (Approx.) classes.
Marks in statistics: Highest marks = 85, Lowest marks = 20, Difference = 65/(i)10width = 7 (Approx.) classes.
Let x and y be marks inmathematics and marks in statistics.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 26

TABULATION

Question 19.
Give a general format of a table.
Solution:
General format of a table
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 27

Question 20.
What are the requisites of a good table? Or What are the General rules to the tabulation?
Solution:

  • Size:- the size of the table should be according to the size of the paper with more rows than columns. Exchange of the data can be done by altering the column and rows. A sufficient space should be provided in a particular to enter any new or to alter any affected figures.
  • Logical order:- The stubs and captions should be arranged logically according to the alphabetical, Chronological, Geographical order and items to be arranged according to the size.
  • Identity:- If desired to locate the particular quantity in the body of the table that should be showed by thick colored inks or shaded off.
  • The table should not be overloaded with number of characteristics, rather can be prepare another table.
  • The table should be complete in all respects by captions and stubs, titles heading and no cell is left blank if left do not put ’o’ but give (—) dash marks or write N.A
  • Miscellaneous column can be provided for the data which do not fit to the data, such as Radius, percentages.
  • Ditto ( “ ) marks should not be used,as they may confuse with the number 11
  • Footnote may contain about errors, omissions, remarks
  • Sources if provided regarding publications, organizations pages Journals etc.

KSEEB Solutions

Question 21.
Elucidate the difference between classification and tabulation.
Solution:
Comparison between Classification and Tabulation:-
The following points may be given as comparison:

  • Classification and Tabulation are not two distinct processes. Before tabulation, data are classified and then displayed under different columns and rows of a table.
  • Classification is the process of arranging the data in to groups or classes according to common characteristics possessed by the items of the data;
    Whereas Tabulation is a process of systematic arrangement of the classified data in rows and columns, in the form of a table.
  • Table contains precise and accurate information, where as classification gives only a classified groups of data.
  • Classification reduces the size of the data and brings the similarities together and tabulation facilitates comparison, reveal the trend and tendencies of the data.

Question 22.
In the college out of total of 1200 applications received for I puc admission, 450
were applied for science and 580 applied for commerce and remaining applied for
arts faculty. Tabulate the above information.
Solution:
Table 1
Table showing the distribution of applicants for I year puc to different faculties
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 28

Question 23.
In the college out of total of 1200 applications received for I puc admission 780 were boys. 450 were applied for science of which 300 were boys and 580 applied for commerce 250 were girls and remaining applied for arts faculty. Tabulate the above information.
Solution
Table 2
Table showing the sex-wise distribution of applicants for I year puc to different faculties
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 29

KSEEB Solutions

Question 24.
In a Sigma multinational accountant consultants there are 180 were accountants, 210 were article helpers, 300 were practitioner trainees. Of all the members 30% were women among accountants, 20% in Articles helpers and 15% among trainees. Tabulate the data.
Solution:
Table 3
Table represents the members of Sigma accountants according to cadre and sex
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 30
Footnote: * 180 × 30%= 54, 210 × 20% = 42

Question 25.
In a trip organized by a college there were 80 persons each of whom paid Rs.15.50 on an average? There were 60 students each of whom paid Rs.16. members of the teaching staff were charged at higher rate. The number of ser ants was 6 (all males) and they were not charged anything. The number of females was 20 percent of the total of which one was a lady staff member. Tabulate the information.
Solution:
Table 4
Table showing the contribution (in Rs.) details made by members of a college trip.
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 31

Question 26.
In a state, there are 30 Medical colleges, 10 Dental colleges and 50 Engineering colleges. Among the Medical colleges, 5 are government colleges, 10 are aided private colleges and remaining are the unaided private colleges. Of the unaided colleges, 5 colleges are run by minority institutions.
Among the Engineering colleges, 10 are government colleges, 20 are aided private colleges and the rest are unaided private colleges. Of the unaided colleges, 10 colleges are run by minority institutions.
Among the dental colleges, 2 are aided private colleges and the rest are the unaided private colleges of which one is run by a minority institutions.
Tabulate the above information.
Solution:
Table showing the Medical, Engineering and Dental colleges run by Government, Private Aided and Private Unaided colleges in a State
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 32

Question 27.
The number of cases filed, hearing made and disposed by different bench judges in a day at High court of Karnataka are as given:
(i) Criminal cases filed 12, hearings made in 8 cases and disposed 3,
(ii) Land dispute cases filed 18, hearings made in 12cases and disposed 5,
(iii) Government service cases filed 6, hearings made in 4 cases and disposed 3,
(iv) Cheating cases filed 15; hearing made in 12 cases and disposed 8.
Tabulate the above information.
Solution:
Table showing the different types of cases filed, heard and disposed in a day at High court of Karnataka
1st PUC Statistics Question Bank Chapter 3 Classification and Tabulation - 33

KSEEB Solutions

2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1

Students can Download Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1

2nd PUC Maths Inverse Trigonometric Functions NCERT Text Book Questions and Answers Ex 2.1

An online inverse function calculator allows you to find the inverse of the given functions with step-by-step calculations in a fraction of a second.

Find the principal values of the following

Question 1.
\(\sin ^{-1}\left(-\frac{1}{2}\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 1

Question 2.
\(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
Answer:

2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 2

KSEEB Solutions

Question 3.
cosec1 (2)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 3

Question 4.
\(\tan ^{-1}(-\sqrt{3})\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 4
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 5

Question 5.
\(\cos ^{-1}\left(-\frac{1}{2}\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 6

Question 6.
tan-1(-1)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 7

KSEEB Solutions

Question 7.
\(\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 8

Question 8.
\(\cot ^{4}(\sqrt{3})\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 9

Question 9.
\(\cos ^{2}\left(-\frac{1}{(\sqrt{2})}\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 10

Question 10.
\({ cosec }^{ -1 }\left( -\sqrt { 2 } \right) \)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 11
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 12

KSEEB Solutions

Find the values of the following:

Question 11.
\(\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 13

Question 12.
\(\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 14

Question 13.
If sin-1 x = y, then
(A) \(0 \leq y \leq \pi\)
(B) \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\)
(C) \(\mathbf{0}<\mathbf{y}<\boldsymbol{\pi}\)
(D) \(-\frac{\pi}{2}<y<\frac{\pi}{2}\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 15
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 16

KSEEB Solutions

Question 14.
\(\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2) \text { is equal to }\)
(A) π
(B) \(-\frac{\pi}{3}\)
(C) \(\frac{\pi}{3}\)
(D) \(\frac{2 \pi}{3}\)
Answer
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Ex 2.1 17

1st PUC Maths Question Bank Chapter 8 Binomial Theorem

Students can Download Maths Chapter 8 Binomial Theorem Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Maths Question Bank Chapter 8 Binomial Theorem

Question 1.
State and prove Binomial theorem.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 1
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 2

Some observations in a binomial theorem:
(1) The expansion of {a + b)n has (n + 1) terms
(2) The coefficients nCr occurring in the binomial theorem are known as binomial coefficients.
(3) The indices of V go on decreasing and that of ‘a’ go on increasing by 1 at each stage.
i.e., for each term: index of a + index of b-n.
(4) Since nCr=nCn_r we have
nC0 =nCn, nCx =nCn_r and so on.
Thus the coefficients of the terms equidistant from the beginning and the end in a binomial theorem are equal.
(5) General term in (a + b)n: Tr+1 – nCr anrbr
(6) Middle terms in (a + b)n

  • When ‘n’ is even, the middle term
    \(=\left(\frac{n}{2}+1\right)^{t h} \text { term }\)
  • When ‘n’ is odd ,the middle term are
    \(\frac{1}{2}(n+1)^{n} \text { term an } \frac{1}{2}(n+3)^{n} \text { term }\)

(7) Taking a = x and b = -y in the expansion, we get (x-y)n =[x + (-y)]n

1st PUC Maths Question Bank Chapter 8 Binomial Theorem 3

KSEEB Solutions

Question 2.
Expand each of the expression
(i)\(\left(x^{2}+\frac{3}{x}\right)^{4}\)
(ii)\((1-2 x)^{5}\)
(iii)\(\left(\frac{2}{x}-\frac{x}{2}\right)^{5}\)
(iv)\((2 x-3)^{6}\)
(v)\(\left(\frac{x}{3}+\frac{1}{x}\right)^{5}\)
(vi)\(\left(x+\frac{1}{x}\right)^{6}\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 4
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 5
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 6

Question 3.
Using binomial theorem, evaluate each of the following:
(i) (98)5
(ii) (96)3
(iii) (102)5
(iv) (101)4
(v) (99)5
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 7
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 8

KSEEB Solutions

Question 4.
Which is longer \((1.01)^{1000000} \text { or } 10,000 ?\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 9

Question 5.
Using Binomial theorem, indicate which number is larger \( (1.1)^{10000} \text { or } 1000 ?\)
Answer:
Spilitting 1.01 and using Binomial theorem write the first few terms, we have.
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 10

Question 6.
Find (a + b)4 – (a -b)4. Hence evaluate
\((\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4}\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 11

Question 7.
Find (x +1)6 +(x -1)6. Hence or otherwise evaluate
\((\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 12
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 13

Question 8.
Evaluate:
\((\sqrt{3}+\sqrt{2})^{6}+(\sqrt{3}-\sqrt{2})^{6}\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 14

Question 9.
Find the value of
\(\left(a^{2}+\sqrt{a^{2}-1}\right)^{4}+\left(a^{2}-\sqrt{a^{2}-1}\right)^{4}\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 15
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 16

KSEEB Solutions

Question 10.
Show that 9n+1 – 8n – 9 is divisible by 64, whenever ‘n’ is a positive integer.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 17

Question 11.
Using binomial theorem, prove that 6n -5n always leaves remainder 1 when
divided by 25.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 18

Question 12.
Prove that
\(\sum_{n=0}^{n} y \cdot c_{n}=4\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 19
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 20

Question 13.
Find the 4th term in the expansion of (x-2 y)12.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 21

Question 14.
Find the 13th term in the expansion of
\(\left(9 x-\frac{1}{3 \sqrt{x}}\right)^{18}, x \neq 0 \)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 22

Question 15.
Write the general term in the expression of
(i) (x2 -y)6
(ii) (x2-yx)12,x≠0
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 23

KSEEB Solutions

Question 16.
Find the coefficient of
(i) x5 in (x + 3)8
(ii) a5b7 in (a – 2b)12
(iii) x6y3 in (x + 2y)9
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 24

Question 17.
Find a, if the 17th and 18th terms of the expansion (2 +a)50 are equal.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 25

Question 18.
In the expansion of (1+ a)m+n, prove that coefficients of am and an are equal.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 26

Question 19.
Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1+x)2n-1
Answer:
In (1+x)2n we have
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 27
From (1) and (2) we get, the coefficient of xn in (1 + x)2n is twice the coefficient of xn in (1 + x)2n-1.

KSEEB Solutions

Question 20.
Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 28

Question 21.
Find the middle terms in the expansions of
(i) \(\left(3-\frac{x^{3}}{6}\right)^{7}\)
(ii)\(\left(\frac{x}{3}+9 y\right)^{10}\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 29
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 30

Question 22.
Show that the middle term in the expansion of \((1+x)^{2 n} \text { is } \frac{1 \cdot 3 \cdot 5 \ldots(2 n-1)}{\lfloor n} 2^{n} x^{n} \)where ‘n’ is a positive integer.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 31

Binomial Expansion Calculator to make your lengthy solutions a bit easier. Use this and save your time. Binomial Theorem & Series Calculator.

Question 23.
The second, third and fourth terms in the binomial expansion (x + a)n are 240, 720 and 1080, respectively. Find x, a and n
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 32
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 33

Question 24.
The coefficients of three consecutive terms in the expansion of (1 + a)n are in the ratio 1:7: 42. Find n
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 34
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 35

KSEEB Solutions

Question 25.
The coefficients (r-1)th, rth,and (r + 1)th, terms in the expansion of (x + 1)th, are in the ratio 1:3:5. Find n and r.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 36
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 37

Question 26.
Find the term independent of x in the expansion of \(\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{6}\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 38

Question 27.
Find the term independent of x in the expansion of \(\left(\sqrt[3]{x}+\frac{1}{2 \sqrt[3]{x}}\right)^{18}, x>0\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 39

Question 28.
Find a, 6 and n in the expansion of (a + b)n if the first three terms of the expansion are 729,7290 and 30375, respectively.
Answer:
Given: Tx = 729, T2 = 7290 and T3 = 30375
∴ an=729……………….(1)
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 40

KSEEB Solutions

Question 29.
Find a if the coefficients of x2 and x3 in the expansion of (3 +ax)9 are equal.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 41

Question 30.
If the coefficients of (r – 5)th and (2r -1)th terms in the expansion of (1 + x)34 are equal, find r.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 42
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 43

Question 31.
If the coefficients of ar-1, ar and ar+1 in the expansion of (1 + a)n are in arithmetic progression, prove that n2 – n(4r +1) + 4r2 -2 = 0.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 44

Question 32.
Show that the coefficient of the middle term in the expansion of (1 + x)2n is equal to the sum of the coefficients of two middle terms in the expansion of (1 + x)2n-1.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 45
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 46

Question 33.
Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of
\(\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{n} \text { is } \sqrt{6}: 1\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 47

Question 34.
Find the rth term from the end in the expansion of (x + a)n.
Answer:
rth term from the end in (x + a)n
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 48

Question 35.
If a and b are distinct integers, prove that a-b is a factor of an -bn, whenever n is a positive integer.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 49

Question 36.
The sum of the coefficients of the first three terms in the expansion of
\(\left(x-\frac{3}{x^{2}}\right)^{m}, x \neq 0, m \)being a natural numbers, is 559. Find the term of the expansion containing x3.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 50

Question 37.
Find the coefficient of x5 in the product (1 + 2x)6(1 – x)7 using binomial theorem.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 51

Question 38.
Find the coefficient of a4 in the product (1 + 2a)4(2-a)s using binomial theorem.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 52

Question 39.
Expand using Binomial theorem
\(\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}, x \neq 0\)
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 53

KSEEB Solutions

Question 40.
Find the expansion of (3x2 – 1ax + 3a2)3 using binomial theorem.
Answer:
1st PUC Maths Question Bank Chapter 8 Binomial Theorem 54

KSEEB Solutions for Class 8 Science Chapter 8 Describing Motion

Karnataka Board Class 8 Science Chapter 8 Describing Motion

KSEEB Class 8 Science Chapter 8 Textual Questions & Answers

I. Four alternatives are given to each of the following incomplete statement/ question, choose the right answer.

Online Circular Velocity Calculator tool makes the calculation faster, and it shows the result in a fraction of seconds.

Question 1.
Uniform circular motion is called continuously accelerated motion mainly because
(a) direction of motion changes
(b) Speed remains the same
(c) Velocity remains the same
(d) direction of motion does not change
Answer:
a) direction of motion changes

Question 2.
A cricketer hits a sixer, the cricket ball moves up with a velocity of 2 ms-1 and falls down. Its initial velocity while falling down will be
a) 1 ms-1
b) 1 ms-2
c) 0 ms-1
d) 2 ms-1
Answer:
c) 0 ms-1

II. Fill in the blanks with suitable words:

Question 1.
S.I. unit of acceleration is ………….
Answer:
meter per second2/ms-2

Question 2.
Velocity has both speed and ……….
Answer:
direction

Question 3.
If an object starts from A and comes back to A. It displacement will be ………….
Answer:
(0) zero

III. Solve

Questions 1.
An object is moving in a circular path of radius 3.5 m. If completes one full cycle, what will be the displacement and what is the distance traveled?
Answer:
radius = 3.5 m
starting point is A and ending point also A.
the displace = 0
Distance traveled = π × 3.5 × 2
KSEEB Solutions for Class 8 Science Chapter 8 Describing Motion iii-1
Question 2.
An object changes its velocity from 30 ms-1 to 40 ms-1 in a time interval of 2 seconds what is its acceleration?
Answer:
u = 30 ms-1
v = 40 ms-1
t = 2 sec
a = ?
KSEEB Solutions for Class 8 Science Chapter 8 Describing Motion iii-2
Question 3.
An object at rest starts moving. It covers a distance of 2 m in one second. It covers a, further distance of 5 m in two seconds in the same direction what is the average velocity and acceleration?
Answer:
distance traveled on 1 sec = 2 mts
distance traveled on 2 sec = 5 mts
total distance traveled = 2 + 5 = 7 mts
total time taken to cover 7 mts = 1+2 = 3 sec
KSEEB Solutions for Class 8 Science Chapter 8 Describing Motion iii-3
= \(\frac { 7 }{ 3 }\) = 2.3 ms-1
Average velocity = 2.3 ms-1
s = ut + \(\frac { 1 }{ 2 }\) at2
7 = 0  3 +\(\frac { 1 }{ 2 }\) a × 32
7 =0 + \(\frac { 1 }{ 2 }\) a × 9
= \(\frac { 9 }{ 2 }\) a = 7
9a =7 × 2 = 14
a =  \(\frac { 14 }{ 9 }\)  = 1.5 ms-2

IV. Answer the following.

Question 1.
If a body is moving with uniform velocity in a given direction its acceleration will be zero, why?
Answer:
The acceleration will be zero in the above situation since there is no change in velocity in unit time.

Question 2.
Distinguish between speed and velocity.
Answer:

  1. Speed
    • Speed is the distance traveled by an object in unit time
    • Speed = \(\frac { distance }{ time }\)
    • Speed is a scalar quantity
    • Speed has the only magnitude
  2. Velocity
    • Velocity is the rate of displacement of the body in unit time.
    • Velocity =\(\frac { displacement }{ time }\)
    • Velocity is a vector quantity
    • It has both magnitude and direction.

Question 3.
Distinguish between distance traveled and displacement.
Answer:

  1. Distance traveled
    • Distance traveled is the length of actual path traveled by a body from one position to another.
    • It is a scalar quantity
    • It has an only numerical value
  2. Displacement
    • Displacement is the shortest path between initial and final position of the moving body.
    • It is a vector quantity
    • If has both numerical value and direction.

Question 4.
What are uniform and non-uniform speed?
Answer:

  1. If an object covers equal distance in equal intervals of time, it is said to be uniform motion.
  2. non-uniform motion: If an object covers unequal distance in equal intervals of time, it is said to be non uniform motion.

Question 5.
While mentioning acceleration the time is mentioned two times, why?
Answer:
Velocity is the rate of displacement of a body in unit time.
V =\(\frac { displacement }{ time }\)
Acceleration is the change in velocity of a body in unit time.
KSEEB Solutions for Class 8 Science Chapter 8 Describing Motion 8-5
While mentioning acceleration time is mentioned two times. (Acceleration occurs twice)

V. Extended activity
Represent the following motion by a graph.
KSEEB Solutions for Class 8 Science Chapter 8 Describing Motion 8-v
Question 1.
Find the acceleration
Ans. Acceleration from the graph
= \(\frac { 40-10 }{ 4-1 }\) = \(\frac { 30 }{ 3 }\) = 10 ms-1

Question 2.
Find the time taken when the velocity in 35 ms-1
Answer:
Time taken to travel a distance of 35 ms-1  is 1.5 seconds.

KSEEB Class 8 Science Chapter 8 Additional Questions & Answers

Question 1.
Define motion?
Answer:
The change in position of a body with time when compared with that of another body is called motion.

Question 2.
Motion is reflective-substantiate this statement with a suitable example.
Answer:
Imagine that you are sitting inside the train when the train starts moving you a fee that the persons on the platform are moving backward you feel that you and the other persons inside the train are not moving.
But for a person standing outside the train, the feeling will be that the train, you and the other members inside the compartment are all moving.

Question 3.
An object travels from A to B and then from B to C as shown in the figure.
1. Find the distance traveled and
2. Find the displacement
3. Write the conclusion
Answer :

KSEEB Solutions for Class 8 Science Chapter 8 Describing Motion add-3
Let ABC is a right-angled triangle with B as shown in the actual figure,

  1. If a person travels from A to B and then to C the total distance traveled is = (3 m + 4 m) = 7 mts.
  2. The displacement is the change in position from A to C
    KSEEB Solutions for Class 8 Science Chapter 8 Describing Motion 8-3
  3. displacement value is lesser than the distance traveled.

Question 4.
An object moves from A to A in a circular path of radius 7 m. What is the distance traveled? What is the displacement?
Answer:
Displacement = Diameter = 2 × radius = 2 × 7= 14 m
Distance traveled = \(\frac { 1 }{ 2 }\) × circumference
= \(\frac { 1 }{ 2 }\) × 2 × π × r
= \(\frac { 1 }{ 2 }\) × 2 × \(\frac { 22 }{ 7 }\) × 7
= 22 m

Question 5.
Define the following
a) Speed
b) Velocity
Answer:

  1. Speed: is the distance traveled by an object in unit time
  2. Velocity: is the rate of displacement of a body in unit time.

Question 6.
An object is moving in a circular path of radius 7 m. To travel from ‘A’ to ‘B’ along the circumference it takes 2 seconds?
Answer:
Data : radius = 7 m
time = 2 s
average speed = ?
speed = \(\frac { distance traveled }{ time }\)
KSEEB Solutions for Class 8 Science Chapter 8 Describing Motion 8-6
Question 7.
When can we say that speed and velocity are synonyms.
Answer:
Speed and velocity will be equal for uniform motions along a straight line. Then it will be one and the same.

Question 8.
What is meant by acceleration?
Answer:
A change in velocity of a body in unit time or the rate of change of velocity is called acceleration.

Question 9.
For an object moving at a uniform speed in a circular path, though the speed is uniform it is continuously accelerated, why? Give reason.
Answer:.
For an object moving in a circular path, the direction is changing at every point even though the speed is the same. Therefore velocity changes and hence it is continuously accelerated.

Question 10.
An object at rest gains an average velocity of 40 ms-1 in 5 seconds. What will be its acceleration?
Answer:
Its initial velocity, u = 0
Its final velocity, v = 40 ms-1
Time taken t = 5 second the
rate of change of velocity
\(\frac { 40 }{ 8 }\) = 8 ms-2
acceleration, a = 8 ms-2

Question 11.
An object at rest starts moving and attains a velocity of 10 ms-1 after 5 seconds. What is the acceleration?
Answer:
Initial velocity, u = 0
Final velocity, v = 10 ms-1
Time interval, t = 5 second
a = \(\frac { u-v }{ t }\) = \(\frac { 10 – 0 }{ 5 }\) = 2 ms-2

Question 12.
An object moving with a uniform velocity of 10 ms-1 comes to rest after 5 seconds. What is the acceleration?
Answer:
Initial velocity, v = 10 ms-1
Final velocity, u = 0
Time interval, t = 5 second
a = \(\frac { u-v }{ t }\) =  \(\frac { 0 – 10 }{ 5 }\) = – 2 ms-2

Question 13.
An object at rest starts moving with a uniform acceleration of 1 ms-2 Calculate the distance traveled by it in 4 seconds.
Answer:
u = 0
a = 1 ms-2
t = 4 second ,
Distance traveled s = ut + \(\frac { 1 }{ 2 }\) at2
S = 0 × 4 + \(\frac { 1 }{ 2 }\) × 1  × 42
=0 + \(\frac { 1 }{ 2 }\) × 1 × 16 = 18 m

Question 14.
An object starts from rest and attains a uniform acceleration of 4 ms-2 what will be its velocity at the end of half a meter?
Answer:
u = 0
a = 4 ms-2
s = \(\frac { 1 }{ 2 }\)
v2 = ?
v2= u2 + 2as
v2 = 02 + 2 × 42 × \(\frac { 1 }{ 2 }\)
v2 = 4
v2 = \(\sqrt { 4 } \) = 2 ms-1

Question 15.
What is the distance-time graph? What are its uses?
Answer:
Motion can be represented by a time graph. If it represents time on the x-axis and distance traveled on the y-axis, it is called a distance-time graph.

Uses of the distance-time graph are,

  1. The distance at which the body is at present from its starting point can be found out.
  2. Time taken to cover a definite distance can be determined
  3. Since it is uniform motion speed of the body can also be calculated.

Question 16.
What do you infer about the velocity when the velocity-time graph is parallel to the x-axis.
Answer:
The body is at rest no acceleration.

Question 17.
Define negative acceleration?
Answer:
It is a Negative acceleration as the moving body reduces its velocity due to applied force, denoted by “-a”.

KSEEB Solutions for Class 8 Science

KSEEB Solutions for Class 9 Science Chapter 8 Motion

Students can Download KSEEB Solutions for Class 9 Science Chapter 8 Motion, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka Board Class 9 Science Chapter 8 Motion

KSEEB Solutions for Class 9 Science Chapter 8 Intext Questions

Our Displacement Calculator helps determine your engine’s size by its bore and stroke. This is a very useful tool for building high-performance racing engines.

Question 1.
An object has moved through a distance can it have zero displacement? If yes, support your answer with an example.
Answer:
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 1
In the figure distance of an object moved from 0 to A is 60 km and distance travelled is 60 km and displacement is 60 km. From 0 to A and back to B, distance travelled is 60 km + 25 km = 85 km. But displacement (35 km) is not equal to distance travelled (85 km). If we observe like this, displacement of a body is zero, but distance travelled is not zero, starting from point ‘0’ and returning back to ‘O’ final position is mixing with initial point. Hence displacement is zero.

Question 2.
A farmer moves along the boundary of a square field of side 10 m in 40s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Answer:
A farmer moves along the boundary of a square field of side 10 m in 40s
2 minutes 20 seconds means 140 seconds.
Distance travelled by farmer
= \(\frac{40}{40}\) × 140 m
= 140 m.
∴ Farmer moves for 2 minutes 20 seconds. It means \(\frac{140}{40}\) = 3.5 rounds
From the original point, he moves for 2 minutes 20 seconds.
Case i)
Original point means any point in the corner of the field. In this case he moves for 2 min 20 seconds from the diagonal of the field.
Displacement is equal to diagonal of the field displacement
= \(\sqrt{10^{2}+10^{2}}\)
= 14.1 m.
Case ii)
Original point means any point in the middle of the side of the field.
∴ Displacement is equal to any side of the field = 10 m.
It means displacement is between any original point ie in between 14.1 m and 10 m.

Question 3.
Which of the following is true for displacement?
a) It cannot be zero.
b) Its magnitude is greater than the distance travelled by the object.
Answer:
b) Its magnitude is greater than the distance travelled by the object.

Question 4.
Distinguish between speed and velocity
Answer:

Speed Velocity
1. Scalar quantity Vector quantity
2. Distance travelled in a given time. Distance travelled along with a path with the given time.
3. It is always is + ve. It is + ve or – ve

Average velocity calculator is really nice tool to use and it will help you solve all your problem related to average velocity.

Question 5.
Under what condition (s) is the magnitude of the average velocity of an object equal to its average speed?
Answer:
If the speed of the object changing uniformly, the magnitude of the average velocity of an object equal to its average speed.

Question 6.
What does the odometer of an automobile measure?
Answer:
Odometer measures the speed of the vehicle.

Question 7.
What does the path of an object look like when it is in uniform motion?
Answer:
When the object is in uniform motion, it covers equal distance at equal intervals of time.

Question 8.
During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, which is 3 × 10 ms-1.
Answer:
Time taken by the signal to reach the ground station from the spaceship = 5 min = 5 × 60 = 300 seconds
Speed of the signal = 3 × 108 m/s.
Speed = \(\frac{\text { Total distance travelled }}{\text { Total time taken }}\)
∴ Distance travelled = speed × time taken
= 3 × 108 × 300
= 9 × 1010m
∴ Distance of change of velocity = 9 × 1010 m

A versatile acceleration calculator with which you can calculate the acceleration given initial and final speed and acceleration time.

Question 9.
When will you say a body is in

  1. Uniform acceleration?
  2. non-uniform acceleration?

Answer:

  1. If the object covers equal distances in equal intervals of time, it is added to be in uniform motion.
  2. If the object covers unequal distances in equal intervals of time, it is said to be in non-uniform motion.

Question 10.
A bus decreases its speed from 80 km h-1 to 60 km h-1 in 5s. Find the acceleration of the bus.
Answer:
Initial velocity of the bus u = 80 kmh-1
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 2
Final velocity, V = 60 km/h
time taken to decrease the velocity of bus, t = 5 seconds
acceleration, a = 5 seconds
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 2.1
= -1.112 ms-2.

Question 11.
A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h-1 in 10 minutes. Find its acceleration.
Answer:
Initial speed of the train u = 0
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 3
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 3.1
∴ acceleration of a train = 0.0185 ms-2.

Question 12.
What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Answer:
(i) The nature of the distance-time graphs for the uniform motion of an object is a straight line.
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 4
(ii) The nature of tire distance time graph for non-uniform motion is curved line.
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 4.2
Distance-time graph for a car moving with non-uniform speed

Question 13.
What can you say about the motion of an object whose distance-time graph is straight line parallel to the time axis?
Answer:
This graph indicates the object is at rest.

Question 14.
What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Answer:
This graph indicates the object is in uniform motion.

Question 15.
What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer:
The quantity which is measured by the area occupied below the velocity-time graph is length = l

Question 16.
A bus starting from rest moves with a uniform acceleration of 0.1 ms-2 for 2 minutes. Find
(a) the speed acquired,
(b) the distance travelled.
Answer:
(a) Initial velocity of the bus u = 0 (Bus is at rest).
acceleration, a = 0.1 ms-2
time t = 2 minutes = 120 seconds
Let the speed of bus be ‘V’
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 5
∴ v = 12m/s
(b) As per third law of motion:
v2 – u2 = 2as
(12)2 – (0)2 = 2(0. 1)s
∴ s = 720 m.
the speed of acquired by bus = 12 m/s.
the distance travelled = 720 m.

Question 17.
A train is travelling at a speed of 90 kmh-1 Brakes are applied so as to produce a uniform acceleration of -0.5 ms-2. Find how far the train will go before it is brought to rest.
Answer:
Let the initial speed of the train be u = 90 Km/h
= 25 m/s.
Final speed of the train, v = 0 (train comes to rest)
acceleration a = 0.5 ms-2
As per 3rd law of motion
v2 = u2 + 2as
(0)2 = (25)2 + 2(0.5)s
s = train travelled distance
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 6
Train will travel 625 km before it is brought to rest.

Question 18.
A trolley, while going down an inclined plane has an acceleration of 2cms-2 what will be its velocity 3s after the start?
Answer:
Initial velocity of a trolley, u = 0 (at rest)
acceleration, a = 2 cms-2 = 0.02 m/s2 time t = 3S
As per 1 st law of motion
v = u + at
Here V means velocity of a trolley after 3 s
V = 0 + 0.02 × 3
= 0.06 m/s.
∴ = 0.06 m/s.

Question 19.
A racing car has a uniform acceleration of 4ms-2, what distance will it cover in 10s after start?
Answer:
Initial velocity of a racing car, u = 0 (at rest)
acceleration, a = 4 m/s2
time t = 10 s
As per second law of motion
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 7
Here S means distance travelled by car after 10S
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 7.1
∴ Distance covered by car after 10 S = 200 m.

Question 20.
A stone is thrown in a vertically upward direction with a velocity of 5ms-1 If the acceleration of the stone during its motion is 10ms-2 in the downward direction, what will he the height attain by the stone and how much time will it take to reach there?
Answer:
The initial velocity of a stone, u = 5 ms-1.
Final velocity of a stone v = 0 (at rest)
If the acceleration of the stone during its motion is = -10ms-2.
Let the maximum height be ‘s’
v = u + at
0 = 5 + (-10)t
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 8
time taken by stone = 0.5s
As per 3rd Law of motion
v2 = u2 + 2as
(0)2 = (5)2 + 2(-10)s
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 8.1
The height attained by the store = 1.25 m.

KSEEB Solutions for Class 9 Science Chapter 8 Textbook Exercises

Question 1.
An athlete completes one round of a circular track of diameter 200 m. in the 40s. What will be the distance covered and the displace¬ment at the end of 2 minutes 20s?
Answer:
Diameter of circular track, d = 200 m.
radius of the circular track,
r = \(\frac{\mathrm{d}}{2}\) = 100m
Circumference = 2πr = 2p(100) = 200πcm.
in 40s, athlete covers distance 200πm
in 1s, distance covered by athlete = \(\frac{200 \pi}{40}\) m
Athlete runs for 2 minutes 20s means 140s
∴ Displacement of all athlete is equal to the diameter of circle
= \(\frac{200 \times 22}{40 \times 7}\) × 140
= 2200 m
∴ Displacement = 200 m.

Question 2.
Joseph jogs from one end A to the other end B of a straight 300 M road in 2 minutes 30 seconds and then turns around and jogs 100 M back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Answer:
a) from A to B distance covered by Joseph = 300 M.
Time taken to cover this distance
= 2 min 30 sec. = 150 seconds
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 9
Displacement = Nearest distance between A and B = 300 m.
Time taken = 150s
Average velocity = ygy = 2 m/s.

b) From A to C :
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 10
Total distance covered = Distance between A to B + distance between B to C
= 300 + 100 = 400 m.
Total time taken to cover from A to B + Time taken to cover from B to C
= 150 + 60= 210s.
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 21
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 11
Displacement from A to C
= AC + AB – BC
= 300 – 100
= 200m
Total time taken = Time taken to travel from A to B + Time taken to travel from B to C
= 210 s
∴ Average velocity = \(\frac{200}{210}\) =0.952 ms-1 .

Question 3.
Abdul, while during to school, computes the average speed for his trip to be 20Kmh-1 on his return trip along the some route, there is less traffic and the average speed is 30 Kmh-1. What is the average speed for Abdul’s trip?
Answer:
i) Average speed of Abdul’s trip while driving to school = 20 Kmh-1
Average speed = \(\frac{\text { Total distance }}{\text { Total time taken }}\)
Total distance = Distance travelled to reach school = t1
∴ 20 = \(\frac{\mathrm{d}}{\mathrm{t}_{1}}\)
t1 = \(\frac{d}{20}\) …………… (i)
While returning from school
Total distance = Distance travelled while returning from school = d Now total time taken = t2.
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 12

Question 4.
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms-2 for 8.0 s. How far does the boat travel during this time?
Answer:
Initial velocity of motor boat, u =0
(at rest) ,
Accieratlon of a motor boat a = 3m/s2
Time taken, t = 85
According to second equation of motion, s
S = 0 +\(\frac{1}{2}\) × (8)2 = 96m.
∴ Motorboat travels = 96m.

Question 5.
A driver of a car travelling at 52kmh-1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5s. Another driver going at 3kmh-1 in another car applies his brakes slowly and stop In 10s. on the same graph paper. Plot the speed versus time graphs for the two cars, which of the two cars travelled farther after the brakes were applied?
Answer:
initial velocity of A car u1 = 52 kmh-1
= 14.4 m/s.
Time taken to stop car t1 = 5 s.
After 5s car comes to rest Initial velocity of B car, u2 = 3Kmh-1
= 0.833 m/s. = 0.83 m/s.
Time taken to stop car t2 = 10s
After 10s car comes to rest.
Graph:
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 13
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 14
Area of ∆OPR = Area of ∆OSQ
When compared car A to car B, A car travelled father. After brake it travels 52 Kmh-1

Question 6.
Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions.
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 15
a) Which of the three is travelled the fastest?
b) Are all three ever at the same point on the road?
c) How far has C travelled when B passes A?
d) How far has B travelled by the time passes C?
Answer:
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 16
Slope of B object is greater than slopes A and C. This is travelling fastest.

b) All three objects never meet at a point, hence these are not at the same point.

c) There are 7 boxes in X-axis = 4 Km
1 box = \(\frac{4}{7}\) Km
In the beginning, C object is 4 boxes away from 0
= \(\frac{16}{7}\) km
Distance to C from origin = Distance of C when it moves from B to A =8 Km
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 18
∴ Distance travelled by C when B passes A = 5.714 km.

d) Distance B travelled by the time it passes C = 9 boxes
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 19
∴ B has travelled 5.143 k.m. by the time when it passes C.

Question 7.
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10ms-1 with what velocity will it strike the ground? After what time will it strike the ground?
Answer:
Distance ball covers, S = 20 m
acceleration, a = 10Kmh-1 = 10 m/s
initial velocity, u = 0 (at rest)
When ball strikes, its final velocity,
According to 3rd law of motion
v2 = u2 + 2as
v2 = 0 + 2(10)(20)
v = 20 m/s.
According to 1 st Law of motion v = u + at
Time taken by ball to strike the ground, t,
20 = 0 + 10 (t)
∴ t = 2s
∴ 2s is required to ball for striking the ground and velocity is 20 m/s

Question 8.
The speed-time graph for a car is shown in figure
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 17
a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by car during the period.
b) Which part of the graph represents uniform motion of the car?
Answer:
a) Shade lightly in the graph about distance travelled by car in 4s.
b) Shade with red colour about 6 cm to 10 cm. It represents the uniform motion of the car.

Question 9.
State which of the following situations are possible and give an example for each of these;
a) an object with a constant acceleration but with zero velocity.
b) an object moving in a certain direction with an acceleration in the perpendicular direction.
Answer:
a) This is possible
Eg: when a ball is thrown to a height its velocity is zero.
(For a ball G = 9.8 m/s2)

b) This is possible
Eg: when a car moves in a curved line, its acceleration is perpendicular to the given direction)

Question 10.
An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Answer:
Satellite is moving in circular orbit radium of satellite r = 42250 km
time t = 24 Hrs = 3600 × 24 S
KSEEB Solutions for Class 9 Science Chapter 8 Motion Q 20
∴ Speed of the satellite when it moves in circular orbit = 3.07 kms-1.

KSEEB Solutions for Class 9 Science Chapter 8 Additional Questions

Fill in the blanks with suitable words.

Question 1.
The two physical quantities required to know final position of an object are _________ and _________
Answer:
Distance travelled, displacement.

Question 2.
S.I. unit of speed is ________
Answer:
m/s

Question 3.
Distance travelled by the object in unit time is ________
Answer:
Speed.

Question 4.
If acceleration is in the direction of velocity it is _________ acceleration.
Answer:
positive.

Question 5.
When an object moves in a circular path with uniform speed, its motion is called ________
Answer:
uniform circular motion.

Answer the following questions:

Question 1.
Give an express10n for centripetal force?
Answer :
\(f=\frac{m v^{2}}{r}\)

Question 2.
What is uniform circular mot10n?
Answer :
Body moving in a circular path arbitrary any instant along a tangential to the posit10n of the body on the circular path at that instant or time.

Question 3.
A body falls freely. What is constant?
Answer :
When the body falls freely. It has a constant acceleration.

Question 4.
Define uniform speed.
Answer :
The speed of an object is said to be uniform speed if it travels equal distances in equal intervals of time.

KSEEB Solutions for Class 9 Science Chapter 3 Atoms and Molecules

Students can Download Kannada Chapter 3 Atoms and Molecules Questions and Answers, Summary, Notes Pdf, Siri KSEEB Solutions for Class 9 Science, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka Board Class 9 Science Chapter 3 Atoms and Molecules

KSEEB Solutions for Class 9 Science Chapter 3 Intext Questions

Question 1.
In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + ethanoic acid ➝ sodium ethanoate + carbon dioxide + water
Answer:
Sodium carbonate + ethanoic acid ➝ sodium ethanoate + carbon dioxide + water
5.3 + 6 ➝ 8.2 + 2.2 + 0.9
= 11.3 g = 11.3 g
Weight of reactants is equal to weight of products. This observation is in agreement with the law of conservation of mass.

Question 2.
Hydrogen and Oxygen combine in the ratio or 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
Answer:
24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas. Because in water the ratio of the mass of hydrogen to the mass of oxygen is always 1 : 8.

Question 3.
Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Answer:
The relative number and kinds of atoms are constant in a given compound. This postulate is the result of the law of conservation of mass.

Question 4.
Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer:
Atoms combine in the ratio of small whole numbers to form compounds. This postulate explains the law of definite proportions.

Question 1.
Define the atomic mass unit.
Answer:
One atomic mass unit is a mass unit equal to exactly one-twelfth (1/12th) the mass of one atom of carbon-12.

Question 2.
Why is it not possible to see an atom with naked eyes?
Answer:
Atoms are very small, they are smaller than anything that we can imagine or compare with. Therefore it is not possible to see an atom with naked eyes.

Question 1.
Write down the formulae of:

  1. sodium oxide
  2. alluminium chloride
  3. sodium sulphide
  4. magnesium hydroxide

Answer:

  1. sodium oxide: Na2O
  2. aluminium chloride: Al2Cl3
  3. sodium sulphide : NaS
  4. magnesium hydroxide: Mg(OH)2

Question 2.
Write down the names of compounds represented by the following formulae:

  1. Al2(SO4)3
  2. CaCl2
  3. K2SO4
  4. KNO3
  5. CaCO3

Answer:

  1. Al2(SO4)3: Aluminium sulphate
  2. CaCl2: Calcium chloride
  3. K2SO4: potassium sulphate
  4. KNO3: potassium nitrate
  5. CaCO3: calcium carbonate

Question 3.
what is meant by the term chemical formula?
Answer:
The chemical formula of a compound is a symbolic representation of its composition.

Question 4.
How many atoms are present in a
i) H2S molecule and
ii) PO43- ion?
Answer:
i) H2S molecule:
KSSEB Solutions for Class 9 Science Chapter 3 Atoms and Molecules Q 3
ii) PO43- ion
KSSEB Solutions for Class 9 Science Chapter 3 Atoms and Molecules Q 4

Question 1.
Calculate the molecular masses of
H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.
Answer:
i) H2
= 2 × H
= 2 × 1
= 2u

ii) O2
= 2 × O
= 2 × 16
= 32u

iii) Cl2
= 2 × Cl
= 2 × 35.5
= 7lu

iv) CO2
= 1 × 12 + 2 × O
= 1 × 12 + 2 × 16
= 12 + 32
= 44u

v) CH4
= 1 × 12 + 4 × 1
= 12 + 4
= 16u

vi) C2H6
= 2 × 12 + 6 × 1
= 24 + 6
= 30u

vii) C2H4
= 2 × 12 + 4 × 1
= 24 + 4
= 28u

viii) NH3
= 14 × 1 + 3 × 1
= 14 + 3
= 17u

ix) CH3OH
= 12 × 1 + 3 × 1 + 16 × 1 + 1 × 1
= 12 + 3 + 16 + 1
= 32u

Question 2.
Calculate the formula unit masses of ZnO, Na2O, K2CO3 given atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u and O = 16u.
Answer:
i) ZnO
(Automic mass of Zn) + (Automic mass of O)
= 65 + 16 = 81u.

ii) Na2O
(2 × Automic mass of Na) + (Automic mass of O)
= 2 × 23 + 1 × 16
= 46 + 16
= 62u.

iii) K2CO3
(2 × Atomic mass of K)+ (Atomic mass of C) + (3 × Atomic mass of oxygen)
= 2 × 39 + 12 × 1 + 16 × 3
= 78 + 12 + 48
= 138u.

Question 1.
If one mole of carbon atoms weighs 12 gms, what is the mass (in gms) of 1 atom of carbon?
Answer:
Number of moles = n
Given mass = m
molar mass = M
Given the number of particles = N
Avogadro number of particles = N0
i) Atomic mass of carbon = 12u.
Atomic mass of one mole of carbon=12g
But Atomic mass of carbon = 12 g mass of 6.022 × 1023 carbon atoms = 12g.
KSSEB Solutions for Class 9 Science Chapter 3 Atoms and Molecules Q 1
∴ Mass of carbon atoms
= 1.9926 x 10-23 gm.

Question 2.
Which has more number of atoms, 100 g, 100 gms of sodium or 100 gms of sodium or 100 gms of iron (given, the atomic mass of Na=23u, Fe = 56u).
Answer:
Atomic mass of sodium = 23u (data)
It means the gram atomic mass of sodium = 23 gm
Now atoms present in 23 gm sodium = 6.022 × 1023
KSSEB Solutions for Class 9 Science Chapter 3 Atoms and Molecules Q 2.1
It means Number of atoms in 100 gm sodium = 1.6753 × 1024
∴ 100 gm sodium has more number of atoms rather than 100 gms of iron.

KSEEB Solutions for Class 9 Science Chapter 3 Textbook Exercises

Question 1.
A 0.24 g sample of a compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Answer:
Mass of Boron = 0.096 g (Data)
Mass of Oxygen = 0.144g (Data)
Given mass = 0.24 g (Data)
∴ The percentage composition of Boron
KSSEB Solutions for Class 9 Science Chapter 3 Atoms and Molecules Q 1.1
∴ The percentage of O2 = \(\frac{0.144}{0.24}\) × 100
= 60%.

Question 2.
When 3.00 g of carbon is burnt in 8.009 of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50,000 g of oxygen? Which law of Chemical combination will govern your answer?
Answer:
3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.
But when 3 g of carbon is burnt in 50 g of oxygen, only 3 g of carbon reacts with 8 g of oxygen.
Remaining 42 gm of oxygen will not react.
11 gm of carbon dioxide is produced.
∴ Our answer obeys law of constant proportion.

Question 3.
What are polyatomic ions? Give examples.
Answer:
A group of atoms carrying a charge is known as a polyatomic ion.

Question 4.
Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Answer:
(a) Magnesium chloride : MgCl2
(b) Calcium oxide : CaO
(c) Copper nitrate : Cu(NO3)2
(d) Aluminium chloride : AlCl3
(e) Calcium carbonate : CaCO3.

Question 5.
Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
Answer:
(a) Quick lime: Calcium, carbon, oxygen
(b) Hydrogen bromide: Hydrogen, Bromine
(c) Baking powder: Sodium, Bicarbonate
(d) Potassium sulphate: Potassium, Sulphur, Oxygen.

Question 6.
Calculate the molar mass of the following substances.
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus Molecule, P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid HNO3.
Answer:
(a) Ethyne C2H2
Molar Mass = 2 × 12 + 2 × 1
= 24 + 2
= 26 g.

(b) Molar mass of Sulphur molecule
= 8 × 32
= 256 g.

(c) Molar Mass of Phosphorus molecule = 4 × 31
(Atomic mass of Phosphorus) = 124 g.

(d) Molar mass of Hydrochloric acid = HCl
= 1 + 35.5
= 36.5 g.

(e) Molar mass of HNO3
= 1 + 14 + 3 × 16
= 15 + 48
= 63 gm.

Question 7.
What is the mass of –
(a) 1 mole of nitrogen atoms?
(b) 4 moles of aluminium atoms (Atomic mass of Aluminium = 27)
(c) 10 moles of Sodium Sulphite (Na2SO3)
Answer:
(a) Mass of 1 mole of nitrogen= 14g.

(b) Mass of 4 moles of Aluminium
= 4 × 27
= 108 g.

(c) Mass of 10 moles of Sodium sulphite
= 10 × [2 × 23 + 32 + 3 × 16]
= 10 × 126
= 1260 gm.

Question 8.
Convert into mole:
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide
Answer:
(a) 32gm of oxygen means 1 mole
12gm oxygen means \(\frac{12}{32}\) mole
∴ 12g of oxygen gas = 0.375 mole
(b) 18 gm of water means = 1 mole
20 gm of water means \(\frac{20}{18}\) mole = 1.11 mole
(c) 22 g of carbon dioxide means
\(\frac{22}{44}\) = 0.5 mole.

Question 9.
What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer:
(a) Mass of 1 mole of oxygen = 16 g
Mass of 0.2 mole of oxygen = 0.2 × 16
= 3.2 g.
(b) Mass of 1 molecule of water= 18 gm.
Mass of 0.5 mole of water = 0.5 × 18
= 9 gm.

Question 10.
Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.
Answer:
One mole of Sulphur (S) = 8 × 32
= 256 gm.
256 g of solid sulphur = 6.022 × 1023 molecules.
KSSEB Solutions for Class 9 Science Chapter 3 Atoms and Molecules Q 10
= 3.76 × 1022
(Approximate)

Question 11.
Calculate the number of aluminium ions present in 0.0519 of aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u).
Answer:
One mole of Aluminium oxide
= 2 × 27+ 3 × 16
= 54 + 48
= 102 g.
102 g of Al2O3 = 6.022 × 1023 moles (aluminium oxide)
KSSEB Solutions for Class 9 Science Chapter 3 Atoms and Molecules Q 11
It means Aluminium present in 0.051 gm
= 3.011 × 1020 Aluminium oxide molecules
Number of Al ions in one mole of Al2O3 = 2
∴ Number of Al Ions In 3.011 × 1026 molecules
0.051 Al2O3
= 2 × 3.011 × Number of Al ions In 3.011 × 1020
= 6.022 × 1020.

KSEEB Solutions for Class 9 Science Chapter 3 Additional Questions

Question 1.
Write the symbols of the following:
a) Iron
b) lead
c) Zinc
d) Oxygen
e) Chlorine
Answer:
a) Iron = Fe
b) lead = Pb
c) Zinc = Zn
d) Oxygen = O
e) Chlorine = Cl

Question 2.
What is a molecule?
Answer:
The smallest particle of an element or a compound that is capable of independent existence and shows all the properties of that substance.

Question 3.
What is an ion?
Answer:
An ion is a charged species present in metals and non-metals.

Question 4.
What is molecular mass?
Answer:
The molecular mass of a substance is the sum of the atomic masses of all the atoms in a molecule of the substance.

Question 5.
What is a mole?
Answer:
One mole of any species (atoms, molecules, ions, or particles) is that quantity in number having a mass equal to its atomic or molecular mass in grams.

KSEEB Class 9 Science Atoms and Molecules Additional Questions and Answers

Question 1.
An element Z forms an oxide with formula Z2O3. What is its valency?
Answer :
Valency is 3+

Question 2.
Mention the elements present in (1) quick line (2) sodium hydrogen carbonate.,
Answer :
(1) Quick line (calcium oxide) (CaO) element present are calcium and oxygen.
(2) Sodium hydrogen carbonate (NaHCO3) elements are sodium, hydrogen, hydrogen, carbon and oxygen.

Question 3.
Calculate the total number of ions in 0.585 g of sodium chloride.
Answer :
Gram formula mass of NaCl = 23 + 35-5 = 58.5 g
58.5 g of NaCl have ions = 2 × NA
58.5 g of NaCl have ions
\(=2 \times \mathrm{N}_{\mathrm{A}} \times \frac{0.585}{58 \cdot 5}=0 \cdot 02 \times \mathrm{N}_{\mathrm{A}}\)
= 0.02 × 6.022 x 1023
= 1.20 × 1022 ions

 

Karnataka Class 10 Hindi Solutions वल्लरी Chapter 12 रोबोट

You can Download रोबोट Questions and Answers Pdf, Notes, Summary KSEEB SSLC Class 10 Hindi Solutions to help you to revise complete Syllabus and score more marks in your examinations.

रोबोट Questions and Answers, Notes, Summary

अभ्यास

I. एक वाक्य में उत्तर लिखिए :

प्रश्न 1.
वर्षों से सक्सेना के परिवार में कौन काम कर रहा था?
उत्तर:
वर्षों से सक्सेना के परिवार में साधोराम सेवक काम कर रहा था।

प्रश्न 2.
धीरज सक्सेना किस कार्यालय में जा पहुँचे ?
उत्तर:
धीरज सक्सेना रोबोटोनिक्स कॉरपोरेशन के कार्यालय में जा पहुँचे।

प्रश्न 3.
एक रोबोट वैक्यूम क्लीनर से क्या साफ कर रहा था?
उत्तर:
एक रोबोट वैक्यूम क्लीनर से दफ्तर के फर्श को साफ कर रहा था।

प्रश्न 4.
रोबोनिल की मुलाकात किससे हुई ?
उत्तर:
रोबोनिल की मुलाकात रोबोदीप से हुई।

प्रश्न 5.
शर्मा परिवार के कुत्ते का ना लिखिए।
उत्तर:
शर्मा परिवार के कुत्ते का नाम शेरू

प्रश्न 6.
रोबोनिल और रोबोदीप किससे मिलने गए ?
उत्तर:
रोबोनिल और रोबोदीप कंपनी के मालिक रोबोजीत से मिलने गए।

प्रश्न 7.
वैज्ञानिक लेखक का नाम लिखिए।
उत्तर:
वैज्ञानिक लेखक का नाम आइजक आसिमोव था।

II. दो-तीन वाक्यों में उत्तर लिखिए:

प्रश्न 1.
साधोराम को क्या हुआ था?
उत्तर:
साधोराम सक्सेना परिवार में वर्षों से काम कर रहा था। एक दिन अचानक चलती बस से गिरकर उसे खतरनाक चोट आई। उसे अस्पताल में भर्ती कराना पडा।

प्रश्न 2.
धीरज सक्सेना को बुद्धिमान रोबोट की जरूरत क्यों थी ?
उत्तर:
साधोराम के अस्पताल जाने से धीरज सक्सेना के परिवार की तकलीफें अधिक बढ़ गई। सक्सेनाजी से परिवारवालों का यह दुःख देखा नहीं गया। उन्हें उनके नाती-पोतों का होमवर्क कराने एवं वर्ड प्रोसेसर पर काम सँभालने के लिए बुद्धिमान रोबोट की जरूरत थी।

प्रश्न 3.
रोबोनिल ने रोबोनिल से क्या कहा ?
उत्तर:
रोबोदीप ने रोबोनिल से कहा कि उनके नियम कुछ भी हो पर हम किसी इन्सान के नुकसान का कारण बने यह सही नहीं है और किसी इन्सान की नौकरी को खतरा पहुंचे यह गलत है।

प्रश्न 4.
रोबोनिल ने रोबोजीत को क्या समझाने की कोशिश की ?
उत्तर:
रोबोनिल ने रोबोजीत को यह समझाया कि उनके कारण अगर किसी को नुकसान पहुँचता है तो वह सही नहीं है। यंत्रमानव और मानव नजरिये से यह गलत है कि किसी भी इन्सान की नौकरी को खतरा पहुँचे। पर यह बात रोबोजित के समझ से बाहर थी।

प्रश्न 5.
कहानी को टाइप करते समय रोबोनिल को क्या हुआ ?
उत्तर:
धीरज सक्सेना ने रोबोनिल को एक विज्ञान कथा वर्ड प्रोसेसर पर टाइप करने के लिए दी। कहानी टाइप करके रोबोनिल की धात्विक और तारों भरे परिपंथवाली खोपड़ी में यकायक मानों नीली रोशनी हो गई। रोबोदीप के साथ रोबोटिक संघ से संपर्क साधने के बारे में सोचने लगा।

III. पाँच-छः वाक्यों में उत्तर लिखिए:

प्रश्न 1.
धीरज सक्सेना ने घरेलू कामकाज के लिए रोबोट रखने का निर्णय क्यों किया ?
उत्तर:
वर्षों से सक्सेना परिवार में साधोराम काम कर रहा था। अचानक एक दिन चलती बस से गिरकर उसे खतरनाक चोट आ गई। वह अस्पताल में भर्ती हो गया। सक्सेना परविार को साधोराम के अस्पताल भर्ती होने से बहुत तकलीफें सहन करनी पड़ी। परिवार बडा था। घरेलु कामकाज के लिए रोबोट रखने का निर्णय लिया।

प्रश्न 2.
रोबोनिल और रोबोदीप की मुलाकात का वर्णन कीजिए।
उत्तर:
रोबोनिल घरेलू कुत्ते शेरू को घुमाने ले जाता था। उसकी मुलाकात एक दिन रोबोदीप से हुई जो शर्मा परिवार के कुत्ते झबरू को घुमाने लाता है। दोनों के बीच दोस्ती हो गयी। एक दिन रोबोदीप ने रोबोनिल को साधोराम का किस्सा सुनाया। और बताया कि अब सक्सेना परिवार उसे काम पर नहीं रखेगा। वे उसकी छुट्टी करनेवाले हैं। रोबोनिल ने कहा कि यह रोबोटिकी के नियम के विरुद्ध है। दोनों साधोराम के लिए कुछ करने का सोचते है।

प्रश्न 3.
विज्ञान कथा का सार लिखिए।
उत्तर:
विज्ञान कथा रोबोट यंत्रमानव की कहानी है। इसमें रोबोनिल और रोबोदीप यंत्रमानव हैं, जो काम पर लग जाते है। अपने अपने परिवार में इन दोनों ने कार्यभाग अच्छी, तरह से सँभला था। एक दिन इन दोनों के बीच मुलाकात होने पर पता चलता है कि उनकी वजह से साधोराम सेवक की नौकरी चली जाएगी। अपनी वजह से किसी की नौकरी चली जाएगी इस बात का उनको बडा दुःख होता है। दोनों अपने संघर्ष दवारा साधोराम की नौकरी वापस दिलाते है। यंत्रमानव को खुशी होती है कि उनसे इंमानों को कोई नुकसान नहीं सेलना पडा।

प्रश्न 4.
रोबोटिक कंपनियों के मालिकों के बीच हलचल क्यों मच गई ?
उत्तर:
सक्सेना परिवार का नौकर साधोराम अचानक एक दिन चलती बस से गिरकर उसे खतरनाक चोट आती है। उसके अस्पताल पहुंच जाने से घर के सभी की तकलीफें बढ़ जाती है। धीरज सक्सेना से परिवारवालों का यह दुख नहीं देखा जाता है। घरेलू कामकाज के लिए साधोराम को निकालकर रोबोट को रख दिया जाता है। रोबोट को जब यह पता चलता है तो वह रोबोटिक संघ से मिलकर रोबोटो की हड़ताल की घोषणा करवा देता है। इस आहवान से रोबोटिक कंपनियों के मालिकों के बीच हलचल मच जाती है।

प्रश्न 5.
सक्सेना परिवार रोबोनिल को पाकर बहुत खुश था। क्यों? विवरण दीजिए।
उत्तर:
रोबोनिल सक्सेना परिवार के सभी लोगों का काम करता था। वह सुबह नाश्ता कराने, मेहमानों के लिए द्वार खोलने, घर के छोटे बच्चों को कहानियाँ सुनाने, उनके होमवर्क कराने, सक्सेना जी के वर्ड प्रोसेसर पर काम करने के अलावा शाम को परिवार के कुत्ते को घुमाने भी ले जाने का काम करता था। इसलिए सक्सेना परिवार रोबोनिल को पाकर बहुत खुश था।

IV. निम्नलिखित कारकों को चुनकर लिखिएः (का, के, में, से, की)

  1. वर्ष 2030 ……… नवंबर का महीना था।
  2. आप साधारण रोबोट ………. भी काम चला सकते है।
  3. रोबोनिल ……… घर में आ जाने से सभी ने राहत की साँस ली।
  4. रोबोटों ………. हडताल की घोषणा हुई।
  5. संघ को यह बात मानने …….. कोई आपत्ति नहीं थी।

उत्तर:

  1. के
  2. से
  3. के
  4. से
  5. में

V. जोडकर लिखिए:

1. यह सुनकर काउंठ पर  गुप्त मंत्रणा हूं।
2. मगर, रोबेदीप, यह तो पाकर फूला नहीं समा रहा था।
3. शाम को रोबोनिल बैठ रघट बोला।
4. दोनों के बीच कुछ रोबोटिकी के नियम के रुिद्ध है।
5. सक्सेना परिवार
रोबोनिल के
और रोबोदीप मिले।

उत्तर – जोडकर लिखना

1. यह सुनकर काउंठ पर  बैठ रघट बोला।
2. मगर, रोबेदीप, यह तो रोबोटिकी के नियम के रुिद्ध है।
3. शाम को रोबोनिल और रोबोदीप मिले।
4. दोनों के बीच कुछ  गुप्त मंत्रणा हूं।
5. सक्सेना परिवार रोबोनिल के  पाकर फूला नहीं समा रहा था।

VI. अन्य वचन रूप लिखिए:

  1. बेटा – बेटे
  2. नाती – नाती
  3. कुत्ता – कुत्ते
  4. दुट्टी – छुट्टियाँ
  5. बेटियाँ – बेटी
  6. पोता – पोते
  7. कंपनियाँ – कंपनी
  8. नौकरियाँ – नौकर

VII. अनुरूपता :

1. शेरू के टहलाना : रेम्बोनिल :: झबरू को घुमाना : ——–
2. मुखिया : धीरज सक्सेना :: सेवक : ——
3. टस से मस न होना : अटल रहना. :: फूले न समाना : ———
उत्तर:

  1. रोबोदीप
  2. साधोराम
  3. बहुत खुशहोना

भाषा ज्ञान :

मुहावरे :
जो वाक्यांश अपने सामान्य अर्थ को छोडकर किसी विशेष अर्थ को प्रकट करता है, वह मुहावरा कहलाता है।
उदाहरण :
आँखें चुराना – अपने आप को छिपाना
आँखें दिखाना – धमकाना, डराना
अक्ल का अंधा – मूर्ख
आस्तीन का साँप – कपटी मित्र
कान भरना – चुगली करना

I. उदाहरण के अनुसार मुहावरे लिखिए :

प्रश्न 1.
शरीर के अंगों से संबंधित – उदा : अँगूठा दिखाना – साफ इनकार करना
उत्तर:
पेट में चूहे दौडना – बहुत भूख लगना
अँगूठा दिखाना – साफ इनकार करना
आँख लगना – नींद आना

प्रश्न 2.
अंको से संबंधित – उदा : नौ दो ग्यारह होना – इधर-उधर भाग जाना ।
उत्तर:

  1. सवासोलह आने – पूरी तरह
  2. चार चाँद लगना – प्रतिष्ठा बढना
  3. छक्के छुडानां – बुरी तरह हराना

II. मुहावरों का अर्थ लिखकर वाक्यों में प्रयोग कीजिए

  1. आँख खुलना – सच्चाई का पता चलना।
  2. ईद का चाँद होना – बहुत दिनों के बाद दिखाई देना।
  3. कान खडे होना – सचेत होना।
  4. वा से बातें करना – तेज भागना।
  5. बात का धनी – बात का पक्का।

III. मुहावरों को सही अर्थ के साथ जोडकर लिखिए :

1. राहत की साँस लेना विचलित न होना।
2. पेट पर लात मारना चैन की साँस लेना/तसल्ली कना
3. टस से मस न होना आश्चर्यचकित होना
4. फूला नहीं समाना नौकरी या सहूलियत छैनना
5. आँच आना वक्त आने पर इनकार करना।
6. अँगूठ दिखा देना हानि पहुँचना
7. हलचल मचाना बहुत सुख होना
8. हाथों के तोते उङमा  शोर मचाना

उत्तर – जोडकर लिखना

1. राहत की साँस लेना चैन की साँस लेना/तसल्ली कना
2. पेट पर लात मारना नौकरी या सहूलियत छैनना
3. टस से मस न होना विचलित न होना।
4. फूला नहीं समाना बहुत सुख होना
5. आँच आना हानि पहुँचना
6. अँगूठ दिखा देना वक्त आने पर इनकार करना।
7. हलचल मचाना शोर मचाना
8. हाथों के तोते उङमा आश्चर्यचकित होना

IV. निम्नलिखित मुहावरों से वाक्य पूरा कीजिए:
(साँस रोके हुए, नौ दो ग्यारह हो जाना, गुस्सा हवा हो जाना, चिंगारियाँ सुलगना, दाँतों तले उँगली दबाना, भूचाल आ जाना)

  1. मास्टर साहब की आँखों में चिंगारियाँ रही थी।
  2. मालिक को देखकर ङ-इवर का गुस्सा हवा हो गया।
  3. अपने सामने शेर को देख मैं काफी देर तक साँस रोके खडा रहा।
  4. अध्यापिका की अनुपस्थिति में बच्चों ने कक्षा में इतना शोर मचा रखा था, ऐसा लग रहा था, मानो | कक्षा में भूचाल हो।
  5. एक व्यक्ति को अपने मुँह से ट-क खींचते देख हम दाँतों तले ऊँगली दबाकर रह गए।
  6. बिल्ली को सामने से आता देख चूहा नौ दो ग्यारह हो गया।

रोबोट Summary in Hindi

रोबोट पाठ का सारांशः
जब वर्षों से सक्सेना परिवार में काम कर रहा साधोराम एक दिन बस से गिरकर चोटिल हो गया तो सक्सेना परिवार पर मुसीबत का पहाड़ टूट पड़ा। सभी की तकलीफें बढ़ गई। धीरज सक्सेना से यह दुःख देखा नहीं गया और वे ‘रोबोटोनिक्स कॉरपोरेशन’ के कार्यालय जा पहुंचे।

यह 2030 के नवंबर की बात है। वैज्ञानिक रोबोट में कृत्रिम बुद्धि के साथ-साथ मानवीय संवेदना के गुण भी पैदा करने में सफल हो चुके थे। काउंटर पर बैठे एक रोबोट ने धीरज सक्सेना का स्वागत किया और उनसे आने का कारण पूछा। सक्सेना बोले – मुझें घरेलू कामकाज के लिए एक बुद्धिमान रोबोट चाहिए जो मेरे नाती-पोतों का होमवर्क करा सकें और मेरा कम्प्यूटर का काम भी कर सकें।

रोबोट Summary in Hindi 1

उस दिन कंपनी का एक कर्मचारी सक्सेना परिवार में काम करने के लिए रोबोनिल को छोड़ गया। रोबोनिल के आने से सभी ने राहत की सांस ली। सुबह नाश्ता कराने, द्वार खोलने, बच्चों को कहानियाँ सुनाने, होमवर्क में मदद और सक्सेना के वर्ड प्रोसेसर पर काम करने एवं शाम को पालतू कुत्ते शेरू को घुमाने का काम वह करने लगा।

रोबोनिल की मुलाकात एक दिन शेरू को घुमाते समय रोबोदीप से हुयी। दोनों पक्के दोस्त बन गए। एक दिन रोबोदीप ने रोबोनिल को साधोराम की बात बताई। उसने बताया साधोराम सक्सेना परिवार का पुराना नौकर था। अब वह एक दुर्घटना के कारण अस्पताल में है। अब सक्सेना जी उसकी छुट्टी करने वाले हैं क्योंकि तुम उन्हें भा गए हो। वे साधोराम को मुआवजा देकर गाँव भिजवा देंगे।

रोबोट Summary in Hindi 2

रोबोनिल इसे रोबोटिकी के नियम के खिलाफ बताता है। यह वैज्ञानिक एवं विज्ञान लेखक आइजक आसिमोव के रोबोटिकी नियम के खिलाफ है कि कोई रोबोट किसी इंसान के नुकसान का कारण बने, इंसान की नौकरी को खतरा पहुंचे।

दोनों दोस्त तय करते है कि वे साधोराम के लिए कुछ करेंगे। लेकिन कंपनी के मालिक अनुबंध की शर्तों को तोड़ने से मना कर देते हैं और मदद करने से मना कर देते है। रोबोनिल बहुत निराश होता है। एक दिन धीरज सक्सेना ने एक विज्ञान कथा टाइप करने को दी। इस कथा में एक इंसान की नौकरी के लिए ‘रोबोटिक संघ’ द्वारा हड़ताल का वर्णन था।

रोबोट Summary in Hindi 3

दूसरे ही दिन रोबोनिल ने रोबोटिक संघ से मिलकर साधोराम के समर्थन में हड़ताल करवा दी। रोबोजीत को संघ की शर्ते माननी पड़ी और उसने सक्सेना परिवार से हुए अनुबंध को रद्द कर दिया। मजबूर होकर धीरज सक्सेना ने संघ के अध्यक्ष से गुजारिश की कि जब तक साधोराम पूरी तरह से ठीक नहीं हो जाता है, तब तक रोबोनिल उसके पास काम करता रहे। साधोराम के ठीक होते ही वह उसे दोबारा काम पर रख लेगा। संघ ने यह बात मान ली। इस प्रकार रोबोनिल और रोबोदीप के संघर्ष से साधोराम की नौकरी बच जाती है।

रोबोट Summary in English

ROBOT Summary in English:

Dr. Pradeep Mukhopadhay ‘Alok’ The writer has given a message that machine cannot replace humans and humans
will not lose their proting given a mess will not lose their ‘work.

Sadhoram was working with the Saksena family for many years. One day he met with an accident falling from a moving bus. He was hospitalized. Saksena’s family was big. With the hospitalization of Sadhorm all the members of the family faced difficulty. Dhiraj Saksena wanted to mitigate the difficulty of the family members and hence contacted the ‘Robotonics Corporation’ for help.

In the year 2030, Various robots were engaged in cleaning activities in Robotonics corporation: The Robots were given artificial intelligence and some of them had human qualities also.

A Robot is the corner offered Saksena help and enquired about his requirement. Saksena wanted an intelligent Robot for household work his own personal work etc. Saksena hired an intelligent Robot for work.

Robonil, an intelligent Robot started working in Saksena household and all the members of the family were very happy as Robonil did all the household work and also assisted Saksena.

Robotnik ‘used to take the dog for a walk in the evenings and met another Robot named Robodeep, who was working in the neighbourhood.

One day Robodeep informed Robonil the shay of Sadhoram, who was hospitalised Robodeep also informed Robonil that Saksena’s family was planning to remove Sadhoram from his employment. Robotnik felt guilty and very bad as it was against the principles of Robotics that any human he coming Jobless became of Robots, they discussed among themselves and come to a conclusion.

Next day both the Robots Robonil and Robodeep went to Robotonics corporation and narrated the entire happenings to Robojeeth, the owner of ti withdraw the services of Robots so that Sadhoram’s Job could be saved. But Robojeeth did not agree as he would lose business and lose profit. Both the Robots when disappointed and returned to their employers.

Robonil was very sad and worried about sadhoram’s losing Saksena gave a scientific strong and asked Robonil to type in the computer the story mentioned that a person was seriously ill and hence replaced by a Robot. Even after recovery, he was not taken back and the services of a Robot continued. The Robot contracted the Robot Association and arranged to give a call for strike. Due to the threat of strike the worker was reemployed. After reading the story Robonil was excited. Now he knew what to do to help Sadhoram.

Next day Robonil and Robodeep approached their Association, briefed them, and requested for help. The Robotic Association gave a call for strike accordingly.

रोबोट Summary in Kannada

रोबोट Summary in Kannada 1
रोबोट Summary in Kannada 2
रोबोट Summary in Kannada 3
रोबोट Summary in Kannada 4
रोबोट Summary in Kannada 5

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