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Students can Download 2nd PUC Physics Model Question Paper 2 with Answers, Karnataka 2nd PUC Physics Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Physics Model Question Paper 2 with Answers

Time: 3 Hrs 15 Min
Max. Marks: 70

General Instructions:

1. All parts are compulsory.
2. Answers without relevant diagram/figure/circuit wherever necessary will not cany any marks
3. Direct answers to the Numerical problems without detailed solutions will not carry any marks.

Part – A

I. Answer all the following questions ( 10 × 1 = 10 )

Question 1.
Write the SI unit of Electric field.
Answer:
Newton per coulomb (N/C)

Question 2.
When will the magnetic force on a movi¬ng charge be maximum in a mag notified?
Answer:
F = qVBsinθ , If θ = 90° (F_m – qυB)
The magnetic force is maximum only when moving charge is ⊥^r to field.

Question 3.
Where on the Earth’s surface is the magnetic dip zero?
Answer:
At equator, dip is zero

Question 4.
State Curie’s law in magnetism.
Answer:
The magnetic susceptability of a para magnetic substance varies inversely to its absolute temperature(T).
i.e x α \(\frac{\mathrm{C}}{\mathrm{T}}\) (x = \(\frac{\mathrm{C}}{\mathrm{T}}\))

Question 5.
What is the significance of Lenz’s law?
Answer:
Law of conservation of energy.

Question 6.
Write the formula for Malus law.
Answer:
i = i₀cos²θ
Where I → is intensity of the light transmitted by the analyser.
I₀ → is intensity of the light incident on the analyser.
θ → is angle between the pass axes of the analyser & polariser.

Question 7.
What is the ratio of the nuclear densities of two nuclei having mass numbers in the ratio 1:3?
Answer:

Nuclear density is independent of mass number & is approximately constant.
Hence \(\frac{P_{1}}{P_{2}}\) = 1

Question 8.
Define current amplification factor in a common – emitter mode of transistor.
Answer:
It is defined as the ratio of change in collector current ( _ΔI_c )to the change in base current (_ΔI_b) at constant collector emitter voltage (V_CE) when the transistor is in active state

Question 9.
Write the truth table of NAND gate.
Truth table of NAND gate

Question 10.
Why sky wave propagation is not possi¬ble for waves having frequency more than 30 MHz?
Answer:
The sky wave range is short & it is used in short wave broad cast service.

Part – B

II. Answer anyfive ofthe following questions. ( 5 × 3 = 15 )

Question 11.
Sketch the electric lines of force due to a point charge q. If i) q<0 and ii) q>0
Answer:

Question 12.
A galvanometer having a coil of resistance 12 Ω gives full scale deflection for a current of 4 mA. How can it be converted into a voltmeter of range 0 to 24V?
Answer:
Given G = 12 Ω, I_g = 4 x 10^-3A. V = 24V
We have,
\(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{B}}}-\mathrm{G}\)

R = 5988Ω
A resistance of 5988 Ω must be connected in series with galvanometer.

Question 13.
Distinguish between paramagnetic and ferromagnetic substances.
Answer:

Paramagnetic substances	Ferromagnetic Substances
1. Paramagnetics are feebly attracted by magnets.	1. They are strongly attracted by magnets.
2. Magnetic permeability is slightly greater than one i.e. Mr>1	2. Relative permeability is greater than 1000
3. Magnetic susceptability is low & positive.	3. Magnetic susceptability is +ve & large.

Question 14.
What is meant by Self Inductance and Mutual Inductance?
Answer:

1. The phenomena in which an emf is induced in a coil due to change of current through the same coil is known as self-induction.
2. The phenomena in which an emf is induced in one coil due to change of current is the neighbouring coil is called mutual induction

Question 15.
What are electromagnetic waves? Write the expression for the velocity of electro magnetic waves in terms of permittivity and magnetic permeability of free space.
Answer:
The waves in which there are a sinusoidal variation of electric & magnetic field vectors at right angles to each other & also right angle to the direction of propagation of the wave is called e.m waves.
∴ The expression for velocity of e.m. wave is
\(C=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\) Where µ₀ → is permeability of free
space, ε₀ → is the permittivity of free space.

Question 16.
Write the relation between the path difference and wavelength of light wave used for constructive and destructive interference of light.
Answer:
a) For constructive interference
∴ Path difference = 2n\(\frac{\lambda}{2}\) = nλ
Wheve λ → is wave length of light used
n → 0, 1, 2,
b) For destructive interference :
∴ Path difference = (2n+l)\(\frac{\lambda}{2}\)
Wheve λ → is wave length of light used n → 0, 1, 2,………

Question 17.
Define :
i) photoelectric work function
ii) electron volt (ev)
i) Photoelectric work fraction : the minimum energy required to remove an electron from the metal surface is called work function, i.e. w = hV₀
Where h → is planck’s constant
V₀ → is threshold frequency
ii) Electron volt is the kinetic energy gained by an electron when it is accelerated through a potential difference of 1 volt.
i.e. lev = 1.602 × 10^-19J.

Question 18.
Draw the block diagram of a AM receiver.
Answer:

x → Intermediate frequency stage
y → power amplifier

Part – C

III. Answer any five of the following questions. ( 2 × 5 = 10 )

Question 19.
Derive an expression for potential energy of a system of two charges in the absence of external electric field.

Consider 2 point charges q₁ & q₂ are separated by a distance ‘r’ are as sho in the fig.
∴ The change q₁ is bringing from ∞ to given point A, No work done i.e. w₁= 0 |||ly the charge q₂ is bringing from ∞ to given point ‘B’ against us field q₂
∴work is done it is given by
i.e. w₂ = vq₂

Question 20.
Arrive at an expression for drift velocity.

Controller a metalic conductor is connected to a battery
Let \(\overrightarrow{\mathrm{E}} \rightarrow\) is E.f. setup inside the conductor
m → is mass of electom
e → charge of free elector
V_d → is drift velovity
τ → is ralaxation time
the force experienced by an electron in the field is given by \(\overrightarrow{\mathrm{F}}=-\mathrm{e} \overrightarrow{\mathrm{E}}\)
-Ve sign shows that divertion of \(\overrightarrow{\mathrm{E}}\) & \(\overrightarrow{\mathrm{F}}\) are in apposite each other

Question 21.
State and explain Gauss law in magnetism.
Answer:
The net magnetic flux through any closed surface is always zero
\(\sum \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{As}}=0\)

consider a closed surface
S in a uniform
\(\mathrm{M} . \mathrm{F} \cdot \overrightarrow{\mathrm{B}}\) Let \(\overrightarrow{\Delta \mathrm{S}}\)
be a small area element of this surface with \(\hat{n}\)
along its normal

Question 22.
Derive the expression for motional emf induced in a conductor moving in a uniform magnetic field.
Answer:
Consider a straight metallic rod PQ of length ‘t’ placed in a uniform M.F. \(\overrightarrow{\mathrm{B}}\). The rod is moved with a velocity \(\overrightarrow{\mathrm{v}}\) is a direction ⊥^r to
\(\overrightarrow{\mathrm{B}}\) Let the rod moved through a distance ‘x’ in time‘t’ them the area covered by the rod is A= 1 xx The magnetic flux linked with the rod is
Φ = B.A Φ = B1x
∴ The included emf in the rod is

Motional emf (e – Blv)
∴ \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\)

Question 23.
With a diagram, explain the working of a transformer.

Answer:
AC voltage is applied to the primary, it creates a varying M.F. & hence a changing magnetic flux in the core. Since the secondary coil is magnetically coupled, due to the mutual induction the changing flux causes induced emf in it. Thus, the power is transferred from primary to secondary.

Question 24.
What is total internal reflection? Mention two applications of optical fibres.
Answer:
The phenomena of complete reflection of light at the interface of two optical media when a ray of light travelling in denser medium is called total internal reflection.

• Optical fibre are used in telecommunication
• It is used to measure rate of flow of blood.

Question 25.
What are the matter waves? Write the expression for De – Broglie wavelength of a particle and explain the terms.
Answer:
The waves associated with material particles in motion is called matter waves. The expression for de Broglie wavelength is \(\lambda=\frac{\mathrm{h}}{\mathrm{mv}}\)
where
h → is Planck’s constant
m → is mass of moving particle
v → is the velocity of moving particle.

Question 26.
Write three differences between n-type and p-type semiconductors.
Answer:
P – type S.C.

1. It is a semiconductor doped with trivalent impurities.
2. Majority charge carriers are holes
3. The impurit atom is called acceptor impurity.

n-type S.c

1. It is a semiconductor doped with pentavalent impurities.
2. Majority charge carriers are electron
3. The impurity atom is called Sonar impurity

Part – D

IV. Answer any two of the following questions  ( 2 × 5 = 10 )

Question 27.
Derive an expression for the electric field at a point due to an infinitely long thin charged straight wire using Gauss law.
Answer:
Consider an infinitely long thin straight wive with uniform linear charge q density λ. Let P be a point at ⊥^r distance r from the wire.

To calculate the E.F \(\overrightarrow{\mathrm{E}}\) at P, imagine a cylindrical Gaussian surface.
∴ The surface area of the] curved part S = 2πrl
Total charge enclosed by the Gaussian surface q = λl
Electric fliix through the end Surfaces of the cylinder is Φ = 0
Electric flux through the curved Surfaces of the cylinder is Φ₂ = Ecosθ.s
Φ₂ = EX1X2πrl
The total electric flux Φ = Φ₁ + Φ₂
Φ = 0 + E2πrl, Φ₂= 2πrlE …………… 1
A/C to Gauss law,

from (a) and (2)

Question 28.
Obtain the bridge balanced condition of Wheatstone’s bridge network by applying Kirchhoff’s rules.
Answer:
Wheatstone’s bridge is a device used to determine unknown resistance.

It resistances R₁, R₂, R₃ & R₄ a galvanometer of resistance G & a battery with key. Let I is the math current & splits into branch current I₁ & I₂ respectively
i.e. I = I₁ +I₂
At B, I₁ = I₃+ I_g………….(1)
A + D, I₂ = ₄ + I_g ………. (2)
Apply KVL to the mesh ABDA
I₁R₁ + i_gG – I₂R₂ = o ………….. (3)
to the mesh BCDB
I₃R₃ – I₄R₄– I_gG = 0 ……………… (4)

The wheatstones bridge is said to be balanced
when no current flowing throught the galvano
meter i.e. I_g 0 ∴ eqn (1) is I₁ = I₃
eqn(2)is I₂ = I₄
eqn (3) is I₁R₁ – I₂R₂ = 0
∴ I₁R₁ = I₂R₂ …………. (5)
eqn (4) is I₃R₃ – I₂R₂
∴I₃R₃ = I₄R₄ …………………. (6)

Question 29.
Two straight parallel conductors are placed at certain distance in free space. The direction of current in both the conductors is same. Find the magnitude and direction of the force between them. Hence define ampere.
Answer:

Consider two infinitely long straight parallel condu ctors x & y carrying a current I₁ & I₂ respectively. Let d is the ⊥^r distance between them.
Let I₁ is the current flowing through x conductor, it produces a M.F. (B₁)

Now the conductor Y carrying current I₂ in the M.F.B, it experiences a mehanical force of length 1 is

The direction of force can be obtained using ampere’s left-hand rule.
∴ The force bn. two ||le conductors carrying currents in the same direction is attractive. ||ly the force bn two ||le conductors carrying current in opposite direction is repulsive.
Definition of ampere If I1 = I2 = IA, d=lm them F = 2 × 10^-7N.

Ampere is the steady current which when flowing through each of 2 infinitely long straight ||le conductors placed in the air at a distance of lm produces a force of 2 × 10^-7 N/m.

V. Answer any two of the following question ( 2 × 5 = 10 )

Question 30.
Derive Lens Maker’s formula for a con vex lens.
Answer:

Consider a thin convex lens of focal length f & R.I (n) placed in air as shown in fig let R₁ & R₂ → are the of curvatur of the surfaces ABC & ADC of the respectively.
o → Luminous point object on the principle axis.
A ray op invident at p, after refraction, emerges along QI & meet at I on the principal axis.
Image formution takes place in two stages. (1) Refraction at the surface ABC In the observe of ADC, the refracted ray is meet at I¹, then

(2) Refraction at the surface ADC the image I1 acts as a virtual object to form a real image at a distance V, then

Question 31.
Assuming the expression for radius of the orbit, derive an expression for total energy of an electron in hydrogen atom.
Answer:
Consider an electron of mass m, charge, e revoking around the nucleus of radius of r. The charge on the nucleus + Te.
∴ T.E = KE + P.E
T. E = K + U ………… (1)
For circulation
Centripetal force = Electrostate force bn nucleus & electron.


The Potential energy of the electron is the field o nucleus is

Question 32.
With the help of circuit diagram, explain the working of NPN transistor as a common emitter amplifier.
Answer:
The circuit diagram of a CE amplifier using NPN transistor is as shown m fit. The input circuit is forward biased & the output circuit is reverse biased when the ac input signal to be amplified is fed to the base-emitter circuit. The output voltage V0 varies in accordance with the relation, V₀=V_CE =V_cc – I_cR_c, These variations is the collector voltage V_CE uppers as amplified output.

During the -t-ve half cycle of ac input signal the forward base of emitter as junction increase Due to this base current IB increase & hence collector current IC increases. As a result of this ICRC increases output voltage VO this indicates that the +ve half cycle of input ac signal voltage is amplified through -ve half cycle.

During the -ve half cycle of ac input signal, the forward bias of emitter-base junction decreases. Due to this base current I_B decreases & hence collector current I_C decreases. As a result of this ICRC decreases the. output voltage V₀ is +ve. This indicates that the -ve half cycle of input ac signal voltage is amplified through +ve half cycle.

Thus, the weak input signal is amplified & output signal is out of phase with the input signal by 180°

VI. Answer any three of the following questions ( 3 × 5 = 15 )

Question 33.
Charges 2 µ C, 4 µ C and 6 µ C are placed at the three corners A, B and C respectively of a square ABCD of side x metre. Find, what charge must be placed at the fourth corner so that the total potential at the center of the square is 0
Answer:

In the Δle ABC
AC² = AB² + BC²
AC² = x² + x²
AC² = 2x²
∴ AC = √2 x,.m

Question 34.
A wire having length 2.0m, diameter 1.0 mm and resistivity 1.963 × 10^-8 Ω m is connected in series with a battery of emf 3 V and internal resistance 1 Calculate the resistance of the wire and current in the circuit.
Answer:
Given,
length (l) = 2m
Diameter (D) = 1mm = 1 × 10^-3 m
∴ radius (r) = \(\frac{\mathrm{D}}{2}\) = 0.5 × 10^-3 m
resistivity (f) = 1.963 × 10^-8Ωm
E = 3V Internal senstarce (r) = 1 Ω
R=? I = ?
We have,

Question 35.
An inductor and a bulb are connected in series to an AC source of 220 V, 50 Hz. A current of 11A flows in the circuit and phase angle between voltage and current is \(\frac{\pi}{4}\) radians. Calculate the impedance and inductance of the circuit.
Answer:
Given,
V_rms = 220v, f = 50 Hz
I_rms = 11A
Φ = \(\frac{\pi}{4}\) radian, Z=?, L =?

Question 36.
In Young’s double slit experiment while using a source of light of wavelength
4500 Å, the fringe width is 5mm. 1f the distance between the screen and the plane of the slits ¡s reduced to half, what should be the wavelength of light to get fringe 4 mm?
Answer:
Given,
wavelength (λ) 45OOA°= 45OO × 10^-10 m
fringe width (w) = 5mm = 5 × 10^-3 m
D = \(\frac { D }{ 2 }\)
wavelength (λ¹) = ?
fringe width (w¹) = 4mm = 4 × 10^-3 m
we have, fringe width (w) = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\)
5 × 10^-3 =4500 × 10¹⁰ × \(\frac { D }{ d }\)

Question 37.
The activity of a radioactive substance is 4700 per minute. Five minutes later the activity is 2700 per minute. Find
a) decay constant and
b) half – life of the radioactive substance.
Answer:
Given A₀=4700 per minute
A = 2700 per minute
t = 5 minute λ = ? T_y2 = ?
we have,

Students can Download 2nd PUC Hindi Previous Year Question Paper June 2017, Karnataka 2nd PUC Hindi Model Question Papers with Answers help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Hindi Previous Year Question Paper June 2017

समय : 3 घंटे 15 मिनट
कुल अंक : 100

I. अ) एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिए : (6 × 1 = 6)

प्रश्न 1.
जो मनुष्य सत्य बोलता है, वह किससे दूर भागता है?

प्रश्न 2.
किसका व्यापारीकरण हो रहा है?

प्रश्न 3.
मन्नू भंडारी को प्रभावित करनेवाली हिन्दी प्राध्यापिका का नाम लिखिए।

प्रश्न 4.
चीफ़ की दावत किसके घर पर थी?

प्रश्न 5.
स्वर्ग या नरक में निवास स्थान अलॉट करनेवाले कौन हैं?

प्रश्न 6.
‘नारा’. का प्रसिद्ध मंदिर कौनसा है?

आ) निम्नलिखित प्रश्नों में से किन्हीं तीन प्रश्नों के उत्तर लिखिएः (3 × 3 = 9)

प्रश्न 7.
घर में सुजान भगत का अनादर कैसे हुआ?

प्रश्न 8.
झूठ की उत्पत्ति और उसके कई रूपों के बारे में लिखिए।

प्रश्न 9.
गंगा मैया का कुर्सी से क्या अभिप्राय है?

प्रश्न 10.
विश्वेश्वरय्या की प्रसिद्धि तथा पदोन्नति देख कुछ इंजीनियर क्यों जलते थे?

प्रश्न 11.
भोलाराम का परिचय दीजिए।

II. अ) निम्नलिखित वाक्य किसने किससे कहे? (4 × 1 = 4)

प्रश्न 12.
“दिन भर एक न एक खुचड़ निकालते रहते हैं।”

प्रश्न 13.
“यू हैव मिस्ड समथिंग।”

प्रश्न 14.
“वह जरूर बना देंगी। आप उसे देखकर खुश होंगे।”

प्रश्न 15.
“महाराज, रिकार्ड सब ठीक है।”

आ) निम्नलिखित में से किन्हीं दो का ससंदर्भ स्पष्टीकरण कीजिए: (2 × 3 = 6)

प्रश्न 16.
“भगवान की इच्छा होगी, तो फिर रुपये हो जायेंगे। उनके यहाँ किस बात की कमी है?”

प्रश्न 17.
“वे बोलते जा रहे थे और पिताजी के चेहरे का संतोष धीरे-धीरे गर्व में बदला जा रहा था।”

प्रश्न 18.
“मेरी माँ गाँव की रहनेवाली हैं। उमर-भर गाँव में रही हैं।”

प्रश्न 19.
“यहाँ तो डेंटिस्ट मक्खी मारते होंगे।”

III. अ) एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिए : (6 × 1 = 6)

प्रश्न 20.
श्रीकृष्ण के अनुसार किसने सब माखन खा लिया?

प्रश्न 21.
बाँसुरी किस रंग की है?

प्रश्न 22.
बेटी सजने-धजने से क्या महसूस करती है?

प्रश्न 23.
देश को किससे बचाना है?

प्रश्न 24.
भारतीयता कहाँ बहु रूप में सँवरती है?

प्रश्न 25.
दीवार किसकी तरह हिलने लगी?

आ) निम्नलिखित प्रश्नों में से किन्ही दो प्रश्नों के उत्तर लिखिए: (2 × 3 = 6)

प्रश्न 26.
श्रीकृष्ण के रूप सौन्दर्य का वर्णन कीजिए।

प्रश्न 27.
‘गहने’ कविता के द्वारा कवि ने क्या आशय व्यक्त किया है?

प्रश्न 28.
पर्यावरण के संरक्षण के संबंध में कवि कुँवर नारायण के विचार लिखिए।

प्रश्न 29.
दक्षिण प्रदेश की महत्ता को अपने शब्दों में लिखिए।

इ) ससंदर्भ भाव स्पष्ट कीजिए: (2 × 4 = 8)

प्रश्न 30.
अब कैसे छूटै राम रट लागी।
प्रभु जी तुम चंदन हम पानी,
जाकी अंग-अंग बास समानी।
अथवा
धनि रहीम जल पंक को, लघु जिय पिअत अघाय।
उदधि बड़ाई कौन है, जगत पिआसो जाय॥

प्रश्न 31.
ऐसा तेरा लोक, वेदना
नहीं, नहीं जिसमें अवसाद,
जलना जाना नहीं, नहीं
जिसने जाना मिटने का स्वाद!
अथवा
युद्धं देहि कहे जब पामर
दे न दुहाई पीठ फेर कर;
या तो जीत प्रीति के बल पर
या तेरा पद चूमे तस्कर।

IV. अ) एक शब्द या वाक्यांश या वाक्य में उत्तर लिखिए : (5 × 1 = 5)

प्रश्न 32.
मिश्रानी कितने वर्षों से मूलराज के परिवार में काम कर रही थी?

प्रश्न 33.
छोटी बहु के मन में किसकी मात्रा जरूरत से ज्यादा है?

प्रश्न 34.
कवि किस पर शासन करता है?

प्रश्न 35.
शास्त्रार्थों में पंडितों को किसने पराजित किया?

प्रश्न 36.
पितृ-हत्या का दण्ड क्या नहीं है?

आ) निम्नलिखित प्रश्नों में से किन्हीं दो प्रश्नों के उत्तर लिखिए: (2 × 5 = 10)

प्रश्न 37.
इन्दु को अपनी भाभी बेला पर क्यों क्रोध आया?
अथवा
बेला की चारित्रिक विशेषताओं पर संक्षेप में प्रकाश डालिए।

प्रश्न 38.
भारवि अपने पिता से क्यों बदला लेना चाहता था?
अथवा
प्रायश्चित को लेकर पिता और पुत्र के बीच हुए संवाद को लिखिए।

V. अ) वाक्य शुद्ध कीजिएः (4 × 1 = 4)

प्रश्न 39.
i) मेरा तो प्राण निकल गया।
ii) संदीप को पूछो।
iii) यह एक इतिहासिक घटना है।
iv) कोयल डाली में बैठी है।
उत्तरः
i) मेरे तो प्राण निकल गये।
ii) संदीप से पूछो
iii) यह एक ऐतिहासिक घटना है।
iv) कोयल डाली पर बैठी है।

आ) कोष्टक में दिये गए उचित शब्दों से रिक्त स्थान भरिए: (4 × 1 = 4)
(पावन, भला, समाज, समय)

प्रश्न 40.
i) आप …………… तो जग भला।
ii) ……… परिवर्तनशील है।
iii) वह सरस्वती देवी का …………… मंदिर है।
iv) साहित्य …………… का दर्पण है।
उत्तरः
i) भला
ii) समय
iii) पावन
iv) समाज।

इ) निम्नलिखित वाक्यों को सूचनानुसार बदलिए: (3 × 1 = 3)

प्रश्न 41.
i) वसुंधरा गाना गाती थी। (वर्तमानकाल में बदलिए)
ii) वह आग चिता पर रखेगा। (भूतकाल में बदलिए)
iii) पृथ्वीराज ने देश की सेवा की। (भविष्यत्काल में बदलिए)
उत्तरः
i) वसुंधरा गाना गाती है।
ii) उसने आग चिता पर रख दी।
iii) पृथ्वीराज देश की सेवा करेगा।

ई) निम्नलिखित मुहावरों को अर्थ के साथ जोड़कर लिखिए: (4 × 1 = 4)

प्रश्न 42.
i) खिल्ली उड़ाना a) रूठ जाना
ii) मुँह फुलाना b) सफल न होना
ii) टोपी उछालना c) हँसी उड़ाना
iv) दाल न गलना d) अपमानित करना
उत्तरः
i – c, ii – a, iii – d, iv – b.

उ) अन्य लिंग रूप लिखिए: (3 × 1 = 3)

प्रश्न 43.
i) श्रीमती
ii) गाय
iii) पुत्रवान।
उत्तरः
i) श्रीमान
ii) बैल
iii) पुत्रवती।

ऊ) अनेक शब्दों के लिए एक शब्द बनाइए : (3 × 1 = 3)

प्रश्न 44.
i) नीचे लिखा हुआ।
ii) जो पुत्र गोद लिया हो।
iii) जो छिपाने योग्य हो।
उत्तरः
i) निम्नलिखित
ii) दत्तक
iii) गुप्त/रहस्य।

ए) निम्नलिखित शब्दों के साथ उपसर्ग जोड़कर नए शब्दों का निर्माण कीजिए : (2 × 1 = 2)

प्रश्न 45.
i) परिवार
ii) शासन।
उत्तर:
i) परिवार = स + परिवार = सपरिवार।
ii) शासन = अनु + शासन = अनुशासनं।

ऐ) निम्नलिखित शब्दों में से प्रत्यय अलग कर लिखिए: (2 × 1 = 2)

प्रश्न 46.
i) बलवान
ii) घुमाव।
उत्तरः
i) बलवान = बल + वान।
ii) घुमाव = घुम + आव।

VI. अ) किसी एक विषय पर निबंध लिखिए : (1 × 5 = 5)

प्रश्न 47.
i) a) इंटरनेट की दुनिया।
b) खेल जगत में क्रिकेट का स्थान।
c) राष्ट्रीय एकता।
अथवा
ii) चरित्र प्रमाण-पत्र प्राप्त करने हेतु अपने कॉलेज के प्राचार्य को आवेदन-पत्र लिखिए।

आ) निम्नलिखित अनुच्छेद पढ़कर उस पर आधारित प्रश्नों के उत्तर लिखिए: (5 × 1 = 5)

प्रश्न 48.
यह कहावत सत्य है कि समय बलवान है, इस पर किसी का वश नहीं चलता। आनेवाले समय में अच्छा-बुरा क्या घट जाए, कोई नहीं जानता। ऐसे अनिश्चित समय के लिए यदि उसके पास कुछ संचित धन है तो उसके काम आ सकता है। कोई भी व्यक्ति दूसरों के सहारे न तो कल रह सका है, न आज रह पा रहा है, न ही भविष्य में रह सकता है। सत्य है। मनष्य का आज का जीवन कई प्रकार की आकस्मिकताओं वाला बन चुका है। अतः उन आकस्मिकताओं का ठीक प्रकार से सामना करने के लिए, प्रत्येक व्यक्ति को चाहिए कि वह प्रतिदिन जितनी भी हो अधिक-से-अधिक बचत करता रहे। इसी में उसकी भलाई है। आज की गई एक-एक पैसे की बचत कल का अनंत सुख सिद्ध हो सकती हैं। धन की कमी से सुखपूर्वक तो क्या सामान्य जीवन कतई संभव नहीं है। मनुष्य के जीवन में हमेशा से धन की आवश्यकता बनी रही है। उसे पूरा करने के लिए बचत करना नितांत आवश्यक है। अतः व्यक्ति अपने सभी तरह के स्रोतों से आज और कल में संतुलन बनाए रखकर ही सुख-चैन से जीवन जी सकता है।
प्रश्नः
i) कौन सी कहावत सत्य है?
ii) कटु सत्य क्या है?
iii) मनुष्य की भलाई किस में है?
iv) धन की कमी से क्या संभव नहीं है?
v) व्यक्ति सुख-चैन से कब जी सकता है?

इ) हिन्दी में अनुवाद कीजिए: (5 × 1 = 5)

प्रश्न 49.

Students can Download Chapter 6 Integers Ex 6.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Maths Chapter 6 Integers Ex 6.3

Question 1.
Find
Solution:
a) 35 – (20)
35 – (20) = 15

b) 72 – (90)
72 – (90) = -18

c) (-15) – (-18)
(-15) – (-18) = -15 + 18 = 3

d) (-20) – (-13)
(-20) – (-13) = -33

e) 23 – (-12)
23 – (-12) = 23 + 12 = 35

f) (-32) – (-40)
(-32) – (-40) = -32 + 40 = 8

Question 2.
Fill in the blanks with >,< or = sign
a) (-3) + (-6) ___ (-3) – (6)
(-3) + (-6) = -9
(-3) – (-6) = -3 + 6 = 3 – 9 < 3
Hence (3) +(-6) < (-3)-(-6) b) (-21) – (-10) ___ (-31) + (-11) -31 + (-11) = -42 – 11 > -42
(9 – 21) – (- 10) ≥ (-31) +(-11)

c) 45 – (-11) ___ 57 + (-4)
57 + (-4) = 57 – 4 = 53
56 > 53

d) (-25) – (-42) ___ (-42) – (-25)
-42 – (-25) = -42 + 25 = -17
117 > -17
(-25) – (-42) ≥ (-42) – (-25)

Question 3.
Fill in the blanks
a) (-8) + 8 = 0
b) 13 + (-13) = 0
c) 12 + (- 12) = 0
d) (-4) + (-8) =-12
e) 5 – 15 = -10

Question 4.
Find
Solution:
a) (-7) -8 – (-25)
(-7) – 8 – (-25) = (-7) – 8 + 25 = -15 + 25 = 10

b) (-13) + 32 – 8 – 1
(-13) + 32 – 8 – 1 = -13 + 32 – 8 – 1 = 32 – 22 = 10

c) (-7) + (-8) +(-90)
(-7) + (-8) + (-90) = -7 – 8 – 90 = -105

d) 50 – (-40) – (-2)
50 – (-40) – (-2) = 50 + 40 + 2 = 92

Students can Download Hindi Lesson 10 मेरा, हमारा, तेरा, तुम्हारा Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 6 Hindi helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 6 Hindi Chapter 10 मेरा, हमारा, तेरा, तुम्हारा

मेरा, हमारा, तेरा, तुम्हारा Questions and Answers, Summary, Notes

अभ्यास

1. उचित शब्द चुनकर खाली :

1. मोहन …………… भाई हैं (मेरी / मेरा)
2. ……………घर बड़ा हैं (हमारा / हमारी)
3. …………. नाम क्या है ? ( तेरी / तेरा )
4. ………. गाँव कहाँ है ? (तुम्हारी / तुम्हारा)

उत्तर:

1. मेरा
2. हमारा
3. तेरा
4. तुम्हारा

2. मिलान करो :

1. सतीश – a. गाँव के पास हैं
2. कुत्ता  – b. सविता का बड़ा भाई हैं
3. खरगोश – c. वफादार होता हैं
4. विद्यालय – d. प्यारा हैं

उत्तर:

1. b सविता का बड़ा भाई हैं
2. c वफादार होता हैं
3. d प्यारा हैं
4. a गाँव के पास हैं

You can Download Chapter 3 Financial Statements of a Company Questions and Answers, Notes, 2nd PUC Accountancy Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Accountancy Question Bank Chapter 3 Financial Statements of a Company

Question 1.
What is public company?
Answer:
A public company is defined as a company that offers a part of its ownership in the form of shares, debentures, bonds, securities to the general public through stock market:

Question 2.
What is private limited company?
Answer:
As defined by the Section 3. (1) (iii) of Companies Act 1956, private limited company is defined by the following characteristics:

• It restricts the right to transfer its shares.
• There must be at least two and a maximum of 50 members (excluding current and former employees) to form a private company.
• It ca not invite application from the general public to subscribe its shares, or debentures.
• It cannot invite or accept deposits from persons other than its members, Directors and their relatives.

Question 3.
Define Government Company?
Answer:
As per the Section 617 of Company Act of 1956, a Government Company means any company in which not less than 51% of the paid up share capital is held by the Central Government, or by any State Government or Governments, or partly the Central Government and partly by one or more State Governments and includes a company which is a subsidiary of a Government Company as thus defined.

Question 4.
What do you mean by a listed company?
Answer:
Those public companies whose shares are listed and can be traded in a recognised stock exchange for public trading like, Tata Moto. s, Reliance, etc are called Listed Company. These companies are also called Quota Cornpanies.

Question 5.
What are the uses of securities premium?
Answer:
As per the Section 78 of the Companies Act of 1956, the amount of securities premium can be used by the company for the following activities:

• For paying up unissued shares of the company to be issued to members of the company as fully paid bonus share.
• For writing off the preliminary expenses of the company.
• For writing off the expenses of, or the commission paid or discount allowed on, any issue of shares or debentures of the company.
• For paying up the premium that is to be payable on redemption of preference shares or debentures of the company.
• Further, as per the Section 77A, the securities premium amount can also be utilised by the company to Buy-back its own shares.

Question 6.
What is buy-back of shares?
Answer:
Buy-back of shares means repurchasing of its own shares by a company from the market for reducing the number of shares in the open market.

Question 7.
Write a brief note on ‘Minimum Subscription’.
Answer:
When shares are issued to the general public, the minimum amount that must be subscribed by the public so that the company can allot shares to the applicants is termed as Minimum Subscription. As per the Company Act of 1956, the Minimum Subscription of share cannot be less than 90% of the issued amount. If the Minimum Subscription is not received, the company cannot allot shares to its applicants and it shall immediately refund the entire application amount received to the public.

2nd PUC Accountancy Financial Statements of a Company Long Answer Type Questions and Answers

Question 1.
Explain the nature of the financial statements.
Answer:
The nature of the financial statements depends upon the following aspects;
1. Recorded facts: The items recorded in the financial statements reflect their original cost i.e. the cost at which they were acquired. Consequently, financial statements do not reveal the current market price of the items. Further, financial statements fail to capture the inflation effects.

2. Conventions: The preparation of financial statements is based on some accounting conventions like, Prudence Convention, Materiality Convention, Matching Concept, etc. The adherence to such accounting conventions makes financial statements easy to understand, comparable and reflects the true and fair financial position of the company.

3. Accounting Assumptions: These basic accounting assumptions like Going Concern Concept, Money Measurement Concept, Realisation Concept, etc are called as postulates. While preparing financial statements, certain postulates are adhered to. The nature of these postulates is reflected in the nature of the financial statements.

4. Personal Judgments: Personal value judgments play an important role in deciding the nature of the financial statements. Different judgments are attached to different practices of recording.transactions in the financial statements. Thus, personal judgments determine the nature of the financial statements to a great extent.

Question 2.
Explain in detail about the significance of the financial statements.
Answer:
The importance of financial statements is mentioned below:
1. Provides Information: Financial statements provide information to various accounting users both internal as well as external users. It acts as a basic platform for different accounting users to derive information according to varying needs. For example, the financial statements on one hand help the shareholders and investors in assessing the viability and return on their investments, while on the other hand, the financial statements help the tax authorities in calculating the amount of tax liability of the company.

2. Cash Flow: Financial statements provide information about the cash flows of the company. The financial statements help the creditors and other investors.in determining solvency of company.

3. Effectiveness of Management: The comparability feature of the financial statements enables management to undertake comparisons like inter-firm and intra-firm comparisons. This not only helps in assessing the viability and performance of the business but also helps in’ designing policies and drafting policies. The financial statements enhance the effectiveness and efficacy of the management.

4. Disclosure of Accounting Policies: Financial statements provide information about the various policies, important changes in the methods, practices and process of accounting by the company. The disclosure of the accounting policies makes financial statements simple, true and enables different accounting users to understand without any ambiguity.

5. Policy Formation by Government: It needs information to determine national income, GDP, industrial growth, etc. The accounting information assist the government in the formulation of various policy measures and to address various economic problems like employment, poverty etc.

6. Attracts Investors and Potential Investors: They invest or plan to invest in the business. Hence, in order to assess the’viability and prospectus of their investment, creditors need information about profitability and solvency of the business.

Question 3.
Explain the limitations of financial statements.
Answer:
The following are the limitations of financial statements.
1. Historical Data: The items recorded in the financial statements reflect their original cost i.e. the cost at which they were acquired. Consequently, financial statements do not reveal the current market price of the items. Further, financial statements fail to capture the inflation effects.

2. Ignorance of Qualitative Aspect: A financial statement does not reveal the qualitative aspects of a transaction. The qualitative aspects like colour, size and brand position in the market, employee’s qualities and capabilities are not disclosed by the financial statements.

3. Biased: Financial statements are based on the personal judgments regarding the use of methods of recording. For example, the choice of practice in the valuation of inventory, method of depreciation, amount of provisions, etc. are based on the personal value judgments and may differ from person to person. Thus, the financial statements reflect the personal value judgments of the concerned accountants and clerks.

4. Inter-firm Comparisons: Usually, it is difficult to compare the financial statements of two companies because of the difference in the methods and practices followed by their respective accountants.

5. Window dressing: The possibility of window dressing is probable. This might be because of the motive of the company to overstate or understate the assets and liabilities to attract more investors or to reduce taxable profit. For example, Satyam showed high fixed deposits- in the Assets side of its Balance Sheet for better liquidity that gave false and misleading signals to the investors.

6. Difficulty in Forecasting: Since the financial statements are based on historical data, so they fail to reflect the effect of inflation. This drawback makes forecasting difficult. Prepare the format of income statement and explain its elements

Prepare the format of balance sheet and explain the various elements of balance sheet.
Vertical form of balance sheet

1. Share Capital: It is the first item on the Liabilities side. It consists of the following items:

• Authorised Capital
• Issued Capital: Equity share and preference share.
• Subscribed Capital less Call in Arrears add Forfeited Shares

2. Reserve and Surplus: As per the Schedule VI,  it consists of the following items:

• Capital Reserve
• Capital Redemption Reserve
• Security Premium
• Other Reserve less Debit balance of P & L A/c
• Credit balance of P & L A/c
• Proposed Additions.
• Sinking Fund

3 . Secured Loans

1. Debentures
2. Loan and advances from bank etc.

4. Unsecured Loans

• Fixed Deposits
• Loan & Advances from subsidiaries

5. Fixed Assets: These are those assets that are used for more than one year, like:

• Goodwill
• Land
• Building
• Plant ec Machinery
• Patents, Trade Marks
• Livestock
• Vehicles, etc.

6. Current Assets: Assets that can be easily converted into cash or cash equivalents are termed as current assets. These are required to run day to day business activities; for example, cash, debtors, stock, etc.

7. Current Liabilities: Those liabilities that are incurred with an intention to be paid or are payable within a year; for example, bank overdraft creditors, bills payable, outstanding wages, short-term loans, etc are called current liabilities.

Question 4.
Explain how financial statements are useful to the various parties who are interested in the affairs of an undertaking?
Answer:
The various parties that are directly or indirectly interested in the financial statements of a company can be categorized into the following two categories:
Internal Parties: The following are the various internal accounting users who are directly related to the company.

a) Owner: The owner/s is/’are interested in the profit earned or loss incurred during an accounting period. They are interested in assessing the profitability and viability of the capital invested by them in the business.

b) Management: The financial statements help the management in drafting various policies measures, facilitating planning and decision making process. The financial statements also enable management to exercise various cost controlling measures and to remove inefficiencies.

c) Employees and workers: They are interested in the timely payment of wages and salaries, bonus and appropriate increment in their wages and salaries. With the help of the financial statements they can know the amount of profit earned by the company and can demand reasonable hike in their wages and salaries.

External Parties: There are various external users of accounting who need accounting information for decision making, investment planning and to assess the financial position of the business. The various external users are given below.

a) Banks and other financial institutions: Banks provide finance in the form of loans and advances to various businesses. Thus, they need information regarding liquidity, creditworthiness, solvency and profitability to advance loans.

b) Creditors: These are those individuals and organisations to whom a business owes money on account of credit purchases of goods and receiving services; hence, the creditors require information about credit worthiness of the business.

c) Investors and potential investors: They invest or plan to invest in the business. Hence, in order to assess the viability and prospectus of their investment, creditors need information about profitability and solvency of the business.

d) Tax “authorities: They need information about sales, revenues, profit and taxable income in order to determine the levy various types of tax on the business.

e) Government: It needs information to determine national income, GDP, industrial growth, etc. The accounting information assist the government in the formulation of various policies measures and to address various economic problems like employment, poverty etc.

f) Researchers: Various research institutes like NGOs and other independent research institutions like CRISIL, stock exchanges, etc. undertake various research projects and the accounting information facilitates their research work.

g) Consumers: Every business tries to build up reputation in the eyes of consumers, which can be created by the supply of better quality products and post-sale services at reasonable and alfordable prices. Business that has transparent financial records, assists the customers to know the correct cost of production and accordingly assess the degree of reasonability of the price charged by the business for its products and, thus, helps in repo building of the business.

h) Public: Public is keenly interested to know the proportion of the profit that the business spends on various public welfare schemes; for example, charitable hospitals, funding schools, etc. This information is also revealed by the profit and loss account and balance sheet of the business.

Question 5.
Financial statements reflect a combination of recorded facts, accounting conventions and personal judgments’ discuss.
Answer:
The financial statements are the end-products of the accounting process. The financial statements not only reveal the true financial position of the company but also help various accounting users in decision making and policy designing process. The nature of the financial statements depends upon the following aspects like recorded facts, conventions, concepts, and personal judgment.

1. Recorded facts: The items recorded in the financial statements reflect their original cost i.e. the cost at which they were acquired. Consequently, financial statements do not reveal the current market price of the items. Further, financial statements fail to capture the inflation effects.

2. Conventions: The preparation of financial statements is based on some accounting conventions like,: Prudence Convention, Materiality Convention, Matching Concept, etc. The adherence to such accounting conventions makes financial statements easy to understand, comparable and reflects the true and fair financial position of the company.

3. Accounting Assumptions: These basic accounting assumptions like Going Concern Concept, Money Measurement Concept, Realisation Concept, etc are. called as postulates. While preparing financial statements, certain postulates are adhered to, The nature of these postulates is reflected in the nature of the financial statements.

4. Personal Judgments: Personal value judgments play an important role in deciding the nature of the financial statements. Different judgments are attached to different practices of recording transactions in the financial statements. Thus, personal judgments determine the nature of the financial statements to a great extent.

Question 6.
Explain the process of preparing income statement and balance sheet.
Answer:
The process of preparing Horizontal Form of Income Statement is explained below in a chronological order:

1. Prepare a Trial Balance on the basis of the balances of various accounts in the ledger.
2. Record Opening Stock, Purchases, Manufacturing Expenses and other direct expenses on the debit side of Trading Account
3. Record Sales and Closing Stock on the credit side of the Trading Account.
4. Ascertain the balancing figure by totalling both the sides of the Trading Account If the credit side exceeds the debit side, then the balancing figure is termed as Gross Profit, but if the debit side exceeds the credit side, then the balancing figure is termed as Gross Loss.
5. Carry forward the Gross Profit (Gross Loss) to the credit (debit) side of the Profit and Loss Account.
6. Record all current year’s operating and non-operating revenue expenditures with their relevant adjustments on the debit side of the Profit and Loss Account.
7. Record ail current year’s operating and non-operating revenue incomes with their relevant adjustments on the credit side of the Profit and Loss Account.
8. Ascertain the balancing figure by totalling both the sides of the Profit and Loss Account. If the credit exceeds the debit side, then the balancing figure is termed as Net Profit, but if the debit side exceeds the credit side, then the balancing figure is termed as Net Loss.

The process of preparing Horizontal Form of Balance Sheet is explained below in a chronological order:

1. Prepare a Trial Balance on the basis of the balances of various accounts in the ledger.
2. Record all the debit balances of Real and Personal Accounts on the left hand side (i.e. Assets side) of the Balance Sheet after making all adjustments for provision and other related items.
3. Record all the credit balances of Real and Personal Accounts on the right hand side (i.e. Liabilities side) of the Balance Sheet after making all adjustments for interest and outstanding items.
4. Add Net Profit to the Opening Capital and deduct Net Loss, if any from the Opening Capital
5. Ascertain the total of two sides, jvhich must be equal.

2nd PUC Accountancy Financial Statements of a Company Numerical Questions and Answers 

Question 1.
Show the following items in the balance sheet as per the provisions of the companies Act, 1956 in (Revised) Schedule VI:

Question 2.
On 1stApril, 2013, Jumbo Ltd., issued 10,000; 12% debentures of₹ 100 each a discount of 20%; redeemable after 5 years. The company decided to write-off discount on issue of such debentures over the life time of the Debentures. Show the items of the company immediately after the issue of these debentures.
Answer:

Question 3.
From the following information prepare the balance sheet of Gitanjali Ltd. Invntories ₹ 14,00,000; Equity Share Capital ₹ 20,00,000; Plant and Machinery ₹ 10,00, 000; Preference Share Capital ₹ 12,00,000; Debenture Redemption Reserve ₹ 6,00,000; Outstanding Expenses ₹ .3,00,000; Proposed-Dividend ₹ 5,00,000; Land and Building ₹ 20,00,000; Current Investment ₹ 8,00,000; Cash. Equivalent ₹ 10,00,000; Short term loan from Zaveri Ltd. (A Subsidiary Company of Twilight Ltd.) ₹ 4,00,000; Public Deposits ₹ 12,00,000.
Answer:

Question 4.
From the following information prepare the balance sheet of Jam Ltd. Inventories ₹ 7,00,000; Equity Share Capital ₹ 16,00,000; Plant and Machinery ₹ 8,00,000; Preference Share Capital ₹ 6,00,000; General Reserves ₹ 6,00,000; Bills payable Rs 1,50,000; Provision for taxation ₹ 2,50,000; Land and Building k 16,00,000; Non-current Investments ₹ 10,00,000; Cash at Bank ₹ 5,00,000; Creditors ₹ 2,00,000; 12% Debentures ₹ 12,00,000.
Answer:

Note to account

Question 5.
Prepare the balance sheet of Jyoti Ltd., as at March 31, 2018 from the following information. Building ₹ 10,00,000; Investments in the shares of Metro Tyers Ltd. ₹ 3,00,000; Stores & Spares ₹ 1,00,000; Discount on issue of 10% debentures of ₹ 10,000; Statement of Profit and Loss (Pr.). ₹ 90,000; 5,00,000 Equity Shares of ₹ 20 each fully paid-up; Capital Redemption Reserve ₹ 1,00,000; 10% Debentures ₹ 3,00,000; Unpaid dividends ₹ 90,000; Share options outstanding account ₹ 10,000.
Answer:

Question 6.
Brinda Ltd., has furnished the following information:
(a) 25,000,10% debentures of ₹ 100 each;
(b) Bank Loan of ₹ 10,00,000 repayable after 5 years;
(c) Interest on debentures is yet to be paid.
Show the above items in the balance sheet of the company as at March 31, 2018.
Answer:

Question 7.
Prepare a Balance sheet of Black Swan Ltd., as at March 31, 2013 from the following information:

Answer:

Karnataka SSLC Time Table 2020: The officials of Karnataka Secondary Education Board (KSEEB) have released the Karanataka SSLC Time Table for the students of Class 10.  According to the Karnakata SSLC Time Table For Class 10th, the exams will commence from 27th March 2020 to 9th April 2020.

Karnataka SSLC Board Exam Time Table 2020 for Class 10

Before getting into the details of Karnataka SSLC Time Table for Class 10, let’s have an overview of the examination:

Description	Details
Name of the Exam	Karnataka SSLC Examinations
Conducting Body	Karnataka Secondary Education Examination Board
Exam Mode	Offline
Exam Start Date	27th March 2020
Exam End Date	9th April 2020
Category	Karnataka SSLC Time Table For Class 10
Official Website	kseeb.kar.nic.in

Karnataka SSLC Time Table for Class 10

The Karnataka SSLC Time Table for Class 10 is tabulated below:

Date and Day	Subject Name
27 Mar 2020 Fri	First Language Kannada (01) Telugu (04) Hindi (06) Marathi (08) Tamil (10) Urdu (12) English (14) Sanskrit (16)
30 Mar 2020 Mon	Core Subjects Science (83) Political Science (97) Karnataka Music/ Hindustani Music (98)
01 Apr 2020 Wed	English (31) Kannada (33)
03 Apr 2020 Fri	Third Language Hindi (61) Kannada (62) English (63) Arabic (64) Persian (65) Urdu (66) Sanskrit (67) Konkani (68) Tulu (69) NSQF Exam Subjects Information Technology (86) Retail (87) Automobile (88) Health Care (89) Brauty and Wellness (90)
04 Apr 2020 Sat	Elements of Electrical and Mechanical Engineering (71) Engineering Graphics (72) Elements of Electronic Engineering (73) Elements of Comp[uter Science (74) Economics (96)
07 Apr 2020 Tue	Mathematics (81) Sociology (95)
09 Apr 2020 Thu	Social Science (85)

Karnataka SSLC Time Table 2020 in English

Karnataka SSLC Time Table in English is given below:

Karnataka SSLC Time Table 2020 in Kannada

Karnataka SSLC Time Table in Kannada is given below:

You can Download Chapter 2 Solutions Questions and Answers, Notes, 2nd PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Chemistry Question Bank Chapter 2 Solutions

Question 21.
Two elements A and B form compounds having formulas AB₂ and AB₄. When dissolved in 20 g of benzene (C₆H₆), 1 g of AB₂ lowers the freezing point by 2.3 K whereas 1.0 g of AB₄ lowers it by 1.3 K. Tin molar depression – constant for benzene is 5.1 K kg mol^-1. Calculate atomic masses of A and B.
Answer:


Suppose atomic masses of A and B are ‘a’ and ‘b’ respectively. Then
Molar mass of AB₂ = a + 2b = 110.87 g mol^-1
Molar mass of AB₄ = a + 4b =196.15 g mol^-1
Equation (ii) – Equation (i) gives
2b = 85.28 orb = 42.64
substituting in equation (i) we get
a + 2 × 42.64 = 110.87 or a = 25.59
Thus atomic mass A = 25.59 u
Atomic mass of B = 4.64 u.

Question 22.
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Answer:
π = CRT
∴ In the first case,
4.98 = \(\frac{36}{180}\) × R × 300 = 60R
In the second case, 1.52 = C × R × 300
Dividing (ii) by (i), we get C = 0.061 M.

Question 23.
Suggest the most important type of intermolecular attractive interaction in the following pairs.

1. n-hexane and n-octane
2. I₂ and CCl₄
3. NaClO₄ and water
4. methanol and acetone
5. acetonitrile (CH₃CN) and acetone (C₃H₆O).

Answer:

1. London’s forces
2. London’s forces
3. Ion-dipole interactions
4. Intermolecular hydrogen bonding
5. Dipole-dipole interactions.

Question 24.
Based on solute-solvent interactions, arrange the following in order of increasing solubility in n*octane and explain. Cyclohexane, KCl, CH₃OH, CH₃CN.
Answer:
(a) Cyclohexane and n-octane both are non-polar. They mix completely in all proportions.
(b) KCl is an ionic compound, KCl will not dissolve in n-octane.
(c) CH₃OH is polar. CH₃OH will dissolve in n-octane.
(d) CH₃CN is polar but lesser than CH₃OH. Therefore, it will dissolve in n-octane but to a greater extent as compared to CH₃OH. Hence, the order is KCl < CH₃OH < CH₃CN < Cyclohexane.

Question 25.
Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol.
Answer:
(i) Partially soluble (because phenol has a polar -OH group but aromatic phenyl, C₆H₅ – group)
(ii) Insoluble because toluene is nonpolar while water is polar
(iii) Highly soluble because formic acid can form hydrogen bonds with water.
(iv) Highly soluble because ethylene glycol can form hydrogen bonds with water
(v) Insoluble chloroform is an organic liquid
(vi) Partially soluble because-OH the group is polar but the large hydrocarbon part (C₅H₁₁) is nonpolar.

Question 26.
If the density of some lake water is 1.25g mL^-1 and contains 92 g of Na⁺ ions per kg of water, calculate the molality of Na+ ions in the lake.
Answer:
Number of moles in 92 g of Na⁺ ions

As these are present in 1 kg of water, by definition, molality = 4m.

Question 27
If the solubility product of CuS is 6 × 10^-16, calculate the maximum molarity of CuS in an aqueous solution.
Answer:
Maximum molarity of CuS in aqueous solution = solubility of CuS in mol L^-1
If S in the solubility of CuS in mol L^-1 , then

Question 28.
Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 g of C₉H₈O₄ is dissolved in 450 g of CH₃CN.
Answer:

Question 29.
Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. The dose of nalorphine generally given is 1.5 mg. Calculate the mass of 1.5 x 10^-3 m aqueous solution required for the above dose.
Answer:
1.5 × 10^-3 m solution means that 1.5 × 10^-3 mole of nalorphine is dissolved in I kg of water.
Molar mass of C₁₉H₂₁NO₃ = 19 × 12 + 21 + 14 + 48 = 3119 mol^-1
∴ 1.5 × 10^-3 mole of C₁₉H₂₁NO₃;
= 1.5 × 10^-3 × 3119 = 0.467 g = 467 mg
∴ Mass of solution = 1000 g + 0.467 g = 1000.467g
Thus, for 467 mg of nalorphene, solution required =1000.467 g for 1.5 mg of nalorphene, solution required =\(\frac{1000.467}{467} \times 1.5 = 3.219\)

Question 30.
Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Answer:
0.15 M solution means that 0.15 mole of benzoic acid is present in 1 L
i.e. 1000 mL of the solution.
Molar mass of benzoic acid (C₆H₅COOH) = 72 + 5 + 12 + 32 + 1 =122 g mol^-1
∴ 0.15 mole of benzoic acid = 0.15 × 122g= 18.39
Thus, 1000 mL of the solution contain benzoic acid = 18.39
∴ 250 ml of the solution will contain benzoic acid = \(\frac{18.3}{1000}\) × 250= 4.575g

Question 31.
The depression in the freezing point of water observed for the same amount of acetic acid, trichloroacetic acid, and trifluoroacetic acid increases in the order given above. Explain briefly.
Answer:
The depression in freezing points are in the order:
acetic acid < trichloroacetic acid

Fluorine, being most electronegative, has the highest electron-withdrawing inductive effect. Consequently, trifluoroacetic acid is the Strongest acid while acetic acid is the weakest acid. Hence, trifluoroacetic acid ionizes to the largest extent while acetic acid ionizes to minimum extent to give ions in their solutions in water. Greater the ions produced, greater is the depression in freezing point. Hence the depression in freezing point is maximum for trifluoroacetic acid and minimum for acetic acid.

Question 32.
Calculate the depression in the freezing point of water when 10 g of CH₃CH₂ CHCICOOH is added to 250 g of water. K_a = 1.4 × 10^-3, K_f= 1.86 K kg mol^-1.
Answer:
Molar mass of CH₃CH₂CHCICOOH = 15 + 14 + 13 + 35.5 + 45 = 122.5 g mol^-1

If α is the degree of dissociation of CH₃CH₂CHCICOOH, then CH₃CH₂CHCLCOOH □ CH3CHXHCICOO^– + H +

To calculate vant’s Hoff factor:
CH₃CH₂CHCICOOH □ CH3CH₂HCICOO^– + H⁺

Question 33.
19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoro acetic acid. K_f of water is 1.86 K kg mol^-1.
Answer:
Here, w₂ = 19.5 g, w₁, = 500g, K_f = 1.86 K kg mol^-1, (AT_f) obs = 1.0°

M₂ (Calculated) for CH₂ FCOOH = 14+19+45 = 78 g mol^-1
Vant Hoff factor (i) =

calculation of dissociation constant. Suppose the degree of dissociation at the given concentration is a.
Then

Taking volume of the solutions as 500ml,

Question 34.
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Answer:
Here, P° = 17.535 mm, w₂ = 25g,
w₁ = 450g
For solute (glucose, C₆H₁₂O₆, M₂ = 180 g mol^-1
For solvent (H₂O), M₁ = 18g mol^-1
Applying Raoult’s law,

substituting the given values, we get

substituting the given values, we get

Question 35.
Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 10s mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
Answer:
Here, K_H = 4.27 × 10⁵ mm
p = 760 mm
Applying Henry’s law
P = K_H x

i.e, mole fraction of methane in benzene = 1.78 x 10^-3

Question 36.
100 g of liquid A (molar mass 140 g mol^-1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^-1). The vapour pressure of pure liquid B was found to be 500 torrs. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Answer:
Number of moles of a liquid

Number of moles of a liquid B (Solvent)

Mole fraction of B in the solution (x_A)

Mole fraction of B in the solution (x_B)
= 1-0.114 = 0.886
Also, given p⁰_B = 500 Torr
Applying Raoult’s law,
P_A = x_AP°_A = 0.114 × P°_A
P_B= x_B p°_B = 0.886 × 500 = 443 Torr
p Total = p_A + p_B
475 = 0.114 P°_A +443 or

Substituting this value in equation (i), we get p_A = 0.114x 280.7 Torr = 32 Torr

Question 37.
Benzene and toluene form ideal solution over the entire range of composition. The vapour pressures of pure benzene and toluene at 300 K are 50. 71mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in the vapour phase if 80 g of benzene is mixed with 100 g of toluene.
Answer:
Molar mass of benzene (C₆H₆ = 78 g mol^-1
Molar mass of toluene (C₆H₅CH₃) = 92 g mol^-1
∴ Number of moles in 80 g of benzene

Number of moles in 100 g of toluene

In the solution, mole fraction of benzene

=0.486
mole fraction of toluene = 1-0,486 = 0.514
p° Benzene = 50.71 mm, p° Toluene = 32.06mm
Applying Raoult’s law,
p Benzene= x_Tolucno × p°_Tolucne =0.514 × 32.06 mm = 16.48mm
Mole fraction of benzene in vapour phase

Question 38.
The air is a mixture of a number of gases. The major components are oxygen and, nitrogen with the approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 10⁷ mm and 6.51 × 10⁷ mm respectively, calculate the composition of these gases in water.
Answer:
Total pressure of air in equilibrium with water = 10 atm
As air contains 20% oxygen and 79% nitrogen by volumes
partial pressure of oxygen (P ) = \(\frac{20}{100}\) x 10 atm = 2atm
= 2 x 760 mm
= 1520 mm
partial pressure of nitrogen \(P_{N_{2}}\)
= \(\frac{79}{100}\) × 10 atm = 7.9 atm = 7.9 x 760 mm
= 6004 mm

K_H(O₂) = 3.30 x 10⁷mm,K_H (N₂) = 6.51 × 10⁷ mm
Applying Henry’s law

Question 39.
Determine the amount of CaCl (i= 2.47) dissolved in 2.51itre of water such that its osmotic pressure is 0.75 atm at 27° C.
Answer:

Molar mass of CaCl₂= 40 + 2x 35.5 = 111 gmol^-1
Amount dissolved = 0.0308 xlllg = 3.42g

Question 40.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄in 2 litre of water at 25° C, assuming that it is completely dissociated.
Answer:
K₂SO₄ dissolved = 0.025g
volume of solution = 2L
T = 250°C = 298 K
Molar mass of K₂SO₄ = 2 x 39 + 32 + 4 × 16= 174gmol^-1
As, K₂SO₄ dissociates completely as K₂SO₄ → 2K+ SO₄^2-
i.e., ions produced = 3 ∴ i =3

2nd PUC Chemistry Solutions Additional Questions and Answers

Question 1.
How does molarity of a solution change with temperature?
Answer:
Molarity decreases with increase in temperature because volume of solution increases with increase in temperature.

Question 2.
State Raoult’s law.
Ans:
It states that for a solution of volatile liquids its vapour pressure of each component is directly proportional to their mole fraction in the solution.
The relative lowering of vapour pressure is equal to mole fraction of solute in case of non-volatile solute.

Question 3.
Why is liquid ammonia bottle first cooled in ice before opening it?
Answer:
At room temperature, the vapour pressure of liquid ammonia is very high. On cooling, vapour pressure decreases. Hence the liquid ammonia will not splash out.

Question 4.
Sodium chloride and calcium chloride are used clear show from the roads. Why?
Answer:
Sodium chloride depresses the freezing point of water to such an extent that it cannot freeze to form ice. Therefore, it melts off easily at the prevailing temperature.

Question 5.
Two liquids A and B boil at 145°C and 190°C
respectively. Which of them has a higher vapour pressure of 80°C? (CBSE 2006)
Answer:
A being more volatile will have higher vapour pressure at 80°C.

Question 6.
Calculate the molality of sulphuric acid solution in which the mole fraction is 0.85.
Answer:

Question 7.
The molar freezing point depression constant for benzene is 4.90 kg mol’1. Selenium exists as a polymer. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112°C lower for pure benzene. Decide the molecular formula of selenium (At Wt. of selenium is 78.8 g mol^-1). (CBSE 2002)
Answer:

Question 8.
CCI₄ and water are immiscible whereas ethanol and water are miscible in all proportions. Correlate this behaviour with molecular structure of these compounds. (CBSE 2003,2001)
Answer:
CCI₄ is a non-polar covalent compound.
Water is a polar compound. CCI₄ can neither form H^– bonds with water molecules nor can it break H^– bonds between water molecules. Therefore, it is insoluble in water.

Ethanol is a polar compound and can form H’ bonds with water, which is a polar solvent, therefore it is miscible with water in all proportions.

Question 9.
The molarity of a solution of sulphuric acid is 1.35 M. Calculate its molarity (The density of the acid solution is 1.02 g cm³ )
Answer:
Let the solution be 1 litre or 1000 cm³
∴ Number of moles of H₂SO₄= 1.35
Wt. of solution = 1000 x 1.02 = 1020 g
Wt. of sulphuric acid = 1.35 x 98= 132.3g :
Wt. of water = 1020 – 132.3 = 887.79
Molality of H₂SO₄= ppp x 1000 = 1.52 m

From (i) M, = 110.82, from (ii) M2 = 196.15 AB4 – AB, = B2 196.15- 110.82 = B2 85.33 = B2 B = 42.665
Molar mass of AB2 = Atomic mass of A+ x 2 atomic mass of B 110.82=Atomic mass of A+85.33 Atomic mass of A = 110.82 – 85.33 = 25.49 Atomic mass of A = 25.499 Atomic mass of B = 42.669

Question 10.
Two elements A and B form compounds having molecular formulas AB2 and AB4. When dissolved 20 g of C6H6, 1 g of AB2 lowers the freezing point by 2.3 K, whereas 1.0 g of AB„4 lower it by 1.3 K. The molar depression constant for benzene is 5.1 kg mol*1. Calculate the mass of A and B. (CBSE2004)
Answer:
Let the molar mass of AB₂ and AB₄ be M₁and M₂
Then, for AB₂,

From (I) M₁ = 110.82, from (ii) M₁ = 196.15
AB₄– AB₂= B₂
196.15 – 110.82 =B₂
85.33 = B₂
B = 42.665
Molar mass of AB2 = Atomic mass of A+ x 2 atomic mass of B
110.82 = Atomic mass of A+ 85.33
∴ Atomic mass of A = 110.82-85.33 = 25.49
Atomic mass of A = 25 .499
Atomic mass of B = 42.669

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

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

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

(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)

The Polynomial Roots Calculator will find the roots of any polynomial with just one click.

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.

Question 11.
State Euclid’s division lemma.

Question 12.
Find the number of zeroes of a polynomial p(x) from the graph given

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

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.

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.

Solutions

Solution 1.
(B) or (D)

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

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.

Solution 14.

Solution 15.
Nos. divisible by 3 = 12,15,18, 21, …….., 99
a = 12, d = 3, T = 99, n = ?
T_n = 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.

Solution 17.

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)

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.

Solution 19.

Solution 20.

Solution 21.

Solution 22.
Let us assume that (3 + √5) is rational

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

Question 25.
3x² – 5x + 2 = 0
Compare it with the standard form

Solution 26.

Solution 27.

Solution 28.

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.

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.

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.

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

Solution 34.

Solution 35.

Solution 36.

Solution 37.

Solution 38.

Solution 39.

Solution 40.

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

(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.

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.

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.

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

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, T_n = 96, n = ?
T_n = 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.

Solution 17.

Solution 18.
Let the present ages of A = x years, B = y years.


Age of A = 40 years
Age of B = 10 years

Solution 19.

Solution 20.

Solution 21.

Solution 22.

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.

Solution 25.
It is in the form ax² + bx + c = 0
a = 5, b = -a, c = 3

Solution 26.

Solution 27.

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.

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

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.

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

Solution 33.

Solution 34.

Solution 35.

Solution 36.

Solution 37.
List of two-digit nos.
divisible by 3 = 12, 15, 18, ……. , 99.
This is an AP
a = 12, d = 3, T_n = 99, n = ?
T_n = 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

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

Solution 38.
data: ΔABC || ΔPQR
To Prove that:

Solution 39.



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 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

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 = …….

(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

(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).

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

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.

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

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)

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.

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.

Solution 14.

Solution 15.
Nos. divisible by 5 are 10, 15, 20, ………, 95
They are in AP
a = 10, d = 5, T_n = 95, n = ?
T_n = 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.

Solution 17.

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)

Present age of the father = 80 years
Age of the son = 30 years.

Solution 19.

Solution 20.

Solution 21.

Solution 22.
Let us assume that 5 + √7 is rational.

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.

Solution 25.
Compare this with ax² + bx + c = 0
a = 2, b = -5, c = 3

Solution 26.

Solution 27.

Solution 28.

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.

Solution 31.

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

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.

ΔABC is required Δ
ΔA’BC’ is required Δ

Solution 33.

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

Solution 35.

Solution 36.

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
a_n = -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, a_n = 5
a_n = 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.

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.

Solution 40.

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 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 =…

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 = ………

Question 10.
Identify the largest chord in the given figure.

Question 11.
Find the HCF of 25 and 15

Question 12.
From the graph find the number of zeros of the polynomial p(x)

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.

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

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

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.

Solution 10.
XY

Solution 11.
25 = 5 × 5
15 = 5 × 3

Solution 12.
2

Solution 13.
(0, 0)

Solution 14.

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, a_n = 99, n = ?
a_n = 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

Solution 17.

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.

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.

Solution 20.
PA and PB are tangents to the circle at the ends of radii

Solution 21.

Solution 22.
Let us assume that (5 + √3)

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

Solution 25.

Solution 26.

Solution 27.

Solution 28.

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

Solution 30.

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

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

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

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

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.

Solution 34.
Because A, B, C are the angles of a ∆ABC
A + B + C = 180°
⇒ B + C = (180 – A)

Solution 35.

Solution 36.

Solution 37.
a = 45
d = a₂ – a₁ = 41 – 45 = -4
Let nth term of the AP be the first negative term then a_n < 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

Solution 38.

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.

Solution 40.

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 =?

(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 = ……..

(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

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

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

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.

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:

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)

Solution 13.

Solution 14.

Solution 15.

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.

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

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

Solution 20.
Draw the rough figure & calculate OA using Pythagora’s theorem. Then construct the tangent.

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

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

Solution 25.
2x² – 5x + 3 = 0
It is in the form ax² + bx + c = 0
a = 2, b = -5, c = 3

Solution 26.

Solution 27.

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)

Solution 28.

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.

Solution 30.
r + h = 37
T.S.A. of the cylinder = 2πrh + 2πr² = 2πr (h + r)
But 2πr (h + r) = 1628

Solution 31.

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.

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)

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)

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°

Solution 35.

Solution 36.

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
T_n = 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. a_n = 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. T_n = (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

Solution 39.

Solution 40.

Karnataka SSLC Maths Model Question Papers

Students can Download English Poem 10 Photograph Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 9 English Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Board Class 9 English Poem Chapter 10 Photograph

Photograph Questions and Answers, Summary, Notes

Comprehension:

C1. Answer the following in a sentence or two each:

Question 1.
How many people are there in the photograph?
Answer:
There are three people in the photograph.

Question 2.
How is the poet related to the people in the photograph?
Answer:
The people in the photograph are the poet’s mother and her cousins, Betty and Dolly.

Question 3.
Who was taking the snapshot?
Answer:
An uncle was taking the snapshot.

Question 4.
Is the mother described in the photo alive?
Answer:
No, she is dead.

Question 5.
Which aspect of the mother does the poet like very much?
Answer:
The poet has sweet memories of the smiling and sweet face of the mother which she likes. This nature of the mother is captured in the photo.

C2. Answer the following:

Question 1.
Why does the writer say
“And of this circumstance
There is nothing to say at all.
Its silence silences”?
Answer:
The poet says that her mother had been dead and now she finds herself in a situation in which there is nothing to be said but only emptiness. The silence of this situation silences her. In other words, she is left speechless. Fate has killed all the feelings in her.

By the phrase ‘of this circumstance’, the poet means the circumstance of the death of her mother. The poet puns upon the. word ‘silence’. Since there are no details, what she is left with is silence and this silence silences her.

There are different possible meanings of the use of the word ‘silence’ as verb: she cannot say anything as she does not know anything about the circumstance of the death; or she does not have words to express her feelings; or the loss of her mother has stifled her voice. If we go one step further, we can even wonder whether the silence is owing to the unnatural nature of the death of the mother.

Question 2.
Does the poet notice any change in the mother after the poet was bom? What do you think could have made the change in the mother’s face?
Answer:
Yes, the poet notices the change in the mother’s face after she was born. This could have been the outcome of sorrowful incidents or hardships in life. Age and ill health also might have made the mother lose the sweetness of her face and smile.

The poet is looking at her mother’s photograph which is indeed an old one. With it she can see how her mother looked when she was a little girl of twelve. The photo shows her on a beach with her two girl cousins who are younger than her, holding her hand.

It might have been windy at that time as their hair was flying on their faces when the uncle took the photograph. All the three smile through their flying hair. Looking at the photograph, the poet says that her mother had a sweet face, but it was a time before the poet was born.

The sea was washing their feet. The poet says that the sea has changed only a little but change has come about the ones whose feet it was washing. After 30 or 40 years, the mother would take out the photograph and take a look at it.

By that time, she was married and had a daughter. She would laugh a little and say “Look at Betty and Dolly, see how they have dressed for the beach”. By now, she can only remember those days. A huge change has come about her and she is no longer that small, innocent girl of twelve. After some years, when the poet’s mother dies, for the poet, her mother’s laughter becomes a thing of the past.

That’s why she says “the sea holiday was her past and mine is her laughter”. In the same way as the mother remembers her old days, the poet can remember her mother. The poem also shows that in due course of time, the two of them learned to live with their losses though the loss had made a permanent impression on their wry faces.

Question 3.
Why are the feet described as ‘transient feet’?
Answer:
The word ‘transient’ means fleeting/ passing or temporary. It is the opposite of permanent. The ‘feet’ are described with an adjective ‘transient’ to drive home the truth that the impression they make on the sand is transitory. They get washed off by the waves.

Everything about life is transient. Single incidents, as well as whole life itself, are transient. Only nature remains as a permanent feature around us. Thus, the poet presents a contrast between human life and the sea.

The poem lends itself to multiple interpretations with regard to the transitory nature of life. If the passage of time is one aspect of the transitory nature of life, sudden unexpected occurrences can also indicate how ephemeral everything is.

We can have a different reading of the poem here. At one point the children are posing for the photograph holding their aunt’s hand. At another, the sea washes their terribly transient feet. The word ‘terribly’ makes us wonder whether the children are swept away by the waves of the sea.

C3. Answer the following questions on your own.

Question 1.
What is the mood of the poet?
Answer:
Melancholy marks the utterances of the poet. The poet has a deep sense of loss on losing her mother and the tone of sadness is all-pervasive.

Question 2.
Which line in the poem do you like the most? Why?
Answer:
I like the line ‘All three stood still to smile through their hair’. On the one hand, it creates a powerful mental picture of the three human figures against the breeze on the seashore posing for the photograph. The golden time of childhood when children are filled with mirth and are free from all worries and complexities of life is powerfully pictured here.

Smiling through the hair is typical of people who pose on the seashore. The description captures the innocuous and blithe spirit of the children which is contrasted against the care and concern of old age. It shows the poet’s power of observation and description.

The line gets connected to the title also. A photograph becomes a metaphor for all that life captures and also loses. Life is transitory and we are likely to lose many things which we will remember only through photographs.

Question 3.
Is there any change in the life of the poet’s mother over the years? What kind of a person, you think, she was? Describe the mother in the poem in your own words.
Answer:
The three stanzas of the poem depict the three stages in the life of the mother – as a child with her cousins, as a mother looking at the old photograph, and as a memory for the daughter after her death. She had a smiling and sweet face in the photograph when she posed for it, holding the hands of her cousins.

However, the poet cannot remember witnessing the same cheerfulness on the mother’s face in her recollections of her mother after she was born. What could have been the reason for the change? Apparently, the poet’s mother had posed for the photograph with her cousins when she was young and was not yet bogged down by the responsibilities and hardships of life.

As people age, along with inevitable physical changes, they also experience a change in their mental make-up because of the challenges in life. The line,’… how they dressed us for the beach,’ indicates that the mother herself was quite young and followed the directions given by others when she accompanied her cousins to the beach.

Additional Questions:

Question 1.
What does the word ‘cardboard’ denote in the poem? Why has this word been used?
Answer:
The word ‘cardboard’ means a very stiff and thick paper. Here the cardboard is a part of the frame that keeps the photograph intact. Its use in the poem is ironical. It keeps the photograph of that twelve-year-old girl safe who herself was ‘terribly transient’. She had died years ago.

Question 2.
What has not changed over the years? What does this suggest?
Answer:
The sea has not changed over the years. It brings out the ‘transient’ nature of man when compared to nature and its objects. Time spares none. The pretty faces and the feet of the three girls are ‘terribly transient’ or mortal when compared to the ageless and the unchangeable sea.

Question 3.
The poet’s mother laughed at the snapshot. What did this laugh indicate?
Answer:
The poet’s mother would laugh at the snapshot as it would revive her memories of the old happy days on the sea beach and the strange way in which they were dressed up for the beach. The laugh indicates her youthful spirit.

Question 4.
What is the meaning of the line “Both wry with the laboured ease of loss”?
Answer:
Both the poet’s mother and the poet suffer a sense of loss. The mother has lost her childlike innocence and joyful spirit that the photograph captured years ago. For the poet the smile of her mother has become a thing of the past. Ironically, both labour to bear this loss with ease.

Question 5.
What does ‘this circumstance’ refer to?
Answer:
‘This circumstance’ refers to the death of the poet’s mother. The photograph of the dead mother brings sad nostalgic feelings in the poet. But the poet has nothing to say at all about this circumstance. The silence of the poet makes the pall of silence prevailing there still deeper.

Question 6.
The three stanzas depict three different phases. What are they?
Answer:
In the first stanza, the poet’s mother is shown as a twelve-year-old girl with a pretty and smiling face. She went paddling with her two cousins. This phase is before the poet’s birth. The second phase describes the middle-aged mother laughing at her own snapshot. The third phase describes the chilling pall of silence that the death of the mother has left in the life of the poet.

Question 7.
“The sea, which appears to have changed less, washed their terribly transient feet.” – How does the poet contrast the girls’ ‘terribly transient feet’ with the sea?
Answer:
All the girls standing on the beach have ‘terribly transient’ existence. They are mortal and suffer physical changes with the passage of time. The mother’s sweet face and her smile has disappeared years later. But the vast sea remains unchanged of seems ‘to have changed less’ in their comparison.

Question 8.
‘Both wry with the laboured ease of loss’ – What is the loss? Describe the irony in the situation.
Answer:
Actually, both the poet and her mother suffer a sense of loss. The mother loses her carefree childhood. She can’t have those moments of enjoyment again that she once experienced at the beach. She can’t be a sweet smiling girl of twelve again.

This is also the poet’s loss. Perhaps she will never see that smiling face and experience her laughter again in life. The irony of the situation is that both of them struggle to bear the loss with tolerable ease.

Multiple Choice Questions:

Question 1.
The poet looks at the photograph of her mother taken when she was
A) ten years old
B) twelve years old
C) an old woman
D) a young lady
Answer:
B) twelve years old

Question 2.
The mother had gone on a sea-holiday
A) with her father
B) with her mother
C) with her cousins
D) with her daughter
Answer:
C) with her cousins

Question 3.
The photograph was taken by her
A) father
B) cousin
C) sister
D) uncle
Answer:
D) uncle

Question 4.
The Phrase ‘Their terribly transient feet’ suggests that
A) their feet were moving
B) they were padding
C) life is not permanent
D) life is everlasting
Answer:
C) life is not permanent

Question 5.
The poet’s mother is dead and the poet has nothing to say about that situation because
A) she does not want to say anything
B) the silence of death has silenced her
C) she cannot relive the past
D) she knows her mother would come back.
Answer:
B) the silence of death has silenced her

Question 6.
The mother saw the photograph after …………. years.
A) twenty or thirty
B) twelve or thirteen
C) four or five
D) one or two.
Answer:
A) twenty or thirty

Question 7.
The names of the cousins are
A) Bertha and Pinky
B) Betty and Dolly
C) Rosa and Mary
D) Laila and Pinky.
Answer:
B) Betty and Dolly

Question 8.
The cardboard shows the pictures of
A) two schoolgirls
B) two real sisters
C) two neighbours
D) narrator’s mother and her two cousins.
Answer:
D) narrator’s mother and her two cousins.

Question 9.
The ‘big girl’ here means
A) the eldest cousin of the mother
B) the tallest of the girls
C) narrator’s mother
D) the heaviest of the girls.
Answer:
C) narrator’s mother

Question 10.
The sweet face the photograph showed was that of the
A) narrator’s cousin
B) narrator’s father
C) narrator’s mother
D) narrator’s brother.
Answer:
C) narrator’s mother

Question 11.
The photograph was taken
A) when the narrator was about twelve
B) about twelve years ago
C) when the narrator was a child
D) when the narrator was not even born.
Answer:
D) when the narrator was not even born.

Question 12.
Who were ‘Betty’ and ‘Dolly’?
A) narrator’s cousins
B) her sisters
C) her mother’s cousins
D) her neighbours.
Answer:
C) her mother’s cousins

Question 13.
‘Its silence silences’ means
A) death’s silence
B) silence only brings out deeper silence
C) poet’s silence
D) silence caused by the mother’s death gives birth to a pall of silence.
Answer:
D) silence caused by the mother’s death gives birth to a pall of silence.

Photograph by Shirley Toulson About The Poet:

Shirley Toulson was born in 1924 in Henley-on-Thames, England. She had a huge passion for writing and was greatly influenced by her father who was a writer too. She secured a B.A. in Literature from Brockenhurst College in London in the year 1953. Shortly, she took writing as a career but meantime also served as editor for many magazines.

Celtic Christianity, influenced her so greatly that most of her major works such as ‘Celtic Alternative’ in 1987 and ‘Celtic Year’ in 1993 were on that topic. But a whole lot of other works, including collections of poetry, essays on education and other publications on diverse topics brought her even more fame.

Photograph Summary in English

The poem ‘Photograph’ is an interesting piece describing how a photograph can set in motion recollections of different kinds. The photograph is also an indicator to the fact that everything in life is transitory. What is captured years ago on the photograph is a contrast to what people turn out to be years later.

The photograph can also be illusory in nature as it might not show exactly how the people being featured in the photograph felt at that particular point of time. The photograph can also refer to our desperate attempts to defy the passage of time and the inevitable changes that take place along with the passage of time.

The speaker begins the poem by saying, ‘A photograph: The cardboard shows me how it was.’ The photograph is of the speaker’s mother with her cousins. The photograph features the two girls holding the hands of the speaker’s mother and the speaker knows that the photograph was taken by an uncle. The speaker is struck by the realisation that her mother was sweet in her looks at the time the photo was taken.

This was even before the speaker was 4}orn. While referring to the sea washing the terribly transient feet of the people who posed for the photograph, the speaker introduces a contrast between the sea and the people.

People change with the passage of time, but the sea seems to have changed very little. This is an indicator to the permanence of nature and transience of man. Unfortunately, the change in human beings, both physically and emotionally, is for the worse.

The smiling, sweet face of the young girl is replaced by the melancholic face of the mother. Just as physical changes are inescapable, even mental deterioration also seems to be inevitable.

Twenty or thirty years passed since that photograph had been taken. Whenever mother looked at the photograph, she would laugh. She would point out to Dolly and Betty how they were dressed up (by their parents) for the sea-holiday.

That sea-holiday had become a thing of the past for the mother. And that sweet laughter of the mother which the photograph had captured had become the poet’s own past. Both of them suffered from a sense of loss. Ironically, both of them were labouring to ease the loss.

The poet says that her mother has been dead for years. She has been dead nearly as many years as she had lived. She (the poet) has nothing to say about the circumstances of her death or about the situation captured in the photo. Silence only brings out deeper silence and makes things mysterious.

The photograph describes three stages. In the first stage, the photograph shows the poet’s mother standing on the beach enjoying her holiday with her two girl cousins. She was 12 or so at that time.

The second stage takes us twenty or thirty years later. The mother would laugh at the way she and her cousins Betty and Dolly were dressed up for the sea-holidays. In the third stage, the poet remembers the mother with a heavy heart as the mother is already dead. The photograph revives a nostalgic feeling in the poet.

Thus, through an everyday, commonplace incident of a photo being clicked on the seashore, the poet is able to drive home philosophical reflections on the very nature of life.

Glossary:

paddling: moving like rowing.
transient: momentary; not lasting for long
wry: distorted.

Students can Download English Lesson 8 The Dinner Party Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 5 English Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Board Class 5 English Prose Chapter 8 The Dinner Party

The Dinner Party Questions and Answers, Summary, Notes

Comprehension:

I. Answer the following questions

Question 1.
Who are the hosts and guests in the story?
Answer:
A colonial official and his wife, army officers and government official and their wives.

Question 2.
Describe the place where the dinner is hosted
Answer:
Dining room, which has a bare marble floor, open rafters and wide glass doors opening onto a veranda.

Question 3.
What is the young girl discussing with the colonel?
Answer:
A woman’s unfailing reaction in any crisis and a man’s nerve control.

Question 4.
According to the colonel Who has greater self-control, the man or the woman?
Answer:
Man.

Question 5.
Why do you think that the American did not join the discussion?
Answer:
The American is a naturalist, who just observes others and studies their reactions. So he did not join the discussion.

Question 6.
What did the hostess want her servants to do? Why?
Answer:
The hostess wants her servant to place a bowl of milk on the Veranda Just outside the open doors. Because a cobra had just crawled over her feet.

Question 7.
Do you think this action was a routine one?
Answer:
No.

Question 8.
What made the naturalist think that there must be a cobra in the room?
Answer:
In India milk in a bowl means only one thing bait for a snake. He realizes there must be cobra in the room.

Question 9.
Where did the American search for the cobra?
Answer:
He looked up at the rafters – the likeliest place, but they were bare. The three of the room were empty, and in the fourth, the servants were waiting to serve the next course. So he deduced that the cobra might be under the table.

Question 10.
What did the American do to test the self-control of the people at the dinner party?
Answer:
When the American Naturalist concluded that the cobra was under the table his first impulse was to jump back and warn others. But he did not want to frighten others and the cobra into striking one of them. In a sobering tone, he announced that he just wanted to know what self – control everyone at the table had.

He challenged them that he will start counting to three hundred – that’s five minutes and asked everyone to not move a muscle, and those who moved will lose fifty rupees. All the twenty guests sit like stone images while he counted. When he was counting two hundred and eight – the cobra emerged from under the table and went to the veranda where the bowl of milk was placed. When the guests saw the cobra, they were shocked and started, screaming. But the American quickly jumped up and slammed the veranda doors safely shut.

Question 11.
Who shows greater self-control in the stray The American guest or the hostess?
Answer:
The Hostess.

Question 12.
How does the American figure out that there is a cobra under the table?
Answer:
The American naturalist, while watching the hostess observes a strange expression come over the face of the hostess. She was staring straight ahead, he muscles contracting slightly.

She gestured to the servant standing behind her chain and whispered something to him. The servant’s eyes widened and he quickly left the room. The American then noticed that the servant placed a bowl of milk on the veranda just outside the open doors. In India, Milk in a bowl means – bait for a snake.

He looked up at the rafters – the likeliest place, but they were bare. The three of the room were empty, and in the fourth, the servants were waiting to serve the next course. So he deduced that the cobra might be under the table.

Question 13.
The colonel believes that women behave very differently from men in a crisis. Do you agree with the colonet’s belief.
Answer:
Yes, I agree that women behave very differently from men in a crisis. Women have more patience though the cobra was crawling across Mrs. Wynnes’s feet she kept her self – control.

Question 14.
How do you describe the followings:
a) The American Naturalist:
He is the one who kept control of the difficult situation in the lesson.

b) Mrs. Wynnes:
She realised that a cobra was in the room. Because it was crawling across her foot. But she kept her self – control.

c) The colonel:
The colonel is a male-centric person. He believed that men have more nerve control than woman.

d) The servants:
The servants were obedient and discrete.

Language Exercise

I. Use the following words in sentences of your own.

• Forfeit: The student cheated in the exams, so he had to for diet an year of schooling.
• Sobar: The children were playing daily in the garden, but they all turned sober when they saw a cobra, crawling about.
• Gesture: The teacher gestured at Nagaraj and called him to solve the Maths problem on the blackboard.
• Stare: It is not right to stare at physically handicapped people.

Grammar:

In one of the earlier lessons you have learnt the functions of adjectives. In this lesson lets learn how the three different forms of degrees of adjective are formed.

The three different forms/degree of adjectives are

1. The Positive degree
2. The Comparative degree
3. The Superlative degree

Usually, the comparative degree is formed by the addition of ‘er’ to the positive degree form ex: tall-taller short – shorter
The superlative degree is usually formed by the addition of ‘est’.

ex: strong stronger strongest broad broader broadest near nearer nearest Now, form the comparative and superlative degrees of the adjectives given below

Positive Comparative Superlative

1. fine – finer – finest
2. sweet – sweeter – sweetest
3. dear – dearer – dearest
4. high – higher – highest
5. fair – fairer – fairest
6. long – longer – longest
7. weak – weaker – weakest
8. fair fairer fairest
9. cool – coller – coolest
10. low – lower – lowest
11. bold – bolder – boldest
12. few – fewer – fewest

II. Match the adjectives given in column ‘A’ with the words given in B’

A	B
1. Brilliant	i) sea
2. Sharp	ii) light
3. loud	iii) shower
4. juicy	iv) knife
5. heavy	v) fruit
6. busy	vi) noise
7. rough	vii) cut
	viii) street

Answer:

A	B
1. Brilliant	ii) light
2. Sharp	iv) knife
3. loud	vi) noise
4. juicy	v) fruit
5. heavy	iii) shower
6. busy	viii) street
7. rough	i) sea

III. Complete the phrases below using appropriate adjectives

1. a delicious meal
2. a rough sea
3. a cute child
4. a gold ring
5. a terrible accident
6. a barking dog
7. an interesting story
8. a tough task
9. a good friend
10. a beautiful hat

IV. Write five sentences of your own using any of the completed phrases above.

1. Nazir read a thrilling story about pirates.
2. My uncle has a harking dog.
3. Francis is a good friend.
4. Arabian sea is a rough sea.
5. My mother prepared a delicious meal.

The Dinner Party Summary In English

The lesson ‘The Dinner Party” written by Mona Gardner explains about the importance of self-control. A colonial official and his wife are giving a large dinner party for the army officers and government officials and their wives, and a visiting American naturalist. A spirited discussion springs up between a young girl and a woman.

The American notices the servant place a bowl of milk on the veranda just outside the open doors. He realizes there must be cobra in the room. The American told all that he would count three hundred, that was five minutes and not one of you is to move a muscle.

Those who move will forfeit fifty rupees. The twenty people sat like stone images by the ti pie the cobra emerged and went to the varanda where the bowl of milk kept. The colonel appreciated the American who has just shown us an example of perfect control.

The Dinner Party Summary In Kannada