2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves

You can Download Chapter 8 Electromagnetic Waves Questions and Answers, Notes, 2nd PUC Physics Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves

2nd PUC Physics Electromagnetic Waves NCERT Text Book Questions and Answers

Question 1.
Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A.
(a) Calculate the capacitance and the rate of charge of potential difference between the plates.
(b) Obtain the displacement current across the plates.
(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 1
Answer:
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 2
(c) Yes, Kirchhoff’s first law will be valid at each plate of the capacitor, provided the electric current means the sum of the conduction and the displacement currents.

Question 2.
A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 Y ac supply with an (angular) frequency of 300 rad s1.
(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of B at a point 0 cm from the axis between the plates.
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 3
Answer:
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 4
(b) Yes the conduction current and the displacement current will be equal. It is true, even if the conduction current is oscillating.
(c) Let us first derive an expression for the amplitude of the magnetic field at a distance x from the axis of the capacitor. To do so, place a circular loop of radius r between the two plates of the capacitor with its face parallel to the plates and its centre on the axis of the plates.
According to modified Ampere’s circuital law,
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 5
The amplitude of the magnetic field at a distance r from the axis of the plates is then, given by.
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 6

Question 3.
What physical quantity is the same for X- rays of wavelength 10-1 m, the red light of wavelength 6800 A, and radio waves of wavelength 500m?
Answer:
All e.m. waves travel with the same speed c = 3 x 108 m s-1 in a vacuum.
Therefore, X-rays, red light, and radio waves have the same speed.

Question 4.
A plane electromagnetic wave travels in a vacuum along the z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
Answer:
The electric and magnetic field vectors are in X Y – plane,
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 7

Question 5.
A radio can tune in to any station in the 7.5 MHz to 12 MHz bands. What is the corresponding wavelength band?
Answer:
Corresponding to v = 7.5 MHz = 7.5 × 106Hz
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 8

Question 6.
A charged particle oscillates about its mean equilibrium position with a frequency of 109. What is the frequency of the electromagnetic waves produced by the oscillator?
Answer:
The frequency of the produced E.M. wave is the same as the frequency of oscillating charged particles i. e. 109 Hz.

KSEEB Solutions

Question 7.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in a vacuum is Bo = 510 nT. What is the amplitude of the electric field part of the wave?
Answer:
B0 = 510 nT = 510 x 10-9T
The amplitude of the electric field part of the wave,
E0 =CB0 =3 x l08 x 510 x l0-9 =153NC-4

Question 8.
Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/ C and that its frequency is v = 50.0 MHz.
(a) Determine, B0,ω, k, and λ,
(b) Find expressions for E and B.
Answer:
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 9

Question 9.
The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hv (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?
Answer:
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 10

2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 11

Question 10.
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 x 1010 Hz and amplitude 48 V m-1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field, [c = 3 x 108 ms-1]
Answer:
u =2 x 1010Hz
E0 = 48 Vm4 C
= 3 x l08ms-1
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 12
Question 11.
Suppose that the electric field part of an electromagnetic wave in a vacuum is E= {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 x 106 rad/s)t]} \(\hat { i } \)

  • What is the direction of propagation?
  • What is the wavelength λ?
  • What is the frequency V?
  • What is the amplitude of the magnetic field part of the wave?
  • Write an expression for the magnetic field part of the wave.

Answer:
The electric field part of the electromagnetic wave is given by,
\(\vec { E } \) = (3.1 NC-1 ) cos [(1,8 rad m-1) y + (5.4x 106 rad s1) t]\(\hat { i } \) it follows that
E0 = 3.1 NC-1
k =1.8 rad m-1
ω = 5.4×106 rad s-1
(a) Since the argument of since in the expression for electric field is of type (ky +cot), the direction of propagation of the electromagnetic wave is along the negative Y-axis.

(b) The wavelength of the wave,
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 13

(e) The electric field and the direction of propagation of the e.m. wave are along negative X-axis and negative Y-axis respectively. Therefore, the magnetic field is along negative Z-axis and expression for it is given by
B = (10.33nT) cos [(1.8 rad m-1) y + (5.4 x 106rad s-1)t] \(\hat { k } \)

KSEEB Solutions

Question 12.
About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
(a) At a distance of lm from the bulb?
(b) At a distance of 10 m? Assume that the radiation is emitted isotropically and neglect reflection.
Answer:
P = 100 W
n = 5%
Since the efficiency of the bulb is 5%, the effective power of the bulb.
\({ P }_{ eff }=p\times n=100\times \frac { 5 }{ 100 } =5w\)
At a distance, ‘r’ the radiation coming from the bulb is distributed over the surface of a sphere of radius r i.e. over an area.
A = 4πr2
Therefore, the intensity of light at a distance of r,
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 14

Question 13.
Use the formula λmT = 29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?
Answer:
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 15

2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 16

These numbers tell us the order of temperature for obtaining radiations in different parts of the electromagnetic spectrum. At these temperatures, the radiation of the corresponding wavelength in a particular case can be emitted at a lower temperature also, but its intensity will not be maximum.

Question 14.
Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).
(c) 7 K [temperature associated with the isotropic radiation fHifag nil space-thought to be a relic of the big-bang origin of the universe}.
(d) 5890 A – 5890 A (double lines of sodium}
(e) 14.4 keV {energy of a particular transition in 57Fe nucleus associated with a famous high-resolution spectroscopic method (Mossbauer spectroscopy)].
Answer:
(a) Given wavelength is of the order of 10-2 m i.e. short radio wave.
(b) Frequency is of the order of 109 Hz e. short radio wave.
(c) Microwave
(d) Visible(Yellow)
(e) X-rays (or soft γ-rays) region

Question 15.
Answer the following questions:

  1. Long-distance radio broadcasts use short-wave bands. Why?
  2. It is necessary to use. satellites for long-distance TV transmission. Why?
  3. Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why?
  4. The small ozone layer on top of the stratosphere is crucial for human survival. Why?
  5. If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?
  6. Some scientists have predicted that a global nuclear war on the earth would be followed by a severe nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction?

Answer:

  1. Long-distance radio broadcasts make use of sky waves. The ionosphere of the earth’s atmosphere reflects the radiations of this range.
  2. For very long-distance TV transmission, a very high frequency is required. Waves of this frequency just pass through the ionosphere and are not reflected back. Therefore, a satellite is required to reflect the signals to the earth.
  3. The atmosphere of the earth can absorb X-rays but visible waves and radio waves pass through it.
  4. The ozone layer absorbs ultraviolet (UV) radiation emitted by the sun and prevents them to reach the earth. Ultraviolet radiations are harmful for the life on earth.
  5. In this case, there will be no greenhouse effect. So the earth will be at a low temperature in the absence of the atmosphere.
  6. In the case of worldwide nuclear war, the sky may get overcast with clouds of nuclear radiation. These clouds will stop the passage of sunlight to earth. Thus, the earth will be as cool as in winter.

2nd PUC Physics Electromagnetic Waves Additional Questions and Answers

Question 1.
What is Maxwell’s displacement current?
Answer:
The displacement current is produced, when in an electric circuit (say across the plates of a charged capacitor), the electric field changes with time. Mathematically,
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 17

Question 2.
Can an electromagnetic wave be deflected by a magnetic or electric field? Explain.
Answer:
No, an electromagnetic wave cannot be deflected by a magnetic or electric field. It is because they do not consist of charged particles. They merely propagate in the form of electric and magnetic fields varying both in space and time.

Question 3.
The dimensions of \(\frac { 1 }{ { \mu }_{ 0 }{ \in }_{ 0 } } \) where symbols have their usual meanings are:
(A) [L1 T]
(B) [L2 T2]
(C) [L2 T-2]
(D) [L T1]
Answer:
The velocity of the electromagnetic wave in free space
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 18

Question 4.
Which of the following are not electromagnetic waves
(A) Cosmic rays
(B) γ-rays
(C) β – rays
(D) X-rays
Answer:
(C) β – rays

Question 5.
If \({ \in }_{ 0 }\) and \({ \mu }_{ 0 }\) are the electric permeability and magnetic permeability in free space, ∈ and µ are the corresponding quantities in a medium, then index of refraction of the medium is
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 19
Answer:
The velocity of the electromagnetic waves in free space is given by
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 20

Question 6.
The speed of electromagnetic waves is independent of
(A) Wavelength
(B) Frequencies
(C) Intensity
(D) Medium in which it travels
Answer:
(C) Intensity

KSEEB Solutions

Question 7.
If an electromagnetic wave propagating through vacuum is described by,
E = E0 sin (Kx – wt); B = B0 sin (Kx -ωt) then,
(A) E0K = Boω
(B) E0B0 = ωK
(C) E0ω=B0K
(D) E0B0=ω/K
Answer:
(A) E0K = Boω

Question 8.
The oscillating magnetic field in a plane electromagnetic wave is given by
By = 8×106sin [2 × 10-6t + 300 πx] (int) calculate the wavelength of the wave
Answer:
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 21

Question 9.
Find the wavelength of the electromagnetic wave of frequency 5x 1019 Hz in free space. Identify the wave.
Answer:
2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 22

Question 10.
Give the ratio of velocities of light rays of wavelength 4000 A in a vacuum.
Answer:
The velocity of light rays of different wavelengths in a vacuum is the same and hence the ratio of their velocities is 1.2nd PUC Physics Question Bank Chapter 8 Electromagnetic Waves 16

Siri Kannada Text Book Class 7 Solutions Gadya Chapter 8 Sankrantiyandu Sukha-Dukha

Students can Download Kannada Lesson 8 Sankrantiyandu Sukha-Dukha Questions and Answers, Summary, Notes Pdf, Siri Kannada Text Book Class 7 Solutions, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Siri Kannada Text Book Class 7 Solutions Gadya Bhaga Chapter 8 Sankrantiyandu Sukha-Dukha

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1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids

You can Download Chapter 9 Mechanical Properties of Solids Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids

1st PUC Physics Mechanical Properties of Solids Textbook Questions and Answers

Question 1.
A steel wire of length 4.7 m and cross-sectional 3.0 × 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10-5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Answer:
Let the Young’s Modulus of steel and copper be Ys and Yc respectively.
Length of steel wire, Ls = 4.7 m
Length of copper wire, Lc = 3.5 m
Area of cross-section of steel wire,
As= 3 × 10-5 m2
Area of cross-section of copper wire,
Ac = 4 × 10-5 m2
Since change in lengths are equal.
∴ ∆Ls = ∆Lc = ∆ L.
As the load is same, Fs = Fc = F
We have, Y = \(\frac{F}{A} \times \frac{L}{\Delta L}\)
\(\therefore \frac{Y_{s}}{Y_{c}}=\frac{L_{s} A_{c}}{A_{s} L_{c}}=\frac{4 \times 4.7 \times 10^{-5}}{3 \times 3.5 \times 10^{-5}}\) = 1.79
The ratio of Young’s Modulus of steel to that of copper is ≃ 1.8 :1.

Question 2.
Figure shows the strain – stress curve for a given material. What are
1. Young’s modulus and
2. approximate yield strength for this material?
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 1
Answer:
1. F from the given graph,
For a stress of 150 × 106 Nm-2
Strain = 0.002
∴ Young’s modulus (Y) = \(\frac{\text { stress }}{\text { strain }}\)
\(=\frac{150 \times 10^{6} \mathrm{Nm}^{-2}}{0.002}\) = 7.5 × 1010 Nm-2

2. 300 × 106 Nm-2 is the maximum stress that the material can sustain without crossing the elastic limit. Thus, the yield strength of the material is 3 × 108 Nm-2 or 300 × 106 Nm-2.

Question 3.
The stress-strain graphs for materials A and B are shown in Figure.
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 2
The graphs are drawn to the same scale.

  1. Which of the materials has the greater Young’s modulus?
  2. Which of the two is the stronger material?

Answer:

  1. Material A has greater Young’s modulus as the slope of the graph of material A is greater than material B.
    Young’s Modulus = Y = Slope = \(\frac{\text { stress }}{\text { strain }}\)
    ∴ Material A has greater Young’s Modulus.
  2. From the graph it can be approximated that the stress at the fracture point for material A is higher than material B, thus material A is stronger.

KSEEB Solutions

Question 4.
Read the following two statements below carefully and state, with reasons, if it is true or false.
1. The Young’s modulus of rubber is greater than that of steel;
2. The stretching of a coil is determined by its shear modulus.
Answer:
1. The Young’s modulus of a -material defined as,
Y = \(\frac{\text { stress }}{\text { strain }}\)
Since, rubber stretches more than steel ∆ L for rubber is greater than that of steel, therefore ( \(\frac{\Delta L}{L}\) 0f rubber) > (\(\frac{\Delta L}{L}\) of steel) Since, Y \(\alpha \frac{1}{\text { strain }}\) and, strain = \(\frac{\Delta L}{L}\)
Since the strain of rubber is greater, its Youngs modulus is less. Hence false,

2. The Stretching of a coil depends on the shear modulus of the material because for the coil to be stretched, a shear force has to applied to change the shape of the coil. Hence the statement is true.

Question 5.
Two wires of diameter 0.25 cm. one made of steel and the other made of brass are loaded as shown in Fig.9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of steel and the brass wires.
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 3
Answer:
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 4
given, d = 0.25 cm
r = \(\frac{\mathrm{d}}{2}\) = 0.125 cm.
As = AB = π × (0.125 × 10-2)2 m2
= 4.91 × 10-6 m2
As = AB because the diameters are same.
Fs = (4 × g + 6 × g) N
= (4 × 9.8 + 6 × 9.8) N
= 10 × 9.8 N =98 N
FB = (6 × 9.8) N = 58.8 N
∆Ls = ?
∆LB = ?
Ls = 1.5 m
LB =1.0m
Ys = 2 × 1011 N m-2 and
YB = 0.91 × 1011 Nm-2
∆Ls = \(\frac{F_{s} \cdot L_{s}}{A_{s} \cdot Y_{s}}=\frac{98 \times 1.5}{4.91 \times 10^{-6} \times 2 \times 10^{11}} \mathrm{m}\)
= 1.49 × 10-4m = 0.149 mm
To find the elongation of Brass,
∆LB = \(\frac{F_{B} L_{B}}{A_{B} Y_{B}}=\frac{(58.8)(1)}{\left(4.91 \times 10^{-6}\right)\left(0.91 \times 10^{11}\right)} \mathrm{m}\)
= 1.32 × 10-4m = 0.132 mm
The elongation of steel wire is 0.149 mm.
The elongation of Brass wire is 0.132 mm.

Question 6.
The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Answer:
Edge length, L = 10 cm = 0.1 m
Weight attached = 100 kg
F = (100 × 9.8) N = 980 N
A = L2 = 0.1 × 0.1 m2 = 0.01 m2
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 5
= 3.92 × 10-7 m
∴ the deflection is 3.92 × 10-7 m

Question 7.
Four identical hollow cylindrical columns of mild steel support a big structure of mass of 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Answer:
Mass of big structure = 50,000 kg
Net force = (5 × 104 × 9.8) N
= 4.9 × 105 N = FB
Since it is supported by four cylinders the weight is equally distributed.
Force (F) on each cylinder is
F = \(\frac{F_{B}}{4}\) = 122500 N
Young’s Modulus = Y = = \(\frac{\text { stress }}{\text { strain }}=\frac{F \times L}{A \times \Delta L}\)
∴ Compressional strain of each column is given by, \(\frac{\Delta L}{L}=\frac{F}{A \times Y}\)
Here, F = 122500 N
A = π(0.6)2 – π(0.3)2 = π(0.36 – 0.09)
= 0.27 π = 8.48 × 10-1 m2
Y = 2 × 1011 N m-2 (Standard Values)
∴ Strain = \(\frac{122500}{(0.27 \pi)\left(2 \times 10^{11}\right)}\) = 7.22 × 10-7
Thus the compressional strain per cylinder is 7.22 × 10-7

KSEEB Solutions

Question 8.
A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
Answer:
F = 44,500 N
Strain =?
Length = 19.1 mm, Breath = 15.2 mm
Area = A = 19.1 × 15.2 × 10-6m2
= 2.98 × 1 O-4 m2
G = 42 × 109 Nm-2
We have, G = \(\frac{F}{A \times \text { strain }}\)
∴ Strain = \(\frac{F}{G \times A}\)
\(=\frac{44,500}{2.98 \times 10^{-4} \times 42 \times 10^{9}}\)
= 3.65 × 10-3

Question 9.
A steel cable with a radius of 1.5 cm supports a chair lift at a ski area. If the maximum stress does not exceed 108 Nm-2, What is the maximum load the cable can support?
Answer:
Maximum load = maximum stress x area of cross-section
= 108 n r2 =108 X (22/7) x (1.5 x 10-2)2 = 7.07 x 104 N

Question 10.
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of Iron. Determine the ratios of their diameters if each is to have the same tension.
Answer:
Since the rigid bar is supported symmetrically by three wires each of length 2 m, the extension is same in all the three wires. Thus strain is same.
\(Y=\frac{F}{\pi r^{2} \times \text { strain }}\) = \(\frac{4 \mathrm{F}}{\pi \mathrm{d}^{2} \times \text { strain }}\)
∴ Y \(\alpha \frac{1}{\mathrm{d}^{2}}\)
Yi = Young’s modulus of Iron
di = diameter of iron wire.
Yc = Young’s modulus of copper
dc = diameter of copper wire.
Yi = 190 × 109 N m-2
Yc = 110 × 109 N m-2
\(\frac{d_{c}}{d_{i}}=\sqrt{\frac{Y_{i}}{Y_{c}}}=\sqrt{\frac{19}{11}}=1.31: 1\)

Question 11.
A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.
Answer:
L = 1 m
W = 2 revolutions/s
A = 0.065 cm2 = 6.5 x 10-6 m2
F = mg + mlw2
= 14.5 × 9.8 +14.5 × 1 × (2)2
= 200.1 N
∆L = \(\frac{F l}{A Y}\)                [∵ Y = \(\frac{\mathrm{FI}}{\mathrm{A} \Delta \mathrm{I}}\)]
Young’s modulus of steel is 2 × 1011 Nm-2
∆L = \(\frac{200.1 \times 1}{6.5 \times 10^{-6} \times 2 \times 10^{11}} \mathrm{m}\) = 1.539 × 10-4m

Question 12.
Compute the bulk modulus of water from the following data: Initial Volume = 100.0 litre. Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
Answer:
VF = final volume = 100.5 I
= 100.5 × 10-3 m3
Vi = initial volume = 100 × 10-3 m3
∆V = VF – Vi =0.5 × 10-3 m3
= 5 × 10-4 m3
∆ P = 100 atm = 100 × 1.013 × 105 Pa
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 6
= 2.03 × 109 Pa
Bulk modulus of air = 1 × 105 Pa
\(\frac{\text { Bulk modulus of water }}{\text { Bulk modulus of air }}=\frac{2.026 \times 10^{9}}{1 \times 10^{5}}\)
= 2.026 × 104
Since air is highly compressible compared to water, the ratio is so high.

Question 13.
What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m-3?
Answer:
Density of water at surface
= \(\rho_{1}\) = 1.03 × 103 kgm-3
density of water at depth h = \(\rho_{2}\)
Pressure at depth h = 80 atm
= 80 × 1.01 × 105 Pa.
V1 = Volume of water of mass m at surface.
V2 = Volume of water at depth h.
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 7
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 8
\(\frac{1}{\mathrm{B}}\) Compressibility of water
= 45.8 × 10-11
∴ \(\frac{\Delta V}{V_{1}}\) =80 × 1.013 × 105 × 45.8 × 10-11
= 3.71 × 10-3
Since,
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 9
= 1.034 × 103 kg m-3

Question 14.
Compute the fractional change in volume of a glass slab, when subjected to hydraulic pressure of 10 atm.
Answer:
Hydraulic pressure exerted = 10 atm
= 10 × 1.013 × 105 Pa
B = \(\frac{P \cdot V}{\Delta V}\)
\(\frac{\Delta V}{V}\) is the fractional change in volume
\(\frac{\Delta V}{V}=\frac{P}{B}=\frac{10 \times 1.013 \times 10^{5}}{37 \times 10^{9}}\)
= 2.73 × 10-5
∴ The fractional change in Volume is 2.73 × 10-5.

Question 15.
Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 106 Pa.
Answer:
P = 7 × 106 Pa
I = 10 cm
V = I3= 10-3 m3
B = 140 × 109
B = \(\frac{P \times V}{\Delta V}\)
∆V = \(\frac{P}{B} \times V=\frac{7 \times 10^{6}}{140 \times 10^{9}} \times 10^{-3} \mathrm{m}^{3}\)
= 5 × 10-8 m3
= 0.05 cm3
∴ The contraction is 0.05 cm3

KSEEB Solutions

Question 16.
How much should the pressure on a litre of water be changed to compress it by 0.10%
Answer:
P =?
V = 1L ,
B = 2.2 × 109Nm-2
\(\frac{\Delta v}{V}\) × 100 = 0.1
⇒ \(\frac{\Delta v}{V}\) = \(\frac{0.1}{100 \times 1}\) = 10-3
Since, B = \(\frac{\mathrm{PV}}{\Delta \mathrm{V}}\)
P = B \(\frac{\Delta v}{V}\)
= 2.2 × 109 Pa × 10-3
= 2.2 × 106 Pa.
∴ The pressure is required is 2.2 × 106 Pa.

1st PUC Physics Mechanical Properties of Solids Additional Exercises Questions and Answers

Question 1.
Anvils made of single crystals of diamond, with the shape as shown in Fig. are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 10
Answer:
Diameter at narrow end
= d =0.5 mm
Compressional Force = F = 50,000 N
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 38
∴ The pressure is 2.546 × 1011 Pa

Question 2.
A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig.9.15. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2 respectively. At what point along the rod should a mass m be suspended in order to produce.

  1. equal stresses.
  2. equal in both steel and aluminium wires.

1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 11
Answer:
Let the initial lengths of the 2 wires be ‘L’
Given that the cross-sectional area of A
= Aa = 1 mm2 = 10-6
Cross-sectional area of
B = Ab = 2 mm2 = 2 × 10-6m2
Length of rod = 1.05 m.
Young’s modulus of wire A
= Ysteel = 2 × 1011 N/m2
Young’s modulus of wire B
YAl = 7 × 1010 N/m2
Let the mass ‘m’ be suspended at a distance ‘x’ from the end with wire A attached.

1. For equal stress on the two wires :
i.e., stress A = stress B
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 12
or Fb = 2 Fa …..(1)
where, Fa is force exerted on steel wire and Fσ is force exerted on aluminium wire. Since the total torque at the point when mass is suspended is zero, we can write
Fa × (x) = Fb × (1.05 – x)
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 13
⇒ 3x = 2 × 1.05 or x = 0.7 m
So, for equal stress on the wires the mass should be suspended at a distance 0.7m, from wire A.

2. For equal strain:
We know that, Strain = \(\frac{\text { Stress }}{Y}=\frac{F}{\text { AY }}\)
we require (strain)a = (strain)b
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 14
\(\frac{F_{a}}{F_{b}}\) = \(\frac{10}{7}\)
……………(3) Since the torque is ‘0’ at the point of suspension equation (2) is valid.
⇒ \(\frac{F_{a}}{F_{b}}\) = \(\frac{10}{7}\) = \(\frac{1.05-x}{x}\)
⇒ 17 x = 7 × (1.05)
x = 0.432 m
So, for equal strain on the wires the mass should be suspended at a distance 0.432 m from wire A.

Question 3.
A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10-2cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the mid-point.
Answer:
Given that length of the wire = 1.0 m
Area of cross-section = 0.50 × 10-2cm2
= 5.0 × 10-7m2
Mass suspended =100g
= 10-1 Kg
Ysteel = 2 × 1011 Pa
The mass causes the steel wire to bend as shown.
So new length of wire = AX + XB
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 15
Change in length is, ∆l = (AX + XB) – AB
Now, AX = \(\sqrt{\mathrm{AM}^{2}+\mathrm{MX}^{2}}\) = XB
[ Pythogorus Theorem]
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 16
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 17
[Expanding and neglecting higher-order terms]
The tension in the steel wire is,
T cos θ + T cos θ = mg
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 18
\(\sqrt[3]{\frac{0.5 \times 0.1 \times 9.8}{4 \times 2 \times 10^{11} \times 5 \times 10^{-7}}}\) = 0.0106 m
So, the depression at the centre is 0.0106 m or 1.06 cm.

Question 4.
Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 107 Pa? Assume that each rivet is to carry one-quarter of the load.
Answer:
Given diameter of metal strips
= 6.0 × 10-3 m
⇒ Area = \(\pi\left(\frac{\mathrm{d}}{2}\right)^{2}=\frac{\pi \times 36 \times 10^{-6} \mathrm{m}^{2}}{4}\)
= 2.827 × 10-5m2
For the maximum stress of 6.9 × 107 Pa
the maximum force is given by
Max Force = \(\frac{\text { Max Force }}{\text { Area }}\)
⇒ Max force = Max Stress × Area
= 6.9 × 107 × 2.827 × 10-5
= 1950.63 N
Since there are 4 rivets and each rivet carries 1/4th of the load,
Max tension that can be exerted by the rivet strip is 4 × 1.95063 KN
= 7.80252 KN.

KSEEB Solutions

Question 5.
The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench Is about 1.1 × 108 Pa. A steel ball of Initial volume 0.32 m3 is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?
Answer:
Pressure at bottom ’of the trench = ρgh = 1.1 × 108 Pa
Initial volume of the ball = 0.32 m3
Bulk modulus of steel = 1.6 × 1011 Pa
Let ∆v the change in volume of Ball when it falls down to the bottom.
We know that, B = \(\frac{P}{(\Delta V / V)}\)
⇒ ∆v = \(\frac{P V}{B}\)
\(=\frac{1.1 \times 10^{8} \times 0.32}{1.6 \times 10^{11}}\) = 2.2 × 10-4 m3
Therefore, change in volume of the ball on reaching the bottom of the trench is 2.2 × 10-4 m3 or 220 cm3

1st PUC Physics Mechanical Properties of Solids One Mark Questions and Answers

Question 1.
What is elasticity?
Answer:
The property of the body to restore the original shape or size on removal of the external force is called elasticity.

Question 2.
Define stress.
Answer:
Stress is defined as the deforming force acting on a unit area of the body.

Question 3.
What is the S.l unit of stress?
Answer:
Nm-2

Question 4.
Define strain.
Answer:
Strain is the ratio of change in dimension of a body due to stress to the original dimension.

KSEEB Solutions

Question 5.
Define Shear Strain.
Answer:
The angular tilt between different layers of a body due to tangential force (Shear Force) is called Shear Strain.

Question 6.
Define Modulus of elasticity.
Answer:
Modulus of elasticity is defined as the ratio of stress to strain.

Question 7.
State Hooke’s law.
Answer:
Within elastic limit, stress is directly proportional to strain.

Question 8.
Mention the expression for Young’s modulus in the case of a stretched string.
Answer:
Young’s modulus in case of a stretched string is Y = \(\frac{\mathrm{FL}}{\pi r^{2}(\Delta \mathrm{L})}\)
F – Force acting on the wire,
L – Initial length of the wire
∆L – change in length of the wire,
r – radius of the wire.

Question 9.
Which is more elastic, rubber or iron?
Answer:
Iron.

Question 10.
What are anisotropic materials?
Answer:
The materials which have different physical properties along different directions are called anisotropic materials.

Question 11.
Name 2 main groups of solids.
Answer:
Amorphous and Crystalline.

KSEEB Solutions

Question 12.
What are glass solids?
Answer:
Glassy solids are amorphous solids.

Question 13.
What is the Bulk modulus of an incompressible object?
Answer:
Since its incompressible strain is 0 and Bulk modulus is infinity.

Question 14.
Show graphically stress v/s strain curve for a Brittle material.
Answer:
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 19

Question 15.
If the length of the wire Is made 2 times, what Is the effect on the Increase in length under same load?
Answer:
∆l ∝ l. So, increase in length is also doubled.

1st PUC Physics Mechanical Properties of Solids Two Marks Questions and Answers

Question 1.
If a wire is replaced by another similar wire but of half the diameter. What is the effect on

  1. Increase in the length for a given load,
  2. Max load It can sustain.

Answer:
We know that, \(\frac{\Delta L}{L}=\frac{F}{A Y}\)
1.
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 20
The increase in the length (∆L)is 4 times the original wire.

2.  We know that, F = F=A \(\left(\frac{\mathrm{Y} \Delta \mathrm{L}}{\mathrm{L}}\right)\) Thus F ∝ A and hence maximum load it can sustain becomes 1/4th of the original wire.

Question 2.
What is a perfectly elastic and inelastic body?
Answer:
If a body completely regains its natural shape or size after the removal of external forces, it is called a perfectly elastic body.
Eg: rubber.
If the body does not regain its original size or shape it is called inelastic body or plastic body.

Question 3.
Identical springs made of steel and aluminium are equally stretched. On which more work will have to be done? Why?
Answer:
Work done for steel spring is more because steel is more elastic than aluminium, (alternatively: Spring constant of steel is more than that of aluminium).

KSEEB Solutions

Question 4.
Why does a cycle tube burst in summer?
Answer:
In summer as the temperature increases, the pressure increases inside the tube This pressure would be enough to cross the breaking point of the tyre making it to burst.

Question 5.
When the pressure on a sphere increases by 50 atm then its volume decreases by 0.02%. Find Bulk modulus.
Answer:
P = 50 atm
P = 50 atm ≅ 50 × 1.01 × 105
\(\frac{\Delta V}{V}=\frac{0.02}{100}=2 \times 10^{-4}\)
Bulk modulus= \(\frac{\mathrm{P}}{(\Delta \mathrm{V} / \mathrm{V})}=\frac{5.05 \times 10^{6}}{2 \times 10^{-4}}\)
= 2.525 × 1010N/m2

Question 6.
Which is more elastic, rubber or Steel? Why?
Answer:
For a given force, rubber extends more than steel, i.e., Steel is more elastic than rubber. This is because for the same amount of extension. Steel requires more force. For the same force ∆l ∝ Y, and F ∝ Y, for same extension Ysteel > Yrubber.

Question 7.
What is elastic Limit?
Answer:
Elastic limit is the value of applied force beyond which the material (body) does not come back to its original shape when the applied force is reduced.

Question 8.
Plot load v/s extension curve for a substance on the graph show,

  1. Yield point
  2. Breaking point
  3. Elastic limit
  4. Crushing point.

Answer:
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 21
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 22

Question 9.
The Young ‘s modulus of a wire of length L and radius ‘r’ is Y. If length is reduced to L/3 and Radius r/4. What will be Its ratio of extension?
Answer:
Young’s modulus is given by
Y = \(\frac{F}{A} \frac{l}{\Delta}\) ⇒ ∆l = \(\frac{F}{A} \frac{l}{Y}\)
So, ∆l ∝ l and ∆l ∝ \(\frac{1}{A}\) and A ∝ r2
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 23

Question 10.
The Young’s modulus of a material is Y. It has an area A and length ‘L’. Now if a is made \(\frac{A}{16}\) and length L/2. What is the new Young’s modulus?
Answer:
Even though the value of Young’s modulus is given by, Y = \(\frac{F}{A} \cdot \frac{l}{\Delta l}\). It is only material dependent. It is constant for the given wire irrespective of its dimension. So, the new value of Young’s modulus is Y.

KSEEB Solutions

Question 11.
The length of a metal is L1 when a force is F1 is applied on it and its length is L2 when applied force is F2. Find the original length of wire.
Answer:
We know that Y = \(\frac{\mathrm{F} l}{\mathrm{A} \Delta l}\)
Let original length be L. Then,
L1 = L + ∆l1
L2 = L + ∆l2
Since, Y is a constant,
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 24

Question 12.
Explain the terms Young’s modulus and elastic fatigue.
Answer:
Young’s modulus is defined as the ratio of longitudinal stress to longitudinal strain.
Y = \(\frac{(F / A)}{(\Delta l / l)}\)
Elastic Fatigue is the time (delay) required for a material to regain its original position on removing the deforming force.

1st PUC Physics Mechanical Properties of Solids Three Marks Questions and Answers

Question 1.
When a wire is stretched by a certain force its elongation is ‘a’. If the second wire having double-radius and 4-time length. Find its elongation.
Answer:
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 25
= 4 × 4 a
∆l2 = 16 a.

Question 2.
Define stress and strain and derive their units. Write one limitation of Hooke’s Law.
Answer:
Stress is the deforming force experience by the body per unit area.
Stress = \(\frac{F}{A}\)
Units = \(\frac{(N)}{\left(m^{2}\right)}\) = Nm-2
Strain is the change in the considered quantity (l,v, A, θ…..) per unit of its original value.
Strain = \(\frac{\Delta x}{x}\)
It is unitless.
Hooke’s Law is valid only for small deformation i.e., in its limit.

1st PUC Physics Mechanical Properties of Solids Five Marks Questions and Answers

Question 1.
The Stress-Strain graph of a metal wire is shown in figure up to pint E. The wire returns to its original state 0 along EPO when it is gradually unloaded point B corresponds to the fracture of wire.
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 26

  1. Up to what point of the curve Hooke’s Law is obeyed?
  2. Which point on the curve corresponds to yield point or elastic limit.
  3. Indicate the Elastic region and plastic region.
  4. Describe what happens when the wire Is loaded up to the stress corresponding to point A and then unloaded gradually. In particular to the dotted line.
  5. What is peculiar about portion of curve from C to B? Upto what stress can be applied without causing fracture?

Answer:

  1. Since the graph is linear till point E. Hooke’s Law is obeyed till point E.
  2. The point E corresponds to the elastic limit because till there the body returns to its original state along EPO.
  3. Elastic – O to E; Plastic – E to B
  4. Up to point E an almost linear increase in strain with stress. After point E, if it is loaded till A, the strain increases faster for some increase in stress. But on unloading the body returns along line AO to a permanent strain (change in shape) of O’.
  5. In the curve C to B, the body undergoes a strain even if it is being unloaded and then fractures. To present fracture the value of stress has to be just less than that of C.

Question 2.
Explain different types of strain.
Answer:
There are 3 types of strain.
1. Longitudinal strain :
It is defined as the ratio of change in length produced (dl) to the original length (L) of a thin rod.
Longitudinal strain = \(\frac{d l}{L}\)

2. Shearing strain :
It is defined as the ratio of displacement of a surface on which stress is acting to the height of the surface.
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 27
Shearing strain θ = \(\frac{x}{h}\)
\(=\frac{\text { displacement }}{\text { heightof the surface }}\)

3. Bulk strain :
It is defined as the ratio of change in volume (dV) to the original volume (V).
Bulk strain = \(\frac{\mathrm{d} \mathrm{v}}{\mathrm{v}}\).

KSEEB Solutions

Question 3.
Explain the different types of elastic constant.
Answer:
1. Young’s modulus (y):
It is defined as the ratio of longitudinal stress to the longitudinal strain.
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 28
where F – Longitudinal force,
L – original length,
dl – change in length,
A – area of cross-section of the wire.

2. Rigidity modulus (η):
It is defined as the ratio of shearing stress to the shearing strain.
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 29

where F – tangential force,
A – area of the surface,
h – height of the surface,
x – displacement of the surface

3. Bulk modulus (K) :
It is defined as the ratio of bulk stress to the bulk strain. Bulk modulus is given by,
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 30
where F/A = P – pressure, V is the original volume and dV is the change in volume.

1st PUC Physics Mechanical Properties of Solids Numerical Problems Questions and Answers

Question 1.
What is the percentage increase in length of a wire of diameter 12.5 mm stretched by a weight of 1000 kg ωt ?
[Given Y = 12.5 × 1011 dyne/sq.cm]
Answer:
Area = \(\frac{\pi d^{2}}{4}=\frac{\pi(1.25)^{2}}{4} \mathrm{cm}^{2}\)
F = 1000 kg ωt
= 1000 × 103 × 980
F = 9.8 × 108 dyne
We know that,
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 31
\(=\frac{9.8 \times 4 \times 10^{-3}}{\pi \times(12.5)(1.25)^{2}}\)
= 0.638 × 10-3
Percentage increase = 0.064%.

Question 2.
A cube of side 3 cm Is subjected to shearing force. The top of cube is sheared through 0.012 cm with respect to bottom face. If cube is made of a material of shear modulus 2.1 × 1010Nm-2 then find

  1. Shearing strain
  2. Shearing stress
  3. Force applied.

Answer:
1. Shearing strain
\(=\frac{\Delta l}{l}=\frac{0.012 \mathrm{cm}}{3 \mathrm{cm}}\)
= 0.004

2. Shearing stress
= Shearing strain × Shear modulus
Shearing stress = 0.004 × 2.1 × 1010Nm-1
= 8.4 × 107Nm-2

3. Shearing stress = \(\frac{\text { S.Force }}{\text { Area }}\)
Shearing force = Shearing stress × Area
= 8.4 × 107 × (3 × 10-1)2
= 7.56 × 104 N
= 75.6 KN.

Question 3.
A composite wire of materials A and B are of lengths 3.2 m and 2.4 m, and areas 1cm2 and 1cm2 respectively. If the extension observed is 1.2 mm. Find the force ‘F’. Given YA = 1.6 × 1011 Nm-2 and YB = 2.4 × 101 Nm-2
Answer:
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 32
We know that Y = \(\frac{F}{A} \times \frac{l}{\Delta L}\)
Let the individual extension be ∆lA and ∆lB
∆LA = \(F\left(\frac{L_{A}}{A_{A} Y_{A}}\right)\); ∆LB = \(F\left(\frac{L_{B}}{A_{B} Y_{B}}\right)\)
Given ∆LA + ∆LB = 1.2 mn
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 33
= 1.2 × 10-4
F(3 × 10-7) = 1.2 × 10-4
F = 400 N

Question 4.
A cube of 5 cm has its upper face displaced by 0.2 cm by a force of 8 N. Find Shear Modulus, Stress and Strain.
Answer:
Shear stress = \(\frac{F}{A}=\frac{8 N}{25 \times 10^{-4} m^{2}}\)
= 3200 Nm-2
Shear Strain = \(\frac{\Delta l}{l}\)
But ∆l = 0.2 cm
l = 5 cm
Shear strain = \(\frac{0.2 \mathrm{cm}}{5 \mathrm{cm}}\)
= 0.04
Shear Modulus = \(\frac{\text { Shear Stress }}{\text { Shear Strain }}\)
\(=\frac{3200}{0.04}\)
= 80 × 103Nm-2 or 80 kPa.

KSEEB Solutions

Question 5.
A rubber string 10 m long is suspended from a rigid support at one end. Calculate extension of the string due to its own weight. The density of rubber is 1.5 × 103 kg/m3 and Yrubber = 5 × 106Nm-2.
Answer:
The weight at any point x from below is, \(\frac{\mathrm{M}}{10}\) (10 – x) where M is mass of string.
So, Force at any point dx is
F = \(\frac{\mathrm{Mg}}{l}\) (l – x)dx
We know that,
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 34
Net extension =
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 35
So, Net extension of the rubber due to its own weight = \(\frac{1.5 \times 10^{3} \times 10 \times 100}{2 \times 5 \times 10^{6}}\)
= 0.15 m.

Question 6.
A metal rod is tied vertically. If the breaking stress is 5.4 × 109 Nm-2 and density 5.4 × 103 kg/m3. Find the maximum length of rod that can be held without breaking.
Answer:
Stress = \(\frac{F}{A}\)
= \(\frac{\mathrm{Mg}}{\mathrm{A}}\) [∵ \(\frac{M}{A L}\) = ρ]
= ρgL
If ρgL > Maximum stress that it can sustain than the rod break.
So, ρgL < 5.4 × 109 Nm-2
l< \(\frac{5.4 \times 10^{9}}{5.4 \times 10^{3} \times\left(10 m s^{-1}\right)}\)
l < 105 m
Maximum length of wire is 105 m or 100 km..

Question 7.
Consider a tapered wire as shown. Find the extension if Y is its Young’s modulus.
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 36
Answer:
Consider an element dx which is at a distance x from the end with r, as radius. Tfhe change in length in this portion is
dl = \(\left(\frac{F}{A}\right) \frac{d x}{Y}\)
From figure we can write radius at a distance x is, rx = r1 + xtanθ
⇒ A = π(rx)2
= π (r1 + xtanθ)2
1st PUC Physics Question Bank Chapter 9 Mechanical Properties of Solids img 37
Total extension = \(\frac{F\left(r_{2}-r_{1}\right)}{Y \cdot r_{1} \cdot r_{2} \tan \theta}\)
\(=\frac{\mathrm{FL} \tan \theta}{\mathrm{Y} \mathrm{r}_{1}-\mathrm{r}_{2} \tan \theta}\)
Total extension = \(\frac{F \cdot L}{Y r_{1} r_{2}}\)
[∵ tanθ = \(\frac{r_{2}-r_{1}}{L}\)]

2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments

You can Download Chapter 9 Ray Optics and Optical Instruments Questions and Answers, Notes, 2nd PUC Physics Question Bank with Answers, Karnataka State Board Solutions help you to revise the complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments

2nd PUC Physics Ray Optics and Optical Instruments NCERT Text Book Questions and Answers

Question 1.
A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature of 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Answer:
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 1
Therefore, the screen should be placed at a distance of 54cm from the mirror on the same side as the object,
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 2
When the candle is moved closer to the mirror, the image will move away from the mirror and likewise, the screen has to be moved away to bring the image on the screen. However, when the candle is at a distance of less than 18 cm, the image formed will be virtual. In that situation, the image cannot be brought to the screen.

KSEEB Solutions

Question 2.
A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
Answer:
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 3
When the needle is moved farther from the mirror, the image also moves away from the mirror but will not go beyond the focus of the mirror.

Question 3.
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Answer:
When the tank is filled with water:
Real depth = 12.5cm
Apparent depth = 9.4cm
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 4
Therefore objective and eyepiece should be placed at 11,67cm apart and the object at a distance of 1,5cm from the objective.

Question 4.
A small telescope has an objective lens of a focal length of 140 cm and an eyepiece of a focal length of 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?
(b) the final image is formed at the least distance of distinct vision (25cm)?
Answer:
Here, f0 = 140cm, fe = 5.0cm
(a) When the telescope is in normal adjustment,
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 5
(b) When the image is formed at least distance of distinct vision, then
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 6

Question 5.
(a) For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25cm?
Answer:
Here, f0 = 140cm; fe = 5.0cm
(a) The separation between the objective and the eye-piece,
L = f + f = 140 + 5 = 145 cm
(b) Let a be the angle formed by the object at the objective lens. Then,
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 7

Let A’ B’ be the height of the image produced by the objective lens at its focus. From the figure, it follows that the angle formed by the image a’ g’ at the objective lens is also equal to a.
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 8

KSEEB Solutions

Question 6.
A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with mirrors 20mm apart. If the radius of curvature of the large mirror is 220mm and the small mirror is 140mm, where will the final image of an object at infinity be?
Answer:
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 9
When the object lies at infinity, the concave mirror will form the image of the object at its focus. As this image lies behind the convex mirror, it acts as a virtual object for the convex mirror.

2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 10

Question 7.
Light incident normally on a plane mirror attached to a galvanometer coil retraces backward as shown in Fig. 9.36. A current in the coil produces a deflection of 3.5o of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 11
Answer:
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 12
When a plane mirror rotates through a certain angle, the reflected ray turns twice the angle of rotation. Therefore, the angle between the incident ray AO and the reflected ray is,
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 13

Question 8.
Figure 9.37 shows a biconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0cm. What is the refractive index of the liquid?
Answer:
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 14
The liquid between the convex lens and the plane mirror behaves as a plano-concave lens. Let f1 and f2 be the focal length of the convex lens and the liquid plano-concave lens. In the arrangement as shown in. Figure (b) as the needle and its image coincide with each other, the rays of light retrace their path. Therefore, the rays of light after refraction through the lens are falling normally on the plane mirror. It can happen so if the needle lies at the focus of the convex lens. Hence, f, = 30cm From the lens maker’s formula, we have

2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 15
In the arrangement as shown in Figure (a). As discussed above, now the needle lies at the focus of the combination of the two lenses. If F is the focal length of the combination of the two lenses, then
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 16

2nd PUC Physics Ray Optics and Optical Instruments Additional Questions and Answers

Question 1.
Why is that sun-glasses (gaggles), which has curved surface, do not have any power?
Answer:
The two surfaces of the goggle lens are parallel i.e. possess the same radii of curvature. Since one surface is convex and other equally concave, the powers of the two surface is equal but of opposite signs. The total power of the goggle lens (p’) is the algebraic sum of the powers of the individual surfaces
p’ =p1 +p2 =p+(-p) = 0

KSEEB Solutions

Question 2.
An illuminated object and a screen are placed 90 cm apart. Determine the focal length and nature of the lens required to produce a clear image on the screen, twice the size of the object.
Answer:
Since the image is formed on the screen, it is a real image. Let the distance of the object from the lens be x. Then, distance of the image from the lens be (90-x)
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 17

Question 3.
Two lenses of power + 15D and -5D are in contact with each other forming a combination lens,
(a) what is the focal length of this combination?
(b) An object of size 3cm is placed at 30cm from this combination of lenses. Calculate the position and size of the image formed.
Answer:
(a) Power of the combination,
P = P1 + P2= 15 +(-5)= 10D
Therefore, focal length of the combination,
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 18

Question 4.
A glass prism has a minimum angle of deviation of 5 in the air. State with reason, how the angle of minimum deviation will change if the prism is immersed in a liquid of refractive index greater than 1.
Answer:
We know that for a glass prism placed in the air,
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 19
when the prism is immersed in a liquid of refractive index greater than 1, its refractive index w.r.t liquid will become less than p. From the above relation, it follows that the angle of minimum deviation will decrease.

Question 5.
Which of the following is used in optical fibres?
(A) Total internal reflection
(B) Scattering
(C) Diffraction
(D) Refraction
Answer:
(A) Total internal reflection

Question 6.
An astronomical telescope has a large aperture to:
(A) Reduce spherical aberration
(B) Have high resolution
(C) Increase span of observation
(D) Have low dispersion.
Answer:
(B) Have high resolution
The resolving power of the telescope is given by
\(R.P=\frac { D }{ 1.22\lambda } \)
∴ The resolving power of the telescope will be high, its objective is of a large aperture.

KSEEB Solutions

Question 7.
The refractive index of the material of an equilateral prism is \(\sqrt { 3 } \). What is the angle of minimum deviation?
(A) 45°
(B) 60″
(C) 37°
(D) 30°
Answer:
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 20

Question 8.
Greenlight of wavelength 5460 is incident on an air-glass interface. If the refractive index of glass is 1.5, the wavelength of light in glass would be
(A) 3,640 Å
(B) 5460 Å
(C) 4861 Å
(D) None of the above
Answer:
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 21

Question 9.
A light wave of frequency V and wavelength X travels from air to glass. Then,
(A) V changes
(B) x does not change
(C) V does not change, X changes
(D) V and X changes
Answer:
(C) V does not change, X changes

Question 10.
Two mirrors are kept at 60° to each other and a body is placed at the middle. Total number of images formed is
(A) six
(B) four
(C) five
(D) three
Answer:
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 22
(C) five

Question 11.
Two thin lenses of power +5D and -3D are in contact. What is the focal length of the combination (CBSE 2001)
Answer:
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 23
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 24

Question 12.
A glass lens of refractive index 1.5 is kept in a liquid. What must be the refractive index of the liquid so that the lens will disappear  (CBSE 2008)
Answer:
To disappear the lens the refractive index of liquid should be the same i.e. 1.5.

KSEEB Solutions

Question 13.
For which wavelength is an eye the most sensitive.
Answer:
5500 A

Question 14.
Use mirror formula to show that for an object lying between the pole and the focus of a conceive mirror, the image formed is always virtual for a concave mirror (CBSE 2008)
Answer:
f is negative and u is negative
Since object lying between the pole and focus u < f
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 25

Question 15.
Using the Lens formula, show that a concave Lens produces a virtual and diminished image independent of the location of the object. (KU – 2009)
Answer:
2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments 26
magnification is less than 1 hence image is virtual and diminished

KSEEB Solutions

Question 16.
What should be the position of the object relative to the biconvex lens so that it behaves like a magnifying glasses.
Answer:
If an object is kept between the optical centre and the focus of the biconvex lens, then the lens behaves like a magnifying glass.

Tili Kannada Text Book Class 6 Solutions Nataka Karnataka Chapter 1 Kodi Nanna Balyava

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Tili Kannada Text Book Class 6 Solutions Nataka Karnataka Chapter 1 Kodi Nanna Balyava

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Siri Kannada Text Book Class 7 Solutions Gadya Chapter 7 Billa Habba

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Siri Kannada Text Book Class 7 Solutions Gadya Bhaga Chapter 7 Billa Habba

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