300+ REAL TIME Engineering Physics Objective Questions & Answers 2023

250+ TOP MCQs on Optical Instruments and Answers

Engineering Physics Multiple Choice Questions on “Optical Instruments”.

1. The inability of a lens to form a white image of a white object is known as ________
a) Spherical Aberration
b) Chromatic Aberration
c) Monochromatic Aberration
d) Coma
Answer: b
Clarification: If a lens forms colored images of an object with white light, it is known as chromatic aberration. Because since that the refractive index of the material of the lens is different for different wavelengths of light.

2. What is the ratio of the focal lengths of the two Plano-convex lenses in Huygens’s Eyepiece?
a) 2:1
b) 3:1
c) 3:2
d) 4:3
Answer: b
Clarification: The focal lengths of the two Plano-convex lenses in Huygens’s Eyepiece are in the ration 3:1(3f and f) and the distance between them is equal to 2f. The focal length and positions of the two lenses are such that the eyepiece is achromatic and free from spherical aberration.

3. In which of the following instruments, the objective has a large focal length and a very large eyepiece?
a) A simple microscope
b) A Compound microscope
c) Telescope
d) Interferometer
Answer: c
Clarification: In telescopes, it is desired to provide angular magnification of distant objects. A very large focal length and the even larger eyepiece is used so that the light from a distant object enters the objective and the image formed is clear.

4. X is an optical defect due to which a comet-like image is formed instead of a point image. X is ___________
a) Coma
b) Astigmatism
c) Curvature
d) Distortion
Answer: a
Clarification: Coma is a type of monochromatic aberration. In this, a comet-like image is formed instead of a point image, of a point object situated away from the lens. It can be reduced by using the aplanatic lens.

5. Huygens’s eyepiece is also known as __________
a) Spherical Eyepiece
b) Positive Eyepiece
c) Negative Eyepiece
d) Double Eyepiece
Answer: c
Clarification: In Huygens’s eyepiece, the real inverted image formed by the objective of the microscope lies behind the field lens and that image is virtual. Due to this, Huygens’s eyepiece is also known as Negative Eyepiece. This eyepiece cannot be used to examine a real image formed by the objective.

6. Which of the following is not a part of an Electron Microscope?
a) Electron Gun
b) Objective
c) Magnetic lens
d) Fluorescent Screen
Answer: b
Clarification: The electron microscope works on the principle that a beam of electrons exhibit wave nature and they can be focused by suitable electric and magnetic fields. Thus, electron microscope has an electron gun for producing electrons, a Magnetic lens to focus the electron beam and the fluorescent screen to obtain the image.

7. The condition in which lines in one direction appear to be well focused while those in perpendicular direction appear distorted is known as ___________
a) Presbyopia
b) Myopia
c) Hypermetropia
d) Astigmatism
Answer: d
Clarification: Astigmatism occurs when the cornea is not spherical. Due to this, lines in one direction seems to be focused while in the perpendicular direction looks distorted. It can be corrected by the use of cylindrical lenses.

8. A thin converging lens and a thin diverging lens of each focal length 10 cm are placed coaxially 5cm apart. What will be the focal length of the combination?
a) +10cm
b) -10cm
c) +20cm
d) -20cm
Answer: c
Clarification: We know, f₁=10 cm, f₂ = -10 cm, d = 5 cm
(frac{1}{f}=frac{1}{f_1}+frac{1}{f_2}-frac{d}{f_1f_2})
= 1/10 – 1/10 + 5/100
f = + 20 cm.

9. Two lenses of focal length 8 cm and 6 cm are placed a certain distance apart. If they form an achromatic combination, the separation between them is ___________
a) 5cm
b) 6cm
c) 7cm
d) 8cm
Answer: c
Clarification: For an achromatic combination, the separation x is given by,
x = (f1 + f2)/2
As we know, f1 = 8cm and f2 = 6cm
x = 14/2
x = 7 cm.

10. The effective focal length of Ramsden’s eyepiece is 3cm. The focal length of a single lens is ___________
a) 3cm
b) 4cm
c) 5cm
d) 6cm
Answer: b
Clarification: Here, F = 3 cm, f =? d =?
In Ramsden’s eyepiece, F = f and d = 2f/3
Therefore, applying the formula
(frac{1}{F}=frac{1}{F}+frac{1}{f}-frac{d}{Ff})
F = 3f/4
f = 4F/3
f = 4 X 3 /3
f = 4 cm.

11. Find the position of the image in the following lens combination:

a) 30 cm to the right of the third lens
b) 20 cm to the right of the third lens
c) 30 cm to the left of the third lens
d) 20 cm to the left of the third lens
Answer: a
Clarification: Image by the first lens,
(frac{1}{V_1}-frac{1}{U_1}=frac{1}{f_1} )
(frac{1}{V_1}-frac{1}{-30}=frac{1}{10})
v₁ = 15 cm
For the second lens,
(frac{1}{V_2}frac{-1}{-10}=frac{1}{10})
V₂ = ∞
For the third,
(frac{1}{V_3}frac{-1}{infty}=frac{1}{30})
V₃ = 30 cm.

12. The far point of a myopic person is 40 cm. What should be the power of the lens that he must use to see clearly?
a) -0.4 D
b) +0.4 D
c) -2.5 D
d) +2.5 D
Answer: c
Clarification: Myopia is the defect when a person can’t see distant object clearly. To see clearly, the person should use a concave lens of focal length -40 cm.
Hence, the power of the lens = 100/-40
= -2.5 Diopters.

13. If the magnification of a lens is dependent on the distance from the principal axis, the aberration that arises is called __________
a) Coma
b) Astigmatism
c) Curvature
d) Distortion
Answer: d
Clarification: Distortion is the defect that arises when the images of equal parts of an object will not be of the same length. In such a case, the magnification of the lens is dependent on the distance from the principal axis.

14. The refractive index of a Plano-convex lens is 1.6 for violet and 1.5 for red light. The radius of curvature is 0.20 cm. The separation between violet and red foci of the lens is ____________
a) 0.05 cm
b) 0.06 cm
c) 0.07 cm
d) 0.08 cm
Answer: c
Clarification: Here, μ_v = 1.6, μ_r = 1.5, R₁ = 0.20 m, R₂ = ∞
Using lens maker’s formula: (frac{1}{F_v}=(μ_v-1)(frac{1}{R_1}+frac{1}{R_2}))
F_v = 0.33 m
Similarly solving for red color: (frac{1}{F_r}=(μ_r-1)(frac{1}{R_1}+frac{1}{R_2}))
F_r = 0.40 m
Separation between red and violet, F_r–F_v = 0.40 – 0.33
= 0.07 m.

15. The component in an optical instrument used to increase the angular object field and to minimize aberrations is called as ___________
a) Objective lens
b) Eye lens
c) Field Lens
d) Plano-concave lens
Answer: c
Clarification: Field lens is placed between the objective and the eye lens. It increases the field of view and brightness of the image. It helps in reducing the aberrations in the lens. It brings the center of the exit pupil near the eye lens. Together, the field lens and the eye lens constitute an eyepiece or ocular.

250+ TOP MCQs on Nicol Prism and Answers

Engineering Physics Multiple Choice Questions on “Nicol Prism”.

1. The working of Nicole prism is based on the phenomenon of ____________
a) Refraction
b) Reflection
c) Diffraction
d) Double Refraction
Answer: d
Clarification: The working of Nicol Prism is based on the phenomenon of Double Refraction. When a beam of unpolarized light passes through a doubly refracting crystal such as calcite, it splits into two plane-polarized beams.

2. In the O-ray is eliminated by ____________
a) Refraction
b) Total internal reflection
c) Dispersion
d) Transmission
Answer: b
Clarification: In a Nicol Prism, the O-ray is eliminated by total internal reflection while the E-ray is transmitted through the Nicol Prism. Thus, a plane polarized light with vibrations in the principal section of the crystal is obtained.

3. In the Nicol Prism experiment, the calcite crystal length is x times its breadth. What is x?
a) 2
b) 3
c) 1/2
d) 1/3
Answer: b
Clarification: In this experiment, a calcite crystal whose length is three times its breadth is taken. The calcite crystal is the doubly refracting crystal that splits the unpolarized light into two plane-polarized beams.

4. For E-ray, the Canada balsam layer is optically rarer.
a) True
b) False
Answer: b
Clarification: Canada Balsam is a transparent substance, whose refractive index lies between the refractive indices of O-ray and E-ray. Thus, for an E-ray, it is optically denser while for an O-ray it is optically rarer.

5. The critical angle for O-ray with respect to Canada balsam is ___________
a) 45.7°
b) 57°
c) 59°
d) 69°
Answer: d
Clarification: We know Canada balsam is optically denser than calcite for O-ray. The refractive index for O-ray is 1.069 while the critical angle turns out to be 69°.

6. The prism does not act as a polarizer when _____________
a) The angle of incidence is less than the critical angle for the O-ray
b) The angle of incidence is greater than the critical angle for the O-ray
c) The angle of incidence is less than the critical angle for the E-ray
d) The angle of incidence is greater than the critical angle for the E-ray
Answer: a
Clarification: If the angle of incidence is less than the critical angle for the O-ray, it is not reflected and is transmitted through the prism and so the prism does not act as a polarizer.

7. Nicol Prism can act as a polarizer as well as an analyzer.
a) True
b) False
Answer: a
Clarification: Nicol Prism can be used as a Polarizer as well as an analyzer. Thus, it can polarize an unpolarized light and also detect a plane-polarized light.

8. What should be the maximum angle between the extreme rays of the incident beam for the Nicol prism to work perfectly as a polarizer?
a) 25°
b) 28°
c) 34°
d) 43°
Answer: b
Clarification: The angle between the extreme rays of the incident beam should be less than 28^o so as to avoid the total internal reflection of E-ray and the transmission of O-ray.

9. The combination as shown in the figure is called as _____________

a) Polarimeter
b) Polaroid
c) Polariscope
d) Polarizer and Analyzer
Answer: c
Clarification: In the figure, two Nicol Prisms are placed together. One acts as a polarizer while the other acts as an analyzer. This combination is called Polariscope.

10. In polariscope, if the two prisms are crossed, what would be the intensity of the transmitted polarized beam?
a) I_o
b) 2 I_o
c) 1/2 I_o
d) 0
Answer: d
Clarification: As the two prisms are crossed, the angle between the principal sections of the two prisms would be 90°. Thus, according to Malus law, the intensity of the transmitted bean would be zero. Thus, no light would come out.

250+ TOP MCQs on X-Rays and Compton Effect and Answers

Engineering Physics Multiple Choice Questions on “X-Rays and Compton Effect”.

1. What is the relation between the interaction parameter, ‘b’, and atomic radius, R, for the Compton Effect?
a) b > R
b) b ≈ R
c) b < R
d) No relation between b and R
Answer: b
Clarification: If b ≈ R, the incident photons are scattered by the electron of the atom and the electron itself gets scattered. This phenomenon is known as the Compton effect.
If b < R, the photon is directly converted into an electron-positron pair, known as pair production.
If b > R, it means the interaction parameter is greater than the atomic radius. In this case, the electron is ejected by the photon and it is known as the Photoelectric effect.

2. The Compton effect can be explained on the basis of __________
a) Wave nature of light
b) Quantum theory of light
c) Ray optics
d) Wave optics
Answer: b
Clarification: The quantum theory of light i.e., the photon concept successfully explained the Compton effect. The wave theory predicts the scattered radiation will have the same wavelength.

3. What kind of photon is required for the Compton effect to occur?
a) Visible Light Photon
b) X-ray Photon
c) Infrared
d) UV Photon
Answer: b
Clarification: When a γ-ray and X-ray Photon passes close to an atomic nucleus, the scattered radiation have radiations of smaller wavelength along with the one of the same wavelength.

4. The expression for Compton shift is _________
a) (frac{h}{m_0 c}costheta)
b) (frac{h}{m_0 c}(1 – costheta))
c) (frac{h}{m_0 c}sintheta )
d) (frac{h}{m_0 c}(1 – sintheta) )
Answer: b
Clarification: The expression (frac{h}{m_0 c}(1 – costheta)) is the expression for the Compton shift. It gives the expression for the change in wavelength that occurs during Compton effect.

5. Visible Light rays can show Compton effect.
a) True
b) False
Answer: b
Clarification: Compton effect cannot be observed for light rays. It is so because the Compton shift observed for visible light rays is about 0.001% which is undetectable.

6. X rays of wavelength 0.15 nm are scattered from a block of carbon. What is the wavelength of X-rays scattered at 00?
a) 0.15 nm
b) 0.154 nm
c) 0.165 nm
d) 0.178 nm
Answer: a
Clarification: As we know, (Deltalambda = frac{h}{m_0 c}(1 – costheta))
Here, as cosθ = 1, Δλ = 0
Hence, λ’ = 0.15 nm.

7. X-rays with wavelength 0.1 nm are scattered from a carbon block. The scattered radiations are viewed at right angles to the direction of incident beam. What is the Compton shift?
a) 0.0014 nm
b) 0.0024 nm
c) 0.0034 nm
d) 0.0044 nm
Answer: b
Clarification: As we know, (Deltalambda = frac{h}{m_0 c}(1 – costheta))
Here, cosθ = 0. Therefore,
Δλ = (frac{2h}{m_0 c})
= 0.0024 nm.

8. In Compton scattering, if the incident photon has a wavelength of 0.2 nm and Φ = 90°, the angle at which recoil electron appears is ___________
a) 30.12°
b) 38.46°
c) 44.57°
d) 53.12°
Answer: c
Clarification: We know, tanθ = (frac{lambda sin⁡Phi}{lambda^{‘}-lambda cos⁡Phi})
= 2 X 10^-10/(2.204 – 2) X 10^-10
= 2/2.024
= 0.9980
Θ = tan^-10.9980
= 44.57°.

9. The following graph is seen for which angle?

a) 30°
b) 45°
c) 90°
d) 120°
Answer: c
Clarification: The given figure shows the variation of intensity with wavelength in Compton scattering when the angle θ is equal to 90°.

10. Identify X in the expression for Kinetic energy of recoil electron: (k_e=hv(frac{x(1-cos⁡theta)}{1+x(1-cos⁡theta)})).
a) hv/m_o
b) hv/m_oc
c) hv/m_oc²
d) hv/c²
Answer: c
Clarification: In the given expression, x = hv/m_oc². This is the expression for the Kinetic Energy of recoil electron. It is derived by Einstein’s mass energy relation and the expression for Compton shift.

250+ TOP MCQs on Classification of Semiconductors and Answers

Engineering Physics Multiple Choice Questions on “Classification of Semiconductors”.

1. What is the energy level below which all levels are completely occupied at Zero Kelvin called?
a) Boson Energy
b) Fermi Energy
c) Stable Energy
d) Ground Energy
Answer: b
Clarification: Fermi energy is said to be the energy of the highest possible occupied energy level at 0 K. Below this level, all the states are completely occupied.

2. What are the current carriers in semiconductors?
a) Electrons and Protons
b) Electrons and Nucleons
c) Electrons and Photons
d) Electrons and Holes
Answer: d
Clarification: Electrons and holes are the two current carriers in semiconductors. Electrons are negatively charged while holes are positively charged. Their movement gives rise to a current in the semiconductor.

3. The concentration of doping is kept below ______________
a) 1 %
b) 5 %
c) 10 %
d) 50 %
Answer: a
Clarification: The concentration of doping in semiconductors is generally kept below 1 %. However, it is enough to bring a huge drop in the energy gap.

4. In N-Type semiconductors, which extra energy level is added?
a) Conduction level
b) Donor Energy Level
c) Acceptor energy level
d) Valence level
Answer: b
Clarification: In N-Type semiconductor level, a new energy level below the conduction band is formed. The energy difference between the two is about 0.045 eV.

5. Which of the following can be used to create a P-Type Semiconductor?
a) P
b) Sb
c) Ga
d) As
Answer: c
Clarification: For a P-Type semiconductor, a material with 3 valence electrons is chosen. Out of the given choices, Ga can be used to create a P-Type Semiconductor.

6. The following graph depicts the I-V characteristics of which instrument?

a) Photodiode
b) Light Emitting Diode
c) Solar Cell
d) Zener diode
Answer: c
Clarification: The generation of EMF by a solar cell is due to three basic processes: generation of electron-hole pair, separation of electrons and holes and collection of electrons by the front contact.
The p-side becomes positive and the n-side becomes negative giving rise to photo voltage.
The I-V characteristics of the solar cell are drawn in the fourth quadrant of the coordinate axis because a solar cell does not draw current but supplies the same to the load.

7. The Hall coefficient of a specimen is 3.66 x 10^-4 m³C^-1. If it’s resistivity is 8.93 x 10^-3 Ωm, what will be its mobility?
a) 0.01 m²V^-1s^-1
b) 0.02 m²V^-1s^-1
c) 0.03 m²V^-1s^-1
d) 0.04 m²V^-1s^-1
Answer: d
Clarification: We know, Mobility = Hall coefficient/resistivity
Therefore, Mobility = 3.66 x 10^-4/8.93 x 10^-3
= 0.04 m²V^-1s^-1.

8. Which one of the following is not an intrinsic semiconductor?
a) Carbon
b) Silicon
c) Germanium
d) Lead
Answer: a
Clarification: There are 4 bonding electrons in all the above materials. However, the energy required to take out an electron will be maximum for carbon as the valence electrons are in the second orbit. Hence, the number of free electrons for conduction is negligibly small in C.

9. Which of the following is n-type semiconductor?
a) CaO
b) MgO
c) ZnO
d) BaO
Answer: c
Clarification: II-VI semiconductors are generally p-type semiconductors except for ZnO and ZnTe. II-VI semiconductors are those which contain atoms of materials that have 2 valence electrons and 6 valence electrons.

10. P-Type semiconductor has a lower electrical conductivity than N-Type semiconductor.
a) True
b) False
Answer: a
Clarification: Due to comparatively lower mobility of holes than electrons for the same level of doing as in an N-Type semiconductor, it has lower electrical conductivity.

11. Pure Si at 300 K has equal electron (ni) and hole concentration (p) of 1.5 X 10¹⁶ m^-3. Doping by indium increases p to 4.5 X 10²² m^-3. What is n in the doped silicon?
a) 4.5 X 10⁹ m^-3
b) 4.5 X 10²² m^-3
c) 5 X 10⁹ m^-3
d)5 X 10²² m^-3
Answer: c
Clarification: Here, n_i = 1.5 X 10¹⁶ m^-3, p = 4.5 X 10²² m^-3
We know, np = n_i²
n = n_i²/p
= 5 X 10⁹ m^-3.

12. Identify the type of material.

a) Intrinsic Semiconductor
b) N-Type semiconductor
c) P-Type semiconductor
d) Conductor
Answer: c
Clarification: In the figure, as we can see there is an acceptor energy level just above the valence band. This happens in the case of P-Type semiconductors.

13. In a semiconductor it is observed that three-quarters of the current is carried by electrons and one quarters by holes. If the drift speed is three times that of the holes, what is the ratio of electrons to holes?
a) 1 : 1
b) 1 : 2
c) 2 : 1
d) 4 : 1
Answer: a
Clarification: In a semiconductor, I = I_e + I_h
Here, I_e = ³⁄₄ I and I_h = ¹⁄₄ I
Now v_e = 3v_h
I_e/I_h = nv_e/nv_h
3 = 3n/p
n = p
Hence the ration is, 1 : 1

14. Holes are the majority carries in Intrinsic Semiconductors.
a) True
b) False
Answer: b
Clarification: A pure semiconductor is called an intrinsic semiconductor. Hence, in this case, the number of electrons and holes are same, as the electron that moves out of its position leaves a hole behind. Hence, the concentration of holes and electrons is the same in an intrinsic semiconductor.

15. If the number of electrons (majority carrier) in a semiconductor is 5 X 10²⁰ m^-3 and μ_e is 0.135 mho, find the resistivity of the semiconductor.
a) 0.0926 Ωm
b) 0.0945 Ωm
c) 0.0912 Ωm
d) 0.0978 Ωm
Answer: a
Clarification: We know, Conductivity, σ = en_e μ_e
= 5 X 1.6 X 0.135 X 10 mho/m
= 10.8 mho/m
Resistivity = 1/σ
= 0.0926 Ωm.

250+ TOP MCQs on Motion and Mechanics and Answers

Engineering Physics Multiple Choice Questions on “Motion and Mechanics”.

1. Rest and motion are relative terms.
a) True
b) False
Answer: a
Clarification: A passenger sitting in a train is in rest with respect to his other passengers but is in motion with respect to the things outside the train. Thus, an object maybe at rest with respect to one object and, at the same time, in motion with respect to another object. Hence rest and motion are relative terms.

2. A train is under a journey of several hundred kilometers. How can it be regarded?
a) An object in motion
b) An object under rest
c) An object under absolute motion
d) A point object
Answer: d
Clarification: If the position of an object changes by distances much greater than its own size in a reasonable time, then the object maybe regarded as a point object.

3. A car is moving along a zigzag path on a level road. This is an example for which of the following?
a) Point object
b) Two dimensional motion
c) Three dimensional motion
d) One dimensional motion
Answer: b
Clarification: The motion of an object is said to be two dimensional if only two of the three coordinates specifying its position changes with time. Hence a car moving in a zigzag path on a level road is an example for two dimensional motions.

4. Which of the following can be regarded as an example for three dimensional motions?
a) Motion of planets around the sun
b) Motion of a train along a straight track
c) Motion of a free falling body
d) A kite flying on a windy day
Answer: d
Clarification: The motion of an object is said to be three dimensional if all the three coordinates specifying the position changes with respect time. Thus kite flying on a windy day is an example for three dimensional motions.

5. Displacement is a scalar quantity.
a) True
b) False
Answer: a
Clarification: Displacement is the change in a potion of an object in a fixed direction. It has both magnitude and direction. Thus displacement is a vector quantity.

6. A body travels from A to B at 40m/s and from B to A at 60m/s. Calculate the average speed.
a) 0
b) 48m/s
c) 240m/s
d) 3.5m/s
Answer: b
Clarification: Total time taken by the body to travel from A to B and then from B to A,
t₁+t₂ = AB/40 + AB/60 = AB/24 s
Total distance covered = AB + BA = 2AB
Average speed = 2AB/(t₁+t₂) = 48m/s.

7. On a 60km track travels the first 30km with a uniform speed of 30km/h. How fast must the train travel the next 30km so as to average 40km.h for the entire trip?
a) 60km.h
b) 90km/h
c) 120km/h
d) 30km/h
Answer: a
Clarification: v_av = (2v₁ v₂)/(v₁+v₂)
40 = (2×30×v₂)/(30+v₂)
v₂ = 60km/h.

8. What is the acceleration of a bus approaching a bus stop?
a) Uniform acceleration
b) Instantaneous acceleration
c) Average acceleration
d) Negative acceleration
Answer: d
Clarification: If the velocity of an object decreases with time, its acceleration is negative. When a bus approaches its stop, its acceleration decreases, hence it has negative acceleration.

9. A jet plane starts from rest with an acceleration of 3m/s² and makes a run for 35s before taking off. What is the minimum length of the runway?
a) 105 m
b) 1837.5 m
c) 2451 m
d) 1204 m
Answer: b
Clarification: Minimum length of the driveway is given by s = ut + 1/2 at² = 0 + 1/2 × 3 × 35 × 35 = 1837.5 m.

10. A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving at a speed of 54km/h and the brakes cause a deceleration of 6.0m/s², find the distance travelled by the car after he sees the need to put the brakes.
a) 18.75m
b) 225 m
c) 21.5 m
d) 12 m
Answer: c
Clarification: Distance covered in 0.20s = 15 × 0.20 = 3 m
For motion with deceleration:
U = 15m/s v = 0 a = -6 m/s²
As v² – u² = 2as
s = 18.75 m
Total distance travelled = 3.0 + 18.75 = 21.75 m.

250+ TOP MCQs on Friction and Forces – 2 and Answers

Engineering Physics Problems focuses on “Friction and Forces – 2”.

1. A body in linear motion can never be in equilibrium.
a) True
b) False
Answer: b
Clarification: A body in linear motion can be in equilibrium provided that the vector sum of two forces acting upon the body is zero.

2. A man jumping out of a moving train ___________
a) Lands on the ground
b) Falls with his head forward
c) Will start rolling on the ground
d) Falls with this head backward
Answer: b
Clarification: As the man jumps out of a moving train, his feet suddenly come to rest on touching the ground while the upper part of his body continues to move forward. That is why he falls with his head forward. In order to save himself, he should run some distance in the forward direction.

3. A stone, when thrown on a glass window, smashes the windowpane to pieces, but a bullet from the gun passes through making a clean hole. Why?
a) Due to the large size of the stone
b) Depends on the material of the window
c) Because of the force
d) Because of the small speed of the stone
Answer: d
Clarification: Because of the small speed, the stone remains in contact with the windowpane for a longer period. It transfers its motion to the pane and breaks it into pieces. But the particles of windowpane near the hole are unable to share the fast motion of the bullet and so remain undisturbed.

4. An astronaut is suddenly thrown out of his small spaceship accelerating in interstellar space at a constant rate of 100/s². What is the acceleration of the astronaut the instant after he is outside the spaceship? Assuming there are no nearby stars.
a) 200m/s²
b) 100 m/s²
c) Zero
d) 400 m/s²
Answer: c
Clarification: The moment he gets out of the ship, there is no external force on him. By the law of motion, the acceleration of the astronaut is zero.

5. A soda-water bottle is falling freely. The bubbles of the gas rise in the water of the bottle.
a) True
b) False
Answer: b
Clarification: The water in the freely falling bottle is in the state of weightlessness. Consequently, the pressure in water does not increase with depth. No upthrust acts on the bubbles and they do not rise.

6. A bird is sitting on the floor of a wire cage and the cage is in the hand of a boy. The bird starts flying in the cage. The boy will ___________
a) Also experience the flying feel
b) Will not feel anything different
c) Feel the weight of the cage has become lighter
d) Feel no change in weight of the cage
Answer: c
Clarification: The air inside the wire cage is in free contact with the atmospheric air. When the bird starts flying inside the cage, the weight of the bird is no more experienced and the cage will appear lighter than before.

7. When a man jumps down for a height of several storeys into a stretched tarpaulin, he receives no injury.
a) True
b) False
Answer: a
Clarification: When the man jumps, the tarpaulin gets depressed at the place of impact. This increases the time of impact. As a result, the tarpaulin exerts a very small force on the man, and he receives no injury.

8. The fuel in the vehicle can be saved by ___________
a) Not using vehicle often
b) Mixing water with fuel
c) Cleaning the vehicle
d) Properly inflating the tyres
Answer: d
Clarification: When the tyres are properly inflated, the area of contact between the tyre and the ground is reduced. This reduces rolling friction. Consequently, the automobile covers greater distance for the same quantity of fuel consumed.

9. Which of the brakes are effective on a bicycle?
a) Small ones
b) Bigger ones
c) Both small and big
d) Neither small nor big
Answer: c
Clarification: Action of brakes is based on friction. But the friction is independent of the area of surfaces in contact so long as the normal reaction remains the same, Hence, larger size brakes and normal size brakes will be equally effective if the material of brakes remains unchanged.

10. It is difficult to put a cycle into motion than to maintain the motion because of ___________
a) Limiting friction
b) Kinetic friction
c) Rolling friction
d) Static friction
Answer: a
Clarification: To put a cycle into motion, one needs to overcome limiting friction while maintaining its motion, one needs to overcome kinetic friction. Limiting friction is greater than kinetic friction. So it is difficult to put a cycle into motion than to maintain its motion.

Engineering Physics Problems,