Category: Class 11 Physics Objective Questions

Physics Multiple Choice Questions on “Simple Projectile Motion”.

1. A body of mass m, projected at an angle of θ from the ground with an initial velocity of v, acceleration due to gravity is g, what is the maximum horizontal range covered?
a) R = v² (sin 2θ)/g
b) R = v² (sin θ)/2g
c) R = v² (sin 2θ)/2g
d) R = v² (sin θ)/g
Answer: a
Clarification: The calculation for range uses the horizontal component of the initial velocity and time. Range = distance x Time taken for the motion. The time can be found by utilising the fact that the vertical velocity becomes zero at the maximum height (i.e. at half of the time period) and using the first equation of motion for this. Hence, on calculating, the range will be obtained as R = v² (sin 2θ)/g.

2. A body of mass 5 kg, projected at an angle of 60° from the ground with an initial velocity of 25 m/s, acceleration due to gravity is g = 10 m/s², what is the maximum horizontal range covered?
a) 54.13 m
b) 49 m
c) 49.16 m
d) 60 m
Answer: a
Clarification: The formula for horizontal range is R = v² (sin 2θ)/g. Here, v = 25, θ = 60°, g = 10. Hence, on solving we will get range as 54.13 m.

3. A body of mass 10 kg, projected at an angle of 30° from the ground with an initial velocity of 5 m/s, acceleration due to gravity is g = 10 m/s², what is the time of flight?
a) 0.866s
b) 1.86 s
c) 1.96 s
d) 1.862 s
Answer: a
Clarification: The formula for time of flight is t = 2(v sinθ/g). Here, v = 5, θ = 30°, g = 10. Hence, on solving we will get the time of flight as 0.866 s.

4. On calculating which of the following quantities, the mass of the body has an effect in simple projectile motion?
a) Velocity
b) Force
c) Time of flight
d) Range
Answer: b
Clarification: All mentioned quantities except force are kinematic quantities. Force is a kinetic quantity. Force = m x a. Hence, the mass of the body has an effect on force calculation.

5. A ball of mass 100 g, projected at an angle of 30° from the ground with an initial velocity of 11 m/s, acceleration due to gravity is g = 10 m/s², what is the maximum height attained?
a) 1.5 m
b) 3.0 m
c) 1.0 m
d) 2.0 m
Answer: a
Clarification: The formula for maximum height is h = (v sinθ)²/2g. Here, v = 11, θ = 30°, g = 10. Hence, on solving we will get the maximum height attained as 1.5125 m. The value can be rounded off to 1.5.

6. When do we get maximum range in a simple projectile motion?
a) When θ = 45°
b) When θ = 60°
c) When θ = 90°
d) When θ = 0°
Answer: a
Clarification: The formula for horizontal range is R = v²(sin 2θ)/g. This will be maximum when sin 2θ = 1, which implies that 2θ = 90°, which in turn implies that θ = 45°. Hence the correct answer is when θ = 45°.

7. When do we get maximum height in a simple projectile motion?
a) When θ = 45°
b) When θ = 60°
c) When θ = 90°
d) When θ = 0°
Answer: c
Clarification: The formula for horizontal range is h = (v sinθ)²/2g. This will be maximum when sin θ = 1, which implies that θ = 90°. Hence the correct answer is when θ = 90°.

8. A football is projected at an angle of 45° from the ground with an initial velocity of 10 m/s, take acceleration due to gravity is g = 10 m/s².What is the time of flight?
a) 1.4142 s
b) 1.5361 s
c) 1.8987 s
d) 1.5651 s
Answer: a
Clarification: The formula for time of flight is t = 2(v sinθ/g). Here, v = 10, θ = 45°, g = 10. Hence, on solving we will get the time of flight as 1.4142 s.

9. A bag of mass 1000 g, projected at an angle of 90° from the ground with an initial velocity of 5 m/s, acceleration due to gravity is g = 10 m/s², what is the maximum height attained?
a) 1.25 m
b) 3.0 m
c) 1.5 m
d) 2.0 m
Answer: a
Clarification: The formula for maximum height is h = (v sinθ)²/2g. Here, v = 5, θ = 90°, g = 10. Hence, on solving we will get the maximum height attained as 1.25 m. The maximum height can alternatively be found by simply using equations of motion as this bag is thrown vertically upwards.

10. A body of mass 55 kg, projected at an angle of 45° from the ground with an initial velocity of 15 m/s, acceleration due to gravity is g = 10 m/s², what is the maximum horizontal range covered?
a) 22.5 m
b) 25 m
c) 16 m
d) 15 m
Answer: a
Clarification: The formula for horizontal range is R = v² (sin 2θ)/g. Here, v = 15, θ = 45°, g = 10. Hence, on solving we will get range as 22.5 m.

11. At what angle of projectile (θ) is the horizontal range minimum?
a) θ = 45°
b) θ = 60°
c) θ = 90°
d) θ = 75°
Answer: c
Clarification: The formula for horizontal range is R = v² (sin 2θ)/g. When θ = 90°, sin 2θ = sin 180° = 0. Hence, the range covered becomes 0. Since range cannot be negative, 0 is the minimum value it can attain.

12. A body of mass 5 kg, projected at an angle of 45° from the ground covers a horizontal range of 45 m, acceleration due to gravity is g = 10 m/s², what is the velocity with which it was projected covered?
a) 21.21 m/s
b) 20 m/s
c) 22 m/s
d) 21.1 m/s
Answer: a
Clarification: The formula for horizontal range is R = v² (sin 2θ)/g. Here, R = 45, θ = 45°, g = 10. Hence, on solving we will get velocity as 21.21 m/s. The result can be cross checked by putting this value of velocity and finding out the range.

13. A big stone of mass 1000 g, projected at an angle of 30° from the ground it covers a maximum vertical distance of 5 m, acceleration due to gravity is g = 10 m/s², what is the velocity with which it was thrown?
a) 11.55 m/s
b) 11.5 m/s
c) 1.155 m/s
d) 12.0 m/s
Answer: a
Clarification: The maximum vertical distance is the maximum height. The formula for maximum height is h = (v sinθ) ²/2g. Here, h = 5, θ = 30°, g = 10. Hence, on solving we will get the initial velocity as 11.55 m/s. The result can be cross checked by putting this value of velocity and finding out the maximum height.

14. A soccer ball is projected at an angle of 60° from the ground.It attains its maximum height in 10s. Considering acceleration due to gravity as g = 10 m/s². What is the velocity with which it was projected?
a) 115.5 m/s
b) 117 m/s
c) 120 m/s
d) 11.55 m/s
Answer: a
Clarification: The time for achieving the maximum height is equal to half of the time of flight. Therefore, the time of flight is 20s. The formula for time of flight is t = 2(v sinθ/g). Here, t = 20, θ = 60°, g = 10. Hence, on solving we will get the initial velocity as 115.5m/s.

Physics Multiple Choice Questions on “Potential Energy”.

1. The potential energy of an object maximises as its velocity increases.
a) True
b) False
Answer: b
Clarification: The potential velocity of an object is maximum when it is at rest. The kinetic energy increases as velocity increases while the potential energy decreases.

2. An object of mass 5kg is taken from a height of 10 m to a height of 30 m. What is the increment in its potential energy? (Assume g = 10m/s²)
a) 500 J
b) 1000 J
c) 1500 J
d) 2000 J
Answer: b
Clarification: PE = m x g x h
m = 5 kg
g = 10 m/s²
Let;
h1 = 10m
h2 = 30m
PE1 = 5 x 10 x 10
= 500 J
PE2 = 5 x 10 x 30
= 1500 J
Increment in PE = PE2 – PE1
= 1500 – 500
= 1000 J.

3. A 10kg object is raised to a height of 20m. What is the magnitude of its potential energy? (Assume g = 10 m/s²)
a) 500 J
b) 1000 J
c) 1500 J
d) 2000 J
Answer: d
Clarification: PE = m x g x h
m = 10 kg
g = 10 m/s²
h = 20 m
PE = 10 x 10 x 20
= 2000 J.

4. An object is thrown from the ground with a velocity of 5 m/s. If the object has a mass of 2kg, what will be its potential energy at the top-most point of its trajectory? (Assume g = 10 m/s²)
a) 5 J
b) 15 J
c) 25 J
d) 50 J
Answer: c
Clarification: Maximum height reached by the object;
h = v²/(2 x g)
= 25/20
= 1.25 m
PE = m x g x h
= 2 x 10 x 1.25
= 25 J.

5. Which of the following is true?
a) Potential energy decreases as altitude increases
b) Potential energy increases as altitude increases
c) Potential energy first increases and then decreases as altitude increases
d) Potential energy first decreases and then increases as altitude increases
Answer: b
Clarification: As altitude increases potential energy increases. From the equation of potential energy (PE);
PE = m x g x h
We can see that the potential energy is directly proportional to height. Hence the statement is justified.

6. Since the potential energy of an object depends on the acceleration due to gravity and since the acceleration due to gravity decreases as altitude increases, we can conclude that the potential energy of an object decreases as altitude increases.
a) True
b) False
Answer: b
Clarification: The potential energy of an object does not depend on acceleration due to gravity at the instantaneous position of the object but depends on the acceleration due to gravity at the reference point (such as the surface of the earth). Hence, potential energy increases as altitude increases.

7. Consider 2 balls A and B of the same mass. The potential energy of ball A is thrice that of ball B. How high is ball A compared to ball B?
a) Same height as ball B
b) Twice as high height as ball B
c) Thrice as high height as ball B
d) Four times as high height as ball B
Answer: c
Clarification: PE_a = (m x g x h) _a
PE_b = (m x g x h) _b
m_a = m_b = m
g_a = g_b = g
PE_a = 3 x PE_b
m_a x g_a x h_a = m_b x g_b x h_b x 3
m x g x h_a = m x g x h_b x 3
h_a = 3 x h_b.

8. The potential energy of an object of constant mass and fixed reference is determined by its _____
a) mass
b) gravitational acceleration
c) position
d) velocity
Answer: c
Clarification: PE = m x g x h
The mass of the given object is constant.
The fixed reference indicates constant acceleration due to gravity.
Hence, variation in height (h) is nothing but a variation in position.
Hence, the potential energy of an object of constant mass and fixed reference is determined by its position.

9. The potential energy of an object cannot be increased by internal forces.
a) True
b) False
Answer: a
Clarification: The potential energy can only be increased by external forces. Energy increase by internal forces manifests as an increase in kinetic energy. However, potential energy can be reduced by internal forces by increasing kinetic energy.

10. Find the ratio of potential energy if an object is raised to thrice of its height and its mass is tripled.
a) 1:1
b) 1:4
c) 1:9
d) 1:3
Answer: c
Clarification: PE1 = m x g x h
PE2 = (3m) x g x (3h)
= 9 x m x g x h
PE1 : PE2 = mgh : 9mgh
= 1:9.

Physics Multiple Choice Questions on “Gravitation – Kepler’s Laws”

1. From Kepler’s law of orbit, we can infer that the sun is located _____ of the planet’s orbit.
a) at the centre
b) at one of the foci
c) at both foci
d) anywhere along the semi-minor axis
Answer: b
Clarification: According to Kepler’s law of orbit, every planet revolves around the sun in an elliptical orbit and the sun is at one of the foci.

2. Kepler’s laws of planetary motion replaced circular orbits with _____
a) elliptical orbits
b) parabolic orbits
c) conical orbits
d) hyperbolic orbits
Answer: a
Clarification: From the first law of Kepler’s laws of planetary motion, we can infer that the orbit of a planet is an ellipse with the sun at one of the foci.

3. Kepler’s laws of planetary motion were proposed only for _____
a) our sun
b) any star in our galaxy
c) any star in the universe
d) stars of other solar systems
Answer: a
Clarification: The Kepler’s laws of planetary motion were published by Johannes Kepler between 1609 and 1619. They are three scientific laws describing the motion of planets around the Sun.

4. What does Kepler’s law of period relate?
a) Time period and semi-minor axis
b) Time period and eccentricity
c) Time period and semi-major axis
d) Time period and area swept by the planet
Answer: c
Clarification: According to Kepler’s law of periods, the square of time period f revolution of a planet is directly proportional to the cube of the semi-major axis of the planet’s elliptical orbit.

5. In the figure shown, what is the point “A” called?

a) Perigee
b) Apogee
c) Foci
d) Center
Answer: a
Clarification: The point closest to the sun is known as “perigee” or “perihelion”. The velocity of the planet is the greatest at this point, in accordance with Kepler’s law of areas.

6. In the figure shown, what is the point “B”, which is the farthest from the sun, called?

a) Perigee
b) Apogee
c) Foci
d) Center
Answer: b
Clarification: The point closest to the sun is known as “apogee” or “aphelion”. The velocity of the planet is the least at this point, in accordance with Kepler’s law of areas.

7. What is the time taken by a planet to sweep an area of 2 million square km if the time taken by the same planet to cover an area of 1 million square km is 36 hours?
a) 18 hours
b) 36 hours
c) 72 hours
d) 144 hours
Answer: c
Clarification:According to Kepler’s law of area, a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
If it takes 36 hours to sweep an area of 1 million square km, it would take (36 + 36) hours to sweep an area of (1 + 1) million square km.
Therefore, the answer is 72 hours.

8. The velocity of a planet is constant throughout its elliptical trajectory in an orbit.
a) True
b)False
Answer: b
Clarification: From Kepler’s law of area, we know that a planet sweeps equal areas at equal intervals of time. Therefore, when the planet is closer to the sun it sweeps a lesser area for a given velocity than when it was farther to the sun for the same given velocity. Hence, a planet travels faster when it is closer to the sun.

9. The velocity of a planet is the greatest at perigee.
a) True
b) False
Answer: a
Clarification: From Kepler’s law of area, we know that a planet sweeps equal areas at equal intervals of time. Perigee is the closest point to the sun and hence, the velocity of the planet is greatest at the perigee because, only then, can the planet sweep an area equal to that when it was farther from the sun for the same interval of time.

10. What is the constant of proportionality in Kepler’s law of periods known as?
a) Universal gravitational constant
b) Escape velocity
c) There is no constant of proportionality
d) Cannot be determined
Answer: c
Clarification: There is no particular constant of proportionality for Kepler’s law of periods. The law of periods only relates the proportionality of the square of the time period of revolution of a planet to the cube of the semi-major axis of the orbit of the planet.

11. Kepler’s laws of planetary motion improved ______
a) the heliocentric theory
b) the geocentric theory
c) the big bang theory
d) the string theory
Answer: a
Clarification: Kepler’s laws of planetary motion improved the heliocentric theory of Nicolaus Copernicus, replacing its circular orbits with epicycles with elliptical orbits.

12. The elliptical orbits of planets were indicated by calculations of the orbit of which astronomical body?
a) Mercury
b) Earth
c) Earth’s moon
d) Mars
Answer: d
Clarification: By observing the motion of Mars in the sky, Kepler inferred that the planets have elliptical orbits around the sun. Kepler discovered that a simple ellipse would clearly define the orbit of Mars and eliminate many complexities. It would also eliminate the need for epicycles.

13. If the eccentricities of the planetary orbits were taken as zero, then the sun is at the centre of the orbit.
a) True
b) False
Answer: a
Clarification: If the eccentricity is taken as zero, then the trajectory would become circular. So, the sun would be at the centre of the circle since both foci would also lie at the centre.

Physics Multiple Choice Questions on “Solids Mechanical Properties – Elastic Moduli”.

1. Which of the following quantities have the same S.I. unit as that of modulus of elasticity?
a) Energy per unit volume
b) Force per unit length
c) Energy
d) Change in length

Answer: a
Clarification: Modulus of elasticity is the ratio of stress to strain. Its unit is, therefore, N/m². Energy per unit volume has the unit Nm/m³ = N/m². Force per unit length has the unit N/m. So, the correct answer is energy per unit volume.

2. A small body having mass 1kg is rotating in a circle of radius 10cm. It is connected to the centre with the help of a string having young’s modulus = 10¹¹N/m². Find change in length of wire while the mass is rotating with constant speed of 2m/s. The cross sectional area of string is 1cm².
a) 0.4μm
b) 4μm
c) 40μm
d) 0.4m

Answer: a
Clarification: Tension ‘T’ in string will be given by T = mv²/R = 1*4/0.1 = 40N.
Stress = Y*Strain, where Y is young’s modulus. T/A=Y*ΔL/L
∴ ΔL = TL/AY = (40*0.1)/(0.0001*10¹¹)
= 4*10^-7m = 0.4μm.

3. Which of the following represents volumetric strain?
a) -ΔV/V
b) ΔV/V
c) Pressure/Volume
d) -P / (ΔV/V)

Answer: b
Clarification: The volumetric strain is defined as ratio of change in volume to original volume. It can be either positive or negative. The option -P / (ΔV/V) is the bulk modulus which is basically the ratio of volumetric stress to volumetric strain.

4. Bulk modulus is defined as -P / (ΔV/V). Therefore it can be negative. True or False?
a) True
b) False

Answer: b
Clarification: Bulk modulus is defined as the ratio of volumetric stress to volumetric strain. And volumetric stress is force applied per unit area which is pressure. Now, when pressure is applied on a body its volume decreases, therefore a minus sign is used to make bulk modulus positive. Hence, it is positive.

5. A 10N force is uniformly distributed over the surface of a sphere of radius 20cm. The force at each point acts inward along the normal at that point. Find the fractional change in volume if the compressibility of the sphere is 10^-11m²/N.
a) 3*10^-10
b) 1.5*10^-11
c) 1.98*10^-10
d) 2*10^-11

Answer: c
Clarification: Bulk Modulus ‘B’ is defined as -P / (ΔV/V). And B is the reciprocal of compressibility.
∴ B = 1/10^-11= 10¹¹N/m².
ΔV/V = P/B.
P = 10/(4π*0.04)
∴ ΔV/V = -P/B = – 10/(4π*0.04*10¹¹) = 1.98*10^-10.

6. Select the correct option. B is bulk modulus.
a) B_gas > B_liquid > B_solid
b) B_liquid > B_gas> B_solid
c) B_solid > B_liquid > B_gas
d) B_liquid > B_gas > B_solid

Answer: c
Clarification: Bulk modulus is the inverse of compressibility. Solids are the least compressible, followed by liquids and then gases being the most compressible. Thus solids have the maximum B, and liquids the least.

7. A lake has a depth of 500m. The bulk modulus of water is 2*10⁹N/m². Calculate the fractional compression, ΔV/V, of water at the bottom of the lake.
a) 2.5*10^-3
b) 7*10^-3
c) 3.5*10^-3
d) 5*10^-3

Answer: a
Clarification: Pressure at the bottom ‘P’ = ρgh = 1000*10*500 = 5*10⁶Pa.
B = -P/(ΔV/V)
∴ ΔV/V = -P/B = 5*10⁶/2*10⁹ = 2.5*10^-3.

9. Which of the following stresses causes change in density, as long as force acts on the body?
a) Shear stress
b) Compressive Stress
c) Tensile stress
d) Volumetric stress

Answer: d
Clarification: Volumetric stress is due to forces applied on the surface of a body from all directions such force is along normal at each point. This causes a change in volume while mass, obviously, remains the same. So, there is a slight change in density. Note that it is negligible but this change of density doesn’t occur in any other type of stresses, as the change in longitudinal length is compensated by change in lateral length in case of longitudinal stresses.

10. A wire has a cross sectional area ‘A’, length ‘L’ and young’s modulus ‘Y’. It is pulled by a force ‘F’ which causes a total extension of length ‘l’. The force F is so adjusted that the wire is only slowly stretched. Find the work done by the force in pulling the string by a length ‘dx’ when extension is x (0 > x > l).
a) ½*stress*strain*volume
b) (AY/L)xdx
c) F*l
d) ½*stress * strain

Answer: b
Clarification: The force F at any instant will be equal to internal force developed as the wire is to be slowly stretched.
∴ F=(AY/L)x
∴ Work done for dx extension = F*dx = (AY/L)xdx.

Physics Multiple Choice Questions on “Thermal Equilibrium”.

1. Two bodies in thermal equilibrium can have different pressures. True or False?
a) True
b) False
Answer: a
Clarification: Two bodies are said to be in thermal equilibrium if they don’t exchange heat energy when they are in physical contact. For this only their temperatures have to be the same. Pressure and volume will depend upon others properties like mass.

2. The variables like pressure, volume, temperature of a thermodynamic system are microscopic. True or False?
a) True
b) False
Answer: b
Clarification: Thermodynamics does not deal with molecular descriptions. The variables it describes like pressure, volume & temperature are for a very large collection of molecules. Hence, the thermodynamic properties are macroscopic.

3. One end of a rod is held in a fire, while the other is held by one’s hand. Just assume his hand doesn’t get burned. After some time the rod-hand system is in thermal equilibrium. True or False?
a) True
b) False
Answer: b
Clarification: In the given situation, after some time a steady state will be reached. That is temperature at individual points will be constant, but there will be a difference of temperature between 2 points along the length of the rod. For, this heat flow will be continuous so we can’t call it thermal equilibrium.