Category: Class 11 Physics Objective Questions

Physics Multiple Choice Questions on “Straight Line Motion – Instantaneous Velocity and Speed”.

1. Which of the following can be used to describe how fast an object is moving along with the direction of motion at a given instant of time?
a) Instantaneous velocity
b) Instantaneous speed
c) Average velocity
d) Average speed

Answer: a
Clarification: Instantaneous velocity describes the velocity of an object at a given time instant. Average speed is the speed at which the object travels throughout the time period and not an instant. Speed is a scalar quantity; hence it cannot show the direction of motion.

2. Which of the following is the correct formula for instantaneous velocity?
a) v = dx/dt
b) v = x/t
c) v = xt
d) v = t/x

Answer: a
Clarification: The correct formula is v = dx/dt. Instantaneous velocity is the velocity of the object at a given time instant. v = x/t is the formula for average velocity.

3. The trajectory of an object is defined as x = (t-4)², what is the velocity at t = 5?
a) 2
b) 5
c) 1
d) 4

Answer: a
Clarification: The function for velocity can be derived by differentiating the equation with respect to t. v = 2(t-4) is the required function. When t = 5, v = 2(5-4) = 2.

4. A ball is thrown up in the sky, at what position will the instantaneous speed be minimum?
a) Initial position
b) Final position
c) Halfway through the whole trajectory
d) After covering one fourth of the whole trajectory

Answer: c
Clarification: When the ball rises up, there will be a point where it will be in the state of instantaneous rest. At the this position the speed of the ball will be 0. Speed is maximum at the initial and final points.

5. A car is moving in a spiral starting from the origin with uniform angular velocity. What can be said about the instantaneous velocity?
a) It increases with time
b) It decreases with time
c) It remains constant
d) It does not depend on time

Answer: a
Clarification: This type of motion can be called circular motion with increasing radius. As the radius increases, the tangential velocity increases (v = rw). Now, as there is only one velocity present, the speed will be equal to the magnitude of the tangential velocity.

6. What happen to the instantaneous velocity in a non-uniformly accelerated motion?
a) It increases
b) It decreases
c) It varies as the acceleration
d) It remains constant

Answer: a
Clarification: The instantaneous velocity will increase with time. If the motion is accelerated, no matter if the acceleration is constant, or variable, the instantaneous velocity will increase. Variation of acceleration describes how to change in velocity is changing.

Physics Multiple Choice Questions on “Newton’s Third Law of Motion”.

1. The forces involved in Newton’s third law act ____
a) On the same object
b) On different objects
c) In same direction
d) On five bodies
Answer: b
Clarification: The two forces involved in Newton’s third law are the cause force and the reaction force. The cause force acts on the contact surface or the contact body by the parent body. Whereas the reaction force acts on the parent body by the contact body.

2. Two bodies in contact experience forces in ________
a) Same direction
b) Opposite directions
c) Perpendicular directions
d) Five different directions
Answer: b
Clarification: The two bodies in contact will follow Newton’s third law of motion. The third law states that for every action there is an equal and opposite reaction. The two bodies will experience each other’s reaction forces in opposite directions.

3. A batsman hits a ball with a force a 5 N. What force does the bat experience?
a) 5 N
b) 10 N
c) 15 N
d) 20 N
Answer: a
Clarification: From Newton’s third law we know that for every for every action, there is an equal and opposite reaction. From this we can say that the bat experiences a force of 5 N.

4. A driver hits a light pole with a force a 100 N. What force does the car experience?
a) 100 N
b) 10 N
c) 150 N
d) 200 N
Answer: a
Clarification: From Newton’s third law we know that for every for every action, there is an equal and opposite reaction. From this we can say that the car experiences a force of 100 N by the light pole.

5. A truck with a mass of 2500 Kg travelling with an acceleration of 5 m/s² hits a scooter. What force does the truck experience?
a) 12500 N
b) 500 N
c) 10000 N
d) 2500 N
Answer: a
Clarification: From Newton’ second law we get force on scooter = 2500 x 5 = 12500 N. From Newton’s third law we know that for every for every action, there is an equal and opposite reaction. From this we can say that the truck experiences a force of 5 N.

6. A man and a kid accidentally hit each other. What is true about the forces experienced by them?
a) They are equal in magnitude
b) They are different in magnitude
c) They are same in direction
d) They are at an angle of 1.57 rad
Answer: b
Clarification: From the third law of motion, we know that the cause force and the reaction forces are equal in magnitude and opposite in direction. Hence, from this knowledge, we can say that the man and the kid will experience forces in opposite directions and in equal magnitude.

7. A man is standing still. The force by the man on the earth is (vec{F1}). The force by the earth on the man is (vec{F2}). Which one of the following is true?
a) (vec{F1} = vec{F2})
b) (vec{F1} = -vec{F2})
c) (vec{F1} = 5(vec{F2}))
d) (2(vec{F1}) = vec{F2})
Answer: b
Clarification: According to the third law of motion, every action has an equal and opposite reaction. Hence every action force is equal in magnitude to its reaction but in opposite direction. Hence (vec{F1} = -vec{F2}) is the correct answer.

Physics Multiple Choice Questions on “System of Particles – Centre of Mass – 1”.

1. The centre of masses of two particles with masses 2 kg and 1 kg located at (1,0,1) and (2,2,0) is located at _____
a) 4/3, 2/3, 2/3
b) 2/3, 4/3, 2/3
c) 2/3, 2/3, 4/3
d) 1/3, 2/3, 2/3

Answer: a
Clarification:Sum of masses = 2+1 = 3
x-coordinate;
(2*1 + 1*2)/3 = 4/3
y-coordinate;
(2*0 + 1*2)/3 = 2/3
z-coordinate;
(2*1 + 1*0)/3 = 2/3.

2. Two bodies of masses 5kg and 15kg are located in the cartesian plane at (1,0) and (0,1). What is the location of their centre of mass?
a) 1/4, 1/4
b) 3/4, 3/4
c) 3/4, 1/4
d) 1/4, 3/4

Answer: d
Clarification: Sum of masses = 5 + 15 = 20
x-coordinate;
(5*1 + 15*0)/20 = 1/4
y-coordinate;
(5*0 + 15*1)/20 = 3/4.

3. The coordinates of the centre of mass of objects of mass 10, 20, 30 kg are (1,1,1) m. Where should an object of mass 40 kg be placed such that the centre of mass of this new system lies at (0,0,0)?
a) 3/2, 3/2, 3/2
b) -3/2, -3/2, -3/2
c) 3/4, 3/4, 3/4
d) -3/4, -3/4, -3/4

Answer: b
Clarification: Sum of masses = 10 + 20 + 30 + 40 = 100
Since initial centre of mass is at (1,1,1), we can assume an object of 60 kg is at (1,1,1).
x-coordinate;
(60*1 + 40*x)/100 = 0
x = -3/2
y-coordinate;
(60*1 + 40*y)/100 = 0
y = -3/2
z-coordinate;
(60*1 + 40*z)/100 = 0
z = -3/2.

4. Masses 1 kg, 1.5 kg, 2 kg, and “M” kg are situated at (2,1,1), (1,2,1), (2,-2,1) and (-1,4,3). What is the value of “M” if their centre of mass is at (1,1,3/2)?
a) 1 kg
b) 1.5 kg
c) 2 kg
d) 2.5 kg

Answer: b
Clarification: Sum of masses = 1 + 1.5 + 2 + M = 4.5 + M
x-coordinate;
(1*2 + 1.5*1 + 2*2 – M)/(4.5 + M) = 1
4.5 + M = 7.5 – M
2M = 3
M = 1.5 kg.

5.Particles of masses 1 kg and 3 kg are at (2i+5j+13k) m and (-6i+4j-2k) m. What is the position of their centre of mass?
a) 1/4 (-16i + 17j + 7k) m
b) 1/4 (-8i + 17j + 7k) m
c) 1/4 (-6i + 17j + 7k) m
d) 1/4 (-6i + 17j + 5k) m

Answer: a
Clarification: Sum of masses = 1 + 3 = 4
x-coordinate;
(1*2-3*6)/4 = -16/4
y-coordinate;
(1*5 + 3*4)/4 = 17/4
z-coordinate;
(1*13-3*2)/4 = 7/4
Position of centre of mass = 1/4 (-6i + 17j + 5k) m.

6. The distance between the centres of carbon and oxygen in the carbon monoxide molecules is 1.13 x 10^-10 m. The distance of the centre of mass of the molecule relative to the oxygen atom is _____
a) 0.48 x 10^-10 m
b) 0.64 x 10^-10 m
c) 0.56 x 10^-10 m
d) 0.36 x 10^-10 m

Answer: a
Clarification: The location of the centre of mass from an object in a two-object system is inversely proportional to the mass of the object.
The atomic mass of oxygen = 16
The atomic mass of carbon = 12
Total atomic mass = 16 + 12 = 28
The location of centre of mass relative to oxygen atom = (1.13 x 10^-10) x 12 / 28
= 0.48 x 10^-10 m.

7. Two particles of masses 2 kg and 3 kg are at rest and are separated by 10 m. If they move towards each other under the mutual force of attraction, they would meet at _____
a) 6 m from 4 kg body
b) 6 m from 6 kg body
c) 4 m from 4 kg body
d) 5 m from 6 kg body

Answer: a
Clarification: The two bodies would meet at their centre of mass.
Sum of masses = 2 + 3 = 5 kg
Let us consider the 2 kg body as a reference;
The centre of mass from the 2 kg body =(3 x 10) / 5
= 6 m.

8. A ball of mass 3 kg and a ball of mass 2 kg roll towards each other on a flat surface. How far is the centre of mass from the 3 kg ball of the 2 balls are separated by 6 m?
a) 1.2 m
b) 2.4 m
c) 3.6 m
d) 4.8 m

Answer: b
Clarification: Sum of masses = 3 + 2 = 5 kg
Centre of mass from 3 kg ball = (6 x 2) / 5
= 2.4 m.

9. A uniform disc of radius R is put over another uniform disc of radius 2R of same thickness and density. The edges of the two discs touch each other. What is the position of their centre of mass?
a) at 5R/3 from the centre of the larger disc
b) at 2R/3 from the centre of the larger disc
c) at 3R/5 from the centre of the larger disc
d) at 2R/5 from the centre of the larger disc

Answer: c
Clarification:Since both discs have the same thickness and density;
M/(pi x R² x t) = M’/((pi x 4 x R² x t)
M = M’/4
M = Mass of a smaller disc
M’ = 4M = Mass of larger disc
Sum of masses = M + 4M = 5M
The distance of centres of discs = 3R
The distance of the centre of mass from larger disc = (3R x M)/5M
= 3R/5.

10. Three identical spheres each of radius R are placed such that their centres lie on a straight line. What is the location of their centre of mass from the centre of the first sphere?
a) R
b) 2R
c) 3R
d) 4R

Answer: b
Clarification: Distance between first and last sphere = R + 2R + R = 4R
Since the spheres are identical and lie in a straight line, the centre of mass will lie exactly in the middle.
Hence the centre of mass lies at a distance of 2R from the centre of the first sphere.

11. Centre of mass can vary upon the application of internal force.
a) True
b) False

Answer: b
Clarification: No, the centre of mass cannot vary upon the application of internal force. This is because internal force do not produce any net force. Hence the state of centre of mass in this scenario will remain unchanged.

Physics Multiple Choice Questions on “Gravitational Potential Energy”.

1. The force of gravity is a conservative force.
a) True
b) False

Answer: a
Clarification: The amount of work done on the body by the force of gravity is independent of the path. Hence, it is a conservative force.

2. Conventionally, the magnitude of gravitational potential energy for an object at infinity from the earth is _____ ((M = Mass of the earth; m = Mass of the object at infinity; R = Radius of the earth).
a) -(G*M)/R²
b) -(G*M)/R
c) -(G*M*m)/R
d) Zero

Answer: d
Clarification: Gravitational Potential Energy (U) = -(G*M*m)/r
At infinity, r -> infinity
Therefore, U -> 0.

3. The maximum value of gravitational potential energy is zero.
a) True
b) False

Answer: a
Clarification: Since the reference for gravitational potential energy is taken at infinity, conventionally, we denote any finite value of gravitational potential energy, except zero, as a negative number. Hence, the greatest value of gravitational potential energy is zero.

4. A dam produces electricity from the gravitational potential energy of the water stored in it. The same dam has 50 cubic km of water stored 50 meters above the ground. What is the work done by gravity relative to the ground? (Assume g = 10 m/s²)
a) 1.5 x 10¹⁶ J
b) 2.5 x 10¹⁶ J
c) 3.5 x 10¹⁶ J
d) 4.5 x 10¹⁶ J

Answer: b
Clarification: Mass of 50 cubic km of water = 5 x 10¹³ kg; Density of water = 1 g/cubic cm
Work done = m x g x h
= (5 x 10¹³ x 10 x 50) Joules
= 2.5 x 10¹⁶ J.

5. A 4kg eagle picks up a 75g snake and raises it 2.5 m from the ground to a branch. What is the work done by the eagle on the snake? (Assume g = 10 m/s²)
a) 100 J
b) 1.875 J
c) 118.75 J
d) 10 J

Answer: b
Clarification: The work done by the eagle on the snake depends only on the mass of the snake.
Hence, the work done on the snake = (0.075 x 10 x 2.5) J
= 1.875 J.

6. A 4kg eagle picks up a 75g snake and raises it 2.5 m from the ground to a branch. What is the work done to raise the bird’s own centre of mass to the branch? (Assume g = 10 m/s²)
a) 100 J
b) 1.875 J
c) 118.75 J
d) 10 J

Answer: a
Clarification: The work done to raise the bird’s own centre of mass to the branch depends only on the mass of the bird.
Hence, the work done = (4 x 10 x 2.5) J
= 100 J.

7. The expression for gravitational potential energy is “-(G*M)/r”.
a) True
b) False

Answer: b
Clarification: The expression for gravitational potential energy is “-(G*M*m)/r”.
“-(G*M)/r” is the expression for ‘gravitational potential’.
Gravitational potential at a point can be defined as the work done in bringing a unit mass from infinity to that point.

8. Gravitational potential energy can be zero but gravitational potential can never be zero.
a) True
b) False

Answer: b
Clarification: Gravitational potential at a point can be defined as the work done in bringing a unit mass from infinity to that point.Hence, for a given point mass at infinite distance, both, the gravitational potential and the gravitational potential energy can be zero.

10. The value of the gravitational potential at the centre of a ring of radius “a” and mass “M” is _____
a) zero
b) -(G*M)/a
c) -(G*M)/a^1/2
d) infinite

Answer: b
Clarification:We know that the gravitational potential at a point “p” at a distance of “r” from the centre of the ring of radius “a” and mass M” is;
V = -(G*M)/(a²+r²)^1/2
At the centre of the ring; r = 0.
Therefore; V = -(G*M)/a.

11. The value of gravitational potential inside a uniform thin spherical shell is the same everywhere.
a) True
b) False

Answer: a
Clarification: The gravitational potential inside a uniform thin spherical shell of radius “a” is;
V = -(G*M)/a, which is a constant. Hence, the value of gravitational potential inside a uniform thinspherical shell is the same everywhere and not necessarily non-zero.

12. The value of gravitational potential outside a uniform thin spherical shell is the same everywhere.
a) True
b) False

Answer: b
Clarification: The gravitational potential outside a thin uniform spherical shell behaves similar to the potential due to a point source form the centre of the spherical shell.The gravitational potential outside a spherical shell is inversely proportional to the distance of the point outside from the centre of the sphere. Hence, it is not the same everywhere.

13. The gravitational potential energy of a body at a distance “r” from the centre of the earth is V. Its weight at a distance “2r” from the centre of the earth is _____
a) V/r
b) V/4r
c) V/2r
d) 4V/r

Answer: b
Clarification: Gravitational potential energy (V) at a distance “r”;
V = -(G*M*m)/r
Gravitational force at a distance “r” = -(G*M*m)/r²
= V/r
Gravitational force at a distance “2r” = -(G*M*m)/4r²
= V/4r.

14. The velocity with which an object should be projected from the surface of the earth such that it reaches a maximum height equal to “n” time the radius of earth “R” is _____ (M = Mass of the earth)
a) [(n*G*M)/(n+1)R] ^1/2
b) [((n+1)*G*M)/(n+1)R] ²
c) [(n*G*M)/(n+1)R]
d) [((n+1)*G*M)/nR] ²

Answer: a
Clarification:Total initial energy = Total final energy
1/2(m x v²) – (G*M*m)/R = -(G*M*m)/(n+1)R
Simplifying the above equation, we get;
v = [(n*G*M)/(n+1)R]^1/2.

15. The value ofthe gravitational field in a region is given by g = 2i + 3j. What is the change in gravitational potential energy of a particle of mass 5kg when it is taken from the origin O(0,0) to a point P(10, -5)? (Letters in bold are vectorial representations)
a) 5 J
b) 10 J
c) 25 J
d) 50 J

Answer: c
Clarification: The gravitational potential energy, in vectorial mathematics, is the dot product of the gravitational field and relative position multiplied by mass.
Gravitational potential energy = (g . P) x m
= [(2i + 3j) .(10i – 5j)] x 5
= (20 – 15) x 5
= 25 J.

Physics Multiple Choice Questions on “Fluids Mechanical Properties – Viscosity”.

1. Viscosity is said to be internal friction between fluid layers. What type of force is it?
a) Electromagnetic
b) Gravitational
c) Weak Nuclear forces
d) Strong Nuclear forces

Answer: a
Clarification: At the contact surface of two bodies (here, fluid layers) molecular bonds are formed. When there is relative motion between them these bonds try to oppose it. This is how friction works. Therefore we can say it is an electromagnetic force as molecular bonds are considered to be electromagnetic forces.

2. What is the unit of coefficient of viscosity?
a) Ns/m²
b) Nm/s
c) Ns/m
d) Nm/s²

Answer: a
Clarification: The viscous force is given by F = -ηAdv/dx. Where A is area of contact, η is coefficient of viscosity & dv/dx is velocity gradient (vel/dist). From this expression we get, η = -F(dx/dv)(1/A). Therefore the dimension of η is N*s*m^-2 = Ns/m².

3. 1 poise = _________ Ns/m². Fill in the blanks.
a) 0.01
b) 0.1
c) 1
d) 10

Answer: b
Clarification: Poise is the C.G.S unit of coefficient of viscosity. poise = dyne-s/cm². 1Ns/m² = 10⁵* 1/10⁴ dyne-s/cm² = 10 dyne-s/cm².
∴ 1 dyne-s/cm² = 0.1 Ns/m² = 1 poise.

4. Coefficient of viscosity doesn’t depend on temperature because it is a property of the fluid. True or False?
a) True
b) False

Answer: b
Clarification: Coefficient of viscosity is a constant of proportionality for the viscous force expression. Viscous force depends on the area of contact and the velocity gradient, and also on the temperature as more temperature of liquid will decrease friction between its layers hence we say that coefficient of viscosity depends on temperature.

5. The coefficient of viscosity is equal to the ratio of shear stress to shear strain. True or False?
a) True
b) False

Answer: b
Clarification: Imagine a fluid element with its bottom layer having zero velocity and upper layer, at a height ‘h’, having velocity ‘v’. As the fluid moves, the bottom layer stays as it is, while the other layers will move relative to the bottom layer. So, the strain (=v*Δt/h) continuously increases with time. So we talk about strain rate (=v*Δt/h*Δt = v/h) and not about strain. And we say that stress is proportional to strain rate. Therefore the coefficient of viscosity is the ratio of stress & strain rate.

6. What is the effect of temperature increase on viscosity of liquids and gases?
a) Increases for both liquids and gases
b) Increases for liquids and decreases for gases
c) Increases for gases and decreases for liquids
d) Decreases for both liquids and gases

Answer: c
Clarification: In the case of liquids when temperature increases, viscosity decreases because it is the result of molecular bonds (electromagnetic forces) between particles which weaken due to random motion brought about by increase in temperature. In the case of gases, the molecules vibrate more randomly, leading to more obstruction in the path of bodies trying to move through it. And we interpret this increased obstruction as increased viscosity of the gas.

7. What will be the terminal velocity of a rain drop of radius (r) 1mm and mass (M) 0.001g falling through air. The coefficient of viscosity of air (η) is 1.8 X 10^-5Ns/m²? Neglect buoyant force.
a) 20 m/s
b) 29.47 m/s
c) 35.8 m/s
d) 40 m/s

Answer: b
Clarification: At terminal velocity, Mg = viscous force. Therefore, 0.001*10^-3*10 = 6πηrv ∴ v = 10^-5/ 6πηr = 10^-5 / ( 6π*1.8 X 10^-5*0.001) = 29.47 m/s.

8. In a closed pipe of radius R, fluid (having some viscosity) is flowing laminarly. Which point along a cross section will have maximum speed?
a) Centre
b) Near the wall of pipe
c) R/2 from centre
d) All points will have same speed

Answer: a
Clarification: We know that the shear strain rate is v/l, where v is velocity of layer and l is distance of that layer from layer having zero speed Here, the layer in contact with pipe wall will have zero speed. Therefore, the velocity will increase as one goes away from the wall and will be maximum at centre.

9. On which these factors does air/fluid resistance force, acting on a spherical body, not depend?
a) Radius of body
b) Velocity of body
c) Coefficient of viscosity
d) Density of body

Answer: d
Clarification: Resistance force by a fluid on a body = 6πηrv, where η is coefficient of viscosity, r is radius of body & v is its velocity. This doesn’t depend on density.
Another way to think of this is to imagine a hollow and solid sphere of the same radius falling through air at the same speed. The resistance offered by air will be the same on both as they have the same surface area exposed and are moving at the same speed.

Physics Multiple Choice Questions on “Thermodynamic Processes”.

1. In which of the following processes is heat transfer equal to zero?
a) Isentropic
b) Isochoric
c) Isothermal
d) Diathermic

Answer: a
Clarification: Entropy is defined as dQ/T. In an isentropic process dQ/T = 0 which means dQ = 0. Diathermic is a process in which heat flow is easily possible. Isochoric process is one in which volume is constant. Isothermal process is one in which temperature stays constant.

2. Isothermal process can be represented by which law?
a) Charle’s law
b) Boyle’s law
c) Gay-Lussac’s law
d) 2nd law of thermodynamics

Answer: b
Clarification: In an isothermal process, PV=const. This is the same as Boyle’s law. Charle’s law is given by: V/T=const. Gay-Lussac’s law is given by: P/T=const and 2nd law of thermodynamics states that in every process total entropy of the universe must increase.

3. In an isothermal process for an ideal gas, change in internal energy is zero. True or False?
a) True
b) False

Answer: a
Clarification: For an ideal gas internal energy depends only on temperature. In an isothermal process change in temperature is zero, hence internal energy also remains the same.

4. Calculate the work done by the gas in an isothermal process from A to B. P_A = 1Pa, V_A = 3m³, P_B = 3Pa.
a) 3.3J
b) 3J
c) -3.3J
d) -4.58J

Answer: c
Clarification: Since the process is isothermal the product PV will be constant.
P_AV_A = P_BV_B.
∴ V_B = 1*3/3 = 1m³.
Work done in an isothermal process is given by:
nRT*ln(V_B/V_A)
= P_AV_Aln(V_B/V_A)
= 3ln(1/3)
= -3.3J.

5. Which of the following variables is zero for a cyclic process?
a) Work done
b) Heat supplied
c) Total heat + Total work
d) Total heat – Total work

Answer: d
Clarification: In a cyclic process, the starting position is the same as the ending position. So, the change in all state variables is zero. So the net change in internal energy is zero. ΔU = ΔQ – ΔW = 0.