Category: Class 11 Physics Objective Questions

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

1. Velocity has to be the same at the same horizontal level in case of steady laminar flow.
a) True
b) False

Answer: b
Clarification: Velocity can be different at the same horizontal level if the area of the cross section at the two points is different. This is according to the equation of continuity, A₁v₁ = A₂v₂, where A is area and v is velocity.

2. A tube of uniform cross section always has water flowing through it. It is kept vertical in such a way that water enters from top and leaves from the bottom. If the speed at a point A below the opening is ‘v’, what will be the speed at a point B vertically below A such that the distance between A & B is ‘2h’?
a) v
b) (sqrt{v^2 + 4gh})
c) (sqrt{v^2 + 2gh})
d) v/2

Answer: a
Clarification: According to the equation of continuity, the velocity at B will be the same as that at A since the area of the cross section is the same at both points. You may be confused by thinking that gravity will increase its velocity but the pressure will also decrease downwards and therefore net effect will be zero change in velocity.

3. Choose the correct option regarding a streamline.
a) Speed, not velocity, at all points of a streamline is same
b) Two streamlines can intersect in case of laminar flow
c) There is no friction between streamlines in case of steady flow
d) In a given streamline, velocity of a point can vary with time

Answer: d
Clarification: If we consider a streamline in a pipe of varying diameter we can say that speed of different points in that streamline are different. The tangent to a streamline gives the direction of velocity at that point, so two streamlines cannot intersect as at the point of intersection we will not be able to draw a single tangent. In case of steady flow across a pipe, there could be streamlines next to each other having different velocities and hence there will be friction between the streamlines which we call viscosity. If flow is unsteady, but not turbulent, the velocity of a point in a streamline can change. For eg: If we slowly increase the speed of a tap flow will be laminar and speed of a point (on a streamline) will increase with time.

4. On which of these following options is the continuity equation based?
a) Work energy theorem
b) Law of conservation of energy
c) Conservation of mass
d) Conservation of momentum

Answer: c
Clarification: The continuity equation is based on the conservation of mass. We consider that fluid is incompressible (constant density) and say that mass of fluid passing through 2 different regions at the same time is the same.
∴ density X Vol₁ = density X Vol₂.
∴ Vol₁ = Vol₂.
∴ A₁v₁Δt = A₂v₂Δt
∴ A₁v₁ = A₂v₂.

5. In which one of the following cases can the equation of continuity be used?
a) Compressible flow
b) Incompressible flow
c) Turbulent flow
d) Viscous flow

Answer: b
Clarification: For using the continuity equation we assume that the flow is incompressible, laminar and non-viscous. The option that says compressible flow implies that fluid’s density can be changed. So it’s incorrect. The option that says turbulent flow is incorrect as we have to assume flow to be laminar for the equation of continuity to hold. The option that says viscous flow is also incorrect because we assume flow to be frictionless for the equation of continuity to hold.

6. Streamlines of steady flows can be curved.
a) True
b) False

Answer: a
Clarification: A streamline is a curve to which tangent at every point gives the velocity of that point. The flow of water through a bent pipe can be steady, and when we observe a streamline in it, it will be curved.

Physics Multiple Choice Questions on “First Law of Thermodynamics”.

1. ΔU=0 means that the process is isothermic. True or False?
a) True
b) False
Answer: a
Clarification: Internal energy is a function of temperature only. So if ΔU=0 implies that the process will have constant temperature, or it is isothermic.

2. First law of thermodynamics is based on?
a) Conservation of energy
b) Conservation of mass
c) Conservation of momentum
d) Conservation of work
Answer: a
Clarification: The first law of thermodynamics is based on the conservation of energy. It deals with work done and heat energy supplied or removed from a system. It basically says that energy supplied to a system is conserved.

3. If 315cal of heat is given to the system, and the system does 20cal of work, find the change in internal energy.
a) 295cal
b) 335cal
c) 0 cal
d) 335J
Answer: a
Clarification: The first law states that ΔU=ΔQ – ΔW
= 315 – 20
= 295cal.
Work is positive as it is done by the system. Heat is positive as it is supplied to the system.

4. What is the relation between the internal energy and heat supplied in the process 1 & 2 shown in the diagram? Both paths start at A and end at B.

a) U₁ > U₂, Q₁ > Q₂
b) U₁ < U₂, Q₁ > Q₂
c) U₁ = U₂, Q₁ = Q₂
d) U₁ = U₂, Q₂ > Q₂
Answer: d
Clarification: The initial and final states are the same for both processes, so the value of internal energy will be the same (U₁ = U₂).
The area under the PV curve gives work done.
In the given diagram, area under 1 is greater than area under 2. So, W₁ > W₂.
We know, ΔQ =ΔU + ΔW, which implies that Q₁ > Q₂ since U is same for both and W₁ > W₂.

5. A block of mass 1kg is given a velocity of 5m/s on a horizontal flat surface. What will be the change in internal energy of the block-surface system, when the block comes to rest. The coefficient of kinetic friction is 0.3 and coefficient of static friction is 0.4.
a) 25J
b) 12.5J
c) 0
d) 30J
Answer: b
Clarification: The kinetic energy of the block will change into internal energy of the system.
Initial kinetic energy = 0.5*1*25 = 12.5J.
Therefore the internal energy change = 12.5J.

6. Consider a gas contained in a rigid container of volume 20m³. What will be the change in internal energy if 50J of heat is provided to it? Assume the gas exerts a pressure of 1atm on the walls.
a) 50J
b) 0
c) 70J
d) 20J
Answer: a
Clarification: All the heat supplied will be equal to change in internal energy.
The work done is zero because the container is rigid hence no volume change.
Thus, ΔU = ΔQ – ΔW = 50 + 0 = 50J.

7. A disc spinning with 24J of kinetic energy is kept in a container containing a fluid having a heat capacity of 2000J/K. When the disc comes to rest what will be the change in internal energy if ΔU = 3.5ΔT.
a) 0
b) 0.042J
c) 0.021J
d) 0.028J
Answer: b
Clarification: When the disc comes to rest all its kinetic energy will have been converted into internal energy of the system.
The fluid has a heat capacity of 2000J/K which means that for 24J the change in temperature will be 24/2000 = 0.012K.
Given: ΔU = 3.5ΔT
⇒ ΔU = 3.5*0.012
= 0.042J.

8. In the process A to B to C, 20J of heat is supplied from A to B. 20.5J of heat has been removed from B to C and 2J of heat has been added from C to A. Calculate the value of ‘v’ from the given conditions in the diagram. The values of pressure and volume given in the graph are in S.I. units.

a) 4m³
b) 1.5m³
c) 3m³
d) 8m³
Answer: a
Clarification: In one cycle the change in internal energy should be zero.
The heat energy supplied in one cycle is 20 – 20.5 + 2 = 1.5J.
Therefore the work done by the system in one cycle should be 1.5J.
Area under PV curve is work done,
thus 0.5*(v-1)*1 = 1.5.
∴ v = 4m³.

9. In the given system, 20J of heat is supplied to the gas molecules. The spring is initially not elongated or compressed. If the spring gets compressed by 1cm. Calculate the change in internal energy of the system. Spring constant = 200N/m. Assume piston moves slowly.

a) 19J
b) -19J
c) 21J
d) -21J
Answer: a
Clarification: The work done by the piston will be stored in spring.
The energy stored in spring = 1/2kx² = 0.5*200*0.01 = 1J.
The work done by piston = 0.01J.
ΔU = ΔQ – ΔW = 20 – 1 = 19J.

10. At constant pressure P the volume of a gas increases from V₁ to V₂ when ‘Q’ amount of heat is removed from the system. What will happen to internal energy?
a) Remain the same if Q = P(V₁ – V₂)
b) Decrease by an amount P(V₂ – V₁) – Q
c) Decrease by an amount Q + P(V₂ – V₁)
d) Increase by some unknown amount
Answer: c
Clarification: The internal energy will definitely decrease as both factors heat removal and work done by gas support the same. ΔU = ΔQ – ΔW = -Q – P(V₂ – V₁) = -(Q + P(V₂ – V₁)). Thus, we can say that internal energy decreases by an amount = Q + P(V₂ – V₁).

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

1. What will be the velocity v/s time graph of a ball falling from a height before hitting the ground look like?
a) A straight line with positive slope
b) A straight line with negative slope
c) A straight line with zero slope
d) A parabola
Answer: a
Clarification: As the ball falls down, its velocity increases. This is because of the acceleration due to gravity. Hence the graph looks like a straight line with positive slope. The first equation of motion verifies this.

2. Which of the following is the correct formula for average velocity?
a) v = dx/dt
b) v = x/t
c) v = xt
d) v = t/x
Answer: b
Clarification: The correct formula is v = x/t. Average velocity is total change in displacement divided by total change in time. v = dx/dt is the formula for instantaneous velocity.

3. A truck requires 3 Hrs to complete a journey of 150 km, what is the average speed?
a) 50 km/hr
b) 25 km/hr
c) 15 km/hr
d) 10 km/hr
Answer: a
Clarification: Average speed is defined as total distance divided by total time. The total distance is 150 km and total time taken is 3 Hrs, therefore average speed = 150/3 = 50 km/hr.

4. Average speed of a car between points A and B is 20 m/s, between B and C is 15 m/s, between C and D is 10 m/s. What is the average speed between A and D, if the time taken in the mentioned sections is 20s, 10s and 5s respectively?
a) 17.14 m/s
b) 15 m/s
c) 10 m/s
d) 45 m/s
Answer: a
Clarification: Average speed is the total distance divided by total time taken. Total displacement (d = vt) = 20×20 + 15×10 + 10×5 = 600 m. Total time = 20 + 10 + 5 = 35 s. Therefore, average speed = 600/35 = 17.14 m/s.

5. Which of the following statement is correct?
a) Average speed > Instantaneous speed
b) Average speed >= Instantaneous speed
c) Average speed <= Instantaneous speed
d) Average speed < Instantaneous speed
Answer: c
Clarification: Average speed is the ratio of total distance to total time. Instantaneous speed is the ratio of instantaneous change in distance to instantaneous change in time. Average speed can be equal to instantaneous speed but is usually less than that.

6. In a uniformly accelerated motion, the speed varies from 0 to 20 m/s in 4 s. What is the average speed during the motion?
a) 10 m/s
b) 20 m/s
c) 0 m/s
d) 15 m/s
Answer: a
Clarification: From first equation of motion we have, v = u + at, which implies that a = 5 m/s². From second equation of motion we have s = ut + (1/2)at², which implies that s = 40 m. Average speed = s/t = 40/4 = 10 m/s.

7. What happens to the average velocity when a body falls under gravity with terminal velocity?
a) It increases
b) It decreases
c) It remains constant
d) It changes exponentially
Answer: c
Clarification: When the body is moving with terminal velocity, the velocity does not change. It means that equal displacement is being covered in equal time intervals. Hence the average velocity remains constant.

8. A ball is thrown up in the sky. After reaching a height, the ball falls back. What can be said about the average velocity?
a) It is non zero
b) It is zero
c) It is greater than zero
d) It is less than zero
Answer: b
Clarification: The average velocity is zero. The ball covers positive displacement when it goes up and negative displacement when it comes down. Hence the total displacement is zero. Therefore, the average velocity is zero.

9. A car is moving around a tree in a circular path. What can be said about the average velocity?
a) It is non zero
b) It is zero
c) It is greater than zero
d) It is less than zero
Answer: b
Clarification: Take any point on the circular path. The car moves in a circle and hence will come back to the same point after a definite time interval. Therefore, the displacement is 0. Hence, the average velocity is zero.

10. The changes in displacement in three consecutive instances are 5 m, 4 m, 11 m, the total time taken is 5 s. What is the average velocity in m/s?
a) 1
b) 4
c) 7
d) 6
Answer: b
Clarification: The total change in displacement = 20 m. Total time taken = 5 s. Average velocity = total change in displacement/total time taken = 20/5 = 4 m/s.

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

1. What is the force applied on a body with 5 kg of mass and an acceleration of 7 m/s²?
a) 35 N
b) 5 N
c) 7 N
d) 0 N
Answer: a
Clarification: According to the second law of motion, the force on a body is equal to the rate change of its momentum. On simplifying we get, F = ma. Therefore, Force = 5 x 7 = 35 N.

2. Unit of force is _____
a) Newton
b) Pascal
c) Byte
d) Gram
Answer: a
Clarification: The unit of force is newton. This unit has been given after the name of the famous scientist Sir Isaac Newton who had given the three laws of motion. The force on anybody can be found with the help of the second law of motion.

3. The unit ‘Newton’ is equivalent to ______
a) Kg-m/s²
b) Kg-m/s
c) Kg/m/s²
d) Kg²-m/s²
Answer: a
Clarification: From the second law of motion we get that F = ma, when the mass is not changing. Putting in the units, we get, Newton = Kg-m/s².

4. If the force acting on a body is 10 N, and the acceleration is 4 m/s², what can be the mass of the body?
a) 2.5 Kg
b) 25 Kg
c) 0.25 Kg
d) 5 Kg
Answer: a
Clarification: From the second law of motion we get that F = ma, when the mass is not changing. Putting in the values, we get, Force = 10 = m x 4. Which will result in m = 2.5 Kg.

5. If the force acting on a body is 50 N, and the mass is 5 kg, what can be the acceleration of the body?
a) 10 m/s²
b) 80 m/s²
c) 8 m/s²
d) 0.8 m/s²
Answer: a
Clarification: From the second law of motion we get that F = ma, when the mass is not changing. Putting in the values, we get, Force = 10 = 5 x a. Which will result in a = 10 m/s².

6. An example of variable mass system is _____
a) A car moving
b) A rocket taking off
c) A bicycle moving
d) A man walking
Answer: b
Clarification: In case of a rocket taking off, the mass of the fuel is large and cannot be neglected. As the rocket starts moving, a large amount of fuel is getting burnt out from the nozzle, hence the total mass decreases. Hence it is a variable mass system.

7. How do we calculate force in a variable mass constant velocity system?
a) F = ma
b) F = mv
c) F = v(dm/dt)
d) F = m/t
Answer: c
Clarification: The general expression for force is F = dp/dt. On solving we get, F = m(dv/dt) + v(dm/dt). Since the system is constant velocity, hence dv/dt = 0. What remains is F = v(dm/dt).