Category: Class 11 Physics Objective Questions

250+ TOP MCQs on Oscillations – Velocity and Acceleration in Simple Harmonic Motion | Class 11 Physics

Physics Quiz Online for Class 11 on “Oscillations – Velocity and Acceleration in Simple Harmonic Motion”.

1. A particle is executing SHM and currently going towards the amplitude. If it is at A/2, what is the relation between the direction of velocity and acceleration?
a) Both vectors point towards the amplitude
b) Velocity is towards amplitude & acceleration is towards mean position
c) Velocity is towards mean position & acceleration is towards the amplitude
d) Both vectors point towards the mean position
Answer: b
Clarification: The force on a particle in SHM is always towards the mean position. The particle is currently going towards an extreme, thus velocity will be towards the amplitude.

2. In a SHM, for what value of w, will the magnitude of maximum acceleration be greater than the magnitude of maximum velocity? Here, w is angular frequency.
a) w > 1
b) w < 1
c) w = 0
d) Not possible for any value of w
Answer: a
Clarification: Let the velocity equation of SHM be: v = Awcos(wt),
acceleration will be given by: a = -Aw2sin(wt).
Maximum magnitude of vel = Aw
Maximum magnitude of acc = Aw2.
For, Aw2 > Aw
w > 1.

3. A particle of mass m starts from the mean position of a SHM, at t=0, and goes towards -A. If the angular frequency of SHM is w, find the force acting on it as a function of time.
a) mAw2sin(wt)
b) mAw2sin(wt+π)
c) -mAw2cos(wt+π)
d) -mAw2cos(wt)
Answer: a
Clarification: The displacement equation will be given by: x = -Asin(wt).
On taking its derivative, we get:
v = -Awcos(wt).
Further, we get: a = Aw2sin(wt).
Thus, force is given by: F(t) = mAw2sin(wt).

4. The graph of acceleration vs time for a SHM is given. FInd the equation of position of the particle as a function of time. Assume that the particle is oscillating on the x-axis about the origin. And consider RHS of origin to be positive & LHS to be negative.

a) 8cos(πt)
b) -8cos(πt)
c) 0.81sin(πt)
d) -0.81sin(πt)
Answer: d
Clarification: The particle has zero acceleration at t=0, so it starts from the mean position. Then the acceleration becomes positive which implies that the particle is moving towards -A ( negative extreme). The equation of acceleration can be given by: a = 8sinwt.
The time period is 2s,
so w = 2π/2 = πs-1.
∴ a = 8sin(πt).
∴ v = -8/π cos(πt)
∴ x = -8/pi2sin(πt) = -0.81sin(πt).

5. A particle is undergoing a SHM of amplitude 10cm. What should be the minimum value of acceleration at an extreme position for maximum speed at centre to be 5m/s?
a) 20m/s2
b) 5m/s2
c) 0
d) 250m/s2
Answer: d
Clarification: vmax = Aw.
5 = 0.1w.
w = 50s-1.
amin at extreme = Aw2
= 0.1*50*50
= 250m/s2.

6. A particle in uniform circular motion is projected on its diameter. The motion of projection will be simple harmonic. Select the correct option regarding speed and acceleration of the particle in circular motion.
a) Speed is constant, acceleration is zero
b) Speed is constant, acceleration is non-zero
c) Speed changes, acceleration is non-zero
d) Speed changes, acceleration is zero
Answer: a
Clarification: For the particle in uniform circular motion, magnitude of velocity is constant, but direction is continuously changing. Thus, speed is constant but acceleration is non-zero.

7. The curves given below are that of position(x), velocity(v) & acceleration(a) vs time. What can we infer about the value of angular frequency ‘f’ from the given curves?

a) f > 1
b) f < 1
c) f = 0
d) We need the time period in order to find ‘f’
Answer: a
Clarification: Amplitude of position = A
Amplitude of velocity = Af
Amplitude of acceleration = Af2.
From the graph we deduce that the amplitude of a is greater than the amplitude of v which in turn is greater than the amplitude of x.
Thus, angular frequency will be greater than 1.

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250+ MCQs on Straight Line Motion-Position, Path Length and Displacement | Class 11 Physics

Physics MCQs on “Straight Line Motion – Position, Path Length and Displacement”.

1. The displacement of a particle is given as function of time as x = t2 + 2t. How much displacement is covered in the first 5 seconds?
a) 5 units
b) 35 units
c) 40 units
d) 0 units

Answer: b
Clarification: The displacement covered in the first five seconds can be obtained by putting t = 5 in the equation. Therefore, x = 55 + 2(5) = 25 + 10 = 35. Hence the answer is 35 units.

2. Path length does not depend on ____
a) Initial point
b) Final point
c) Path taken
d) Coordinate system

Answer: d
Clarification: The path length depends on the final and initial point. It also depends on the path taken. But it does not depend on the coordinate system. The coordinate system merely defines thee path and does not affect its total length.

3. Which one of the following relations is true?
a) Distance > Displacement
b) Distance < Displacement
c) Distance >= Displacement
d) Distance <= Displacement

Answer: c
Clarification: Displacement is the shortest distance between two points. Hence displacement <= distance or vice versa. If the path followed is the path of shortest distance or displacement, then displacement = distance.

4. A person is standing at -2 location on the number line. He runs to and fro from -2 to +5 location 5 times. How much distance has he covered if he comes back to -2 location at the end?
a) 35 units
b) 7 units
c) 30 units
d) 15 units

Answer: a
Clarification: In one turn the person covers 5 – (-2) units of distance, i.e. 7. Therefore in 5 turns, he will cover 5 x 7 = 35 units of distance.

5. What is the path length of the following path?
A (0, 0) to B (5, 0) to C (5, 5) to D (0, 5)
a) 20 units
b) 25 units
c) 15 units
d) 10 units

Answer: c
Clarification: Total path length = Sum of path lengths of intermediate paths. Therefore, Path length AD = Sum of Path lengths of AB, BC, CD = 5 + 5 + 5 = 15 units.

6. When person moves in the coordinate system from A (0, 0) to B (5, 10), to C (8, 6), what is the displacement covered?
a) 10 units
b) 5 units
c) 7 units
d) 15 units

Answer: a
Clarification: The displacement is the distance between the final and the initial location. Here the final location is C and the initial is A. We can solve this by using distance between two points method. AC = Square root ((82 – 02) + (62 – 02)) = Square root (100) = 10 units.

7. Displacement between two points is ___
a) The shortest path
b) The longest path
c) Equal to distance
d) Greater than distance

Answer: a
Clarification: Displacement between two points is the shortest path between them. It is always less than or equal to the distance between them. Displacement can never be greater than distance.

8. Distance does not depend on _______
a) Initial point
b) Final point
c) Path taken
d) Speed

Answer: d
Clarification: The distance depends on the final and initial points as these points define the path. Distance also depends on the path chosen, the distance between same initial and final point with different paths can be different. It does not depend on speed as whatever the speed may be, if the initial and final points and path remains same, the distance remains same.

9. How many variables are required to define the position of a body in space?
a) 3
b) 2
c) 1
d) 0

Answer: a
Clarification: In space we require a minimum of 3 variables to describe the position of a body, namely x, y, and z (in Cartesian system). There are also systems other than Cartesian Coordinate system to do this like Cylindrical system, Spherical or Radial system.

10. In which coordinate system do we use distance from origin and to angles to define the position of a point in space?
a) Cartesian
b) Cylindrical
c) Spherical
d) 2-D Cartesian

Answer: c
Clarification: In Spherical system, distance from the center, the angle with the X axis, and the angle with the Z axis are used to define the position of a point. These are respectively represented by R, θ and ф.

11. Which of the following is the correct formula for finding distance (d) between two points (x1, y1) and (x2, y2)?
a) d2 = (x2-x1)2 + (y2-y1)2
b) d4 = (x2-x1)2 + (y2-y1)2
c) d3 = (x2-x1)2 + (y2-y1)2
d) d = (x2-x1)2 + (y2-y1) 2

Answer: c
Clarification: The correct answer is d2 = (x2-x1)2 + (y2-y1)2. This expression can be found out using Pythagoras theorem in Cartesian coordinate system. Build a right-angle triangle with sides parallel to the axes and the hypotenuse joining the two points to construct the right-angle triangle.

250+ TOP MCQs on Newton’s First Law of Motion | Class 11 Physics

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

1. What causes the motion of a body which is initially in the state of rest?
a) Force
b) Displacement
c) Speed
d) Velocity
Answer: a
Clarification: The first law of motion states that a body remains in the state of motion or in the state of rest until and unless an external force is applied on it. Hence, the answer is force. This is also known as the law of inertia.

2. People sitting in a moving bus experience a jerk when the bus stops. This is due to _____
a) Inertia of motion
b) Inertia of rest
c) Inertia of turning
d) Inertia of acceleration
Answer: a
Clarification: When the bus stops, the people sitting in it experience a jerk. This is because, while the bus was moving, the people were moving with it. Hence, they were having an inertia of motion. When the bus stops, the inertia of the bus changes from that of motion to that of rest. But the inertia of the people still remains that of motion. Hence, they experience a jerk.

3. Passengers sitting in a stationary car experience a jerk when the car suddenly starts. This is due to _____
a) Inertia of motion
b) Inertia of rest
c) Inertia of turning
d) Inertia of acceleration
Answer: b
Clarification: When the car starts, the people sitting in it experience a jerk. This is because, while the car was stationary, the people were stationary with it. Hence, they were having an inertia of rest. When the car starts, the inertia of the car changes from that of rest to that of motion. But the inertia of the people still remains that of rest. Hence, they experience a jerk.

4. In the following figure, what will happen to the ball hanging from the roof of the car if the car suddenly moves towards left (direction shown by arrow)?

a) Ball moves towards right
b) Ball moves towards left
c) Ball does not move
d) Ball moves out of the paper
Answer: a
Clarification: Since the system was initially at rest, the inertia of the ball is that of rest. When the car suddenly moves the inertia of the car changes but not that of the ball. Hence the ball moves towards right.

5. A straight moving bus takes a sharp right turn. What will happen to the passengers sitting inside the bus?
a) They will tilt rightwards
b) They will tilt leftwards
c) They will stay the way they were
d) They will start jumping
Answer: b
Clarification: When the bus takes a sharp right turn, the inertia of the bus changes. But the inertia of the passengers is still towards the straight direction. Hence, they will relatively move towards the left.

6. Why do we not experience any leaning when a train takes a turn?
a) Because the train is powerful
b) Because of large turning radius
c) Because the train does not turn
d) Because the driver is smart
Answer: b
Clarification: The train’s turning radius is large as compared to the turning radii of normal vehicles. Hence, there is no sharp or sudden turning. Therefore, the passengers do not experience any leaning.

7. Inertia is _____
a) Property of mass to remain unchanged
b) Property of mass to change continuously
c) Property of mass to accelerate
d) Tendency of mass to accelerate
Answer: c
Clarification: Inertia is the property of mass to remain unchanged. Inertia is because of which a mass continues to remain in its original state in absence of any external force. Inertia of a body can be changed only by an external force.

250+ TOP MCQs on Power | Class 11 Physics

Physics Multiple Choice Questions on “Power”.

1. What are the units of power?
a) Newton
b) Joule
c) Watt
d) No units
Answer: c
Clarification: The SI unit of power is “Watt”. It is named after the famous inventor “James Watt” who is widely remembered for his improvements on the steam engine.
1 Watt = 1 Joule per second.

2. A machine gun fires 360 bullets per minute. Each bullet has a mass of 5 grams and travels at 600 m/s. What is the power of the gun? (Assume no loss of energy and 100% power transmission)
a) 300 W
b) 600 W
c) 900 W
d) 1800 W
Answer: c
Clarification: No. of bullets fired per second = 360/60 = 6
Power of the gun is completely transmitted to bullets.
Power (P) = Power of bullets in 1 second
= [(1/2 x m x v2) x 6] / 1
m = 5 x 10-3 kg
v = 600 m/s
P = (1/2 x 5 x 10-3 x 6002 x 6) / 1
= 900 W.

3. A motor is used to deliver water through a pipe. Let the motor have a power P initially. To double the rate of flow of water through the pipe, the power is increased to P’. What is the value of P’/P?
a) 2
b) 4
c) 6
d) 8
Answer: d
Clarification: Power = Force x Velocity
= (Mass/Time) x Velocity2
= Rate of flow x Velocity2
Rate of flow = Area x Velocity
Doubling the rate of flow will double the velocity.
P = Rate of flow x Velocity2
P’ = (2 x Rate of flow) x (2 x Velocity) 2
= 8P
P/P’ = 8.

4. A motor is used to pump water from a depth of 5 m to fill a volume of 10 cubic meters in 5 minutes. If 50% of the power is wasted, what is the power of the motor? (Assume g = 10m/s2)
a) 1000/3 W
b) 3000/3 W
c) 5000/3 W
d) 10000/3 W
Answer: d
Clarification: Density of water = 1000 kg/m3
Mass of water in 10m3 = 1000 x 10
= 10,000 kg
Volume of water filled in 1s = 10/(5×60)
= 1/30 m3
Mass filled in 1s = 1/30 x 1000
= 100/3 kg
Power = 100/3 x 10 x 5; [g = 10m/s2]
= 5000/3 W
Since 50% of power is wasted, power of motor = 5000/3 x 2
= 10000/3 W.

5. A machine has a power of 20kW. How long will it take for it to lift a body of mass 10kg from the ground to a height of 100m? (Assume g = 10m/s2)
a) 1 second
b) 2 seconds
c) 3 seconds
d) 4 seconds
Answer: b
Clarification: Power (P) = Energy (E) x Time (t)
20,000 = (m x g x h) x t
= (10 x 10 x 100) x t
t = 2 seconds.

6. A 2-ton vehicle travels at 20m/s on a road where the frictional force is 10% of the weight of the vehicle. What is the power required? (Assume g = 10m/s2)
a) 220 kW
b) 240 kW
c) 420 kW
d) 440 kW
Answer: d
Clarification:Weight of car = m x g
= 2,000 x 10
= 20,000 N
Frictional force = (10/100) x (20,000)
= 2,000 N
Total force (F) = 20,000 + 2,000
= 22,000 N
Velocity (v) = 20 m/s
P = F x v
= 22,000 x 20
= 4,40,000 W
= 440 kW.

7. 1000 kW of power is supplied to a motor. 90% of this is transmitted to a machine operating at 80% efficiency. The machine lifts an object of 10-tons with a velocity _____ m/s. (Assume g = 10m/s2)
a) 2
b) 4
c) 6
d) 8
Answer: c
Clarification: Power transmitted to machine = (90/100) x (1000) kW
= 900 kW
Power used to lift object = (80/100) x 900 KW
= 720 kW
7,20,000 = F x v
= (1/2 x m x v2) x v
= 10,000/2 x v3
v3 = 216
v = 6 m/s.

8. A 200 kg wagon climbs up a hill of slope 30-degrees in 1 minute at a speed of 10 m/s. How much power is delivered by the engine? (Assume g = 10m/s2)
a) 10 kW
b) 20 kW
c) 30 kW
d) 40 kW
Answer: b
Clarification: Weight of car = m x g
= 2,00 x 10
= 2,000 N
Total force (F) = 2,000
Velocity (v) = 10 m/s
P = F x v
= 2000 x 10
= 20,000 W
= 20 kW.

9. A 200 kg wagon climbs up a hill of slope 30-degrees in 1 minute at a speed of 36 km/h. How much power is delivered by the engine if the frictional force is 25% of the weight of the car? (Assume g = 10m/s2)
a) 25 kW
b) 50 kW
c) 100 kW
d)125 kW
Answer: a
Clarification: Weight of car = m x g
= 200 x 10
= 2,000 N
Frictional force = (25/100) x (2,000)
= 500 N
Total force (F) = 2000 + 500
= 2,500 N
Velocity (v) = 36 km/h
= 36 x (1000/3600)
= 10 m/s
P = F x v
= 2500 x 10
= 25 kW.

10. A body of mass 3 kg starts from rest with a uniform acceleration of unknown magnitude. If the body has a velocity of 30 m/s in 6 seconds, what is the power consumed in 3 seconds?
a) 125 W
b) 225 W
c) 325 W
d) 425 W
Answer: b
Clarification: v = u + at
30 = a x 6
a = 5 m/s2
Force (F) = m x a
= 3 x 5
= 15 N
v’ = u + at’
v’ = 5 x 3
= 15 m/s
Power consumed in 3 seconds = F x v’
= 15 X 15
= 225 W.

11. The force required to tow a vehicle at constant velocity is directly proportional to the magnitude of velocity raised to the first power. It requires 1000 W to tow with a velocity of 10 m/s. How much power is required to tow at a velocity of 8 m/s?
a) 320 W
b) 640 W
c) 160 W
d) 80 W
Answer: b
Clarification: Since force is directly proportional to velocity, the power required will be directly proportional to the square of the velocity.
Power = Force x Velocity
Hence, power is a multiple of 82 and an integer such that;
1000/102 = P/82
P = 640 W.

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250+ TOP MCQs on Acceleration due to Gravity of the Earth | Class 11 Physics

Physics MCQs on “Acceleration due to Gravity of the Earth”.

1. The acceleration due to gravity on the surface of the earth is different at different points on the surface.
a) True
b) False
Answer: a
Clarification: Since the earth is not a perfect sphere and has many irregularities, the acceleration due to gravity is different at different points on the earth’s surface.

2. The acceleration due to gravity on the surface of the earth is _____
a) greater towards the equator and lesser towards the poles
b) lesser towards the equator and greater towards the poles
c) same at all points on the surface of the earth
d) same everywhere except at the poles
Answer: b
Clarification: The earth is not a perfect sphere. The radius of the earth at the equator is greater than that at the poles, hence, the acceleration due to gravity is lesser towards the equator and greater towards the poles.

3. The dimensions of acceleration due to gravity are _____
a) [M0L1T-2]
b) [M1 L-1T-2]
c) [M-1 L2 T-1]
d) [M0L-1T2]
Answer: a
Clarification: The unit of acceleration is m/s2.
m/s2 = [L1 T-2]
= [M0 L1 T-2].

4. For an object on the surface of the earth, the magnitude of the acceleration due to the gravity of the earth it experiences depends also depends on the mass of that object.
a) True
b) False
Answer: b
Clarification: g = (G*M1)/R2;
M1 = Mass of the earth
The acceleration due to the gravity of the earth experienced by any object on the surface of the earth depends only on the mass of earth and the square of the distance between the object and the centre of the earth.

5. What would be the magnitude of the acceleration due to gravity on the surface of the earth if the radius of the earth were reduced by 20%?
a) 9.81 m/s2
b) 12.26 m/s2
c) 15.33 m/s2
d) 49.05 m/s2
Answer: c
Clarification: g = (G*M1)/R2;
M1 = Mass of the earth
R = Radius of the earth
We know; g = 9.81 m/s2
If the radius is reduced by 20% then the new radius is 80% of the original one;
New radius = 0.8R
Therefore, new acceleration;
g / (0.8 x 0.8) = 9.81 / 0.64
= 15.33 m/s2.

6. What would be the magnitude of the acceleration due to gravity on the surface of the earth if the density of the earth increased by 3 times and the radius remained the same?
a) 9.81 m/s2
b) 12.26 m/s2
c) 15.33 m/s2
d) 29.43 m/s2
Answer: d
Clarification: g = (G*M1)/R2;
M1 = Mass of the earth
R = Radius of the earth
Since the radius is the same, the volume would remain constant.
Density = mass/volume
Since density is increased 3 times and volume is the same, this implies that mass is increased 3 times.
We know; g = 9.81 m/s2
New mass = 3 x M1
Therefore, new acceleration;
3 x (G*M1)/R2 = 3x g
= 3 x 9.81
= 29.43 m/s2.

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250+ TOP MCQs on Fluids Mechanical Properties – Streamline Flow | Class 11 Physics

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.