Category: Class 11 Physics Objective Questions

Physics Problems on “Acceleration due to Gravity below and above the Surface of Earth”.

1. Acceleration due to gravity increases as we go towards the centre of the earth.
a) True
b) False

Answer: b
Clarification: Acceleration due to gravity decreases as we go towards the centre of the earth because the volume enclosed by the radius vector from the centre of the earth decreases as depth increases. This reduction in volume leads to a reduction in enclosed mass and hence, the acceleration due to gravity is reduced.

2. Acceleration due to gravity increases as we move away from the surface of the earth radially towards the sky.
a) True
b) False

Answer: b
Clarification: The volume or mass enclosed by the radius vector from the centre of the earth remains the same as we move higher from the earth’s surface. However, the increase in radius decreases the acceleration due to gravity. The mathematical relationship is as follows;
g = (G*M1)/R²;
M1 = Mass of the earth
R = Radius vector from the centre of the earth.

3. Assume that the earth is a perfect sphere but of non-uniform interior density. Then, acceleration due to gravity on the surface of the earth _____
a) will be towards the geometric centre
b) will be different at different points on the surface
c) will be equal at all points on the surface and directed towards the geometric centre
d) cannot be zero at any point

Answer: d
Clarification: Since we assumed the earth to have a non-uniform density, the acceleration due to gravity will not be directed towards the geometric centre. Furthermore, for a perfect sphere, the acceleration due to gravity will be equal at all points on the surface and will be non-zero.

4. Which of the following is the variation of acceleration due to gravity at a height “h” above the earth’s surface? Let “R” be the radius of the earth and “M1” the mass of earth.
a) g = (G*M1)/R²
b) g = (G*M1)/h²
c) g = (G*M1)/(R + h)²
d) g = (G*M1)/(h/R)²

Answer: c
Clarification: If “r” is the distance between the centre of the earth and the object at a height “h” above the earth’s surface, then;
R = R + h
And; g = (G*M1)/r²
= (G*M1)/(R + h)².

5. Which of the following is the variation of acceleration due to gravity at a height “h” above the earth’s surface? Let “R” be the radius of the earth and “M1” the mass of earth. (Assume h < < R)
a) g = (G*M1)/R²
b) g = (G*M1)/h²
c) g = (G*M1)/(R/h)²
d) g = [(G*M1)/R²] x [1 – (2h)/R)]

Answer: d
Clarification: If “r” is the distance between the centre of the earth and the object at a height “h” above the earth’s surface, then;
R = R + h
And; g = (G*M1)/r²
= (G*M1)/(R + h)²
= (G*M1)/R² x (1 + h/R)^-2
Since h < < R;
Using binomial expansion and neglecting higher order terms, we can write the above expression as;
g = [(G*M1)/R²] x [1 – (2h)/R)].

6. Which of the following is the variation of acceleration due to gravity at a depth “d” below the earth’s surface? Let “R” be the radius of the earth and “M1” the mass of earth. (Assume the density of the earth to be constant)
a) g = (G*M1)/(R – d)
b) g = [(G*M1) x density]/d
c) g = (G x M1/R³) / (R – d)
d) g = (G x M1/R³) x (R – d)

Answer: d
Clarification: g = (G*M)/r²;
M = Mass contained in an enclosed volume
r = distance from centre of the earth to the depth “d”
Therefore; r = R – d; where, R = Radius of the earth
M = (density) x (volume)
= (density) x [(4/3) x (pi) x (r³)]
= (density x 4 x pi x r³) / 3
= (M1 x r³) / R³
Therefore;
g = (G x density x 4 x pi x r³) / (3 x r²)
= (G x density x 4 x pi x r¹) / 3
Therefore;
g = [(G x density x 4 x pi) / 3] x (R – d)
= (G x M1/R³) x (R – d).

7. What is the relationship between height “h” above the earth’s surface and a depth “d” below the earth’s surface when the magnitude of the acceleration due to gravity is equal? (Assume h < < R; where, R = Radius of the earth)
a) h = d
b) h = 2d
c) 2h = d
d) 3h = 2d

Answer: c
Clarification: Acceleration due to gravity above the surface of the earth;
g’ = [(G*M1)/R²] x [1 – (2h)/R)]
Acceleration due to gravity below the surface of the earth;
g’’ = (G x M1/R³) x (R – d)
g’ = g’’
Therefore;
[(G*M1)/R²] x [1 – (2h)/R)] = (G x M1/R³) x (R – d)
2h = d.

8. At what height above the surface of the earth, the acceleration due to the gravity of the earth becomes 5% of that of the surface?
a) h = 0.5 R
b) h = 1.5 R
c) h = 2.5 R
d) h = 3.5 R

Answer: d
Clarification:Acceleration due to gravity on the surface of the earth;
g = (G*M1)/R²
Acceleration due to gravity above the surface of the earth;
g’ = (G*M1)/(R + h)²
g’ = (5/100) x g
(G*M1)/(R + h)² = (5/100) x (G*M1)/R²
R² x 100 = (R + h)² x 5
Taking square root on both sides;
R x 10 = (R + h) x 2.24
7.76 x R = 2.24 x h
h = 3.5 x R.

9. The time period of a simple pendulum on the surface of the earth is “T”. What will be the time period of the same pendulum at a height of 2 times the radius of the earth?
a) T
b) 2T
c) 3T
d) 4T

Answer: c
Clarification: Time period of the simple pendulum on the surface of the earth;
T = (2 x pi) x (l / g)^1/2
l = Length of the simple pendulum
The time period of the simple pendulum at a certain height above the earth’s surface;
T’ = (2 x pi) x (l / g’)^1/2
g’ = Acceleration due to gravity at a certain height above the surface of the earth
g’ for a height of 2R (R = Radius of the earth);
g’ = (G*M1)/(R + 2R)²
= (G*M1)/(3R)²
= (1/9) x g
T’ = (2 x pi) x (l / (1/9)g) ^½
= 3 x [(2 x pi) x (l / g)^½]
= 3T.

10. The net acceleration due to gravity is zero at all points inside a uniform spherical shell.
a) True
b) False

Answer: a
Clarification: The point inside the spherical shell experiences gravitational pull by all points of point of the shell. However, the net gravitational force is zero due to vector addition. Hence, the net acceleration due to gravity is also zero.

11. Let the radius of the earth be R. Now, assume that the earth shrunk by 20% but the mass is the same. What would be the new value of acceleration due to gravity at a distance R from the centre of the earth if the value at the same distance in the previous case was “g’”?
a) g’
b) 2g’
c) 3g’
d) 4g’

Answer: a
Clarification: The value of acceleration due to gravity before shrinking at a distance R from the centre of the earth, i.e., on the surface of the earth is;
g’ = (G*M)/R²
Where; M = Mass of the earth
Now, the earth shrunk by 20%, however, the mass remains the same. This implies an increase in density.
The value of acceleration due to gravity from an object at a point outside the object is dependent only on the distance between the centre of gravity of the object and the distance between the point and the object. It is independent of density.
Hence, the new value of acceleration due to gravity at the same distance will remain unchanged, i.e., g’.

12. The acceleration of the moon towards the earth is approximately 0.0027 m/s². The moon revolves around the earth once approximately every 24 hours. What would be the acceleration due to gravity of the earth of the moon towards the earth if it were to revolve once every 12 hours?
a) Become half in magnitude
b) double in magnitude
c) Change direction but remain the same in magnitude
d) Remains unchanged

Answer: d
Clarification: The value of acceleration due to gravity does not depend on the speed of revolution but only the distance between both the centres of gravity. Hence, the magnitude and direction would remain unchanged.

Physics Multiple Choice Questions on “Fluids Mechanical Properties – Bernoulli’s Equation”

1. Consider a tank of height 20m filled with liquid of density 100kg/m³. The area of tank is 10m². If the tank has a hole of area 2m² at the bottom, find the speed of the liquid flowing out through the hole when the height of liquid in the tank is 10m . Assume speed of liquid descending at top of tank is 5m/s.
a) 20m/s
b) 14.14m/s
c) 15m/s
d) 20.615m/s

Answer: c
Clarification: We can’t consider the speed of efflux to be (sqrt{2gh}) as the areas are comparable. So, we use Bernoulli theorem between the top of the tank and the hole. Pressure at the top of tank and hole will be same, equal to P₀, since both are exposed to the atmosphere.
P₀ + ρgH + 1/2ρ (v_{1}^{2}) = P₀+ 1/2ρ(v_{2}^{2})
∴ v₂ = (sqrt{2(gh + v_{1}^{2}/2)})
= (sqrt{2(10 * 10 + 25/2)})
= 15m/s.

2. In which of the following conditions can the Bernoulli equation not be used?
a) Viscous flow
b) incompressible fluid
c) steady flow
d) laminar flow

Answer: a
Clarification: Bernoulli’s equation can be used for non-viscous, incompressible and steady laminar flow.

3. The speed of efflux, in case of a tank with a hole at the bottom, depends upon which of the following factors? Assume that the area of the tank is > > area of hole.
a) area of tank
b) density of liquid
c) height of hole from liquid
d) atmospheric pressure value

Answer: c
Clarification: Given that the area of the tank is > > area of hole, we can use the formula v = (sqrt{2gh}). This shows that speed depends only on acceleration due to gravity and height of hole from the surface.

4. A cylindrical tank of Height H has a hole on its side. It is kept on a flat surface. Assuming that hole’s area is much smaller than the area of the tank, what should be the distance of the hole below the top surface so that water coming out of the hole travels the maximum horizontal distance at the instant when the height of water is H?
a) H
b) H/3
c) H/2
d) same for all positions

Answer: c
Clarification: Let the hole be at a height x below the top surface.
The distance of hole from the ground will be H-x.
Speed of efflux ‘v’ = (sqrt{2gx}).
Let the time taken for water to reach the ground be ’t’. H-x = 1/2gt².
∴ t = (sqrt{2/g(H – x)}).
For maximum range, v*t should be maximum.
∴ d((sqrt{2/g(H – x)})*(sqrt{2gx}))/dx = 0
∴ d((sqrt{Hx – x^2}))/dx = 0
∴ (H-2x)/2 (sqrt{Hx – x^2}) = 0
The numerator should be zero but denominator should be non-zero.
∴ H ≠ x (for denominator to be non-zero)
∴ H = 2x OR x = H/2.

5. A cylindrical tank, filled with water, has an area of 10m². A piston covers its entire top surface. A uniformly distributed load of 1000N is applied on the piston from top. A hole of area 0.001m²is at the bottom of the tank. Find the speed of efflux when the height of water level is 0.1m .
a) 24.5m/s
b) 1.48m/s
c) 0.2m/s
d) 2m/s

Answer: b
Clarification: We should apply Bernoulli theorem between the top surface of water and the hole. The area of hole is very small when compared to the area of the tank. So, 1/2ρv² at the top surface of the tank can be neglected.
P₀ + F/AP + ρgh = P₀ + 1/2ρv²
∴ 1000/10 + 1000*10*0.1 = 0.5*1000 *v²
∴ v = 1.48m/s

6. A cylindrical tank of area 5m² contains water filled to a height of 10cm. A hole of area 0.005m² is present at the bottom. What is the time required for the water level to become half?
a) 100s
b) 10s
c) 1min
d) 10min

Answer: a
Clarification: The speed of efflux when height of water is ’h’ = (sqrt{2gh}). The volume of water flowing out in time dt is Avdt = 0.005*(sqrt{2gh})*dt. Therefore the decrease in height of water level (dh) in time dt is vol flowing out / area of tank = 0.005(sqrt{2gh})dt/5.
dh = (sqrt{2gh}) (0.005)dt/5
(int_{0}^{0.05}dh/sqrt{h} = int_{0}^{t}sqrt{2g})0.001 dt
∴ 2(sqrt{0.05}) = (sqrt{20})0.001 t
∴ t = (sqrt{0.2})*1000/(sqrt{20}) = 100s.

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.