Previous year questions (2014-21): Work, Energy & Power Notes | Study Physics Class 11 - NEET

Q.1. A particle is released from height S from the surface of the earth. At a certain height, its kinetic energy is three times its potential energy. The height from the surface of the earth and the speed of the particle at that instant are respectively:          [2021]
(a)
(b)
(c)
(d)
Ans: B
U + KE = E
4U = E = mgS
4mgh = mgS
h = S/4


Q.2. Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine?(g = 10 m/s²)        [2021]
(a) 12.3 kW
(b) 7.0 kW
(c) 10.2 kW
(d) 8.1 kW
Ans: D
Mass of water falling/second = 15 kg/s h = 60 m, g = 10 m/s², loss = 10 % i.e., 90% is used. Power generated = 15 x 10 x 60 x 0.9 = 8100 W = 8.1 kW.

Q.3. The energy required to break one bond in DNA is 10^–20J. This value in eV is nearly:        [2020]
(a) 0.06 
(b) 0.006 
(c) 6 
(d) 0.6
Ans: a

1.6 × 10^–19 Joule = 1 eV

Q.4. Body A of mass 4m moving with speed u collides with another body B of mass 2m, at rest. The collision is head on and elastic in nature. After the collision the fraction of energy lost by the colliding body A is:    [2019]
(a) 1/9
(b) 8/9
(c) 4/9
(d) 5/9
Ans: b

Fractional loss of KE of colliding body

Q.5. A disc of radius 2 m and mass 100 kg rolls on a horizontal floor. Its centre of mass has speed of 20 cm/s. How much work is needed to stop it?    [2019]
(a) 3 J
(b) 30 kJ
(c) 2 J
(d) 1 J
Ans: a

Work required = change in kinetic energy Final KE = 0
Initial KE

|ΔKE| = 3 J

Q.6. A force F = 20 + 10 y acts on a particle in y direction where F is in newton and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is    [2019]
(a) 30 J
(b) 5 J
(c) 25 J
(d) 20 J
Ans: c

Work done by variable force is

Q.7. The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is:    [2019]
(a) mgR
(b) 2mgR
(c)
(d)
Ans: c

Initial potential energy at earths surface is

Final potential energy at height h = R

As work done = Change in PE∴ W = U_f – U_i
 (∵ GM = gR²)

Q.8. A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be :-    [2018]
(a) 0.5
(b) 0.25
(c) 0.8
(d) 0.4
Ans: b

Q.9. A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to    [2018]

(a) 3/2 D
(b) D
(c) 7/5 D
(d) 5/4 D
Ans: d

Q.10. The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are KA, KB and KC respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then    [2018]

(a) KA < KB < KC
(b) KA > KB > KC
(c) KB < KA < KC
(d) KB > KA > KC
Ans: b

Q.11. A spring of force constant k is cut into lengths of ratio 1: 2 : 3. They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k''. Then k' : k'' is    [2017]
(a) 1 : 6
(b) 1 : 9
(c) 1 : 11
(d) 1 : 14
Ans: c

Force constant of spring

after cutting the spring in the ratio 1:2:3, the force constants will be in the ratio 6:3:2
Force constant in series K' is given by
Force constant in parallel K" : 6k+3k+2k = 11k;
hence K'/K" = 1/11

Q.12. Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take 'g' constant with a value 10 m/s². The work done by the    [2017]
(i) gravitational force and the
(ii) resistive force of air is :-
(a) (i) 1.25 J (ii) – 8.25 J
(b) (i) 100 J (ii) 8.75 J
(c) (i) 10 J (ii) – 8.75 J
(d) (i) – 10 J (ii) – 8.25 J
Ans: c

Q.13. A body of mass 1 kg begins to move under the action of a time dependent force  , whenare unit vectors along x and y axis. What power will be developed by the force at the time t ?    [2016]
(a) (2t³+ 3t⁵)W
(b) (2t² + 3t³)W
(c) (2t² + 4t⁴)W
(d) (2t³ + 3t⁴)W
Ans: a

Q.14. Two particles of masses m₁, m₂ move with initial velocities u₁ and u₂. On collision, one of the particles get excited to higher level, after absorbing energy ε. If final velocities of particles be v₁ and v₂ then we must have :    [2015]
(a)
(b)
(c)
(d)
Ans: d

Q.15. Two similar springs P and Q have spring constants K_P and K_Q, such that K_P > K_Q. They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs W_P and W_Q are related as, in case (a) and case (b), respectively:    [2015]
A: W_P < W_Q; W_Q < W_P
B: W_P = W_Q; W_P > W_Q
C: W_P = W_Q; W_P = W_Q
D: W_P > W_Q; W_Q > W_P

Ans: d

When a mass moves in a circle of radius R₀ with velocity v₀, its kinetic energy is given by

The tension in the string is gradually increased and the radius of the circle decreased to R₀/2. When the radius of the circle is R   the tension in the string is the same as the centripetal force.

where L = mRv is the angular momentum which is conserved.
Work done in reducing the radius of the circle from R₀ to R₀/2 is

The block of mass M = 10 kg is moving in the x - direction with a speed v = 10 m/s. Its initial kinetic energy is

It is subjected to a retarding force F = 0.1 x J/m during its travel from x = 20 m to 30 m.
Work done is given by

Final kinetic energy is, KE_f = KE_i + W = 500 - 25 = 475