JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Work, Power & Energy

If momentum of a body is increased by

20%

, then its kinetic energy increases by :

(A) $36%$
(B) $40%$
(C) $44%$ ✓ Correct
(D) $48%$

Answer: (C)

Solution

When momentum increases by 20%, the new momentum is:

P^{'} = 1.2 P

Since kinetic energy

K = \frac{P ^{2}}{2 m}

, the percentage change in kinetic energy is:

\frac{K ^{'} - K}{K} \times 100 = \frac{\frac{( P ^{'} ) ^{2}}{2 m} - \frac{P ^{2}}{2 m}}{\frac{P ^{2}}{2 m}} \times 100 = [(\frac{P ^{'}}{P})^{2} - 1] \times 100

Substituting

P^{'} = 1.2 P

= [(1.2)^{2} - 1] \times 100 = [1.44 - 1] \times 100 = 44%

The answer is (C) 44%.

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Single correct option

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