Physics · Work, Power & Energy

JEE Main 2024 — 6 April, Shift 2 — Question 45

When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be :

Option A:
$500 \%$
Correct
Option B:
$600 \%$
Option C:
$6 \%$
Option D:
$60 \%$

Answer: A

Step-by-step solution

Kinetic energy $(K)=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}}$ $\Rightarrow \mathrm{P}=\sqrt{2 \mathrm{mK}}$ If $\mathrm{K}_{\mathrm{f}}=36 \mathrm{~K}_{\mathrm{i}}$ So, $\mathrm{P}_{\mathrm{f}}=6 \mathrm{P}_{\mathrm{i}}$

$\%$ increase in momentum $=\frac{P_{f}-P_{i}}{P_{i}} \times 100 \%$

$=\frac{6P_i-P_i}{P_i}\times100\%\\=500\%$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Physics
Chapter: Work, Power & Energy
Topic: Kinetic Energy and Work-Energy Theorem