Physics · Work, Power & Energy

JEE Main 2025 — 28 January, Evening Shift — Question 58

A body of mass 4 kg is placed on a plane at a point $P$ having coordinate $(3,4) \mathrm{m}$ . Under the action of force $\vec{F}=(2 \hat{i}+3 \hat{j}) N$ , it moves to a new point $Q$ having coordinates $(6,10) \mathrm{m}$ in 4 sec . The average power and instantaneous power at the end of 4 sec are in the ratio of :

Option A:
$13: 6$
Option B:
$6: 13$
Correct
Option C:
$1: 2$
Option D:
$4: 3$

Answer: B

Step-by-step solution

\text{Displacement} = (6 - 3)\hat{i} + (10 - 4)\hat{j} = 3\hat{i} + 6\hat{j}

\text{Work} = \vec{F} \cdot \vec{s} = 2(3) + 3(6) = 24\text{ J}

\text{Average power} = \frac{24}{4} = 6\text{ W}

\text{Acceleration } \vec{a} = \frac{\vec{F}}{m} = \frac{2\hat{i} + 3\hat{j}}{4} = 0.5\hat{i} + 0.75\hat{j}\text{ m/s}^2

\text{Velocity at } t = 4\text{ s (starting from rest)} = \vec{a}t = (0.5\hat{i} + 0.75\hat{j})4 = 2\hat{i} + 3\hat{j}\text{ m/s}

\text{Instantaneous power} = \vec{F} \cdot \vec{v} = 2(2) + 3(3) = 13\text{ W}

\text{Ratio avg:inst} = 6:13

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Physics
Chapter: Work, Power & Energy
Topic: Power