JEE Main 2025 Physics - Class 11 JEEM_2025_3 Solution: A body of mass is placed on a plane at a point with coordinates …

QJEEM_2025_3: JEE Main 2025 Physics - Class 11 JEEM_2025_3 Solution: A body of mass is placed on a plane at a point with coordinates …

A body of mass

4 \text{ kg}

is placed on a plane at a point

P

with coordinates

(3, 4) \text{ m}

. Under the action of a constant force

\vec{F} = (2\hat{i} + 3\hat{j}) \text{ N}

, the body moves to a new point

Q

with coordinates

(6, 10) \text{ m}

4 \text{ seconds}

.Determine the ratio of the average power to the instantaneous power at the end of the

4 \text{ s}

interval.

Step-by-Step Verified Solution

\textbf{Given: } m = 4\,\text{kg},\quad \vec{F} = 2\hat{i} + 3\hat{j}\,\text{N},\quad P(3,4),\quad Q(6,10),\quad t = 4\,\text{s}

% Displacement

\vec{s} = (6-3)\hat{i} + (10-4)\hat{j} = 3\hat{i} + 6\hat{j}

% Work done

W = \vec{F}\cdot \vec{s}

W = (2\hat{i}+3\hat{j})\cdot (3\hat{i}+6\hat{j}) = 6 + 18 = 24\ \text{J}

% Average power

P_{\text{avg}} = \frac{W}{t} = \frac{24}{4} = 6\ \text{W}

% Acceleration

\vec{a} = \frac{\vec{F}}{m} = \frac{2}{4}\hat{i} + \frac{3}{4}\hat{j} = 0.5\hat{i} + 0.75\hat{j}

% Velocity after 4 seconds

\vec{v}(4) = \vec{a}t = (0.5\hat{i}+0.75\hat{j})(4) = 2\hat{i} + 3\hat{j}

% Instant power

P_{\text{inst}} = \vec{F}\cdot \vec{v} = (2\hat{i}+3\hat{j})\cdot (2\hat{i}+3\hat{j}) = 4 + 9 = 13\ \text{W}

% Ratio

\text{Ratio} = \frac{P_{\text{avg}}}{P_{\text{inst}}} = \frac{6}{13}

\boxed{\frac{6}{13}}