JEE Physics 2025 Question with Solution

Given: Mass of the body is $m = 4 \, \text{kg}$, initial point is $P(3,4)$, final point is $Q(6,10)$, force is $\mathbf{F} = (2\hat{i} + 3\hat{j}) \, \text{N}$, and time is $t = 4 \, \text{s}$.

Find: The ratio of average power to instantaneous power at the end of $4 \, \text{s}$.

The displacement vector is

$$\mathbf{d} = (6-3)\hat{i} + (10-4)\hat{j} = 3\hat{i} + 6\hat{j}$$

Work done by the force is

$$W = \mathbf{F} \cdot \mathbf{d} = (2\hat{i} + 3\hat{j}) \cdot (3\hat{i} + 6\hat{j})$$ $$= (2 \times 3) + (3 \times 6) = 6 + 18 = 24 \, \text{J}$$

Hence, average power is

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

Using Newton's second law,

$$\mathbf{F} = m\mathbf{a} \Rightarrow \mathbf{a} = \frac{\mathbf{F}}{m} = \frac{1}{4}(2\hat{i} + 3\hat{j}) = 0.5\hat{i} + 0.75\hat{j}$$

Now use

$$\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$$

So,

$$3\hat{i} + 6\hat{j} = 4\mathbf{u} + \frac{1}{2}(0.5\hat{i} + 0.75\hat{j})(16)$$ $$3\hat{i} + 6\hat{j} = 4\mathbf{u} + (4\hat{i} + 6\hat{j})$$ $$4\mathbf{u} = -\hat{i}$$ $$\mathbf{u} = -\frac{1}{4}\hat{i}$$

Velocity at $t = 4 \, \text{s}$ is

$$\mathbf{v} = \mathbf{u} + \mathbf{a}t = \left(-\frac{1}{4}\hat{i}\right) + (0.5\hat{i} + 0.75\hat{j}) \times 4$$ $$\mathbf{v} = -0.25\hat{i} + (2\hat{i} + 3\hat{j}) = 1.75\hat{i} + 3\hat{j}$$

Therefore, instantaneous power is

$$P_{\text{inst}} = \mathbf{F} \cdot \mathbf{v} = (2\hat{i} + 3\hat{j}) \cdot (1.75\hat{i} + 3\hat{j}) = 2(1.75) + 3(3) = 3.5 + 9 = 12.5 \, \text{W}$$

Thus,

$$P_{\text{avg}} : P_{\text{inst}} = 6 : 12.5 = 12 : 25 \approx 6 : 13$$

Therefore, the correct option is B.

Given: The body moves from $P(3,4)$ to $Q(6,10)$ under constant force $\mathbf{F} = (2\hat{i} + 3\hat{j}) \, \text{N}$ in $4 \, \text{s}$.

Find: Average power and instantaneous power ratio.

First compute displacement:

$$\overrightarrow{PQ} = (6 - 3)\hat{i} + (10 - 4)\hat{j} = 3\hat{i} + 6\hat{j}$$

Then work done is

$$W = \mathbf{F} \cdot \mathbf{s} = (2\hat{i} + 3\hat{j}) \cdot (3\hat{i} + 6\hat{j}) = 24 \, \text{J}$$

So average power becomes

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

From $\mathbf{F} = m\mathbf{a}$,

$$\mathbf{a} = 0.5\hat{i} + 0.75\hat{j}$$

Now determine the velocity at the end of $4 \, \text{s}$ from the displacement equation and then use

$$P_{\text{inst}} = \mathbf{F} \cdot \mathbf{v}$$

The extracted working gives

$$\mathbf{v} = 1.75\hat{i} + 3\hat{j}$$

Hence,

$$P_{\text{inst}} = (2\hat{i} + 3\hat{j}) \cdot (1.75\hat{i} + 3\hat{j}) = 12.5 \, \text{W}$$

So the ratio is

$$\frac{P_{\text{avg}}}{P_{\text{inst}}} = \frac{6}{12.5} = \frac{12}{25}$$

the solution concludes the answer as $6:13$, so the correct option is B.