JEE Physics 2023 Question with Solution

Given: Mass $m = 0.4 \, \text{kg}$, time of fall $n = 8$ seconds, and $g = 10 \, \text{m/s}^2$.

Find: The loss of potential energy in the last second of fall.

The distance travelled in the $n^{\text{th}}$ second of free fall is

$$h_n = u + \frac{1}{2} g(2n - 1)$$

Here $u = 0$, so for $n = 8$,

$$h_8 = \frac{1}{2} \times 10 \times (2 \times 8 - 1)$$ $$h_8 = \frac{1}{2} \times 10 \times 15 = 75 \, \text{m}$$

The loss of potential energy in the last second is

$$\Delta PE = mgh_8$$

Substituting the values,

$$\Delta PE = 0.4 \times 10 \times 75 = 300 \, \text{J}$$

Therefore, the loss of potential energy in the last second of fall is $300 \, \text{J}$.

Given: The body is dropped from rest, so $u = 0$.

Find: Distance covered in the eighth second, then the corresponding potential energy loss.

For uniformly accelerated motion, the distance covered in the $n^{\text{th}}$ second is used directly because the decrease in height during that one-second interval determines the loss in gravitational potential energy.

Using

$$h_n = u + \frac{1}{2} g(2n - 1)$$

we get for the eighth second,

$$h_8 = 0 + \frac{1}{2} \times 10 \times 15 = 75 \, \text{m}$$

Now the gravitational potential energy lost in that interval is

$$\Delta PE = mg \times 75$$ $$\Delta PE = 0.4 \times 10 \times 75 = 300 \, \text{J}$$

Hence, the numerical value of the answer is 300.