JEE Physics 2026 Question with Solution

Given: String length $L_1 = 30 \, \text{cm}$, number of oscillations $n_1 = 20$ in time $t = 10 \, \text{s}$. For the new pendulum, $n_2 = 40$ in the same time. Find: Required string length $L_2$.

For a simple pendulum,

$$T = 2\pi\sqrt{\frac{L}{g}}$$

Also, in a fixed total time,

$$T = \frac{t}{n}$$

So,

$$\frac{t}{n} \propto \sqrt{L}$$

For constant $t$,

$$n \propto \frac{1}{\sqrt{L}}$$

Therefore,

$$\frac{n_1}{n_2} = \sqrt{\frac{L_2}{L_1}}$$

Substituting the given values,

$$\frac{20}{40} = \sqrt{\frac{L_2}{30}}$$ $$\frac{1}{2} = \sqrt{\frac{L_2}{30}}$$

Squaring both sides,

$$\frac{1}{4} = \frac{L_2}{30}$$ $$L_2 = \frac{30}{4} = 7.5 \, \text{cm}$$

Therefore, the required length of the string is $7.5 \, \text{cm}$. The correct option is A.

Given: The number of oscillations doubles from $20$ to $40$ in the same time. Find: New length $L_2$.

In the same total time, doubling the number of oscillations means the frequency doubles, so the time period becomes half. For a simple pendulum,

$$T \propto \sqrt{L}$$

Hence,

$$L \propto T^2$$

If $T$ becomes half, then length becomes one-fourth:

$$L_2 = \frac{L_1}{4} = \frac{30}{4} = 7.5 \, \text{cm}$$

Therefore, the required length is $7.5 \, \text{cm}$, so the correct option is A.