NEET (UG) 2024 — Question 38

The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{\ell}{g}}$, which is independent of the mass of the bob. The new length is $\ell' = \frac{\ell}{2}$. The new time period is $T' = 2\pi \sqrt{\frac{\ell'}{g}} = 2\pi \sqrt{\frac{\ell/2}{g}} = 2\pi \sqrt{\frac{\ell}{2g}} = \frac{2\pi \sqrt{\frac{\ell}{g}}}{\sqrt{2}} = \frac{T}{\sqrt{2}}$. Given $T' = \frac{x}{2} T$, we have $\frac{T}{\sqrt{2}} = \frac{x}{2} T$. Cancelling $T$ (non-zero) gives $\frac{1}{\sqrt{2}} = \frac{x}{2}$. Multiplying both sides by 2 yields $x = \frac{2}{\sqrt{2}} = \sqrt{2}$. Thus, the value of $x$ is $\sqrt{2}$, which corresponds to option B.