Q 11-13-006NEETNEET 2024Top questionEasy
If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is $\dfrac{x}{2}$ times its original time period. Then the value of $x$ is:
Answer: (D) $\sqrt{2}$
$T = 2\pi\sqrt{\dfrac{l}{g}}$ does not depend on the mass of the bob.
$$T' = T\sqrt{\frac{1}{2}} = \frac{T}{\sqrt{2}} = \frac{\sqrt{2}}{2}T$$
So $\dfrac{x}{2} = \dfrac{\sqrt{2}}{2}$, giving $x = \sqrt{2}$.
Solution by Sreeraj P, M.Sc Physics