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) $\sqrt{2}$
Concept. $T = 2\pi\sqrt{\dfrac{l}{g}}$ does not depend on the mass of the bob.
Step 1. New period. $T' = T\sqrt{\dfrac12} = \dfrac{\sqrt2}{2}T$.
Step 2. Compare. $\dfrac{x}{2} = \dfrac{\sqrt2}{2} \Rightarrow x = \sqrt2$.
Answer. $\sqrt2$ (Option D).
Solution by Sreeraj P, M.Sc Physics