MHTCET 10 May 2024 Shift 2 Paper

Concept:The period of a simple pendulum depends only on its length and the acceleration due to gravity, not on the mass or density of the bob.Explanation:The period of a simple pendulum is given by:$$T = 2\pi \sqrt{\frac{l}{g}}$$Here, $T$ is the period, $l$ is the length, and $g$ is the acceleration due to gravity.Initially, with length $l$, the period is $T$.Replacing the brass bob with a steel bob of the same size but $x$ times the density does not change the period, because $T$ is independent of the bob's mass or density.Now, the length is changed so that the new period becomes $2T$.Let the new length be $l'$.Then:$$2T = 2\pi \sqrt{\frac{l'}{g}}$$Substitute the original value of $T$:$$2\left(2\pi \sqrt{\frac{l}{g}}\right) = 2\pi \sqrt{\frac{l'}{g}}$$Cancel $2\pi$ from both sides:$$2\sqrt{\frac{l}{g}} = \sqrt{\frac{l'}{g}}$$Square both sides:$$4 \cdot \frac{l}{g} = \frac{l'}{g}$$Multiplying through by $g$, we get:$$l' = 4l$$Answer:The new length is $4l$.Therefore, the correct option is C: $4l$.