MHTCET 11 Apr 2026 Shift 2 Paper

Concept:For a pendulum, the time period depends on the effective acceleration due to gravity, which changes when the lift accelerates upward or downward.Explanation:The time period of a simple pendulum is $T = 2\pi\sqrt{\frac{l}{g_{\text{eff}}}}$.When the lift is at rest, $g_{\text{eff}} = g$.So, $T_{1} = 2\pi\sqrt{\frac{l}{g}}$.This gives $\frac{1}{T_{1}^{2}} = \frac{g}{4\pi^{2}l}$.When the lift moves upward with acceleration $a$, the effective gravity becomes $g + a$.So, $T_{2} = 2\pi\sqrt{\frac{l}{g+a}}$.This gives $\frac{1}{T_{2}^{2}} = \frac{g+a}{4\pi^{2}l}$.When the lift moves downward with acceleration $a$, the effective gravity becomes $g - a$.So, $T_{3} = 2\pi\sqrt{\frac{l}{g-a}}$.This gives $\frac{1}{T_{3}^{2}} = \frac{g-a}{4\pi^{2}l}$.Adding the expressions for $T_{2}$ and $T_{3}$:$\frac{1}{T_{2}^{2}} + \frac{1}{T_{3}^{2}} = \frac{(g+a) + (g-a)}{4\pi^{2}l} = \frac{2g}{4\pi^{2}l}$.Since $\frac{g}{4\pi^{2}l} = \frac{1}{T_{1}^{2}}$, we get:$\frac{1}{T_{2}^{2}} + \frac{1}{T_{3}^{2}} = \frac{2}{T_{1}^{2}}$.Rearranging:$\frac{1}{T_{1}^{2}} = \frac{1}{2}\left(\frac{1}{T_{2}^{2}} + \frac{1}{T_{3}^{2}}\right)$.Simplifying further:$T_{1}^{2} = \frac{2T_{2}^{2}T_{3}^{2}}{T_{2}^{2} + T_{3}^{2}}$.Taking the square root on both sides:$T_{1} = \frac{\sqrt{2}\, T_{2} T_{3}}{\sqrt{T_{2}^{2} + T_{3}^{2}}}$.Answer:The correct relation is $T_{1} = \frac{\sqrt{2}\, T_{2} T_{3}}{\sqrt{T_{2}^{2} + T_{3}^{2}}}$, which corresponds to option B.