MHTCET 26 Apr 2025 Shift 2 Paper

Concept:The time period of a simple pendulum depends on effective acceleration due to gravity. When the lift accelerates upward, effective gravity increases.Explanation:For a pendulum of length $l$, the time period when the lift is at rest is:$$T = 2\pi \sqrt{\frac{l}{g}}$$When the lift accelerates upward with acceleration $a$, the effective gravity becomes:$$g' = g + a$$So the new time period is:$$T' = 2\pi \sqrt{\frac{l}{g + a}}$$Given that the period is reduced to $T/2$, we write:$$2\pi \sqrt{\frac{l}{g + a}} = \frac{1}{2} \left( 2\pi \sqrt{\frac{l}{g}} \right)$$Cancel $2\pi$ from both sides:$$\sqrt{\frac{l}{g + a}} = \frac{1}{2} \sqrt{\frac{l}{g}}$$Squaring both sides gives:$$\frac{l}{g + a} = \frac{1}{4} \cdot \frac{l}{g}$$Cancel $l$ from both sides:$$\frac{1}{g + a} = \frac{1}{4g}$$Cross-multiplying:$$g + a = 4g$$$$a = 3g$$Thus, the required upward acceleration is $3g$.Answer:$$a = 3g$$Hence, the correct option is B. $3g$.