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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{	ext{eff}}}}$,where $g_{	ext{eff}}$ is the effective acceleration d

Vedclass · Accepted Answer

$T_0 = T_1 < T_2$

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{	ext{eff}}}}$,where $g_{	ext{eff}}$ is the effective acceleration due to gravity.$1$. For a stationary lift,$g_{	ext{eff}} = g$,so $T_0 = 2\pi \sqrt{\frac{l}{g}}$.$2$. When the lift descends at a steady speed,the acceleration is zero. Thus,$g_{	ext{eff}} = g$,and the period $T_1 = 2\pi \sqrt{\frac{l}{g}}$. Therefore,$T_0 = T_1$.$3$. When the lift descends with a constant downward acceleration $a$,a pseudo force $ma$ acts upwards on the pendulum bob. The effective acceleration becomes $g_{	ext{eff}} = g - a$. Since $g - a Thus,$T_0 = T_1 < T_2$.

$A$ simple pendulum suspended from the ceiling of a stationary lift has period $T_0$. When the lift descends at steady speed,the period is $T_1$,and when it descends with constant downward acceleration,the period is $T_2$. Which one of the following is true?

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