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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.On the surface of the earth,$T = 2\pi \sqrt{\frac{l}{g}}$,where $g

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$3T/2$

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.On the surface of the earth,$T = 2\pi \sqrt{\frac{l}{g}}$,where $g = \frac{GM}{R^2}$.At a height $h = R/2$,the acceleration due to gravity $g'$ is given by $g' = g \left( \frac{R}{R+h} \right)^2$.Substituting $h = R/2$,we get $g' = g \left( \frac{R}{R + R/2} \right)^2 = g \left( \frac{R}{3R/2} \right)^2 = g \left( \frac{2}{3} \right)^2 = \frac{4}{9}g$.The new time period $T'$ is $T' = 2\pi \sqrt{\frac{l}{g'}} = 2\pi \sqrt{\frac{l}{(4/9)g}} = \frac{3}{2} \left( 2\pi \sqrt{\frac{l}{g}} \right) = \frac{3}{2}T$.

The time period of a simple pendulum on the surface of the earth is $T$. If the pendulum is taken to a height equal to half of the radius of the earth,then its time period is

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