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Q.

The period of a simple pendulum on the surface of earth is T. At an attitude of half of the radius of earth from the surface, its period will be

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a

\sqrt{\frac{2}{3}} T

b

\frac{2 T}{3}

c

\sqrt{\frac{3}{2}} T

d

\frac{3 T}{2}

answer is B.

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Detailed Solution

T \propto \frac{1}{\sqrt{g}} g^{|} = g {[\frac{R}{R + h}]}^{2}

$g^{|} = g {[\frac{R}{R + R/2}]}^{2} = \frac{4 g}{9}$

$\frac{T}{T^{|}} = \sqrt{\frac{g^{|}}{g}} \Rightarrow T^{|} = T \sqrt{\frac{g}{g^{|}}}$

$T^{|} = T \sqrt{\frac{g}{4g} \times 9}$

$T = T \times \frac{3}{2}$

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