The mass and diameter of a planet are twice those of earth. The period of oscillation of pendulum on this planet will be (If it is a second's pendulum on earth)
IIT 1973, Medium
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(b) As we know $g = \frac{{GM}}{{{R^2}}}$
$\Rightarrow \frac{{{g_{{\rm{earth}}}}}}{{{g_{{\rm{planet}}}}}} = \frac{{{M_e}}}{{{M_p}}} \times \frac{{R_\rho ^2}}{{R_e^2}} $
$\Rightarrow \frac{{{g_e}}}{{{g_p}}} = \frac{2}{1}$
Also $T \propto \frac{1}{{\sqrt g }} $
$\Rightarrow \frac{{{T_e}}}{{{T_p}}} = \sqrt {\frac{{{g_p}}}{{{g_e}}}}$
$\Rightarrow \frac{2}{{{T_p}}} = \sqrt {\frac{1}{2}} $
$ \Rightarrow $ ${T_p} = 2\sqrt 2 \,\sec $.
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