The mass and diameter of a planet are twice t | vedclass

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(B) The acceleration due to gravity is given by $g = \frac{GM}{R^2}$.Given: $M_p = 2M_e$ and $R_p = 2R_e$.Therefore,the ratio of gravity on earth to t

Vedclass · Accepted Answer

$2\sqrt{2} \, 	ext{s}$

Vedclass · Answer

(B) The acceleration due to gravity is given by $g = \frac{GM}{R^2}$.Given: $M_p = 2M_e$ and $R_p = 2R_e$.Therefore,the ratio of gravity on earth to the planet is:$\frac{g_e}{g_p} = \frac{M_e}{M_p} 	imes \left(\frac{R_p}{R_e}ight)^2 = \frac{1}{2} 	imes (2)^2 = \frac{4}{2} = 2$.So,$g_p = \frac{g_e}{2}$.The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g}}$,which implies $T \propto \frac{1}{\sqrt{g}}$.For a second's pendulum on earth,$T_e = 2 \, 	ext{s}$.Thus,$\frac{T_p}{T_e} = \sqrt{\frac{g_e}{g_p}} = \sqrt{2}$.$T_p = T_e 	imes \sqrt{2} = 2\sqrt{2} \, 	ext{s}$.

The mass and diameter of a planet are twice those of earth. The period of oscillation of a pendulum on this planet will be (if it is a second's pendulum on earth).

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