The length of a seconds pendulum is 1 ,m on t | vedclass

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(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.For a seconds pendulum, the time period $T = 2 \,s$ on both the

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$0.5$

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(D) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.For a seconds pendulum, the time period $T = 2 \,s$ on both the earth and the planet.Thus, $T_e = T_p = 2 \,s$.This implies $\frac{\ell_e}{g_e} = \frac{\ell_p}{g_p}$, so $\ell_p = \ell_e \left( \frac{g_p}{g_e} ight)$.The acceleration due to gravity is $g = \frac{GM}{R^2}$.Given $M_p = 2M_e$ and $R_p = 2R_e$ (since diameter is double, radius is also double).Therefore, $g_p = \frac{G(2M_e)}{(2R_e)^2} = \frac{2GM_e}{4R_e^2} = \frac{1}{2} g_e$.Substituting this into the length equation: $\ell_p = 1 \,m 	imes \left( \frac{g_e/2}{g_e} ight) = 1 	imes 0.5 = 0.5 \,m$.

The length of a seconds pendulum is $1 \,m$ on the earth. If the mass and diameter of a planet are double that of the earth, then the length of the seconds pendulum on the planet will be: (in $\,m$)

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