The length of seconds pendulum is $1 \mathrm{~m}$ on the earth. If the mass and diameter of the planet is double than that of the earth, then the length of the seconds pendulum on the planet will be

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Aicharya · Accepted Answer

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{\ell_{e}}{g_{e}} 	imes \sqrt{\frac{g_{p}}{\ell_{p}}}} \quad$ But $T_{e}=T_{p}$ $	herefore \quad 1=\sqrt{\frac{\ell_{e} 	imes g_{p}}{g_{e} 	imes \ell_{p}}}$ But $g_{p}=\frac{G 	imes 2 M}{(2 R)^{2}}=\frac{G M 	imes 2}{4 R^{2}}=\frac{g}{2}$ $g_{e} \ell_{p}=\ell_{e} g_{p}$ $\ell_{p}=\ell_{e} \frac{g_{p}}{g_{e}}=1 	imes \frac{1}{2}=0.5 \mathrm{~m}$

The length of seconds pendulum is $1 \mathrm{~m}$ on the earth. If the mass and diameter of the planet is…

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