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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
Options:
the planet is double than that of the earth, then the length of the seconds pendulum
on the planet will be
Solution:
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Verified Answer
The correct answer is:
$0.5 \mathrm{~m}$
$\frac{T_{e}}{T_{p}}=\sqrt{\frac{\ell_{e}}{g_{e}} \times \sqrt{\frac{g_{p}}{\ell_{p}}}} \quad$ But $T_{e}=T_{p}$
$\therefore \quad 1=\sqrt{\frac{\ell_{e} \times g_{p}}{g_{e} \times \ell_{p}}}$
But $g_{p}=\frac{G \times 2 M}{(2 R)^{2}}=\frac{G M \times 2}{4 R^{2}}=\frac{g}{2}$
$g_{e} \ell_{p}=\ell_{e} g_{p}$
$\ell_{p}=\ell_{e} \frac{g_{p}}{g_{e}}=1 \times \frac{1}{2}=0.5 \mathrm{~m}$
$\therefore \quad 1=\sqrt{\frac{\ell_{e} \times g_{p}}{g_{e} \times \ell_{p}}}$
But $g_{p}=\frac{G \times 2 M}{(2 R)^{2}}=\frac{G M \times 2}{4 R^{2}}=\frac{g}{2}$
$g_{e} \ell_{p}=\ell_{e} g_{p}$
$\ell_{p}=\ell_{e} \frac{g_{p}}{g_{e}}=1 \times \frac{1}{2}=0.5 \mathrm{~m}$
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