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The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be :
Option 1)
$\frac{\sqrt{3}}{2} s$
Option 2)
$\frac{3}{2} s$
Option 3)
$\frac{2}{\sqrt{3}} s$
Option 4)
$2\sqrt{3} s$

Answers (1)

best_answer

Time period of oscillation of simple pendulum -

$T=2\pi \sqrt{\frac{l}{g}}$

- wherein

l = length of pendulum

g = acceleration due to gravity.

$M_{P}=3M_{e}$

$D_{P}=3D_{e}$

$\Rightarrow r_{P}=3r_{e}$

$T_{e}=2s$

$T_{P}=?$

$T\propto \frac{1}{\sqrt{9}}$

& $g=\frac{GM}{R^{2}}$

$\Rightarrow \frac{T_{p}}{T_{e}}=\sqrt{\frac{g_{e}}{g_{p}}}=\sqrt{\frac{M_{e}}{\left ( R_{e} \right )^{2}}\times \frac{\left ( R_{p} \right )^{2}}{M_{p}}}$

$\frac{T_{p}}{T_{e}}=\sqrt{\left ( \frac{M_{e}}{M_{p}} \right )\times \left ( \frac{r_{p}}{r_{e}} \right )^{2}}$

$=\sqrt{\frac{1}{3}\times \left ( 3 \right )^{2}}$

$\frac{T_{p}}{T_{e}}=\sqrt{3}$

$T_{p}=\sqrt{3}T_{e}$

$T_{p}=\sqrt{3}\times 2$

$T_{p}=2\sqrt{3}s$

Option 1)

$\frac{\sqrt{3}}{2} s$

Option 2)

$\frac{3}{2} s$

Option 3)

$\frac{2}{\sqrt{3}} s$

Option 4)
$2\sqrt{3} s$

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