The time period of oscillations of a second's pendulum on the surface of a planet having mass and radius double those of earth is?

4

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1

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\sqrt{2}

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2 \sqrt{2}

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Solution

The correct option is C $2 \sqrt{2}$ s
Time period of second's pendulum is given by $T = 2 π \sqrt{\frac{l}{g}}$
where; $g = \frac{G M_{p}}{R_{p}^{2}}$

$⟹ T = 2 π \sqrt{\frac{l R_{p}^{2}}{G M_{p}}}$
So, $T \propto \frac{R_{p}}{\sqrt{M_{p}}}$

On earth : $T_{e} = 2$
Given : $R_{p} = 2 R_{e}$ $M_{p} = 2 M_{e}$

Thus $\frac{T_{p}}{T_{e}} = \frac{R_{p}}{R_{e}} \sqrt{\frac{M_{e}}{M_{p}}}$

$⟹$ $\frac{T_{p}}{2} = 2 \times \frac{1}{\sqrt{2}}$

So, $T = 2 \sqrt{2} s$
Option D is correct.

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The time period of oscillations of a second's pendulum on the surface of a planet having mass and radius double those of earth is?

The correct option is C 2√2 sTime period of second's pendulum is given by T=2π√lgwhere; g=GMpR2p⟹ T=2π√lR2pGMpSo, T∝Rp√MpOn earth : Te=2Given : Rp=2Re Mp=2MeThus TpTe=RpRe√MeMp⟹ Tp2=2×1√2So, T=2√2 sOption D is correct.