Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationField & ForceEasy2 minPYQ_2020

PhysicsEasysingle choice

The mass and the diameter of a planet are three times of the respective values for the earth. If the time period of oscillation of a simple pendulum on earth is $2 s$ , then the time period of oscillation of the same pendulum on the surface of the given planet would be

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Answer:

Solution:

We know that,

$g = \frac{G M}{R^{2}}$

$\Rightarrow \frac{g_{p}}{g_{e}} = \frac{M_{p}}{M_{e}} \times {(\frac{R_{e}}{R_{p}})}^{2} = 3 \times \frac{1}{9} = \frac{1}{3}$

Also,
$T = 2 π \sqrt{\frac{1}{g}}$

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

Time period at earth for seconds pendulum $= 2 s$

$∴ T_{p} = 2 \sqrt{3} s$

Subject:PhysicsTopic:GravitationSubtopic:Field & Force

⏱ 2mℹ️ Source: PYQ_2020

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