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The mass and diameter of a planet are twice those of earth. The period of oscillation of pendulum on this planet will be (If its second's pendulum on earth)

  • Option 1)

    \frac{1}{\sqrt{2}}\;s

  • Option 2)

    2\sqrt{2}\;s

  • Option 3)

    2 \;s

  • Option 4)

    \frac{1}{}2 \;s

 

Answers (1)

best_answer

As we know   g = \frac{GM}{R^{2}}

\Rightarrow \frac{g_{earth}}{g_{planet}} =\frac{M_{e}}{M_{p}} \times \frac{R_{p}^{2}}{R_{e}^{2}} \Rightarrow \frac{g_{earth}}{g_{planet}} = 2

Also,

 T\;\alpha\;\frac{1}{\sqrt{g}} \Rightarrow \frac{T_{e}}{T_{p}} = \sqrt{\frac{g_{p}}{g_{e}}} \Rightarrow \frac{2}{T_{p}} =\sqrt{\frac{1}{2}} \\*\Rightarrow T_{p} = 2\sqrt{2}s

 

Time period of a ball through tunnel in earth -

T= 2\pi \sqrt{\frac{R}{g}}\\=84.6\: min

- wherein

R= Radius of earth

g= acceleration due to gravity

 

 

 


Option 1)

\frac{1}{\sqrt{2}}\;s

This is incorrect.

Option 2)

2\sqrt{2}\;s

This is correct.

Option 3)

2 \;s

This is incorrect.

Option 4)

\frac{1}{}2 \;s

This is incorrect.

Posted by

divya.saini

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