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A pendulum has time period T. If it is taken on another planet having acceleration due to gravity half and mass 9 times that of earth then its time period on the other planet will be

  • Option 1)

    \sqrt{T}

  • Option 2)

    T

  • Option 3)

    T^{\frac{1}{3}}

  • Option 4)

    \sqrt{2}T

 

Answers (1)

best_answer

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

 

Time period of pendulum of large length but small amplitude -

T= 2\pi \sqrt{\frac{1}{g\left ( \frac{1}{l}+\frac{1}{R} \right )}}

- wherein

l= length of pendulum

g= acceleration due to gravity.

R= Radius of earth

 

 

 


Option 1)

\sqrt{T}

This is incorrect.

Option 2)

T

This is incorrect.

Option 3)

T^{\frac{1}{3}}

This is incorrect.

Option 4)

\sqrt{2}T

This is correct.

Posted by

Avinash

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