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The metallic bob of a simple pendulum has the relative density P. The time period of this pendulum is T. If the metallic bob is immersed in water, then the new time period is given by

  • Option 1)

    T \frac{P-1}{P}

  • Option 2)

    T \frac{P}{P-1}

  • Option 3)

    T \sqrt{\frac{P-1}{P}}

  • Option 4)

    T \sqrt{}\frac{P}{P-1}

 

Answers (1)

best_answer

When the bob is immersered in water its effective weight 

\\*= \left[mg - \frac{m}{\rho}g \right ] = mg \left (\frac{\rho -1}{\rho} \right ) \\*\therefore g_{eff} = g\left (\frac{\rho -1}{\rho} \right ) \\* \frac{T'}{T} = \sqrt{\frac{g}{g_{eff}}} \Rightarrow T' =T\sqrt{\left (\frac{\rho -1}{\rho} \right )}

 

Time period of pendulum in a liquid -

T= 2\pi \sqrt{\frac{l}{g\left ( 1-\frac{\rho }{\sigma } \right )}}
 

- wherein

\rho = density of liquid

\sigma = density of  bob

l= length of pendulum.

 

 

 


Option 1)

T \frac{P-1}{P}

This is incorrect.

Option 2)

T \frac{P}{P-1}

This is incorrect.

Option 3)

T \sqrt{\frac{P-1}{P}}

This is incorrect.

Option 4)

T \sqrt{}\frac{P}{P-1}

This is correct.

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