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A reactangluar block of mass m and area of cross-section A, floats in a liquid of density \rho. It is given a small vertical displacement from equillibrium it undergoes oscillation with time period T. Then 

  • Option 1)

    T \alpha \frac{1}{\rho }

  • Option 2)

    T \alpha \frac{1}{\sqrt{m} }

  • Option 3)

    T \alpha \sqrt{\rho }

  • Option 4)

    T \alpha \frac{1}{\sqrt{A}}

 

Answers (1)

best_answer

T = 2\pi\sqrt{\frac{m}{A\rho g}} \Rightarrow T\;\alpha\;\frac{1}{\sqrt{A}}

 

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 \alpha \frac{1}{\rho }

This is incorrect.

Option 2)

T \alpha \frac{1}{\sqrt{m} }

This is incorrect.

Option 3)

T \alpha \sqrt{\rho }

This is incorrect.

Option 4)

T \alpha \frac{1}{\sqrt{A}}

This is correct.

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prateek

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