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If the length of second's pendulum is decreased by 2%, how many seconds it will lose per day

 

 

 

 

  • Option 1)

    3927 s

  • Option 2)

    3727 s

  • Option 3)

    3427 s

  • Option 4)

    864 s

 

Answers (1)

best_answer

T\; \alpha\; \sqrt{l} \Rightarrow \frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l}=\frac{0.02}{2} = 0.01 \Rightarrow \Delta T = 0.01T

Loss of time per day = 0.01 \times 24\times 60\times 60 = 864 s

 

Time period of oscillation of simple pendulum -

T=2\pi \sqrt{\frac{l}{g}}

- wherein

l = length of pendulum 

g = acceleration due to gravity.

 

 

 


Option 1)

3927 s

This is incorrect.

Option 2)

3727 s

This is incorrect.

Option 3)

3427 s

This is incorrect.

Option 4)

864 s

This is correct.

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Aadil

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