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The period of oscillation of a simple  pendulum isT=2\pi \sqrt{\frac{L}{8}} Measured value of L is 20.0 cm known to 1 mm accuracy   and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resolution.  The accuracy in the determination of g is

  • Option 1)

    2%

  • Option 2)

    3%

  • Option 3)

    1%

  • Option 4)

    5%

 

Answers (1)

best_answer

Aw we learnt in

Time period of oscillation of simple pendulum -

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

- wherein

l = length of pendulum 

g = acceleration due to gravity.

 

 T=2\pi\sqrt{\frac{l}{g}}\ \; \Rightarrow\ \;g=\frac{4\pi^{2}L}{T^{2}}

Taking log on both side

l\ ng=ln(4\pi^{2})+ln\ L-2\ lnT

Taking differential 

\frac{\Delta g}{g}=\frac{\Delta L}{L}-\frac{2\Delta T}{T}      or     \frac{\Delta g}{g}=\frac{\Delta L}{L}+\frac{2\Delta T}{T}

Now, \frac{\Delta g}{g}\times100=\frac{\Delta L}{L}\times100+\frac{2\Delta T}{T}\times100

Given L = 20 cm \Delta L =0.1 cm\ \;\Rightarrow\ \; \frac{\Delta L}{L}=\frac{0.1}{20}

\Delta T=\frac{1\ s}{100}=0.01\ sec

T=\frac{90}{100}=0.9\ sec

\frac{\Delta g}{g}\times100=\frac{0.01}{20}\times100+\frac{2\times}{0.9}\times100

\simeq(0.5 +2.2)% = 2.7% \simeq 3%

Correct option is 2.

 


Option 1)

2%

This is an incorrect option.

Option 2)

3%

This is the correct option.

Option 3)

1%

This is an incorrect option.

Option 4)

5%

This is an incorrect option.

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