Physics And Measurement Question 164

Question: The period of oscillation of a simple pendulum is given by $ T=2\pi \sqrt{\frac{l}{g}} $ where l is about 100 cm and is known to have 1mm accuracy. The period is about 2s. The time of 100 oscillations is measured by a stop watch of least count 0.1 s. The percentage error in g is

Options:

A) 0.1%

B) 1%

C) 0.2%

D) 0.8%

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Answer:

Correct Answer: C

Solution:

$ T=2\pi \sqrt{l/g} $

$ \Rightarrow T^{2}=4{{\pi }^{2}}l/g $

$ \Rightarrow g=\frac{4{{\pi }^{2}}l}{T^{2}} $

Here % error in l = $ \frac{1mm}{100cm}\times 100=\frac{0.1}{100}\times 100=0.1% $

and % error in T = $ \frac{0.1}{2\times 100}\times 100=0.05% $

% error in g = % error in l + 2(% error in T)

$ =0.1+2\times 0.05 $ = 0.2 %



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