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A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is :
Option: 1 2.40%
Option: 2 5.40%
Option: 3 4.40%
Option: 4 3.40%

Answers (1)





  • The time period of oscillation of simple pendulum (T)-


     T=2\pi \sqrt{\frac{l}{g}}


m=mass of the bob 

l = length of pendulum 

g = acceleration due to gravity.


  \begin{array}{l}{\frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{1}{2}\left(\frac{\Delta \mathrm{g}}{\mathrm{g}}+\frac{\Delta \mathrm{L}}{\mathrm{L}}\right)} \\ {\frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{2 \Delta \mathrm{T}}{\mathrm{T}}+\frac{\Delta \mathrm{L}}{\mathrm{L}} ; \quad=2\left(\frac{1}{50}\right)+\frac{0.1}{25.0}} \\ {=4.4 \%}\end{array}


Note - We are adding the error just for calculating maximum possible error. 

Hence the correct option is (3).

Posted by

vishal kumar

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