A simple pendulum is being used to determine the value of gravitational acceleration g at a certain place.
Question:

A simple pendulum is being used to determine the value of gravitational acceleration $g$ at a certain place. The length of the pendulum is $25.0 \mathrm{~cm}$ and a stop watch with $1 \mathrm{~s}$ resolution measures the time taken for 40 oscillations to be $50 \mathrm{~s}$. The accuracy in $g$ is:

  1. (1) $5.40 \%$

  2. (2) $3.40 \%$

  3. (3) $4.40 \%$

  4. (4) $2.40 \%$


Correct Option: 3

Solution:

(3) Given, Length of simple pendulum, $l=25.0 \mathrm{~cm}$

Time of 40 oscillation, $T=50 \mathrm{~s}$

Time period of pendulum

$T=2 \pi \sqrt{\frac{\ell}{g}}$

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

$\Rightarrow$ Fractional error in $g=\frac{\Delta g}{g}=\frac{\Delta l}{l}+\frac{2 \Delta T}{T}$

$\Rightarrow \frac{\Delta g}{g}=\left(\frac{0.1}{25.0}\right)+2\left(\frac{1}{50}\right)=0.044$

$\therefore$ Percentage error in $g=\frac{\Delta g}{g} \times 100=4.4 \%$

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