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In an experiment to find acceleration due to gravity $(\mathrm{g})$ using simple pendulum, time period of $0.5 \mathrm{~s}$ is measured from time of 100 oscillation with a watch of $1 \mathrm{~s}$ resolution. If measured value of length is $10 \mathrm{~cm}$ known to $1 \mathrm{~mm}$ accuracy, The accuracy in the determination of $\mathrm{g}$ is found to be $\mathrm{x %}$ . The value of $\mathrm{x }$ is__________.
Option: 1
5

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{T_1=0.5 \mathrm{~s}}$ $\mathrm{\text { for } 100 \text { oscillations, }}$

$\mathrm{\Delta T=1 s}$ $\mathrm{\quad T=100 T_1=50\: s .}$

$\mathrm{\Delta \ell=1 mm=0.1\: cm}$

$\mathrm{\ell=10 \mathrm{~cm}}$

$\mathrm{\frac{\Delta g}{g} \times 100=x }$

$\mathrm{\frac{\Delta g}{g}=\frac{x}{100} }$

$\mathrm{T=2 \pi \sqrt{\frac{\ell}{9}} }$

$\mathrm{g \: \alpha \: \frac{\ell}{T^2}}$

$\mathrm{\frac{\Delta g}{g} =\frac{\Delta \ell}{\ell}+\frac{2 \Delta T}{T} }$

$\mathrm{\frac{x}{100} =\frac{0.1}{10}+2 \times \frac{1}{50} }$

$\mathrm{\frac{x}{100} =\frac{1}{100}+\frac{4}{100} }$

$\mathrm{x =5}$

The value of $\mathrm{x \: is\: 5}$

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