Time period of a simple pendulum is given by T =2 π √{frac{ell}{ g }} . measured value of length

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1.Units, Dimensions and Measurement

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The time period of a simple pendulum is given by $T =2 \pi \sqrt{\frac{\ell}{ g }}$. The measured value of the length of pendulum is $10\, cm$ known to a $1\, mm$ accuracy. The time for $200$ oscillations of the pendulum is found to be $100$ second using a clock of $1s$ resolution. The percentage accuracy in the determination of $'g'$ using this pendulum is $'x'$. The value of $'x'$ to the nearest integer is ...........$\%$

A

$2$

B

$3$

C

$5$

D

$4$

(JEE MAIN-2021)

Solution

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

$\frac{\Delta g}{g}=\frac{\Delta \ell}{\ell}+2 \frac{\Delta T}{T}=\frac{0.1}{10}+2\left(\frac{\frac{1}{200}}{0.5}\right)$

$\frac{\Delta g}{g}=\frac{1}{100}+\frac{1}{50}$

$\frac{\Delta g}{g} \times 100=3 \%$

Standard 11

Physics

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