In the expression for time period $T$ of simple pendulum $T=2 \pi \sqrt{\frac{l}{g}}$, if the percentage error in time period $T$ and length $l$ are $2 \%$ and $2 \%$ respectively then percentage error in acceleration due to gravity $g$ is equal to ......... $\%$

  • A

    $8$

  • B

    $2$

  • C

    $4$

  • D

    $6$

Similar Questions

A student determined Young's Modulus of elasticity using the formula $Y=\frac{M g L^{3}}{4 b d^{3} \delta} .$ The value of $g$ is taken to be $9.8 \,{m} / {s}^{2}$, without any significant error, his observation are as following.

Physical Quantity Least count of the Equipment used for measurement Observed value
Mass $({M})$ $1\; {g}$ $2\; {kg}$
Length of bar $(L)$ $1\; {mm}$ $1 \;{m}$
Breadth of bar $(b)$ $0.1\; {mm}$ $4\; {cm}$
Thickness of bar $(d)$ $0.01\; {mm}$ $0.4 \;{cm}$
Depression $(\delta)$ $0.01\; {mm}$ $5 \;{mm}$

Then the fractional error in the measurement of ${Y}$ is

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An experiment measures quantities $a, b$ and $c$, and quantity $X$ is calculated from $X=a b^{2} / c^{3}$. If the percentage error in $a$, $b$ and $c$ are $\pm 1 \%, \pm 3 \%$ and $\pm 2 \%$, respectively, then the percentage error in $X$ will be