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(C) The formula for Young's modulus is $Y = \frac{4MgL}{\pi D^2 l}$.The relative error in $Y$ is given by $\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \

Vedclass · Accepted Answer

$6.5$

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(C) The formula for Young's modulus is $Y = \frac{4MgL}{\pi D^2 l}$.The relative error in $Y$ is given by $\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta g}{g} + \frac{\Delta L}{L} + 2\frac{\Delta D}{D} + \frac{\Delta l}{l}$.Given values are $M = 3.00 \, kg$ $(\Delta M = 0.01 \, kg)$,$L = 2.820 \, m$ $(\Delta L = 0.001 \, m)$,$l = 0.087 \, cm$ $(\Delta l = 0.001 \, cm)$,$D = 0.041 \, cm$ $(\Delta D = 0.001 \, cm)$.Substituting these values into the error formula:$\frac{\Delta Y}{Y} 	imes 100 = \left( \frac{0.01}{3.00} + \frac{0.001}{2.820} + 2 	imes \frac{0.001}{0.041} + \frac{0.001}{0.087} ight) 	imes 100$.Note: The term $\frac{\Delta g}{g}$ is usually neglected if $g$ is taken as a constant. Calculating the values:$\frac{\Delta Y}{Y} 	imes 100 \approx (0.00333 + 0.00035 + 0.04878 + 0.01149) 	imes 100 \approx 0.06395 	imes 100 \approx 6.4\%$.Rounding to the nearest provided option,the maximum permissible error is $6.5\%$.

In an experiment,the following observations were recorded: $L = 2.820 \, m, M = 3.00 \, kg, l = 0.087 \, cm$,Diameter $D = 0.041 \, cm$. Taking $g = 9.81 \, m/s^2$ and using the formula $Y = \frac{4MgL}{\pi D^2 l}$,the maximum permissible error in $Y$ is ......... $\%$.

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