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In the experiments, the following observations were recorded: $L=2.820\ m,$ $I=0.087\ cm$ , $M=3\ kg$ . Diameter, $D=0.041\ cm$ . Taking $g=9.81\ \frac{m}{s^2}.$ Using the formula,

$Y=\frac{4MgL}{{\pi D}^2l},$ the allowable error Y is:
Option: 1
$7.96%$

Option: 2
$4.56%$

Option: 3
$6.50%$

Option: 4
$8.42%$

Answers (1)

best_answer

$Y=\frac{4MgL}{{\pi D}^2l}$

So maximum permissible error in $Y=\frac{\Delta Y}{Y}\times 100$

$\left ( \frac{\Delta M}{M} +\frac{\Delta g}{g}+\frac{\Delta L}{L}+\frac{2\Delta D}{D}+\frac{\Delta l}{l}\right )\times 100$

$\left ( \frac{1}{300}+\frac{1}{981}+\frac{1}{2820} +2\times \frac{1}{41}+\frac{1}{87}\right )\times 100$

$0.065\times100=6.5$

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shivangi.bhatnagar

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