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(B) Given the relation $T = \frac{rhg}{2} 	imes 10^3$.Since $r = D/2$,the relative error in $r$ is the same as the relative error in $D$,i.e.,$\frac{

Vedclass · Accepted Answer

$1.5$

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(B) Given the relation $T = \frac{rhg}{2} 	imes 10^3$.Since $r = D/2$,the relative error in $r$ is the same as the relative error in $D$,i.e.,$\frac{\Delta r}{r} = \frac{\Delta D}{D}$.The least count for both $D$ and $h$ is $\Delta D = \Delta h = 0.01 	imes 10^{-2} \; m$.The relative error in $T$ is given by $\frac{\Delta T}{T} = \frac{\Delta r}{r} + \frac{\Delta h}{h}$.Substituting the values: $\frac{\Delta T}{T} = \frac{0.01 	imes 10^{-2}}{1.25 	imes 10^{-2}} + \frac{0.01 	imes 10^{-2}}{1.45 	imes 10^{-2}} = \frac{0.01}{1.25} + \frac{0.01}{1.45}$.Percentage error $= \left( \frac{0.01}{1.25} + \frac{0.01}{1.45} ight) 	imes 100 = \frac{1}{1.25} + \frac{1}{1.45} = 0.8 + 0.6896 \approx 1.5 \%$.Thus,the possible error in surface tension is $1.5 \%$.

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