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(A) Given the relation $m = \pi 	an 	heta$.Taking the differential of both sides,we get $dm = \pi \sec^2 	heta \, d	heta$.The relative error is gi

Vedclass · Accepted Answer

$45$

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(A) Given the relation $m = \pi 	an 	heta$.Taking the differential of both sides,we get $dm = \pi \sec^2 	heta \, d	heta$.The relative error is given by $\frac{dm}{m} = \frac{\pi \sec^2 	heta \, d	heta}{\pi 	an 	heta}$.Simplifying the expression: $\frac{dm}{m} = \frac{\sec^2 	heta}{	an 	heta} d	heta = \frac{1}{\cos^2 	heta} \cdot \frac{\cos 	heta}{\sin 	heta} d	heta = \frac{1}{\sin 	heta \cos 	heta} d	heta$.Multiplying the numerator and denominator by $2$,we get $\frac{dm}{m} = \frac{2 d	heta}{2 \sin 	heta \cos 	heta} = \frac{2 d	heta}{\sin 2	heta}$.For the percentage error to be minimum,the denominator $\sin 2	heta$ must be maximum.The maximum value of $\sin 2	heta$ is $1$,which occurs when $2	heta = 90^\circ$.Therefore,$	heta = 45^\circ$.

The percentage error in the measurement of a physical quantity $m$ given by $m = \pi \tan \theta$ is minimum when $\theta = \dots \dots \dots \dots \dots ^\circ$ (Assume that the error in $\theta$ remains constant).

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