Solution:
Volume of wire is \(V = \pi \left(\frac{d}{2}\right)^2 L ⇒ \frac{\Delta V}{V} = 2\frac{\Delta d}{d} + \frac{\Delta L}{L} = 2\left(\frac{0.001}{0.500}\right) + \frac{0.1}{75.0} \approx 0.53%\).
Dashrath measures the length of a wire using a meter scale with a least count of 1 mm and finds it to be L = 75.0 cm. He also measures diameter of thin wire using a screw gauge with a least count of 0.01 mm and finds it to be d = 0.500 cm. He uses these measurements to calculate the volume of wire. The maximum percentage error in volume of wire is nearly
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