Dashrath measures the length of a wire using a meter scale with a least count of \( 1\text{ mm} \) and finds it to be \( L = 75.0\text{ cm} \). He also measures diameter of thin wire using a screw gauge with a least count of \( 0.01\text{ mm} \) and finds it to be \( d = 0.500\text{ cm} \). He uses these measurements to calculate the volume of wire.
The maximum percentage error in volume of wire is nearlySolution:
Volume \( V = \frac{\pi d^2 L}{4} ⇒ \frac{\Delta V}{V} = 2 \frac{\Delta d}{d} + \frac{\Delta L}{L} \). Here, \( \Delta d = 0.001\text{ cm} \) and \( \Delta L = 0.1\text{ cm} \). Thus, \( \frac{\Delta V}{V} = 2\left(\frac{0.001}{0.500}\right) + \frac{0.1}{75} = 0.4% + 0.13% = 0.53% \).
Leave a Reply