If the percentage error in measuring the radius of a sphere is 2%, what is the percentage error in its calculated volume (assuming small errors)?
6%
The volume of a sphere is given by \(V = \frac{4}{3}\pi r^3\), so the percentage error in volume relates to the percentage error in radius through the power rule for error propagation.
Since \(V \propto r^3\), the relative error is \(\frac{\Delta V}{V} = 3 \times \frac{\Delta r}{r}\).
Substituting the given percentage error in radius, \(\frac{\Delta V}{V} \times 100 = 3 \times 2\% = 6\%\).
Hence, the percentage error in the calculated volume is 6%.
Important Questions from Measuring Instruments
Which one of the following weather conditions indicates a sudden fall in barometer reading?
______ is used to measure the pressure inside the eyes of a person.
A ‘gnomon’ is a part of a ________.
An instrument called the ______ measures the electric current in a circuit.
Seismograph is used to measure:
