If the error in the measurement of the radius of the sphere is 1%, then the error in the measurement in its volume is
3%
Understanding how errors propagate in measurements is crucial in physics and engineering. This problem asks us to determine the percentage error in the volume of a sphere when there is a known percentage error in the measurement of its radius.
The formula for the volume (\(V\)) of a sphere with radius (\(r\)) is given by:
\(V = \frac{4}{3}\pi r^3\)
Here, \(\frac{4}{3}\) and \(\pi\) are constants, meaning they have no error in their values for this calculation. The only quantity subject to measurement error is the radius \(r\).
When a physical quantity \(Y\) depends on another measured quantity \(X\) raised to a power, such as \(Y = kX^n\) (where \(k\) is a constant and \(n\) is the power), the fractional error in \(Y\) is related to the fractional error in \(X\) by the following principle:
\(\frac{\Delta Y}{Y} = n \frac{\Delta X}{X}\)
Where \(\Delta Y\) is the absolute error in \(Y\), \(\Delta X\) is the absolute error in \(X\), \(\frac{\Delta Y}{Y}\) is the fractional error in \(Y\), and \(\frac{\Delta X}{X}\) is the fractional error in \(X\).
To convert fractional error to percentage error, we simply multiply by 100%:
\(\frac{\Delta Y}{Y} \times 100\% = n \left(\frac{\Delta X}{X} \times 100\%\right)\)
Let's apply this principle to the volume of the sphere:
Therefore, the error in the measurement of its volume is 3%.
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