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Chapter 1: Problem 12

Error in the measurement of radius of sphere is \(2 \%\). Then error in the calculation of volume will be (A) \(1 \%\) (B) \(5 \%\) (C) \(3 \%\) (D) \(6 \%\)

Short Answer

Expert verified
The error in the calculation of the volume of the sphere will be \(6\%\) (option D).

Step by step solution

01

Find the formula for the volume of a sphere

The formula for the volume (V) of a sphere with radius (r) is given as follows: \[V = \frac{4}{3}\pi r^3\]
02

Differentiate the volume with respect to the radius

To find how the volume changes with the change in radius, differentiate V with respect to r: \[\frac{dV}{dr} = \frac{d}{dr} \left(\frac{4}{3}\pi r^3\right) = 4\pi r^2\]
03

Find the error in the volume

Let \(\Delta r\) denote the absolute error in the radius and \(\Delta V\) denote the absolute error in the volume. We can approximate the change in volume due to the change in radius as follows: \[\Delta V = \frac{dV}{dr} \Delta r \approx 4\pi r^2 \Delta r\]
04

Calculate the percentage error in the volume

The percentage error in the volume can be found by dividing the absolute error in the volume by the volume and multiplying by 100: \[% \Delta V = \frac{\Delta V}{V} \times 100 = \frac{4\pi r^2 \Delta r}{\frac{4}{3}\pi r^3} \times 100\] Simplify the expression: \[% \Delta V = \frac{3 \Delta r}{r} \times 100\] Since the error in the measurement of the radius is \(2\%\), we have: \[% \Delta V = \frac{3 \times 2}{1} \times 100\] \[% \Delta V = 6\%\] Therefore, the error in the calculation of the volume will be \(6\%\) (option D).

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