If the error in the measurement of the radius of the sphere is 1%
| If the error in the measurement of the radius of the sphere is 1%, then the error in the measurement in its volume is
A. 1%
B. 3%
C. 6%
D. 9%
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Concept:
The volume of a sphere is given by:
\(V=\frac{4}{3}\pi R^3\)
R = Radius of the sphere
The change in Volume is obtained by differentiating the above w.r.t. the radius R as:
\(dV=\frac{4}{3} \pi (3R^2)dR\)
Multiplying both the RHS and LHS by R, we get:
\(dV.R=3.\frac{4}{3} \pi (R^3)dR.\)
\(dV.R=3.V.dR\)
The above is rearranged as:
\(\frac{dV}{V}=3\frac{dR}{R}\)
Calculation:
Given \(\frac{dR}{R}\times 100=1\)%
The percentage change in the volume will be given by:
\(\frac{dV}{V}=3\frac{dR}{R}\times 100=3\times1\)%
\(\frac{dV}{V}=3\)%