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# While measuring the volume of sphere an error of $1.2 \%$ is committed in the measurement of radius. What percentage error is introduced in the measurement of volume ?

$\begin{array}{1 1}(A)\;2.4\% \\(B)\;3.6 \% \\(C)\;9.8\% \\(D)\;2 \% \end{array}$

$V= \large\frac{4}{3} $$\pi R^3$$ =kR^3$
Where $k\bigg(= \large\frac{4}{3} $$\pi \bigg) is a constant \Delta V= (V+ \ \Delta V)-V \qquad= k(R+\Delta R)^3 -kR^3 \qquad= kR^3 \bigg(1+ \large\frac{3 \Delta R}{R} \bigg)$$-k R^3$
$\qquad= kR^3 .\large\frac{3 \Delta R}{R}$
So,$\large\frac{\Delta V}{V} = \frac{3 \Delta R}{R}$
$\qquad= 3 \times 1.2 =3.6\%$
Hence B is the correct answer.