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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} $

1 Answer

$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.
answered Jul 11, 2014 by meena.p

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