Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Units and Measurement
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
$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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App