If the error in the measurement of radius of a sphere is 2 %, then the error in the determination of volume of the sphere will be

Question

If the error in the measurement of radius of a sphere is 2 %, then the error in the determination of volume of the sphere will be (A) 8 % (B) 2 % (C) 4 % (D) 6 %

Tardigrade Admin · Accepted Answer

V=(4/3) π R3 ; ln V= ln ((4/3) π)+ ln R3 Differentiating, (dV/V)=3(dR/R) Error in the determination of the volume =3 × 2 % =6 %

Q. If the error in the measurement of radius of a sphere is $2%$ , then the error in the determination of volume of the sphere will be

$8%$

$2%$

$4%$

$6%$

$V = \frac{4}{3} π R^{3}; ln V = ln (\frac{4}{3} π) + ln R^{3}$
Differentiating, $\frac{d V}{V} = 3 \frac{d R}{R}$
Error in the determination of the volume
$= 3 \times 2%$
$= 6%$

Q. If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be

8%

2%

4%

6%