Q151 / 200 · MHT CET - Medical 2026 (24 April Shift 2) · Physics
The error in measurement of radius of sphere is $0.2\ \%$, then the percentage error in its volume is
- Correct answerA$0.6\ \%$
- B$0.8\ \%$
- C$1.2\ \%$
- D$1.0\ \%$
Solution
The volume of a sphere is given by $V = \dfrac{4}{3}\pi r^3$
Taking relative error on both sides, we get $\dfrac{\Delta V}{V} = 3 \dfrac{\Delta r}{r}$
Percentage error in volume is $\dfrac{\Delta V}{V} \times 100 = 3 \left(\dfrac{\Delta r}{r} \times 100\right)$
Given percentage error in radius is $0.2\ \%$
Therefore, percentage error in volume $= 3 \times 0.2\ \% = 0.6\ \%$
Answer: $0.6\ \%$
Taking relative error on both sides, we get $\dfrac{\Delta V}{V} = 3 \dfrac{\Delta r}{r}$
Percentage error in volume is $\dfrac{\Delta V}{V} \times 100 = 3 \left(\dfrac{\Delta r}{r} \times 100\right)$
Given percentage error in radius is $0.2\ \%$
Therefore, percentage error in volume $= 3 \times 0.2\ \% = 0.6\ \%$
Answer: $0.6\ \%$
Hence, the correct answer is option: 'A'.