JEE Main 2026 — Units and Dimensions Question with Solution

The volume of a sphere in terms of its diameter $D$ is given by $V=\frac{\pi }{6}{D}^3$

Taking the natural logarithm on both sides, we get $\ln V=\ln \left(\frac{\pi }{6}\right)+3\ln D$

Differentiating for small errors, we obtain $\frac{\Delta V}{V}=3\frac{\Delta D}{D}$

Multiplying by $100$ to find the percentage error, $\frac{\Delta V}{V}\times 100=3\left(\frac{\Delta D}{D}\times 100\right)$

Given that the percentage error in the diameter is $2\%$, we have $\frac{\Delta D}{D}\times 100=2$

Therefore, the percentage error in the volume is $3\times 2=6\%$

Answer: $6$