CUET Preparation Today
Approximate percentage increase in the volume of sphere if its radius is increased by 3% is |
3% 9% 4% 6% |
9% |
The correct answer is Option (2) - 9% r → radius, v → volume $v=\frac{4}{3}πr^3$ so $\frac{dv}{dr}=4πr^2$ so $Δv=\frac{dv}{dr}×Δr$ $=4πr^2×\frac{3}{100}×r$ $=\frac{9}{100}×\frac{4}{3}πr^3=\frac{9}{100}×v$ or 9% |
