JEE Main 2025 Jan 22 Shift 2, Physics Q3: Units And Dimensions

Density of a cylindrical wire: $\rho =\frac{m}{\pi R^2\ell }$.

For maximum percentage error, errors add directly (with the power of each quantity as the multiplier):

$\frac{\Delta \rho }{\rho }=\frac{\Delta m}{m}+2\frac{\Delta R}{R}+\frac{\Delta \ell }{\ell }$.

Substituting:

$\frac{\Delta \rho }{\rho }\times 100=\left(\frac{0.003}{0.60}+2\cdot \frac{0.01}{0.50}+\frac{0.05}{10.00}\right)\times 100$

$=(0.5+4.0+0.5)\%=5\%$.