QJEEM25Phbf1b: JEE Main 2025 Physics - Class 11 JEEM25Phbf1b Solution: The maximum percentage error in the measurment of density of a…

To determine the maximum percentage error in the density of the wire, follow these steps:

The density of the wire is given by the formula for a cylinder:

$\rho = \frac{m}{\pi r^2 l}$

When calculating the maximum percentage (relative) error, you add the relative errors from each variable, taking into account the power to which each variable is raised. Thus, for density:

$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$

Here:

$\frac{\Delta m}{m}$ is the relative error in the mass.

$2\frac{\Delta r}{r}$ is due to the radius being squared.

$\frac{\Delta l}{l}$ is the relative error in the length.

Now plug in the given values:

Mass: $m = 0.60 \, \text{g}$ with an error $\Delta m = 0.003 \, \text{g}$

$\frac{\Delta m}{m} = \frac{0.003}{0.60} = 0.005$ or 0.5%.

Radius: $r = 0.50 \, \text{cm}$ with an error $\Delta r = 0.01 \, \text{cm}$

$\frac{\Delta r}{r} = \frac{0.01}{0.50} = 0.02$ or 2%. Since the radius is squared in the formula, multiply by 2:

$2\frac{\Delta r}{r} = 2 \times 0.02 = 0.04$ or 4%.

Length: $l = 10.00 \, \text{cm}$ with an error $\Delta l = 0.05 \, \text{cm}$

$\frac{\Delta l}{l} = \frac{0.05}{10.00} = 0.005$ or 0.5%.

Sum these contributions to find the overall maximum relative error:

$\frac{\Delta \rho}{\rho} = 0.005 + 0.04 + 0.005 = 0.05$

Converting this relative error into a percentage gives:

$0.05 \times 100\% = 5\%$

Thus, the maximum percentage error in the density of the wire is 5%, which corresponds to Option C.