QJEEM24Phb460: JEE Main 2024 Physics - Class 11 JEEM24Phb460 Solution: Young's modulus is determined by the equation given by where is …

To determine the percentage error in Young's modulus, we need to first understand the propagation of errors in the given formula.

Given the equation:

$\mathrm{Y}=49000 \frac{\mathrm{M}}{\mathrm{l}} \frac{\mathrm{dyne}}{\mathrm{cm}^2}$

where:

• $M$ is the mass (with its value given as $500 \mathrm{~g}$ )
• $l$ is the extension (with its value given as $2 \mathrm{~cm}$ )

The errors in the measurements are determined by the smallest scale divisions on the graph paper, which are:

• $5 \mathrm{~g}$ for the load axis
• $0.02 \mathrm{~cm}$ for the extension axis

To find the percentage error in Young's modulus ( $Y$ ), we need to compute the relative errors in the measurements $M$ and $l$ , and then propagate these errors through the given formula.

The relative error in $M$ is:

$\frac{\Delta M}{M} = \frac{5 \mathrm{~g}}{500 \mathrm{~g}} = 0.01$

The relative error in $l$ is:

$\frac{\Delta l}{l} = \frac{0.02 \mathrm{~cm}}{2 \mathrm{~cm}} = 0.01$

Since $Y$ is proportional to $M$ and inversely proportional to $l$ , the overall percentage error in $Y$ is the sum of the percentage errors in $M$ and $l$ :

$\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta l}{l} = 0.01 + 0.01 = 0.02$

To express this as a percentage, we multiply by 100:

$\text{Percentage error in } Y = 0.02 \times 100 = 2\%$

Thus, the percentage error in Young's modulus $Y$ is:

Option A: 2%