JEE Physics 2023 Question with Solution

Step-by-step solution

Standard Method

Given: The extension-load graph is a straight line through the origin and makes an angle of $45^\circ$ with the load axis. The wire length is $62.8 \, \text{cm}$ and diameter is $4 \, \text{mm}$.

Find: The value of $x$ in $Y = x \times 10^{10} \, \text{N m}^{-2}$.

From graph:

$$F = \Delta L$$

Young's modulus is

$$Y = \frac{FL}{A\Delta L}$$

Using $F = \Delta L$,

$$Y = \frac{L}{A}$$

Now, radius of wire $r = 2 \times 10^{-3} \, \text{m}$, so

$$A = \pi r^2 = \pi (2 \times 10^{-3})^2$$

Also,

$$L = 62.8 \times 10^{-2} \, \text{m}$$

Therefore,

$$Y = \frac{62.8 \times 10^{-2}}{\pi (2 \times 10^{-3})^2}$$ $$Y = 5 \times 10^{10} \, \text{N/m}^2$$

Therefore, the value of $x$ is $5$.

Common mistakes

• Using diameter in place of radius while calculating area. Since diameter is $4 \, \text{mm}$, the radius is $2 \, \text{mm}$. Use $A = \pi r^2$ with $r = 2 \times 10^{-3} \, \text{m}$.

• Not interpreting the $45^\circ$ slope correctly. Because the graph is between extension and load, slope $= \tan 45^\circ = 1$, so numerically $\Delta L = F$ from the graph relation used in the solution.

• Forgetting to convert centimetres and millimetres into SI units. Young's modulus must be calculated in $\text{N/m}^2$, so convert $62.8 \, \text{cm}$ to metres and $4 \, \text{mm}$ to metres before substitution.