JEE Physics 2026 Question with Solution

Given: Two wires $A$ and $B$ have lengths $L_A = 6.0 \, \text{cm}$ and $L_B = 5.4 \, \text{cm}$, and cross-sectional areas $A_A = 3.0\times10^{-5}$ and $A_B = 4.5\times10^{-5}$. The same load acts on both wires and the extension is the same.

Find: The value of $x$ if $\frac{Y_A}{Y_B} = \frac{x}{3}$.

Young’s modulus is defined as:

$$Y=\frac{\text{Stress}}{\text{Strain}}=\frac{FL}{A\Delta L}$$

For the same load $F$ and the same extension $\Delta L$,

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

Therefore,

$$\frac{Y_A}{Y_B}=\frac{L_A/A_A}{L_B/A_B}=\frac{L_AA_B}{L_BA_A}$$

Substituting the given values,

$$\frac{Y_A}{Y_B}=\frac{6.0\times4.5}{5.4\times3.0}=\frac{27}{16.2}=\frac{5}{3}$$

Given that

$$\frac{Y_A}{Y_B}=\frac{x}{3}$$

So,

$$x=5$$

the solution states option B, but the working shown gives $\frac{Y_A}{Y_B}=\frac{5}{3}$, hence the defensible answer from the working is option A.

Given: Same load and same extension for both wires.

Find: Compare Young’s moduli using geometry.

From

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

when $F$ and $\Delta L$ are identical for both wires,

$$Y_A : Y_B = \frac{L_A}{A_A} : \frac{L_B}{A_B}$$

Hence,

$$Y_A : Y_B = L_AA_B : L_BA_A$$

Now substitute:

$$Y_A : Y_B = 6.0\times4.5 : 5.4\times3.0 = 27 : 16.2$$

Dividing both terms by $5.4$,

$$Y_A : Y_B = 5 : 3$$

So the ratio is $5:3$. Comparing with $x:3$ gives $x=5$.

Therefore, the correct value is $5$, corresponding to option A.