Two slabs with square cross section of different materials (1,2) with equal sides (l) and thickness d_1 and d_2 such that d_2=2 d_1 and l>d_2 . Considering lower edges of these slabs are fixed to the floor, we apply equal shearing force on the narrow faces. The angle of deformation is \theta_2=2 \theta_1 . If the shear moduli of material 1 is 4 \times 10^9 ~N / m^2 , then shear moduli of material 2 is x \times 10^9 ~N / m^2 , where value of x is \underline\hspace2cm .

Properties of Matter PYQ — JEE Main 2025 · 4 Apr · Shift 1 · Q73 with Solution

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Correct answer: 2

For a slab under shearing force,

$G = \frac{\text{shear stress}}{\text{shear strain}} = \frac{F/A}{\theta}$

where for small deformation, shear strain $\approx \theta$ .

Thus,

$\theta = \frac{F}{A G}$

Equal shearing force is applied: $F_1 = F_2$
Square cross section with equal side $l$ , so area of narrow face is same for both: $A_1 = A_2 = l^2$

Hence,

$\theta \propto \frac{1}{G}$

Given,

$\theta_2 = 2\theta_1$

So,

$\frac{\theta_2}{\theta_1} = 2 = \frac{G_1}{G_2}$

Therefore,

$G_2 = \frac{G_1}{2}$

$G_1 = 4 \times 10^9\ \text{N/m}^2$

So,

$G_2 = \frac{4 \times 10^9}{2} = 2 \times 10^9\ \text{N/m}^2$

Hence,

$x = 2$

Stored correct answer = $1$

But from the standard shear modulus relation, the correct value is

$x = 2$