JEE Main 2025 Apr 4 Shift 1, Physics Q22: Properties of Solids

JEE Main2025Apr 4, Shift 1Physics

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 $θ_{2} = 2 θ_{1}$ . If the shear moduli of material 1 is $4 \times 1 0^{9}$ N/m $^{2}$ , then shear moduli of material 2 is $x \times 1 0^{9}$ N/m $^{2}$ , where value of $x$ is ____.

Your answer

Solution

For small shear, the relation is $tan θ \approx θ = \frac{Δ x}{d} = \frac{σ}{η}$ where $σ = F / A$ is the shear stress and $η$ is the shear modulus.

Using

θ = F / (l d η)

and equating

θ_{2} = 2 θ_{1}

$\frac{F}{l d _{2} η _{2}} = 2 \cdot \frac{F}{l d _{1} η _{1}}$

With $d_{2} = 2 d_{1}$ : $\frac{1}{2 d _{1} η _{2}} = \frac{2}{d _{1} η _{1}}$ , hence $η_{2} = \frac{η _{1}}{4} = \frac{4 \times 1 0 ^{9}}{4} = 1 \times 1 0^{9}$ N/m $^{2}$ .

Therefore $x = 1$ .

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JEE Main 2025 Apr 4 Shift 1, Physics Q22: Properties of Solids

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