JEE Main 2025PhysicsMechanical Properties of SolidsHardNumerical

JEE Main 2025Mechanical Properties of Solids Question with Solution

JEE Main 2025 (4 Apr Shift 1)

Question

Two slabs with square cross section of different materials with equal sides and thickness and such that and . 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 . If the shear moduli of material 1 is , then shear moduli of material 2 is , where value of is ________.

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Step-by-step explanation

Deformation angle
$\begin{aligned}
& 2 \theta_1=\theta_2 \\ & \Rightarrow 2 \frac{\sigma_1}{\eta_1}=\frac{\sigma_2}{\eta_2}
\end{aligned}\begin{aligned} & \Rightarrow 2\left(\frac{\mathrm{~F}}{\ell \mathrm{~d}_1 \eta_1}\right)=\frac{\mathrm{F}}{\ell \mathrm{d}_2 \eta_2} \\ & \Rightarrow \eta_2=\frac{\eta_1}{4}=1 \times 10^9 \Rightarrow x=1\end{aligned}$

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About this question

This is a previous-year question from JEE Main 2025, covering the Mechanical Properties of Solids chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.