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JEE Main 20254 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main 2025 · 4 April · Morning Shift · Physics · Mechanical Properties of Solids

Two slabs with square cross section of different materials (1,2)...

Two slabs with square cross section of different materials (1,2)(1,2) with equal sides (l)(l) and thickness d1\mathrm{d}_1 and d2\mathrm{d}_2 such that d2=2 d1\mathrm{d}_2=2 \mathrm{~d}_1 and l>d2l \gt \mathrm{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 θ2=2θ1\theta_2=2 \theta_1. If the shear moduli of material 1 is 4×109 N/m24 \times 10^9 \mathrm{~N} / \mathrm{m}^2, then shear moduli of material 2 is x×109 N/m2\mathrm{x} \times 10^9 \mathrm{~N} / \mathrm{m}^2, where value of xx is ________.Two slabs with square cross section of different materials (1,2) with equal sides (l) and thickness d_1 and d_2 such

Correct answer

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Step-by-step solution

Deformation angle
<br/>2θ1=θ2⇒2σ1η1=σ2η2<br/>\begin{aligned}<br/>& 2 \theta_1=\theta_2 \\ & \Rightarrow 2 \frac{\sigma_1}{\eta_1}=\frac{\sigma_2}{\eta_2}<br/>\end{aligned}
Two slabs with square cross section of different materials (1,2) with equal sides (l) and thickness d_1 and d_2 such
⇒2( Fℓ d1η1)=Fℓd2η2⇒η2=η14=1×109⇒x=1\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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