JEE Mains · Physics · STD 11- Mechanical properties of solids
- JEE Mains 2025
- Hard
- Numerical
Two slabs with square cross section of different materials \((1,2)\) with equal sides \((l)\) and thickness \(\mathrm{d}_1\) and \(\mathrm{d}_2\) such that \(\mathrm{d}_2=2 \mathrm{~d}_1\) and \(l \gt \mathrm{d}_2\). 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 \(\theta_2=2 \theta_1\). If the shear moduli of material 1 is \(4 \times 10^9 \mathrm{~N} / \mathrm{m}^2\), then shear moduli of material 2 is \(\mathrm{x} \times 10^9 \mathrm{~N} / \mathrm{m}^2\), where value of \(x\) is ________.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation
Deformation angle
\(\begin{aligned}
& 2 \theta_1=\theta_2 \\
& \Rightarrow 2 \frac{\sigma_1}{\eta_1}=\frac{\sigma_2}{\eta_2}
\end{aligned}\)
\(\begin{aligned} & \Rightarrow 2\left(\frac{\mathrm{~F}}{\ell \mathrm{~d}_1 \eta_1}\right)=\frac{\mathrm{F}}{\ell \mathrm{d}_2 \eta_2} \\ & \Rightarrow \eta_2=\frac{\eta_1}{4}=1 \times 10^9 \Rightarrow x=1\end{aligned}\)
\(\begin{aligned}
& 2 \theta_1=\theta_2 \\
& \Rightarrow 2 \frac{\sigma_1}{\eta_1}=\frac{\sigma_2}{\eta_2}
\end{aligned}\)
\(\begin{aligned} & \Rightarrow 2\left(\frac{\mathrm{~F}}{\ell \mathrm{~d}_1 \eta_1}\right)=\frac{\mathrm{F}}{\ell \mathrm{d}_2 \eta_2} \\ & \Rightarrow \eta_2=\frac{\eta_1}{4}=1 \times 10^9 \Rightarrow x=1\end{aligned}\)
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