JEE MAIN - Physics (2025 - 4th April Morning Shift - No. 23)
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 $4 \times 10^9 \mathrm{~N} / \mathrm{m}^2$, then shear moduli of material 2 is $x \times 10^9 \mathrm{~N} / \mathrm{m}^2$, where value of $x$ is ________.
Answer
1
Explanation
$$\begin{aligned} &\text { Deformation angle }\\ &\begin{aligned} & 2 \theta_1=\theta_2 \\ & \Rightarrow 2 \frac{\sigma_1}{\eta_1}=\frac{\sigma_2}{\eta_2} \end{aligned} \end{aligned}$$
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$$\begin{aligned} & \Rightarrow 2\left(\frac{F}{\ell d_1 \eta_1}\right)=\frac{F}{\ell d_2 \eta_2} \\ & \Rightarrow \eta_2=\frac{\eta_1}{4}=1 \times 10^9 \Rightarrow x=1 \end{aligned}$$
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