Mechanical Properties of Solids

JEE Main level, numerical answer, medium

105 of 118

Question 105 of 118

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 ________.

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Answer: 1

Deformation angle
$\begin{aligned}

& 2 \theta_1=\theta_2 \\

& \Rightarrow 2 \frac{\sigma_1}{\eta_1}=\frac{\sigma_2}{\eta_2}

\end{aligned}$


⇒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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