A steel plate of face area $1 \,cm ^2$ and thickness $4 \,cm$ is fixed rigidly at the lower surface. A tangential force $F=10 \,kN$ is applied on the upper surface as shown in the figure. The lateral displacement $x$ of upper surface w.r.t. the lower surface is .............. $m$ (Modulus of rigidity for steel is $8 \times 10^{11} \,N / m ^2$ )
Medium
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(b)
$\text { Modulus of rigidity }(G)=\frac{\text { Force } \times \text { Length }}{\text { Area } \times \text { Lateral displacement }}=\frac{F L}{A \times \Delta x}$
$F=10 \,kN =10 \times 10^3 \,N$
$L=4 \,cm =0.04 \,m$
$A=1 \,cm ^2=1 \times 10^{-4} \,m ^2$
$G=8 \times 10^{11}\,N / m ^2$
Substituting values
$8 \times 10^{11}=\frac{10 \times 10^3 \times 0.04}{1 \times 10^{-4} \times \Delta x}$
$\Delta x=\frac{10 \times 10^3 \times 0.04}{1 \times 10^{-4} \times 8 \times 10^{11}}=5 \times 10^{-6} \,m$
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