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(C) The shear modulus $\eta$ is given by the formula: $\eta = \frac{F/A}{x/L}$,where $F$ is the shearing force,$A$ is the area of the face on which th

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$48$

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(C) The shear modulus $\eta$ is given by the formula: $\eta = \frac{F/A}{x/L}$,where $F$ is the shearing force,$A$ is the area of the face on which the force is applied,$x$ is the displacement,and $L$ is the side length of the slab.Rearranging for displacement $x$: $x = \frac{F \cdot L}{A \cdot \eta}$.Given: $F = 18.0 	imes 10^{4} \, N$,$L = 60 \, cm = 0.6 \, m$,thickness $t = 15 \, cm = 0.15 \, m$,and $\eta = 25 	imes 10^{9} \, N m^{-2}$.The area $A$ of the narrow face is $L 	imes t = 0.6 \, m 	imes 0.15 \, m = 0.09 \, m^2$.Substituting the values: $x = \frac{18.0 	imes 10^{4} 	imes 0.6}{0.09 	imes 25 	imes 10^{9}}$.$x = \frac{10.8 	imes 10^{4}}{2.25 	imes 10^{9}} = 4.8 	imes 10^{-5} \, m = 48 	imes 10^{-6} \, m = 48 \, \mu m$.

$A$ square aluminium (shear modulus is $25 \times 10^{9} \, N m^{-2}$) slab of side $60 \, cm$ and thickness $15 \, cm$ is subjected to a shearing force (on its narrow face) of $18.0 \times 10^{4} \, N$. The lower edge is riveted to the floor. The displacement of the upper edge is $....... \mu m$.

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