Mechanical Properties of Solids

JEE Main level, numerical answer, medium

106 of 118

Question 106 of 118

A square aluminium (shear modulus is 25×109Nm−225\times {10}^{9}{Nm}^{-2}) slab of side 60cm60cm and thickness 15cm15cm is subjected to a shearing force (on its narrow face) of 18.0×104N18.0\times {10}^{4}N. The lower edge is riveted to the floor. The displacement of the upper edge is _____ μm\mu m.

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Answer: 48

Shearing stress is given by FA=ηxl\dfrac{F}{A}=\eta \dfrac{x}{l}
Thus, displacement is x=FlAηx=\dfrac{Fl}{A\eta }
⇒x=18×104×60×10−260×10−2×15×10−2×25×109\Rightarrow x=\dfrac{18\times {10}^{4}\times 60\times {10}^{-2}}{60\times {10}^{-2}\times 15\times {10}^{-2}\times 25\times {10}^{9}}
=48×10−6m=48μm=48\times {10}^{-6}m=48\mu m

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