A square aluminium (shear modulus is $25 \times 10^{9}\,Nm ^{-2}$ ) 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 $.......\mu\,m$.
JEE MAIN 2022, Easy
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$\frac{ F }{ A }=\eta \frac{ x }{\ell} \Rightarrow \frac{ F \ell}{ An }= x$
$x =\frac{18 \times 10^{4} \times 60 \times 10^{-2}}{60 \times 10^{-2} \times 15 \times 10^{-2} \times 25 \times 10^{9}}$
$=48 \times 10^{-6}\,m =48\,\mu m$
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