JEE Mains · Physics · STD 11- 8. mechanical properties of solids
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 \(.......\mu\,m\).
- A \(24\)
- B \(12\)
- C \(48\)
- D \(96\)
Answer & Solution
Correct Answer
(C) \(48\)
Step-by-step Solution
Detailed explanation
\(\frac{ F }{ A }=\eta \frac{ x }{\ell} \Rightarrow \frac{ F \ell}{ An }= x\) \(x =\frac{18 \times 10^{4} \times 60 \times 10^{-2}}{60 \times 10^{-2} \times 15 \times 10^{-2} \times 25 \times 10^{9}}\) \(=48 \times 10^{-6}\,m =48\,\mu m\)
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