TS EAMCET · Physics · Mechanical Properties of Solids
A horizontal aluminium rod of diameter \(4 \mathrm{~cm}\) projected \(6 \mathrm{~cm}\) from a wall. An object of mass \(400 \pi \mathrm{kg}\) is suspended from the end of the rod. The shearing modulus of aluminium is \(3.0 \times 10^{10} \mathrm{~N} / \mathrm{m}^2\). The vertical deflection of the end of the rod is \(\left(\because g=10 \mathrm{~m} / \mathrm{s}^2\right)\)
- A \(0.01 \mathrm{~mm}\)
- B \(0.02 \mathrm{~mm}\)
- C \(0.03 \mathrm{~mm}\)
- D \(0.04 \mathrm{~mm}\)
Answer & Solution
Correct Answer
(B) \(0.02 \mathrm{~mm}\)
Step-by-step Solution
Detailed explanation
Modulus of rigidity of a horizontal rod with hanging mass is given as \( \eta=\frac{F_L}{A x}=\frac{m g L}{A x} \) Where, \(x=\) verticle deflection of the end of the rod. Here, \(L=6 \mathrm{~cm}=6 \times 10^{-2} \mathrm{~m}\), radius \(R=2 \mathrm{~cm}\)…
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