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JEE Mains · Physics · STD 11- 8. mechanical properties of solids

A uniform heavy rod of mass \(20\,kg\). Cross sectional area \(0.4\,m ^{2}\) and length \(20\,m\) is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is \(x \times 10^{-9} m\). The value of \(x\) is (Given. Young's modulus \(Y =2 \times 10^{11} Nm ^{-2}\) અને \(\left.g=10\, ms ^{-2}\right)\)

  1. A \(28\)
  2. B \(25\)
  3. C \(24\)
  4. D \(23\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(25\)

Step-by-step Solution

Detailed explanation

\(Y =\frac{ T }{ A } \frac{ dx }{ dy }\) \(m =20\,kg\) \(A =0.4\,m^{2}\) \(1=20\,m\) let extension is \(dy\) in length \(dx\) \(Y =\frac{\text { stress }}{\text { strain }}\) \(Y =\frac{\frac{ T }{ A }}{\frac{ d }{ dx }}=\frac{ T }{ A } \cdot \frac{ dx }{ dy }\)…
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