Question

Elongation of a wire under its own weight is independent of

• A. length
• B. area of cross section
• C. density
• D. yong's modulus

Answer 1

Answer: (B)

Mass $=M$
Force $=\rho ALg$
Let weight of the rod is $W$.
Consider a small length $dx$ of the rod at distance $x$ from the fixed end. The tension $T$ in element equals the weight of rod below it.
$T=(L-x)\frac{W}{L}$
Elongation $=\frac{L\times \text{ Stress }}{y}=\frac{(L-x)Wdx}{LAy}$
Total elongation $={\int }_O^L\frac{(L-x)Wdx}{LAy}=\frac{W}{LAy}{\left(Lx-\frac{x^2}{2}\right)}_0^L=\frac{WL}{2Ay}$
Now, $W=\rho ALg$
Putting it in there: $\Delta L=\frac{\rho A{L}^{2}g}{2Ay}=\frac{\rho L^{2}g}{2y}$