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A uniform rod of mass $m$, length $L$, area of cross-section $A$ and Young's modulus $Y$ hangs from the ceiling. Its elongation under its own weight will be
Zero
$\frac {mgL}{2AY}$
$\frac {mgL}{AY}$
$\frac {2mgL}{AY}$
Solution

Mass of section $\mathrm{BC}=\frac{\mathrm{M}}{\mathrm{L}}(\mathrm{L}-\mathrm{y})$
$\text { Tension at } \mathrm{B}=\mathrm{T}=\frac{\mathrm{m}}{\mathrm{L}}(\mathrm{L}-\mathrm{y}) \mathrm{g}$
$\therefore$ Elongation of element dy at $\mathrm{B}$.
$\mathrm{dx}=\mathrm{dy} \frac{\mathrm{T}}{\mathrm{AY}}=\frac{\mathrm{m}}{\mathrm{L}}(\mathrm{L}-\mathrm{y}) \mathrm{g} \frac{\mathrm{dy}}{\mathrm{AY}}$
Total elongation
$\mathrm{x}=\int \mathrm{dx}=\frac{\mathrm{mg}}{\mathrm{LAY}} \int_{0}^{\mathrm{L}}(\mathrm{L}-\mathrm{y}) \mathrm{dy}=\frac{\mathrm{mgL}}{2 \mathrm{YA}}$
Similar Questions
Column$-II$ is related to Column$-I$. Join them appropriately :
Column $-I$ | Column $-II$ |
$(a)$ When temperature raised Young’s modulus of body | $(i)$ Zero |
$(b)$ Young’s modulus for air | $(ii)$ Infinite |
$(iii)$ Decreases | |
$(iv)$Increases |