Question

A uniform rod of length L, has a mass per unit length $\lambda$ and area of cross-section A. The elongation in the rod is l due to its own weight, if it is suspended from the ceiling of a room.The Young's modulus of the rod is

• A. $\frac{\lambda g{L}^2}{2Al}$
• B. $\frac{3\lambda g{L}^2}{Al}$
• C. $\frac{2\lambda gL}{Al}$
• D. $\frac{\lambda g{L}^2}{Al}$

Answer 1

Answer: (A)

$\frac{\lambda g{L}^2}{2Al}$

Given, a ned of length $=L$

area of cross - sction $=A$

Yourng's modulues $=y$

Now , we can comert this

Sugstem into an equivalent spring - systerm of sprung equivalint $K$.

$\begin{array}{c}\text{Let}dx\text{be the elomgation in}\\ \text{sinall element}dz\text{.}\\ T_1=\left(\frac{M}{L}\right)\left(z\right)\left(g\right)\\ T_1=k\left(dx\right)\\ T_1=\left(\frac{yA}{dz}\right)\left(dx\right)\\ \frac{Mg}{L}{\int }_0^{t}zdz=yA{\int }_0^{x}dx\end{array}$

$\frac{MgL}{2AY}=x$

$\begin{array}{cc}\text{Now ,}& dm=\lambda dx\\ M& =\lambda {\int }_0^{0}dx\\ M& =\lambda L\end{array}$

$\begin{array}{c}\therefore \text{Total elongation}\\ \begin{array}{cc}l& =\frac{\lambda g{L}^2}{2AY}\\ \Rightarrow & y=\frac{\lambda g{L}^2}{2Al}\end{array}\end{array}$