Question

The extension in a uniform rod of length $l$, mass $m$, cross sectional area $A$ and young's modulus $Y$ when it is suspended vertically at one end is

• A. $\frac{mgl}{AY}$
• B. $\frac{mgl}{2AY}$
• C. $\frac{2mgl}{AY}$
• D. $\frac{mgl}{4AY}$

Answer 1

The correct option is B $\frac{mgl}{2AY}$

Let us take a small part of rod of length $dx$ at height $x$ from lower end.


Tension at point $P$ in the rod have to support the mass $m^{\prime }$ of lower portion of rod below point $P$.
So, Tension at point $P$,
$T=m^{{}^{\prime }}g=\frac{mgx}{l}$ .........(1)
[ Since rod is uniform ]
Now, extension of this small part of rod of width $dx$ due to tension at $P$,
$dl=\frac{Tdx}{AY}$
$\Rightarrow dl=\frac{mgxdx}{AYl}$ [ from (1) ]
Therefore, total extension in rod,
$L=\underset{0}{\overset{l}{\int }}\frac{mgxdx}{YAl}=\frac{mgl}{2AY}$