Question

A wire of length $L$, area of cross section $A$ is hanging from a fixed support. The length of the wire changes to $L_1$ when mass $M$ is suspended from its free end. The expression for Young’s modulus is:

• A. $\frac {MgL_1}{AL}$
• B. $\frac {Mg(L_1-L)}{AL}$
• C. $\frac {MgL}{AL_1}$
• D. $\frac {MgL}{A(L_1-L)}$

Answer 1

Answer: (D)

When the new mas is hanged to the wire, the force exerted on the wire is:
$F = mg$
The initial length of the wire is $L$ and the new length is $L_{1}$.
We know that the Young's modulus is given as:
$Y =\frac{ FL _{ o }}{ A \Delta l }$
Here, $L_{o}$ is the initiallength of wire and $\Delta l$ is the chnage in length of wire.
Substitute the values:
$Y =\frac{ mg \times L }{ A \times\left( L _{1}- L \right)}$
So, $Y=\frac{m g L}{A\left(L_{1}-L\right)}$