Question

A load of 4 kg is suspended from a ceiling through a steel wire of length 20 m and radius 2 mm. It is found that the length of the wire increases by 0.031 mm, as equilibrium is achieved. If g = 3.1 $\times \pi m{s}^{-2}$ the value of

Young’s modulus is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Length of the wire is l = 20 m

Radius of wire is r = $2\times {10}^{-3}m$

Increase in length is $\Delta l=0.031\times {10}^{-3}m$

g = $3.1\times \pi m{s}^{-2}$

Load is F = mg = $4\times 3.1\times \pi N$

Young’s ,modulus of material of wire is Y = $\frac{Fl}{ADl}=\frac{Fl}{\pi r^2\Delta l}$

$\therefore Y=\frac{4\times 3.1\times \pi \times 20}{\pi (4\times {10}^{-6}\times 0.031\times {10}^{-3})}=2\times {10}^{12}N{m}^{-2}$