8.Mechanical Properties of Solids
medium

If one end of a wire is fixed with a rigid support and the other end is stretched by a force of $10 \,N,$ then the increase in length is $0.5\, mm$. The ratio of the energy of the wire and the work done in displacing it through $1.5\, mm$ by the weight is

A

$\frac{1}{3}$

B

$\frac{1}{4}$

C

$\frac{1}{2}$

D

$1$

Solution

(c) Work done in stretching a wire 

$W = \frac{1}{2}Fl = \frac{1}{2} \times 10 \times 0.5 \times {10^{ – 3}}$= $2.5 \times {10^{ – 3}}J$

Work done to displace it through $1.5 \,mm$

$W = F \times l = 5 \times {10^{ – 3}}J$

The ratio of above two work $= 1 : 2$

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.