Question

A steel wire of length $1m$ and cross sectional area $1.5{mm}^2$ is hung from a rigid support, with a stone of volume $2000{cm}^2$ hanging from the other end. Find the decrease in the length of the wire, when the stone is completely immersed in water. ($Y_{s}teel=2\times {10}^{11}N{m}^{-}2,{\rho }_{water}$=${10}^{3}Kg{m}^{-3}$

• A. $0.06863mm$
• B. $0.02533mm$
• C. $0.04533mm$
• D. $0.01533mm$

Answer 1

The correct option is A $0.06863mm$

It is given that

${\gamma }_{steel}=2\times {10}^{11}N{m}^{-2}$

$\rho ={10}^{3}Kg{m}^{-3}$ .

$V=2000c{m}^3$

$L=1m$

$A=1.5m{m}^2=1.5\times {10}^{-6}m$

Using the relation as

$Y=\frac{MgL}{A\Delta l}$

$\Delta l=\frac{MgL}{YA}$

$=\frac{\rho VgL}{YA}$

$=\frac{10\times 0.2\times 10\times 1}{2\times {10}^{11}\times 1.5\times {10}^{-6}}$

$=68.6\times {10}^{-5}m$

$=0.0686mm$