A steel wire of length 7 m and cross section 1`mm^(2)` I suspended from a rigid support, with a steel weight of volume ` 10^(3) cm^(3)` hanging from its other end. Find the decrease in the length of the wire when the steel weight is completely immersed in water.
` Y_("steel") = 2xx 10^(11) N//m^(2) ` density of water `= 10^(3) kg//m^(3)` ]

Explanation

Data : L = 7m A= 1 `mm^(2) = 10^(-6) m^(2) , Y = 2 xx 10^(11) N//m^(2)`
`V=10^(3) cm^(3) = 10^(3)xx10^(-6) m^(3) = 10^(-3) m^(3) , p_("water") = 10^(3) "kg/m"^(3) , g= 9.8 m//s^(2)`
` Y = (F//A)/(DeltaL/L) = (FL)/(ADeltaL) " " therefore DeltaL = (FL)/(AY)`
Intitally with the suspended steel weight in air.
`DeltaL_(1) = (F_(1)L)/(AY)`
with the suspended steel weight immersed in water,
`DeltaL_(2) = (F_(2)L)/(AY)`
and teh upthrust due to water is
` F_(1) = F_(2) = V_(rhog)`
The decrease in the length of the wire is
` DeltaL_(1) - DeltaL_(2) = ((F_(1)-F_(2))L)/(AY) = (VrhogL)/(AY)`
`= (10^(-3) xx 10^(3) xx 9.8 xx 7 xx 10^(-5))`
` 3.43 xx 10^(-4) m` or 0.343 mm