A brass wire of length 5m and cross section `1mm^(2)` is hung from a rigid support, with a brass weight of volume 1000 `cm^(3)` hanging from the other end. Find the decrease in the length of the wire, when the brass weight is completely immersed in water.
`(Y_("brass")=10^(11)Nm^(-2), g=9.8ms^(-2), rho_("water")=1g cm^(-3))`

Explanation

When a weight is hung in air from the other end of a wire, F = Mg. The increase in length of the wire, e = ?
Young.s modulus, `Y=(F)/(A)(L)/(e ) e=(MgL)/(AY)`.
When weight hung in a liquid,
Weight of the body in the liquid `=Mg-V rho g`
where V is the volume of the body
This is the force, F acting on the wire i.e.,
`F=Mg-V rho g`
Increase in length of the wire, `d=((Mg-v rhog)L)/(AY)`
which is less than the increase in length of the wire when the weight is in air.
Decrease in length = e in air - e. in liquid
`=(MgL)/(AY)-((Mg-V rhog)L)/(AY)=(V rhogL)/(AY)`
Here, `V=1000cm^(3)=1000xx10^(-6)m^(3)`
`rho= 1 g cm^(-3)=1xx10^(3)kgm^(-3), g=9.8ms^(-2), L=5m`
`A=1mm^(2)=1xx10^(-6)m^(2), Y=1xx10^(11)Nm^(-2)`
The decrease in length
`=(1000xx10^(-6)xx1xx10^(3)xx9.8xx5)/(1xx10^(-6)xx1xx10^(11))`
`=49xx10^(-5)m=0.49` mm