Consider a wire suspended vertically from a rigid support. A load is now attached to the wire and the extension in the wire is 6.2 mm. If the extension is reduced to 4.1 mm, when the body attached is immersed completely in the water, then what is the relative density of the material of the body?

Answer

`02.95`

Explanation

When in air, for wire, tension in wire
T=mg when wire is completely immersed in water, tension in wire
`T.-mg-V rho g`
Where `rho=` density of water
`:.T.=mg(1-( V rho)/m)`
`=T(-(rho)/d)`
Where d=density of material of wire.
`:.(T.)/t=1-(rho)/d`
for a given wire as extension produced in wire is directly proportional to tension in wire i.e.
`F propl`
`:.4.1/6.2=1-(rho)/d`
`:.4.1/6.2=1-1/(rho_("rel"))`
Where `rho_("rel")=` Relative density of material of wire
`:.1/(rho_("rel")=` Relative density of materal of wire
`:.1/(rho_("rel"))=1-4.1/6.2=(6.2-4.1)/6.2=2.1/6.2`
`:.rho_("rel")=6.2/2.1=2.95`