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))

A brass wire of length 5m and cross section 1mm^(2) is hung from a rig

1.	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))
Answer» Solution : 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 - `Vrho 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, e. ` = ((mg -v rho g)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^1` in liquid ` = (MgL)/(AY) - ((Mg - V rho g)L)/(AY) = (Vrho gL)/(AY)` Here , ` V = 1000 cm^3 = 1000 xx 10^(-6) m^3` `A = 1mm^2 = 1 xx 10^(-6) m^2 ,Y = 1 xx 10^11 Nm^(-2)` Thedecrease in length ` = (1000 xx 10^(-6) xx 1 xx 10^3 xx 9.8 xx 5)/(1 xx 10^(-6) xx 1 xx 10^11)` ` = 49 xx 10^(-5) m = 0.49 mm`