A steel wire having cross-sectional area `1.5 mm^(2) ` when stretched by a load produces a lateral strain `1.5 xx 10^(-5)`. Calculate the mass attached to the wire.
(`Y_("steel") = 2xx 10^(11) N//m^(2)`, Poisson's ration `sigma = 0.291, g = 9.8 m//s^(2)` )

Explanation

Given : `A = 1.5mm^(2)`, lateral strain `= 1.5 xx 10^(-5) `, `Y_("steel") = 2 xx 10^(11) N//m^(2) , sigma = 0.291` and `g = 9.8 m//s^(2)`
Poisson's ratio, `sigma = ("Lateral strain")/("Longitudinal strain")`
`0.291 = (1.5 xx 10^(-5))/("Longitudinal strain")`
`:.` Longitudinal strain `= (1.5 xx 10^(-5))/(0.291) = 5.15 xx 10^(-5)`
Longitudinal stress `= Y xx` Longitudinal strain
`= 2 xx 10^(11) xx 5.15 xx 10^(-5)`
`:.` Longitudinal stress `= 10.3 xx 10^(6)`
But Longitudinal stress `= (Mg)/(A)`
`rArr 10.3 xx 10^(6) = (M xx 9.8)/(1.5 xx 10^(-6))`
`rArr M= (10.3 xx 10^(6) xx 1.5 xx 10^(-6))/(9.8)`
`= (15.45)/(9.8)`
`:. M = 1.58 kg`.