Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in figure. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires .Young's modulus of steel is `2.0 xx 10^(11)`Pa and that of brass is `9.1 xx 10^(11)` Pa.

Explanation

For steel wire , total force on steel wire,
`F_(1) = 4 + 6 =10kg f = 10 xx 9.8 N`,
`l_(1) = 1.5 m . Delta l_(1)=? , 2r_(1) = 0.25 cm`
or `r_(1) = (0.25//2) cm = 0.125 xx 10^(-2)m`, `Y_(1) = 2.0 xx 10^(11)Pa`
for brass wire, `F_(2) = 6.0 kg f= 6 xx 9.8 N , 2 r_(2) = 0.25 cm`
or `r_(2) = (0.25//2)cm = 0.125 xx 10^(-2)m`,
`Y_(2) = 0.91 xx 10^(11) Pa, l_(2) = 1.0 m, Delta l_(2)=?`
As, `Y_(1) = (F_(1) xx l_(1))/(A_(1) xx Delta l_(1)) = (F_(1) xx l_(1))/(pi r_(1)^(2) xx Delta l_(1))`
or `Delta l_(1) = (F_(1) xx l_(1))/(pi r_(1)^(2) xx Y_(1)) = ((10 xx 9.8) xx 1.5 xx 7)/(22 xx (0.125 xx 10^(-2))^(2) xx 2 xx10^(11))=1.49 xx 10^(-4)m`
and `Delta l_(2) = (F_(2) xx l_(2))/(pi r_(2)^(2) xx Y_(2)) = ((6 xx 9.8) xx 1.0 xx 7)/(22 xx (0.125 xx 10^(-2))^(2) xx(0.91 xx 10^(11)))=1.3 xx 10^(-4)m`.