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

(i)For steel, `Y_(S)=("Stress")/("Strain")`
`"Stress"=(4Mg)/(piD^(2))=(4W_(s))/(piD_(s)^(2))`
`"Strain"=(DeltaL_(S))/(L_(S))thereforeDeltaL_(S)=(4W_(S)L_(S))/(piY_(S)D_(S)^(2))`
Putting values and simplifying we get, `DeltaL_(S)=(4xx10xx9.8xx1.5)/(3.14xx2.0xx10^(11)xx(2.5xx10^(-3))^(2))`, where `W_(S)=(4+6)9.8N`
or `DeltaL_(S)=1.5xx10^(-4)m`
(ii) Similarly, for brass
`DeltaL_(b)=(4xx6xx98xx1.0)/(3.14xx(2.5xx10^(-3))^(2)xx9.1xx10^(10))`, were `W_(S)=6xx9.8N orDeltaL_(b)=1.3xx10^(-4)m`