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.

Explanation

`r _(A) = r _(B) =(0.25 xx 10 ^(-2))/( 2 ) = 0.125 xx 10 ^(-2) m `
For steel wire,
`= (F _(S) L _(S))/( pi r _(S) ^(2) Y _(S)) = (98 xx 1.5 )/( 3.14 xx (0.125 xx 10 ^(-2)) xx 21 xx 10 ^(11))`
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For brass wire, `F _(B) = 6 xx 9.8 N`
`l _(R) = (F _(B) L _(B))/( pi r _(B) ^(2) Y _(B))`
`= (6 xx 9.8 xx 1 )/( 3.14 xx (0.125 xx 10 ^(-2)) ^(2) xx 0.91 xx 10 ^(11))`
`l _(B) = 1217 xx 10 ^(-7)`
`therefore l _(B) = 1.317 xx 10 ^(-4) m`
`therefore l _(B) ~~ 1.3 x 10 ^(-4) m`