A steel wire of length 4.7 m and cross-sectional area `3.0 xx 10 ^(-5) m ^(2)` stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of `4.0 xx 10 ^(-5) m ^(2)` under a given load. What is the ratio of the Young's modulus of steel to that of copper ?

Explanation

Here, `L _(S) = 4.7 m, A _(S) = 3 xx 10 ^(-5) m ^(2), Delta L _(S) = l, F_(S) = F`
and `L _(CU) = 3.5 m, A _(CU) = 4 xx 10 ^(-5) m ^(2) , Delta L _(CU)=l, F _(CU) = F`
`T _(S) = ((F _(S))/( A_(S)))/( ( Delta L _(S))/( L _(S)) ) = (F_(S) L _(S))/( A _(S) Delta L _(S)) = (FL _(S))/( A _(S) l)`
`Y_(CU ) = ((F _(CU))/( A _(CU)))/((Delta L _(CU))/(L _(CU)))= (F _(CU) L _(CU ))/( A _(CU) Delta L _(CU)) = (F L _(CU))/(A _(CU) l )`
`therefore (Y_(S))/(Y_(CU)) = (L _(S))/(A_(S)) xx (A _(CU))/(L _(CU))`
`=(4.7)/(3 xx 10 ^(-5)) xx (4 xx 10 ^(-5))/( 3.5)= (18.8)/(10.5) = 1.79 ~~1.8`