A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension. `Y_(Cu)=1.2 xx 10 ^(11) N m ^(-2), Y_(Fe) = 1.9 xx 10 ^(11) N, ^(-2)`

Explanation

Suppose cross sectional area of copper and iron wire are `A _(Cu) and A _(Fe)` respectively and their diameter are `D _(Cu) and D_(Fe)`
`therefore A _(Cu ) = (pi D _(Cu ) ^(2))/( 4) and A _(Fe ) = (pi D _(Fe ) ^(2))/( 4)`
`therefore (A _(Cu ))/( A _(Fe)) = ((D _(Cu ))/( D _(Fe)))" "...(1)`
Now `Y= (F//A)/("Strain")`
`therefore` Strain in copper `= (F)/(A _(Cu) Y_(Cu)) and `
strain in iron `= (F)/(A _(Fe) Y _(Fe))`
SInce each wire is to have same tension therefore each wire has same strain,
`therefore (F)/(A _(Cu ) Y _(Cu )) = (F )/( A _(Fe )Y _(Fe))`
`therefore (Y _(Fe)\)/( Y _(Cu )) = (A _(Cu ))/( A _(Fe)) =((D _(Cu))/( D _(Fe))) ^(2) [because ` From result (1)]
`therefore ( D _(Cu ))/( D _(Fe)) = sqrt (( Y _(Fe))/( Y _(Cu )) ) = sqrt (( 1. 9 xx 10 ^(11))/( 1.2 xx 10 ^(11)))`
`therefore (D _(Cu))/( D _(Fe)) = sqrt ((19)/(12)) = sqrt (1.58)`
`therefore (D_(Cu ))/( D _(Fe)) = 1.256 :1`