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)

A rigid bar of mass 15 kg is supported symmetrically by three wires ea

1.	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)
Answer» SOLUTION :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`