A rod of length 1 and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in figure. The cross-sectional areas of wires A and B are `1.0 mm^(2) and 2.0 mm^(2)` respectively. `Y_("steel”) = 200 xx 10 ^(9) Nm ^(-2) and Y_("auminium”) = 70 xx 10 ^(9) Nm ^(-2)`

Options

A. Mass m should be suspended close to wire A to have equal stresses in both the wires.
B. Mass m should be suspended close to B to have equal stresses in both the wires.
C. Mass m should be suspended at the middle of the wires to have equal stresses in both the wires.
D. Mass m should be suspended close to wire A to have equal strain in both wires.

Answer

B, D

Explanation

Suppose, mass m hangs at distance x from wire B.

Let 1 is for steel wire and 2 is for aluminium wire. `T_(1) and T_(2)` are the tension in steel and aluminium wire respectively and `S_(1) and S_(2)` are stress and `in _(1) and in _(2)`are strain `a_(1) and a_(2)` are cross sectional areas.
For balance of wire.
`sum T = 0`
`therefore T_(2) x - T_(1) (l -x) =0`
`therefore T _(2) x = T (l -x)`
`therefore (T_(2))/( T _(1)) = (l-x)/( x) " "...(1)`
Stress:
For steel wire `S _(1) = (T_(1))/(a_(1)) " "...(2)`
For aluminium wire `S_(2) = (T_(2))/( a _(2)) " "...(3)`
If stress is equal,
`S_(1) = S_(2)`
`therefore (T_(1))/( a _(1)) = (T_(2))/(a_(2))`
`therefore (T_(1))/( T _(2)) = (a_(1))/( a _(2)) = (1.0)/( 2.0 ) = 1/2 `
`therefore (x)/( l -x) = 1/2 [ because ` From eq. (1)]
`therefore 2x =l -x`
`therefore 3x =l`
`therefore x = l /3` is the distance from end B
and `l-x =l - l/3 = (2l)/(3)` is the distance from end A.
`therefore (l)/(3) lt (2l)/(3)` hence option (B) is correct.
Now, if strain are same,
`Y_(1) = (S_(1))/( in _(1)) .and Y_(2) = (S_(2))/( in _(2))`
but ` in _(1) = in _(2)`
`therefore ( Y_(1))/( Y _(2)) = (S_(1))/( S _(2)) `
`therefore (200 xx 10 ^(9))/( 70 xx 10 ^(9)) = (T_(1))/( a _(1)) xx (a _(2))/( T_(2))`
[`because` From equation (2) and (3)]
`therefore (20)/(7) =(T_(1))/( T _(2)) xx (a _(2))/( a _(1))`
`therefore (20 )/(7) = (x)/(l -x) xx (2.0)/(10) [because ` From eq. (1)]
`therefore ( 10 )/(7) = (x)/( l -x)`
`therefore 10 l -10 x = 7x`
`therefore 10l = 17x`
`therefore x = (10 l )/(17)` is the distance from end B
`and l - x =l - (10l)/(17) = (7l)/(17)` is distance from end A.
`therefore ( 7l)/(17) lt (10 l)/( 17)` hence option (D) is correct.