A rod of length 1.05 m having negliaible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in fig. The cross-sectional area of wire A and B are `1 mm^(2)` and 2` mm^(2)`, respectively . At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires. Given,
`Y_(steel) = 2 xx 10^(11) Nm^(-2) and Y-(alumi n i um) = 7.0 xx 10^(10)N^(-2)`

Options

A. mass m should be suspended close to the wire A
B. mass m should be suspended close to the wire B
C. mass m should be suspended at the middle of the wires
D. mass m should be suspended from a point at a distance of 1/10 from the wire A

Answer

B

Explanation

Suppose that the mass 'm' is kept at C at a distance x from the end P and let `T_(A)` and `T_(B)` be the tension in the wires A and B respectively. Let `a_(A)` and `a_(B)` be their cross sectional areas.
For the rotational equilibrium of the rod, the total torque acting on the rod should be zero.
`:. T_(A)x - T_(B) (l-x) = 0`
`:. (T_(A))/(T_(B)) = (l-x)/(x)` ...(1)
Stress in the wire `A = S_(A) = (T_(A))/(a_(A))`
and stress in the wire `B = S_(B) = (T_(B))/(a_(B))`
To have `S_(A) = S_(B)` (equal stress)
`(T_(A))/(a_(A)) = (T_(B))/(a_(B)) :. (T_(A))/(T_(B)) = (a_(A))/(a_(B)) = (1)/(2)` ....(2)
`:. (l-x)/(x) = (1)/(2)`
`:. 2l - 2x = x :. x = (2l)/(3)` and `l-x = (l)/(3)`
Thus the mass 'm' should be suspended closer to the wire B, to have equal stresses in both the wires.