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 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

Let the mass is suspended at `x` from the end `B`, which develop equal stress in wires.
Let `T_(A)` and `T_(B)` be the tension in wire `A` and wire `B` respectively.
Stess in steel wire `A, S_(A) = (T_(A))/(A_(B)) = (T_(A))/(10^(-6))`
Stress in `Al` wire , `S_(B) = (T_(B))/(A_(B)) = (T_(B))/(2xx10^(-6))`
where `A_(A)` and `A_(B)` are cross-sectional areas of wire `A` and `B` respectively. Also from rotational equilibrium, net torque is zero, i.r,
`T_(B)x - T_(A) (l-x) = 0`
`rArr (T_(B))/(T_(A)) = (l-x)/(x)`
For equal stress, `S_(A) = S_(B)`
`rArr S_(A) = S_(B) rArr (T_(A))/(10^(-6)) = (T_(B))/(2xx10^(-6))`
`rArr (l-x)/(x) = 2 rArr (l)/(x) - 1 = 2 rArr x = (l)/(3)`
`:. l - x = l - (l)/(3) = (2l)/(3)`
Hence, mass `m` should be suspended close to wrie `B`(`Al` wire)
We know, Strian `= ("stress")/(Y)`
So, for equal strain in the wires,
`rArr (S_(A))/(Y_("steel")) = (S_(B))/(Y_(Al)`
`rArr (Y_("steel"))/(T_(A)//a_(A)) = (Y_(Al))/(T_(B)//a_(B))`
`rArr (Y_("steel"))/(Y_(Al)) = (T_(A))/(T_(B)) xx (a_(B))/(a_(A)) = ((x)/(l-x)) ((2a_(A))/(a_(A)))`
`rArr (200xx10^(9))/(70xx10^(9)) = (2x)/(l-x) rArr (20)/(7) = (2x)/(l-x)`
`rArr 17x = 10l rArr x = (10l)/(17)`
`rArr l-x = l - (10l)/(17) = (7l)/(17)`
Hence, mass `m` should be suspended close to wire `A`(steel wire).