A light rod of length 2 m is suspended from the ceiling horizontally by mean.s of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross-section `10^(-3)m^(2)` and the other is of brass of cross -section `2xx10^(-3)m^(2)`. Find out the position along of corss-section `2xx10^(-3)m^(2)`. Find out the position along the rod at which a weight may be hung to produce, (i) equal stress in both wires (ii) equal strains in both wires
Young.s modulus of brass `=1xx10^(11)N//m^(2)`
Young.s modulus of steel `=2xx10^(11)N//m^(2)`

Explanation

Suppose `a_(1) and a_(2)` are the corss - sectional areas, and `Y_(1) and Y_(2)` are the Young.s moduli of steel and brass wire respectively. Let `T_(1) and T_(2)` are tensions in the steel and brass wires respectively.
Let x is distance of the position of the hanging weight from the steel wire.
(i) First case : For equal stress in both wires, we have
`(T_(1))/(a_(1))=(T_(2))/(a_(2)) (or) (T_(1))/(10^(-3))=(T_(2))/(2xx10^(-3))(or) T_(2)=2T_(1)" "......(i)`
As the whole system is in equilibrium, so `Sigma bartau=0`.
Taking moment of all the forces acting on the rod about C,
we have `T_(1)x-T_(2)(2-x)=0" "...( ii)`
Solving equations (i) and (ii), we get `x=(4)/(3)m`
(ii) Second case: For equal strains in both the wires `e_(1)=e_(2)`
`(T_(1)l)/(a_(1)Y_(1))=(T_(2)l)/(a_(2)Y_(2))(or) (T_(1))/(10^(-3) xx 2xx10^(11))=(T_(2))/(2xx10^(-3)xx10^(11))`
(or) `T_(1)=T_(2)" "...(iii)`
From equations (ii) and (iii), we get x = 1m