A light rod AC of length 2.00 m is suspended from the ceiling horizontally by means 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)`. The position of point D from end A along the rod at which a weight may be hung to produce equal stress in both the wires is [Young's modulus of steel is `2xx10^(11) Nm^(2)` and for brass is `1xx10^(11) Nm^(-2)`]

Options

A. `1.00 m`
B. `(2//3)m`
C. `(4//3)m`
D. `(5//3)m`

Answer

C

Explanation

Given , stress in steel = stress in brass
`:. (T_(s))/(A_(s))=(T_(B))/(A_(B))` or `(T_(s))/(T_(B))=(A_(s))/(A_(B))=(10^(-3))/(2xx10^(-3))=(1)/(2)`

As the system is is equilibrium, taking moments of forces about point D, we have
`T_(s)x=T_(B)(2-x)`
`:. (T_(s))/(T_(B))=(2-x)/(x)` or `(1)/(2)=(2-x)/(x)`
On soliving, `x=4//3 m`