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. `1.33m, 1m`
B. `1m, 1.33m`
C. `1.5m, 1.33m`
D. `1.33m, 1.5m`

Answer

A

Explanation

Given, stress in stell = stress in brass
`therefore (T_(S))/(A_(S)) = (T_(B))/(A_(B)), (10^(-3))/(2xx10^(-3)) = (1)/(2).....(i)`
As the system is in equilibrium, taking moments about `D`, we have
`T_(S)x = Y_(B) (2-x) rArr (T_(S))/(T_(B)) = (2-x)/(x)`.......(ii)
from equation (i) and (ii), we get `x = 1.33m`
strain `= ("stress")/(Y)`
given strain in steel = strain in brass
`therefore (T_(S)//A_(S))/(Y_(S)) = (T_(B)//A_(B))/(Y_(B)), therefore (T_(S))/(T_(B)) = (A_(S) Y_(S))/(A_(S) Y_(S)) = 1....(iii)`
from equations (ii) and (iii) we have `X = 1.0m`