A light rod of length `L=2 m` is suspended horizontally from the ceiling by two wires `A` and `B` of equal lengths. The wire `A` is made of steel with the area of cross section `A_(S)=1xx10^(-5)m^(2),` while the wire `B` is made of brass of cross sectional area `A_(b)=2xx10^(-5)m^(2).` A weight `W` is suspended at a distance `x` from the wire `A` as shown in figure.
Take, Young's modulus of steel and brass as `Y_(s)=2xx10^(11)Nm^(-2)` and `Y_(b)=1xx10^(11)Nm^(-2)`.

Determine the value of `x` so that equal stresses are produced in each wire.

Options

A. `1.33m`
B. `2.5m`
C. `3.6m`
D. `2.1m`

Answer

A

Explanation

Let `F_(s)` and `F_(b)` be the forces in wire `A` and `B`, respectively. The free body diagram of the rod is shown in the figure.

Apply the condition of rotational equilibrium about the point of application of load, we get
`F_(S)X=F_(b)(L-x)`
`(F_(S))/(F_(b))=(L-x)/x`............i
If the stress are equal in two wires, we have
`(F_(S))/(F_(b))=(A_(s))/(A_(b))`
Here `A_(S)=1xx10^(-5)m^(2)` and `A_(b)=2xx10^(-5)m^(2)`
`(F_(S))/(F_(b))=1/2` ............ii
From eqn i and ii we get
`(L-x)/x=1/2` or `x=(2L)/3`
Since `L=2m, x=(4/3)=1.33m`