A light rod of length 200cm 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 `0.1cm^(2)` and the other of brass of cross-section `0.2 cm^(2)`. A distance long the rod at which a weight may be hung to produce equal stresses in both the wires?

Options

A. `(4)/(3)` m from steel wire
B. `(4)/(3)` m from brass wire
C. 1m from steel wire
D. `(1)/(4)` m from brass wire

Answer

A

Explanation

As stresses are equal, so `(T_(1))/(A_(1)) = (T_(2))/(A_(2))`
or `(T_(1))/(T_(2)) = (A_(1))/(A_(2)) = (0.1)/(0.2) or T_(2) = 2T_(1)`
Now for translatory equilibrium of the rod,
`T_(1) + T_(2) = W`
From (i) and (ii), we get `T_(1) = (W)/(3), T_(2) = (2W)/(3)`
Now, if x is the distance of weight W from steel wire, then for rotational equilibrium of rod, `T_(1) x = T_(2) (2-x) or (W)/(3) x = (2W)/(3) (2-x) :. x= (4)/(3)m`