Question

A light rod of length 2 m is suspended ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is ade of steel and is of cross-section 0.1 cm and the other of brass of cross-section 0.2 cm. Along the rod at what distance a weight may be hung to produce stresses in the wires?
`(Y_("Steel") = 2 xx 10^(11) Nm^(-2), Y_("Bass") = 1 xx 10^(11) Nm^(-2))`
A. `(4)/(3)m` from steel wire
B. `(4)/(3)m` from brass wire
C. 1 m from steel wire
D. `(1)/(4)m` from brass wire

Answer 1

Correct Answer - A
The situation is as shown in the figure.
Let a weight W be suspended at a distance x from steel wire. Let `T_(s)` and `T_(b)` be tension in the steel and brass wires respectively.

`therefore` Stress in steel wire `= (T_(B))/(A_(B))`
Stress in brass wire `= (T_(B))/(A_(B))`
For equal stress in both the wire
`(T_(s))/(T_(B)) = (T_(B))/(A_(B))`
`(T_(S))/(T_(B)) = (A_(S))/(A_(B)) = (0.1 cm^(2))/(0.2 cm^(2))=(1)/(2) " "....(i)`
For the rotational equilibrium of the rod,
`T_(S)x = T_(B) (2-x)`
`(2x-x)/(x) = (T_(S))/(T_(B)) = (1)/(2) " "["Using (i)"]`
`4-2x = x or 3 xx = 4 or x=(4)/(3) m`