Question

A light rod of length 200 cm 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 $\text{0.1 c}\text{m}^{2}$ and the other of brass of cross-section $\text{0.2 c}\text{m}^{2}.$ Along the rod at what distance a weight may be hung to produce equal stresses in both the wires?

$(Y_{\text{Steel}} = 2 \times 10^{11}\text{ N }\text{m}^{- 2}\text{, }\text{Y}_{\text{Brass}} = 1 \times \text{1}\text{0}^{\text{11}}\text{ N }\text{m}^{- 2})$

• A. $\frac{4}{3}$m from steel wire
• B. $\frac{4}{3}$ m from brass wire
• C. 1 m from steel wire
• D. $\frac{1}{4}$m from brass wire

Answer 1

Answer: (A)

$\frac{4}{3}$m from steel wire

Answer 2

: The situation is as shown in the figure.

Let a weight W be suspended at a distance x from steel wire. Let $T_{S}andT_{B}$be tensions in the steel and brass wires respectively.

$\therefore$Stress in steel wire$= \frac{T_{S}}{A_{S}}$

Stress in brass wire $= \frac{T_{B}}{A_{B}}$

For equal stress in both the wires

$\frac{T_{S}}{A_{S}} = \frac{T_{B}}{A_{B}}$

$\frac{T_{S}}{T_{B}} = \frac{A_{S}}{A_{B}} = \frac{0.1cm^{2}}{0.2cm^{2}} = \frac{1}{2}$ ….(i)

For the rotational equilibrium of the rod.

$T_{S}x = T_{B}(2 - x)$

$\frac{2 - x}{x} = \frac{T_{S}}{T_{B}} = \frac{1}{2}$ [Using (i)]

$4 - 2x = x$or $3x = 4$ or $x = \frac{4}{3}m$