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 which distance a weight may be hung along the rod, in order to produce equal strains 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: (C)

1 m from steel wire

Answer 2

: As strain =$\frac{Stress}{Young'sModulus}$

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

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

For equal strain in both the wires,

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

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

$= \frac{0.1cm^{2}}{0.2cm^{2}} \times \frac{2 \times 10^{11}Nm^{- 2}}{1 \times 10^{11}Nm^{- 2}} = 1$ ……(ii)

For the rotational equilibrium of the rod,

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

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

2 – x = x or x = 1 m