Question

A copper wire of length 2.4 m and a steel wire of length 1.6 m, both of diameter 3 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.7 mm. The load applied is

$(Y_{copper} = 1.2 \times \text{1}\text{0}^{\text{11}}\text{ N }\text{m}^{- 2}\text{, }Y_{steel} = 2 \times \text{1}\text{0}^{\text{11}}\text{ N }\text{m}^{- 2})$

• A. $1.2 \times 10^{2}\text{ m}$
• B. $1.8 \times 10^{2}\text{ m}$
• C. $2.4 \times 10^{2}\text{ m}$
• D. $3.2 \times 10^{2}\text{ m}$

Answer 1

Answer: (B)

$1.8 \times 10^{2}\text{ m}$

Answer 2

: Let the load applied be W.

Since both the wires have the same tension (equal to the load W) and same area of cross-section A, hence both wires have the same tensile stress.

As stress = Young’s modulus ×Strain

$\therefore\frac{W}{A} = Y_{C} \times \frac{\Delta L_{C}}{L_{C}} = Y_{S} \times \frac{\Delta L_{S}}{L_{S}}$ ……(i)

Where the subscripts C and S refer to copper and steel respectively.

$\therefore\frac{\Delta L_{C}}{\Delta L_{S}} = \frac{L_{C}}{L_{S}} \times \frac{Y_{S}}{Y_{C}}$

Here,

For copper wire, $L_{C} = 2.4m$

$Y_{C} = 1.2 \times 10^{11}Nm^{- 2}$

For steel wire, $L_{S} = 1.6m$

$Y_{S} = 2 \times 10^{11}Nm^{- 2}$

$\therefore\frac{\Delta L_{C}}{\Delta L_{S}} = \left( \frac{2.4m}{1.6m} \right) \times \frac{(2 \times 10^{11}Nm^{- 2})}{(1.2 \times 10^{11}Nm^{- 2})} = \frac{5}{2}$…..(ii)

The total elongation is

$\Delta L_{C} + \Delta L_{S} = 0.7mm = 7 \times 10^{- 4}m$ ……(iii)

Solving Eq. (ii) and (iii), we get

$\Delta L_{C} = 5 \times 10^{- 4}mand\Delta L_{S} = 2 \times 10^{- 4}m$

Form Eq. (i)

$W = A \times Y_{C} \times \frac{\Delta L_{C}}{L_{C}}$

$= \pi \times (1.5 \times 10^{- 3}m)^{2} \times 1.2 \times 10^{11}Nm^{- 2}$

$\times \frac{5 \times 10^{- 4}m}{2.4m}$

=$1.8 \times 10^{2}N$