Question

A brass wire 1.8 m long at 27°C is held taut with negligible tension between two rigid supports. Diameter of the wire is 2 mm, its coefficient of linear expansion, $\alpha_{Brass} = 2 \times 1{0^{- 5}}^{o}C^{- 1}$ and its Young’s modulus, $Y_{Brass} = 9 \times 10^{10}Nm^{- 2}$ If the wire is cooled to a temperature $- 39^{o}C,$ tension developed in the wire is

• A. $2.7 \times 10^{2}N$
• B. $3.7 \times 10^{2}N$
• C. $4.7 \times 10^{2}N$
• D. $5.7 \times 10^{2}N$

Answer 1

Answer: (B)

$3.7 \times 10^{2}N$

Answer 2

Let F be tension developed in the wire,

$\therefore Y = \frac{F/A}{\Delta L/L}$

As $\Delta L = L\alpha\Delta T$

$\therefore = \frac{F}{A\alpha\Delta T}orF = YA\alpha\Delta T$

Hence, $\alpha = 2.0 \times 10^{- 5}{^\circ}C^{- 1},Y = 9 \times 10^{10}Nm^{- 2}$

$A = \pi(1 \times 10^{- 3})^{2}m^{2},$

$\Delta T = 66{^\circ}C$

$\therefore$ $F = 9 \times 10^{10} \times \pi(1 \times 10^{- 3})^{2} \times 2.0 \times 10^{- 5} \times 66$

$= 3.7 \times 10^{2}N$