Question

A brass wire of length $2\,\text{m}$ and radius $1\,\text{mm}$ at $27^\circ\text{C}$ is held taut between two rigid supports. Initially it was cooled to a temperature of $-43^\circ\text{C}$ creating a tension $T$ in the wire. The temperature to which the wire has to be cooled in order to increase the tension in it to $1.4T$ is ___$^\circ$C.

• A. $-71$
• B. $-65$
• C. $-80$
• D. $-86$

Hint:

For wires with fixed ends, thermal stress (and hence tension) is directly proportional to temperature change.

Answer 1

The Correct Option is A

To solve this problem, we need to understand the relationship between temperature and tension in the context of thermal expansion or contraction of the brass wire. The problem indicates that the wire contracts as it is cooled, increasing tension due to restricted movement by the rigid supports.

Let's explore the relevant concepts and solve this numerically:

1. The thermal expansion (or contraction) of materials is given by the formula: 
   \(\Delta L = \alpha L_0 \Delta T\) 
   where \(\alpha\) is the coefficient of linear expansion, \(L_0\) is the original length, and \(\Delta T\) is the change in temperature.
2. The tension, \(T\), created in the wire due to a temperature change \(\Delta T\) can be expressed as proportional to the strain induced in the wire, which depends on its thermal contraction. Assuming the material has a Young's Modulus \(Y\), we express this tension as: 
   \(T = Y \cdot \frac{\Delta L}{L_0} \cdot A\) 
   where \(A\) is the cross-sectional area of the wire.
3. The cross-sectional area \(A\) for a circular wire is: 
   \(A = \pi r^2\) 
   with \(r = 1 \text{ mm} = 0.001 \text{ m}\).
4. Initially, the wire is cooled from \(27^\circ\text{C}\) to \(-43^\circ\text{C}\), creating tension \(T\). The temperature change is: 
   \(\Delta T_1 = 27 - (-43) = 70^\circ\text{C}\).
5. To increase tension to \(1.4T\), we need a new temperature change \(\Delta T_2\) such that: 
   \(1.4T = \text{Y} \cdot \frac{\alpha L_0 (\Delta T_2)}{L_0} \cdot A\) 
   Simplifying the proportional change (since we want the same terms to cancel out), we have: 
   \(1.4 \cdot 70 = \Delta T_2\).
6. Solving gives: 
   \(\Delta T_2 = 98^\circ\text{C}\)
7. Thus, the new temperature \(T_\text{new}\) is: 
   \(T_\text{new} = 27^\circ\text{C} - 98^\circ\text{C} = -71^\circ\text{C}\).

Therefore, the correct answer is \(-71^\circ\text{C}\).