Question

To measure the temperature coefficient of resistivity $\alpha$ of a semiconductor, an electrical arrangement shown in the figure is prepared. The arm BC is made up of the semiconductor. The experiment is being conducted at $25^\circ \text{C}$ and the resistance of the semiconductor arm is $3 \, \text{m}\Omega$. Arm BC is cooled at a constant rate of $2^\circ \text{C/s}$. If the galvanometer G shows no deflection after 10 s, then $\alpha$ is:

• A. $-2 \times 10^{-2} \, \degree \text{C}^{-1}$
• B. $-1.5 \times 10^{-2} \, \degree \text{C}^{-1}$
• C. $-1 \times 10^{-2} \, \degree \text{C}^{-1}$
• D. $-2.5 \times 10^{-2} \, \degree \text{C}^{-1}$

Answer 1

The Correct Option is C

To address this issue, the initial setup and conditions are examined:

The Wheatstone bridge is in a balanced state. Arm BC, composed of a semiconductor, possesses an initial resistance of \(R_0 = 3 \, \text{m}\Omega\) at a temperature of \(25^\circ \text{C}\). This arm undergoes cooling at a rate of \(2^\circ \text{C/s}\). If the galvanometer registers zero deflection after 10 seconds, the bridge remains balanced. The condition for resistance change can be expressed as follows:

The temperature variation over 10 seconds is calculated as:

\(\Delta T = \text{cooling rate} \times \text{time} = 2^\circ \text{C/s} \times 10 \, \text{s} = 20^\circ \text{C}\)

Let \(\alpha\) denote the temperature coefficient of resistivity. The corresponding change in resistance is given by:

\(\Delta R = R_0 \times \alpha \times \Delta T\)

For the bridge to maintain its balance:

The initial configuration includes resistances \(R_{DA} = 1 \, \text{m}\Omega\) and \(R_{AB} = 0.8 \, \text{m}\Omega\). When balanced, the following relationship holds:

\(\frac{R_{AB}}{R_{BC}} = \frac{R_{DA}}{R_{CD}}\)

The resistance change \(\Delta R\) must satisfy:

\(R_0 + \Delta R = R_{BC} + \Delta R_{BC}= R_{CD}\)

Initially, \(R_{BC} = 3 \, \text{m}\Omega\). After 10 seconds, \(R_{CD} = 3 \, \text{m}\Omega\).

Substituting the known values yields:

\(0.8 \, \text{m}\Omega / (3 \, \text{m}\Omega + \alpha \times 3 \, \text{m}\Omega \times 20^\circ \text{C}) = 1 \, \text{m}\Omega / 3 \, \text{m}\Omega\)

Rearranging the equation provides:

\(0.8 \, \text{m}\Omega \times 3 \, \text{m}\Omega = 3 \, \text{m}\Omega \times (3 \, \text{m}\Omega + \alpha \times 3 \, \text{m}\Omega \times 20^\circ \text{C})\)

This equation simplifies to:

\(0.8 = 1 + 60 \times \alpha\)

\(60 \times \alpha = -0.2\)

\(\alpha = -\frac{0.2}{60} = -\frac{1}{300} \, \text{C}^{-1} = -1 \times 10^{-2} \, \text{C}^{-1}\)

Consequently, the temperature coefficient of resistivity \(\alpha\) is:

\(\alpha = -1 \times 10^{-2} \, \degree \text{C}^{-1}\), which corresponds to option \(C\).