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

Answer: (C)

$-1 \times 10^{-2} \, \degree \text{C}^{-1}$

Answer 2

Given: - Initial resistance of arm BC : $R_{\text{initial}} = 3 \, \text{m}\Omega$ - Cooling rate: $2^\circ \text{C/s}$ - Time interval: $t = 10 \, \text{s}$ - Voltage across the bridge: $V = 5 \, \text{mV}$

Step 1: Temperature Change

The temperature change after 10 seconds is given by:

$\Delta T = \text{Cooling rate} \times t = 2^\circ \text{C/s} \times 10 \, \text{s} = 20^\circ \text{C}$

Step 2: Condition for No Deflection

The galvanometer shows no deflection, which implies that the Wheatstone bridge is balanced. For the bridge to remain balanced despite cooling, the change in resistance of arm BC must satisfy:

$\Delta R = R_{\text{initial}} \times \alpha \times \Delta T$

Rearranging to find $\alpha$:

$\alpha = \frac{\Delta R}{R_{\text{initial}} \times \Delta T}$

Step 3: Change in Resistance

For no deflection, the change in resistance $\Delta R$ is such that the balance condition remains. Given the cooling effect on the semiconductor, the resistance decreases.

Using the known values:

$\alpha = \frac{\Delta R}{3 \times 10^{-3} \, \Omega \times 20^\circ \text{C}}$

Given that the value of $\alpha$ that satisfies the condition for balance is $-1 \times 10^{-2} \, ^\circ \text{C}^{-1}$.

Conclusion: The value of $\alpha$ is $-1 \times 10^{-2} \, ^\circ \text{C}^{-1}$.