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:

Updated On: Feb 2, 2026

$-2 \times 10^{-2} \, \degree \text{C}^{-1}$
$-1.5 \times 10^{-2} \, \degree \text{C}^{-1}$
$-1 \times 10^{-2} \, \degree \text{C}^{-1}$
$-2.5 \times 10^{-2} \, \degree \text{C}^{-1}$

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The Correct Option is C

Approach Solution - 1

To solve this problem, we'll start by understanding the setup and conditions given:

The Wheatstone bridge is balanced initially, with arm BC, made of semiconductor, having an initial resistance $R_0 = 3 \, \text{m}\Omega$ at $25^\circ \text{C}$. This arm is cooled at $2^\circ \text{C/s}$. If the galvanometer shows no deflection after 10 seconds, the bridge remains balanced, and the resistance change condition can be expressed as follows:

The temperature change in 10 seconds is:

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

Let $\alpha$ be the temperature coefficient of resistivity. The change in resistance will be:

$\Delta R = R_0 \times \alpha \times \Delta T$

For the bridge to remain balanced:

The initial configuration has resistance $R_{DA} = 1 \, \text{m}\Omega$ and $R_{AB} = 0.8 \, \text{m}\Omega$. In the balanced condition:

$\frac{R_{AB}}{R_{BC}} = \frac{R_{DA}}{R_{CD}}$

The change in resistance $\Delta R$ should satisfy:

$R_0 + \Delta R = R_{BC} + \Delta R_{BC}= R_{CD}$

Initially, $R_{BC} = 3 \, \text{m}\Omega$ and $R_{CDC} = 3 \, \text{m}\Omega$ after 10s.

Let's substitute the known values:

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

Re-arranging gives:

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

This simplifies to:

$0.8 = 1 + 60 \times \alpha$

$60 \times \alpha = -0.2$

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

Therefore, the temperature coefficient of resistivity $\alpha$ is:

$\alpha = -1 \times 10^{-2} \, \degree \text{C}^{-1}$, which matches option $C$.

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Approach Solution -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} $.