In the balanced condition, the values of the resistances of the four arms of a wheatstone bridge are shown in figure below. The resistance $R_{3}$ has temperature coefficient $α = K T ° C^{- 1}$ where $K = 0.0008$ and T is temperature in $° C$ . If the temperature of only $R_{3}$ is increased by $10 ° C$ , then [Take initial temperature $0 ° C$ ; Resistance of voltmeter V is very large]

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The voltage developed between C and D will be 0.5 volt at $T = 0^{0} C$

The voltage developed between C and D will be approx 0.47 volt at $T = 10^{0} C$

The voltage developed between C and D will be 0 volt at $T = 0^{0} C$

The voltage developed between C and D will be approx 0.27 volt at $T = 10^{0} C$

answer is A, C.

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Detailed Solution

$R_{3} = 300 (1 + .0008 \times 50) = 312 Δ V_{3} = \frac{50}{372} \times 312 - 50 \times \frac{5}{6} = 0.27 V$

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