Q194 / 200 · MHT CET - Medical 2026 (24 April Shift 1) · Physics
A potentiometer wire has length 4 m and resistance $8\ \Omega$. The resistance that must be connected in series with the wire and the driving battery of e.m.f. 2 V, so as to get a potential gradient of 1 mV per cm on the wire is
- A$50\ \Omega$
- B$42\ \Omega$
- C$40\ \Omega$
- Correct answerD$32\ \Omega$
Solution
Potential gradient $k = 1\text{ mV/cm} = \dfrac{10^{-3}\text{ V}}{10^{-2}\text{ m}} = 0.1\text{ V/m}$
Length of the potentiometer wire $L = 4\text{ m}$
Potential difference across the potentiometer wire $V = k \times L = 0.1 \times 4 = 0.4\text{ V}$
Current in the circuit $I = \dfrac{V}{R_p} = \dfrac{0.4}{8} = 0.05\text{ A}$
Total resistance of the circuit $R_{eq} = \dfrac{E}{I} = \dfrac{2}{0.05} = 40\ \Omega$
Resistance to be connected in series $R = R_{eq} - R_p = 40 - 8 = 32\ \Omega$
Answer: $32\ \Omega$
Length of the potentiometer wire $L = 4\text{ m}$
Potential difference across the potentiometer wire $V = k \times L = 0.1 \times 4 = 0.4\text{ V}$
Current in the circuit $I = \dfrac{V}{R_p} = \dfrac{0.4}{8} = 0.05\text{ A}$
Total resistance of the circuit $R_{eq} = \dfrac{E}{I} = \dfrac{2}{0.05} = 40\ \Omega$
Resistance to be connected in series $R = R_{eq} - R_p = 40 - 8 = 32\ \Omega$
Answer: $32\ \Omega$
Hence, the correct answer is option: 'D'.