A potentiometer wire of length 300,cm is connected in series with a resistance 780,Omega and a standard cell of emf 4,V. A constant current flows

Q: A potentiometer wire of length $300\,cm$ is connected in series with a resistance $780\,\Omega$ and a standard cell of emf $4\,V$. A constant current flows through potentiometer wire. The length of the null point for cell of emf $20\,mV$ is found to be $60\,cm$. The resistance of the potentiometer wire is$...\Omega$

d Let resistance of potentiometers wire is $R$ $i=\frac{4}{R+780}$ Potential difference across $AB$ $=\frac{4 R}{R+780}$ Potential difference across $AC$ $=\frac{4 R \times 60}{(R+780) \times 300}=\frac{4 R}{5(R+780)}$ This should be equal to $20\,mV$ $\frac{4 R}{5(R+780)}=20 \times 10^{-3}=2 \times 10^{-2}$ $4 R=10^{-1}(R+780)$ $4 R=\frac{R}{10}+78$ $4 R-\frac{R}{10}=78$ $\frac{39 R}{10}=78$ $R=20\,\Omega$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

A potentiometer wire of length $300\,cm$ is connected in series with a resistance $780\,\Omega$ and a standard cell of emf $4\,V$. A constant current flows through potentiometer wire. The length of the null point for cell of emf $20\,mV$ is found to be $60\,cm$. The resistance of the potentiometer wire is$...\Omega$

A$78$
B$200$
C$2$
D$20$

JEE MAIN 2022, Diffcult

STD 11 Science
NEET
STD 12 - 3. current electricity
Physics

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