A potentiometer wire has a length of 100 cm | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) In a potentiometer,the potential drop across a length $\ell$ of the wire is given by $V = k\ell$,where $k$ is the potential gradient.Given the tot

Vedclass · Accepted Answer

$\frac{30E}{100}$

Vedclass · Answer

(D) In a potentiometer,the potential drop across a length $\ell$ of the wire is given by $V = k\ell$,where $k$ is the potential gradient.Given the total length $L = 100 \ cm$ and the total $emf \ E$ applied across the wire,the potential gradient is $k = \frac{E}{L} = \frac{E}{100}$.At the balance point,no current flows through the secondary circuit (the circuit containing the battery with $emf \ E_0$). Therefore,the internal resistance of the battery does not affect the potential difference measured at the balance point.The $emf \ E_0$ is equal to the potential drop across the length $\ell = 30 \ cm$.Thus,$E_0 = k 	imes \ell = \left(\frac{E}{100}ight) 	imes 30 = \frac{30E}{100}$.

Similar Questions

$A$ potentiometer wire has a length of $5 \, m$ and a resistance of $5 \, \Omega$. If the null point is obtained at $300 \, cm$,what is the $emf$ $E$ of the cells (connected in parallel) in $V$?

$A$ cell is connected to a potentiometer,and the balancing length is obtained at $125 \ cm$. When a resistor of $2 \ \Omega$ is connected in parallel with the cell,the balancing length is obtained at $100 \ cm$. What is the internal resistance of the cell in $\Omega$?

The balancing length for a cell is $560 \, cm$ in a potentiometer experiment. When an external resistance of $10 \, \Omega$ is connected in parallel to the cell, the balancing length changes by $60 \, cm$. The internal resistance of the cell in ohms, is

Vedclass Test Series

Exam Paper Generator

Online Exam Module