The adjoining figure shows the connections of a potentiometer experiment to determine the internal resistance of a Leclanché cell. When the cell is on open circuit,the balancing length of the potentiometer wire is $3.4 \, m$,and on closing the key $K_2$,the balancing length becomes $1.7 \, m$. If the resistance $R$ through which current is drawn is $10 \, \Omega$,then the internal resistance of the cell is .............. $\Omega$.

  • A
    $0.1$
  • B
    $1$
  • C
    $10$
  • D
    $1.1$

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Similar Questions

In a potentiometer experiment,cells of e.m.f. $E_{1}$ and $E_{2}$ are connected in series $(E_{1} > E_{2})$,and the balancing length is $64 \ cm$. If the polarity of $E_{2}$ is reversed,the balancing length becomes $32 \ cm$. The ratio $\frac{E_{1}}{E_{2}}$ is:

When a potentiometer is connected between the points $A$ and $B$ as shown in the circuit,the balance point is obtained at $64 \ cm$. When it is connected between $A$ and $C$,the balance point is $8 \ cm$. If the potentiometer is connected between $B$ and $C$,the balance point will be: (in $cm$)

$A$ cell of internal resistance $1.5\,\Omega$ and of $e.m.f.$ $1.5\,V$ balances at $500\,cm$ on a potentiometer wire. If a wire of $15\,\Omega$ is connected between the balance point and the cell,then the balance point will shift:

Two cells of e.m.f. $E_1$ and $E_2$ $(E_1 > E_2)$ are connected as shown in the figure. When a potentiometer is connected between points $A$ and $B$, the balancing length of the potentiometer wire is $412 \text{ cm}$. When the same potentiometer is connected between points $A$ and $C$, the balancing length is $103 \text{ cm}$. The ratio $E_1 : E_2$ is:

$A$ potentiometer circuit is set up as shown. The potential gradient across the potentiometer wire is $k \, V/cm$ and the ammeter present in the circuit reads $1.0 \, A$ when the two-way key is switched off. The balance points,when the key between the terminals $(i)$ $1$ and $2$ and $(ii)$ $1$ and $3$ is plugged in,are found to be at lengths $l_1$ and $l_2$ respectively. The magnitudes of the resistors $R$ and $X$ in ohms are equal to:

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