For a cell of $e.m.f.$ $2\,V$,a balance is obtained for $50\, cm$ of the potentiometer wire. If the cell is shunted by a $2\,\Omega$ resistor and the balance is obtained across $40\, cm$ of the wire,then the internal resistance of the cell is ............. $\Omega$.

  • A
    $0.25$
  • B
    $0.50$
  • C
    $0.80$
  • D
    $1$

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Similar Questions

$AB$ is a potentiometer wire of length $100\, cm$ and its resistance is $10\,\Omega$. It is connected in series with a resistance $R = 40\,\Omega$ and a battery of $e.m.f.$ $2\,V$ and negligible internal resistance. If a source of unknown $e.m.f.$ $E$ is balanced by $40\, cm$ length of the potentiometer wire,the value of $E$ is ................. $V$. (in $,V$)

In a potentiometer circuit,a cell of $2\,V$ $e.m.f.$ and $5\,\Omega$ internal resistance is connected to a uniform wire of length $1000\,cm$ and resistance $15\,\Omega$. The potential gradient of the wire is:

When a cell of e.m.f. $E_1$ is connected to a potentiometer wire,the balancing length is $\ell_1$. Another cell of e.m.f. $E_2$ $(E_1 > E_2)$ is connected such that the two cells oppose each other,and the balancing length is $\ell_2$. The ratio $E_1 : E_2$ is:

When two cells are connected in series in a potentiometer circuit to assist each other,the balancing length is $6 \ m$. When they are connected in series to oppose each other,the balancing length is $2 \ m$. What is the ratio of the $EMF$ of the two cells?

$A$ potentiometer wire of length $1\,m$ and resistance $10\,\Omega$ is connected in series with a cell of $emf$ $2\,V$ with internal resistance $1\,\Omega$ and a resistance box including a resistance $R$. If the potential difference between the ends of the wire is $1\,mV$,the value of $R$ is ............. $\Omega$.

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