For a cell,the balancing length is $0.60 \, m$. For another cell having an electromotive force $(emf)$ $0.1 \, V$ less than the first one,the balancing length is $0.55 \, m$. What are the $emf$ values of the two cells?

  • A
    $1.2 \, V, 1.1 \, V$
  • B
    $1.2 \, V, 1.3 \, V$
  • C
    $-1.1 \, V, -1.0 \, V$
  • D
    None of these

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Similar Questions

In an experiment to measure the internal resistance of a cell by a potentiometer,it is found that the balance point is at a length of $2\,m$ when the cell is shunted by a $5\,\Omega$ resistance; and is at a length of $3\,m$ when the cell is shunted by a $10\,\Omega$ resistance. The internal resistance of the cell is,then ................ $\Omega$.

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The length of a potentiometer wire is $l$. $A$ cell of emf $E$ is balanced at a length $\left(\frac{l}{3}\right)$ from the positive end of the wire. If the length of the wire is increased by $\left(\frac{l}{2}\right)$,the distance at which the same cell gives the balancing point is (The cell in the primary circuit is ideal and no series resistance is present in the primary circuit.)

$A$ wire of length $10 \ m$ and resistance $30 \ \Omega$ is connected to a battery of $emf$ $2.5 \ V$ and internal resistance $5 \ \Omega$ through an external resistance $R$. If the potential gradient along the wire is $50 \ \mu V/mm$,then $R = $ ................. $\Omega$.

$A$ potentiometer wire is $100 \, cm$ long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $50 \, cm$ and $10 \, cm$ from the positive end of the wire in the two cases. The ratio of emfs is:

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