$A$ $10 \, m$ long wire of $20 \, \Omega$ resistance is connected with a battery of $3 \, V$ $e.m.f.$ (negligible internal resistance) and a $10 \, \Omega$ resistance is joined to it in series. The potential gradient along the wire in $V/m$ is:

  • A
    $0.02$
  • B
    $0.3$
  • C
    $0.2$
  • D
    $1.3$

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Similar Questions

$A$ potentiometer wire of length $4 \, m$ and resistance $5 \, \Omega$ is connected in series with a resistance of $992 \, \Omega$ and a cell of e.m.f. $4 \, V$ with internal resistance $3 \, \Omega$. The length of $0.75 \, m$ on the potentiometer wire balances the e.m.f. of (in $ \, mV$)

In the given potentiometer circuit,the resistance of the potentiometer wire $AB$ is $R_0$. $C$ is a cell of internal resistance $r$. The galvanometer $G$ does not give zero deflection for any position of the jockey $J$. Which of the following cannot be a reason for this?

In a potentiometer arrangement,a cell gives a balancing point at $75 \, cm$ length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf's of two cells respectively is $3:2$,the difference in the balancing length of the potentiometer wire in above two cases will be ......... $cm$.

In a potentiometer (see figure) a balance is obtained at a length of $400 \ mm$ when using a known battery of emf $1.6 \ V$. After removing this battery,another battery of unknown emf is used and balance is obtained at a length of $650 \ mm.$ The emf of the unknown battery is ............. $V$. (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 $\Omega$?

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