$A$ potentiometer is used for the comparison of $e.m.f.$ of two cells $E_1$ and $E_2$. For cell $E_1$,the null point is obtained at $20 \ cm$ and for $E_2$,the null point is obtained at $30 \ cm$. The ratio of their $e.m.f.$s will be:

  • A
    $2/3$
  • B
    $1/2$
  • C
    $1$
  • D
    $2$

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$AB$ 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 $,V$)

The sliding contact of a potentiometer is in the middle of the potentiometer wire having a total resistance $R_p = 1 \Omega$,as shown in the figure. An external resistance of $R_e = 2 \Omega$ is connected via the sliding contact. Find the total current drawn from the $0.9 \text{ V}$ battery. (in $\text{ A}$)

In an experiment with a potentiometer,$V_B = 10 \, V$ and the variable resistance is adjusted to $R = 50 \, \Omega$ (figure). $A$ student wanting to measure the voltage $E_1$ of a battery (approx. $8 \, V$) finds no null point. He then diminishes $R$ to $10 \, \Omega$ and is able to locate the null point on the last $(4^{th})$ segment of the potentiometer wire. Find the resistance of the potentiometer wire and the potential drop per unit length.

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$A$ cell of $emf$ $2 \,V$ and internal resistance $5 \,\Omega$ is connected to a wire of length $100 \,cm$ and resistance $15 \,\Omega$. What is the potential gradient along the wire (in $,V/cm$)?

$A$ 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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