$A$ potentiometer wire has length $4\, m$ and resistance $8\, \Omega$. The resistance that must be connected in series with the wire and an accumulator of e.m.f. $2\, V$,so as to get a potential gradient of $1\, mV$ per $cm$ on the wire is ............. $\Omega$.

  • A
    $32$
  • B
    $40$
  • C
    $44$
  • D
    $48$

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The length of a wire of a potentiometer is $100 \, cm$,and the $emf$ of its standard cell is $E \, volt$. It is employed to measure the $emf$ of a battery whose internal resistance is $0.5 \, \Omega$. If the balance point is obtained at $l = 30 \, cm$ from the positive end,the $emf$ of the battery is (where $i$ is the current in the potentiometer wire).

$A$ potentiometer wire of length $100 \, cm$ has a resistance of $10 \, \Omega$. It is connected in series with a resistance $R$ and a cell of $emf$ $2 \, V$ and of negligible internal resistance. $A$ source of $emf$ $10 \, mV$ is balanced against a length of $40 \, cm$ of the potentiometer wire. What is the value of external resistance $R$?

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$A$ potentiometer consists of two wires $AC$ and $CB$ of the same material and of equal lengths but diameters in the ratio $3 : 1$. Then the potential gradients on the two wires will be in the ratio :-

The arrangement as shown in the figure is called:

In the circuit shown, a four-wire potentiometer is made of a $400\, cm$ long wire, which extends between $A$ and $B$. The resistance per unit length of the potentiometer wire is $r = 0.01\, \Omega /cm$. If an ideal voltmeter is connected as shown with jockey $J$ at $50\, cm$ from end $A$, the expected reading of the voltmeter will be: ............... $V$

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