$A$ potentiometer balances at $44 \ cm$ when a cell of internal resistance $1 \ \Omega$ is in the secondary circuit. To obtain the balancing point at $40 \ cm$,the resistance to be connected in parallel to the cell is: (in $\Omega$)

  • A
    $20$
  • B
    $10$
  • C
    $30$
  • D
    $5$

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To compare the $EMF$ of two cells using a potentiometer, the balancing lengths obtained are $200 \ cm$ and $150 \ cm$. The least count of the scale is $1 \ cm$. The percentage error in the ratio of the EMFs is . . . . . . .

In the arrangement shown in the figure,when the switch $S_2$ is open,the galvanometer shows no deflection for $l = L/2$. When the switch $S_2$ is closed,the galvanometer shows no deflection for $l = 5L/12$. The internal resistance $(r)$ of the $6\, V$ cell and the $emf$ $E$ of the other battery are respectively:

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With a potentiometer,null points were obtained at $140 \, cm$ and $180 \, cm$ with cells of $emf$ $1.1 \, V$ and one unknown $X \, V$. The unknown $emf$ is .............. $V$.

$A$ potentiometer wire has a length of $5 \, m$ and a resistance of $16 \, \Omega$. The driving cell has an e.m.f. of $5 \, V$ and an internal resistance of $4 \, \Omega$. When two cells of e.m.f.s $1.3 \, V$ and $1.1 \, V$ are connected so as to assist each other and then oppose each other, the balancing lengths are respectively:

In the determination of the internal resistance of a cell with a potentiometer,the error in the measurement of the balancing length is $\pm 1 \text{ mm}$. When the cell alone is connected in the circuit,the balancing length is obtained at $60 \text{ cm}$ and when the cell is shunted with a resistance of $10 \Omega \pm 2 \%$,the balancing length is obtained at $50 \text{ cm}$. The error in the determination of the internal resistance of the cell is (in $\%$)

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