When the right gap of a meter bridge consists of two equal resistors in series,the balancing point is at $50 \ cm$. When one of the resistors in the right gap is removed and is connected in parallel to the resistor in the left gap,the balancing point is at: (in $cm$)

  • A
    $60$
  • B
    $33.3$
  • C
    $25$
  • D
    $40$

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Similar Questions

Resistance in the two gaps of a meter bridge are $10 \, \Omega$ and $30 \, \Omega$ respectively. If the resistances are interchanged, the balance point shifts by .............. $cm$.

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Reason $(R):$ The resistance of a metal increases with an increase in temperature.

In a meter bridge experiment,resistances are connected as shown in the figure. Initially,resistance $P = 4\,\Omega$ and the null point $N$ is at $60\,cm$ from $A$. Now,an unknown resistance $R$ is connected in series to $P$,and the new position of the null point is at $80\,cm$ from $A$. The value of unknown resistance $R$ is

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Shown in the figure below is a meter-bridge set up with null deflection in the galvanometer. The value of the unknown resistor $R$ is ............. $\Omega$.

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