In a meter bridge,the wire of length $1\, m$ has a non-uniform cross-section such that the variation $\frac{dR}{d\ell}$ of its resistance $R$ with length $\ell$ is $\frac{dR}{d\ell} \propto \frac{1}{\sqrt{\ell}}$. Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point $P$. What is the length $AP$?

  • A
    $0.2$
  • B
    $0.3$
  • C
    $0.25$
  • D
    $0.35$

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When two resistances $R_1$ and $R_2$ are connected in series and introduced into the left gap of a meter bridge and a resistance of $10 \ \Omega$ is introduced into the right gap,a null point is found at $60 \ cm$ from the left side. When $R_1$ and $R_2$ are connected in parallel and introduced into the left gap,a resistance of $3 \ \Omega$ is introduced into the right gap to get a null point at $40 \ cm$ from the left end. The product $R_1 R_2$ is $............. \ \Omega$.

The figure shows a meter-bridge circuit,with $AB = 100\, cm$,$X = 12\,\Omega$ and $R = 18\,\Omega$,and the jockey $J$ in the position of balance. If $R$ is now made $8\,\Omega$,through what distance will $J$ have to be moved to obtain balance? .............. $cm$

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The figure shows a rough sketch of a meter bridge. The galvanometer $(G)$ shows zero deflection at a length $l \, cm$. Now,$R_1$ and $R_2$ are interchanged,and the balancing length increases by $25 \, cm$. Find the ratio $R_1/R_2$.

In the gaps of a meter bridge,two unknown resistances $A$ and $B$ are connected. $A$ null point is obtained when the wire is divided in the ratio $3:4$. When each resistance is increased by $40 \Omega$,the null point divides the wire in the ratio $7:8$. The values of resistances $A$ and $B$ are respectively:

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Resistances in the left gap and right gap of a meter bridge are $10 \Omega$ and $30 \Omega$ respectively. If the resistances in the two gaps are interchanged, the balance point will shift to the right by: (in $\text{cm}$)

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