In a meter bridge,two gaps in the metallic strip are connected by $3 \Omega$ and $9 \Omega$ resistors. What should be the value of the shunt that needs to be added in parallel to the $9 \Omega$ resistor to shift the balancing point by $25 \text{ cm}$ (in $\Omega$)?

In an experiment of a meter bridge, a null point is obtained at the centre of the bridge wire. When a resistance of $10\, \Omega$ is connected in one gap, the value of resistance in the other gap is ............. $\Omega$.

In a meter-bridge experiment,a resistance of $18 \Omega$ is connected in the left gap and an unknown resistance $R$ is connected in the right gap. The null point is obtained at $\ell_{1}$ from the left end. If the unknown resistance is replaced by $(\frac{R}{3}) \Omega$,the null point is obtained at $1.5 \ell_{1}$. The unknown resistance $R$ is: (in $\Omega$)

Two wires $A$ and $B$ of equal lengths are connected in left and right gap respectively of a metre bridge, null point is obtained at $40 \text{ cm}$ from left end. Diameters of the wires $A$ and $B$ are in the ratio $3:1$ respectively, the ratio of specific resistance of $A$ to that of $B$ is (in $: 1$)

In the circuit shown, a meter bridge is in its balanced state. The meter bridge wire has a resistance of $0.1 \, \Omega/cm$. The value of the unknown resistance $X$ and the current drawn from the battery of negligible internal resistance are:

$A$ meter bridge is set up as shown,to determine an unknown resistance '$X$' using a standard $10 \ \Omega$ resistor. The galvanometer shows a null point when the tapping key is at the $52 \ cm$ mark. The end corrections are $1 \ cm$ and $2 \ cm$ respectively for the ends $A$ and $B$. The determined value of '$X$' is (in $\Omega$)