In a metre bridge experiment, the balance point is obtained if the gaps are closed by $2 \text{ } \Omega$ and $3 \text{ } \Omega$. $A$ shunt of $X \text{ } \Omega$ is added to $3 \text{ } \Omega$ resistor to shift the balancing point by $22.5 \text{ cm}$. The value of $X$ is

In the meter bridge experiment, the length $AB$ of the wire is $1 \, m$. The resistors $X$ and $Y$ have values $5 \, \Omega$ and $2 \, \Omega$ respectively. When a shunt resistance $S$ is connected in parallel to $X$, the balancing point is found to be $0.625 \, m$ from $A$. Then, the resistance of the shunt $S$ is

The resistance in the left and right gaps of a meter bridge are $10 \Omega$ and $30 \Omega$ respectively. If the bridge is balanced,then the distance of the null point from the center of the wire is (in $cm$)

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 a meter bridge,the null point is found at a distance of $25\, cm$ from $A$. If now a resistance of $10\,\Omega$ is connected in parallel with $S$,the null point occurs at the midpoint of $AB$. The value of $R$ is ................. $\Omega$. (in $.67$)

$A$ 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 $\Omega$)