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

When a wire is connected in the left gap of a metre bridge,the balancing point is at $40 \ cm$ from the left end of the bridge wire. If the wire in the left gap is stretched so that its length is doubled and again connected in the same gap,then the balancing point from the left end of the bridge wire 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 (in $Omega$)

$A$ resistance of $2 \Omega$ is connected across one gap of a metre-bridge (the length of the wire is $100 \text{ cm}$) and an unknown resistance,greater than $2 \Omega$,is connected across the other gap. When these resistances are interchanged,the balance point shifts by $20 \text{ cm}$. Neglecting any corrections,the unknown resistance is (in $Omega$)

In a meter bridge,the balancing length from the left end is found to be $25 \ cm$. The value of the unknown resistance is (assume,standard resistance of $1 \ \Omega$ is in the right gap). (in $Omega$)

The resistances in the left and right gaps of a metre bridge are $40 \Omega$ and $60 \Omega$ respectively. When the bridge is balanced, the distance of the null point from the centre of the wire towards the left is (in $\text{ cm}$)