In a meter bridge experiment,the ratio of the left gap resistance to the right gap resistance is $2:3$. The balance length from the left end is (in $\,cm$)

The resistance per centimeter of a meter bridge wire is $r$. $A$ resistance of $X \ \Omega$ is placed in the left gap and a resistance of $25 \ \Omega$ is placed in the right gap. The balancing length from the left end is $40 \ cm$. Now,the wire is replaced by another wire of $2r$ resistance per centimeter. The new balancing length for the same settings will be at: (in $cm$)

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 a metre-bridge experiment,the balance point is obtained when the gaps are closed by $2 \Omega$ and $3 \Omega$ resistors. $A$ shunt of $x \Omega$ is added in parallel to the $3 \Omega$ resistor to shift the balance point by $22.5 \ cm$. The value of $x$ is: (in $\Omega$)

$A$ resistance of $2\,\Omega$ is connected across one gap of a meter-bridge and an unknown resistance,greater than $2\,\Omega$,is connected across the other gap. When these resistances are interchanged,the balance point shifts by $20\,cm$. Neglecting any end correction,the unknown resistance is ................ $\Omega$.

In a meter bridge,the point $D$ is the neutral point (null point) as shown in the figure.