In a metre bridge experiment, the resistance in the left gap is $15 \text{ } \Omega$ and in the right gap is $45 \text{ } \Omega$. The bridge is balanced. The distance of the null point from the centre of the wire is (in $\text{ cm}$)

In a meter bridge experiment,when resistance $X$ is in the left gap and $Y$ is in the right gap,the null point is found at $20 \ cm$ from the left end. If $X < Y$,what will be the new position of the null point from the same end if a resistance of $4X$ is placed in the left gap instead of $X$?

In a meter bridge experiment, two resistances $X$ and $Y$ in the two gaps give a null point dividing the wire in the ratio $2 : 3$. When each resistance is increased by $30 \text{ } \Omega$, the null point divides the wire in the ratio $5 : 6$. The resistances $X$ and $Y$ are respectively:

$A$ balanced bridge is shown in the circuit diagram. The metre bridge wire has a resistance of $1 \Omega \text{ cm}^{-1}$. The current drawn from the battery is (Internal resistance of the battery is negligible). (in $\text{ A}$)

In a meter bridge (Figure), the null point is found at a distance of $33.7 \; cm$ from $A$. If now a resistance of $12 \; \Omega$ is connected in parallel with $S$, the null point occurs at $51.9 \; cm$. Determine the values of $R$ and $S$.

$A$ wire of length $10 \ cm$ and radius $\sqrt{7} \times 10^{-4} \ m$ is connected across the right gap of a meter bridge. When a resistance of $4.5 \ \Omega$ is connected in the left gap using a resistance box,the balance length is found to be at $60 \ cm$ from the left end. If the resistivity of the wire is $R \times 10^{-7} \ \Omega \ m$,then the value of $R$ is: