The resistances in the two gaps of a balanced meter bridge are $X \ \Omega$ and $3X \ \Omega$ respectively. If the resistances are interchanged, the balance point shifts by: (in $\text{ cm}$)

Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length $l_{1}$ is $40\,cm$. Now an unknown resistance $x$ is connected in series with $P$ and the new balancing length is found to be $80\,cm$ measured from the same end. Then the value of $x$ will be $.......\Omega$.

In a meter bridge experiment, the balance point is obtained at length $l_1$ cm from the left end when resistances in the left gap and right gap are $15 \text{ } \Omega$ and $R \text{ } \Omega$ respectively. When the resistance $R$ is shunted with an equal resistance, the new balance point is at $1.6 l_1$. The resistance $R$ in ohm 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 $\Omega$)

In a meter-bridge,if the left and right gaps are connected with $2 \Omega$ and $3 \Omega$ resistances,respectively,then the bridge is balanced. What resistance should be connected with the $3 \Omega$ resistance to get the balancing point at the midpoint of the bridge wire?

In the shown arrangement of the experiment of the metre bridge,if $AC$ corresponding to null deflection of the galvanometer is $x$,what would be its value if the radius of the wire $AB$ is doubled?