When a metal conductor connected to the left gap of a meter bridge is heated,the balancing point

In the gaps of a meter bridge,two unknown resistances $A$ and $B$ are connected. $A$ null point is obtained when the wire is divided in the ratio $3:4$. When each resistance is increased by $40 \Omega$,the null point divides the wire in the ratio $7:8$. The values of resistances $A$ and $B$ are respectively:

Explain how the value of an unknown resistor can be obtained by using a meter bridge.

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 $cm$)

Consider a $72\, cm$ long wire $AB$ as shown in the figure. The galvanometer jockey is placed at $P$ on $AB$ at a distance $x\, cm$ from $A$. The galvanometer shows zero deflection. The value of $x$,to the nearest integer,is ..... $cm$.

Two wires $A$ and $B$ of equal lengths are connected in left and right gap respectively of a metre bridge, null point is obtained at $40 \text{ cm}$ from left end. Diameters of the wires $A$ and $B$ are in the ratio $3:1$ respectively, the ratio of specific resistance of $A$ to that of $B$ is (in $: 1$)