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:

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 $\Omega$)

Explain the construction of a meter bridge used in the laboratory.

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

Shown in the figure below is a meter-bridge set up with null deflection in the galvanometer. The value of the unknown resistor $R$ is ............. $\Omega$.

Two resistors $2 \Omega$ and $3 \Omega$ are connected in the gaps of a meter bridge as shown in the figure. The null point is obtained with the contact of the jockey at some point on the wire $XY$. When an unknown resistor $R$ is connected in parallel with the $3 \Omega$ resistor, the null point is shifted by $22.5 \text{ cm}$ toward $Y$. The resistance of the unknown resistor $R$ is . . . . . . $\Omega$.