In a meter bridge,the two gaps contain resistances of $10 \, \Omega$ and $30 \, \Omega$ respectively. If these resistances are interchanged,the null point shifts by ..... cm.

In a meter bridge,the gaps are enclosed by resistances of $2 \ \Omega$ and $3 \ \Omega$. The value of shunt to be added to the $3 \ \Omega$ resistor to shift the balancing point by $22.5 \ cm$ is (in $\Omega$)

In a meter bridge,the resistances $R$ and $S$ are such that the null point is found at a distance of $40 \text{ cm}$ from one end. If a resistance of $10 \Omega$ is connected in parallel with shunt $S$,then the null point occurs at $90 \text{ cm}$ from the same end. The values of the two resistances $R$ and $S$ respectively are:

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 $: 1$)

In a meter bridge,as shown in the figure,it is given that resistance $Y = 12.5 \, \Omega$ and the balance point is obtained at a distance $l_1 = 39.5 \, cm$ from end $A$ (by jockey $J$). After interchanging the resistances $X$ and $Y$,a new balance point is found at a distance $l_2$ from end $A$. What are the values of $X$ and $l_2$?

To find the value of an unknown resistance by using a meter bridge,why are the positions of $R$ and $S$ interchanged?