In a meter bridge experiment,a null point is obtained at $40 \ cm$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$,then the new position of the null point from the same end,if one decides to balance a resistance of $3X$ against $Y$,will be close to .............. $cm$.

$A$ meter bridge with two resistances $R_{1}$ and $R_{2}$ as shown in the figure was balanced (null point) at $40 \text{ cm}$ from the point $P$. The null point changed to $50 \text{ cm}$ from the point $P$, when a $16 \ \Omega$ resistance is connected in parallel to $R_{2}$. The values of resistances $R_{1}$ and $R_{2}$ are . . . . . . .

In a meter bridge experiment,$S$ is a standard resistance and $R$ is a resistance wire. It is found that the balancing length is $l = 25 \; cm$. If $R$ is replaced by a wire of half the length and half the diameter of $R$ but of the same material,then the new balancing length $l^{\prime}$ (in $cm$) will be:

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

Two resistances are connected in the two gaps of a meter bridge. The balancing point is obtained at $20 \ cm$. When a resistance of $15 \ \Omega$ is connected in series with the smaller resistance of the two,the balancing point shifts to $40 \ cm$. The value of the smaller resistance 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?