In a meter bridge experiment,when resistance $X$ is in the left gap and $Y$ is in the right gap,the null point is found at $20 \ cm$ from the left end. If $X < Y$,what will be the new position of the null point from the same end if a resistance of $4X$ is placed in the left gap instead of $X$?

When two resistances $R_1$ and $R_2$ are connected in series and introduced into the left gap of a meter bridge and a resistance of $10 \ \Omega$ is introduced into the right gap,a null point is found at $60 \ cm$ from the left side. When $R_1$ and $R_2$ are connected in parallel and introduced into the left gap,a resistance of $3 \ \Omega$ is introduced into the right gap to get a null point at $40 \ cm$ from the left end. The product $R_1 R_2$ is $............. \ \Omega$.

$A$ 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 the meter bridge experiment,for given values of $R_1$ and $R_2$,the balance point is $20 \ cm$ from point $A$. When $R_2$ is shunted by a $20 \ \Omega$ resistance,the balancing point shifts to $50 \ cm$ from point $A$. Find the value of $R_2$ in $\Omega$.

In a metre bridge experiment,the ratio of the left gap resistance to the right gap resistance is $2:3$. The balance point from the left is: (in $cm$)

Resistances in the left gap and right gap of a meter bridge are $10 \Omega$ and $30 \Omega$ respectively. If the resistances in the two gaps are interchanged, the balance point will shift to the right by: (in $\text{cm}$)