Two resistances are connected in two gaps of a meter bridge. The balance point is $20 \ cm$ from the zero end. $A$ resistance of $15 \ \Omega$ is connected in series with the smaller of the two. The null point shifts to $40 \ cm$. The value of the smaller resistance in $\Omega$ is

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 $X$?

In a meter bridge experiment, the balance point from the left is $37.5 \ cm$. The ratio of the right gap resistance to the left gap resistance is

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 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 meter bridge experiment,the balance point is obtained if the gaps are closed by $2 \Omega$ and $3 \Omega$ resistors. $A$ shunt of $X \Omega$ is added to the $3 \Omega$ resistor to shift the null point by $22.5 \text{ cm}$. The value of $X$ is: (in $\Omega$)