The unknown resistances are connected in two gaps of a metre bridge. The null point is at $20 \ cm$ 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 smaller resistance is (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 $cm$)

In a meter bridge,the balancing length from the left end is found to be $25 \ cm$. The value of the unknown resistance is (assume,standard resistance of $1 \ \Omega$ is in the right gap). (in $\Omega$)

Two known resistances are connected in two gaps of a meter bridge. The null point is obtained at $20 \ cm$ 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 smaller resistance is (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 $\Omega$)

In a meter bridge experiment,a null point is obtained at $20 \ cm$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$,then where will be the new position of the null point from the same end,if one decides to balance a resistance of $4X$ against $Y$ (in $cm$)?