$A$ resistance wire connected in the left gap of a meter bridge balances a $10\, \Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $3: 2$. If the length of the resistance wire is $1.5\, m$,then the length of $1\, \Omega$ of the resistance wire is $....... \times 10^{-2}\, m$.

An unknown resistance $R_1$ is connected in series with a resistance of $10 \,\Omega$. This combination is connected to one gap of a meter bridge while a resistance $R_2$ is connected in the other gap. The balance point is at $50 \, cm$. Now,when the $10 \,\Omega$ resistance is removed,the balance point shifts to $40 \, cm$. The value of $R_1$ is (in $\Omega$):

In a metre bridge experiment,the null point is obtained at $20 \,cm$ from the left end of the wire,when resistance $X$ is balanced against another resistance $Y$ $(X < Y)$. To balance a resistance $4X$ against $Y$,the new position of the null point from the same end will be: (in $\,cm$)

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,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?

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