$A$ 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 $....... \times 10^{-2}\, m$.

In a meter bridge experiment,resistances are connected as shown in the figure. Initially,resistance $P = 4\,\Omega$ and the null point $N$ is at $60\,cm$ from $A$. Now,an unknown resistance $R$ is connected in series to $P$,and the new position of the null point is at $80\,cm$ from $A$. The value of unknown resistance $R$ is

In the circuit shown, a meter bridge is in its balanced state. The meter bridge wire has a resistance of $0.1 \, \Omega/cm$. The value of the unknown resistance $X$ and the current drawn from the battery of negligible internal resistance are:

When a wire is connected in the left gap of a metre bridge,the balancing point is at $40 \ cm$ from the left end of the bridge wire. If the wire in the left gap is stretched so that its length is doubled and again connected in the same gap,then the balancing point from the left end of the bridge wire is

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 $\Omega$)

$A$ resistance of $2 \Omega$ is connected across one gap of a metre-bridge (the length of the wire is $100 \text{ cm}$) and an unknown resistance,greater than $2 \Omega$,is connected across the other gap. When these resistances are interchanged,the balance point shifts by $20 \text{ cm}$. Neglecting any corrections,the unknown resistance is (in $\Omega$)