$A$ 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,the ratio of the left gap resistance to the right gap resistance is $2:3$. The balance length from the left end is (in $\,cm$)

When the two known resistances $R$ and $S$ are connected in the left and right gaps of a meter bridge respectively,the null point is found at a distance $l_1$ from the zero end of the meter bridge wire. An unknown resistance $X$ is now connected in parallel with $S$,and the null point is found at a distance $l_2$ from the zero end of the meter bridge wire. The unknown resistance $X$ is:

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 a balanced meter bridge,the segment of wire opposite to a known resistance of $70 \Omega$ is $70 \text{ cm}$. The unknown resistance is (in $Omega$)

The resistances in the two arms of a meter bridge are $5\,\Omega$ and $R\,\Omega$,respectively. When the resistance $R$ is shunted with an equal resistance,the new balance point is at $1.6\,l_1$. The resistance $R$ is ................. $\Omega$.