The resistances in the two arms of the 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 .................. $\Omega$.

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:

The resistance of the meter bridge wire $AB$ in the given figure is $4\,\Omega$. With a cell of emf $\varepsilon_1 = 0.5\,\text{V}$ and rheostat resistance $R_{h1} = 2\,\Omega$,the null point is obtained at point $J$. When the cell is replaced by another one of emf $\varepsilon_2$,the same null point $J$ is found for $R_{h2} = 6\,\Omega$. The emf $\varepsilon_2$ is ................. $\text{V}$.

To find the value of an unknown resistance by using a meter bridge,why are the positions of $R$ and $S$ interchanged?

$A$ wire of length $3 \, m$ connected in the left gap of a meter-bridge balances $8 \, \Omega$ resistance in the right gap at a point, which divides the bridge wire in the ratio $3:2$. The length of the wire corresponding to a resistance of $1 \, \Omega$ is: (in $ \, m$)

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