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 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

In a meter bridge experiment,to minimize an error due to contact resistance,

In a meter bridge experiment,$S$ is a standard resistance and $R$ is a resistance wire. It is found that the balancing length is $l = 25 \; cm$. If $R$ is replaced by a wire of half the length and half the diameter of $R$ but of the same material,then the new balancing length $l^{\prime}$ (in $cm$) will be:

In a meter-bridge experiment,a resistance of $18 \Omega$ is connected in the left gap and an unknown resistance $R$ is connected in the right gap. The null point is obtained at $\ell_{1}$ from the left end. If the unknown resistance is replaced by $(\frac{R}{3}) \Omega$,the null point is obtained at $1.5 \ell_{1}$. The unknown resistance $R$ is: (in $\Omega$)

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