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 $\,cm$)

In a meter bridge,the point $D$ is the neutral point (null point) as shown in the figure.

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

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 $\Omega$)

The figure shows a meter-bridge circuit,with $AB = 100\, cm$,$X = 12\,\Omega$ and $R = 18\,\Omega$,and the jockey $J$ in the position of balance. If $R$ is now made $8\,\Omega$,through what distance will $J$ have to be moved to obtain balance? .............. $cm$

In a metre bridge experiment,with a standard resistance in the right gap and a resistance coil dipped in water (in a beaker) in the left gap,the balancing length obtained is $l$. If the temperature of water is increased,the new balancing length is: