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

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 meter bridge experiment,resistances are connected as shown in the figure. The balancing length $l_1$ is $55 \, cm$. Now,an unknown resistance $x$ is connected in series with $P$ and the new balancing length is found to be $75 \, cm$. The value of $x$ 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 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:

Two resistances are connected in two gaps of a meter bridge. The balance point is $20 \ cm$ from the zero end. $A$ resistance of $15 \ \Omega$ is connected in series with the smaller of the two. The null point shifts to $40 \ cm$. The value of the smaller resistance in $\Omega$ is