$A$ balanced bridge is shown in the circuit diagram. The metre bridge wire has a resistance of $1 \Omega \text{ cm}^{-1}$. The current drawn from the battery is (Internal resistance of the battery is negligible). (in $\text{ A}$)

$A$ 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$)

During an experiment with a meter bridge,the galvanometer shows a null point when the jockey is pressed at $40.0 \ cm$ using a standard resistance of $90 \ \Omega$,as shown in the figure. The least count of the scale used in the meter bridge is $1 \ mm$. The unknown resistance is:

In a metre-bridge,when a resistance in the left gap is $2 \ \Omega$ and an unknown resistance is in the right gap,the balance length is found to be $40 \ cm$. On shunting the unknown resistance with $2 \ \Omega$,the balance length changes by: (in $cm$)

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

In a meter-bridge,if the left and right gaps are connected with $2 \Omega$ and $3 \Omega$ resistances,respectively,then the bridge is balanced. What resistance should be connected with the $3 \Omega$ resistance to get the balancing point at the midpoint of the bridge wire?