In the experimental setup of a meter bridge shown in the figure,the null point is obtained at a distance of $40\,cm$ from $A$. If a $10\,\Omega$ resistor is connected in series with $R_1$,the null point shifts by $10\,cm$. The resistance that should be connected in parallel with $(R_1 + 10)\,\Omega$ such that the null point shifts back to its initial position is .............. $\Omega$.

$A$ wire of length $10 \ cm$ and radius $\sqrt{7} \times 10^{-4} \ m$ is connected across the right gap of a meter bridge. When a resistance of $4.5 \ \Omega$ is connected in the left gap using a resistance box,the balance length is found to be at $60 \ cm$ from the left end. If the resistivity of the wire is $R \times 10^{-7} \ \Omega \ m$,then the value of $R$ is:

$(a)$ In a meter bridge,the balance point is found to be at $39.5\; cm$ from the end $A$,when the resistor $Y$ is of $12.5\; \Omega$. Determine the resistance of $X$. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?
$(b)$ Determine the balance point of the bridge above if $X$ and $Y$ are interchanged.
$(c)$ What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?

In a meter bridge experiment,when resistance $X$ is in the left gap and $Y$ is in the right gap,the null point is found at $20 \ cm$ from the left end. If $X < Y$,what will be the new position of the null point from the same end if a resistance of $4X$ is placed in the left gap instead of $X$?

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 balanced meter bridge,the segment of wire opposite to a known resistance of $70 \Omega$ is $70 \text{ cm}$. The unknown resistance is (in $\Omega$)