In a meter bridge experiment,a null point is obtained at $20 \ cm$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$,then where will be the new position of the null point from the same end,if one decides to balance a resistance of $4X$ against $Y$ (in $cm$)?

When two resistances $R_1$ and $R_2$ are connected in series and introduced into the left gap of a meter bridge and a resistance of $10 \ \Omega$ is introduced into the right gap,a null point is found at $60 \ cm$ from the left side. When $R_1$ and $R_2$ are connected in parallel and introduced into the left gap,a resistance of $3 \ \Omega$ is introduced into the right gap to get a null point at $40 \ cm$ from the left end. The product $R_1 R_2$ is $............. \ \Omega$.

If the balance point is obtained at the $35^{th} cm$ in a meter bridge,the resistances in the left and right gaps are in the ratio of:

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 $cm$)

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

$(a)$ 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?