In a metre bridge experiment,the null point is obtained at $20 \,cm$ from the left end of the wire,when resistance $X$ is balanced against another resistance $Y$ $(X < Y)$. To balance a resistance $4X$ against $Y$,the new position of the null point from the same end will be: (in $\,cm$)

In a meter bridge,the null point is located at $20 \ cm$ from the left end of the wire when resistances $R$ and $S$ are connected in the left and right gaps respectively. If the resistance $S$ is shunted with $60 \ \Omega$ resistance,the null point shifts by $5 \ cm$. The values of $R$ and $S$ are respectively:

$A$ meter bridge setup is shown in the figure. It is used to determine an unknown resistance $R$ using a given resistor of $15\,\Omega$. The galvanometer $(G)$ shows null deflection when the tapping key is at the $43\,cm$ mark from end $A$. If the end correction for end $A$ is $2\,cm$ and for end $B$ is $1\,cm$,then the determined value of $R$ will be . . . . . . $\Omega$.

$n$ identical resistors are taken,in which $\frac{n}{2}$ resistors are joined in series in the left gap and the remaining $\frac{n}{2}$ resistors are joined in parallel in the right gap of a metre bridge. The balancing length in $cm$ is

In a meter bridge,two gaps in the metallic strip are connected by $3 \Omega$ and $9 \Omega$ resistors. What should be the value of the shunt that needs to be added in parallel to the $9 \Omega$ resistor to shift the balancing point by $25 \text{ cm}$ (in $\Omega$)?

When a wire is connected in the left gap of a metre bridge,the balancing point is at $40 \ cm$ from the left end of the bridge wire. If the wire in the left gap is stretched so that its length is doubled and again connected in the same gap,then the balancing point from the left end of the bridge wire is