In the meter bridge experiment,the length $AB$ of the wire is $1 \ m$. The resistors $X$ and $Y$ have values $5 \ \Omega$ and $2 \ \Omega$ respectively. When a shunt resistance $S$ is connected in parallel to $X$,the balancing point is found to be $0.625 \ m$ from $A$. Then,the resistance of the shunt $S$ is (in $\Omega$)

$A$ meter bridge is set up as shown to determine the end corrections at $A$ and $B$. When a resistance of $15\,\Omega$ is used in the left gap and $20\,\Omega$ in the right gap,the null point is obtained at a distance of $42\,cm$ from $A$. When these resistances are interchanged,the null point is obtained at a distance of $57\,cm$ from $A$. The values of the end corrections are:

In a meter bridge,as shown in the figure,it is given that resistance $Y = 12.5 \, \Omega$ and the balance point is obtained at a distance $l_1 = 39.5 \, cm$ from end $A$ (by jockey $J$). After interchanging the resistances $X$ and $Y$,a new balance point is found at a distance $l_2$ from end $A$. What are the values of $X$ and $l_2$?

In a meter bridge experiment, the balance point is obtained at length $l_1$ cm from the left end when resistances in the left gap and right gap are $15 \text{ } \Omega$ and $R \text{ } \Omega$ respectively. When the resistance $R$ is shunted with an equal resistance, the new balance point is at $1.6 l_1$. The resistance $R$ in ohm is:

$A$ 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 . . . . . . $\Omega$.

In the given figure of a meter bridge experiment,the balancing length $AC$ corresponding to null deflection of the galvanometer is $40 \, cm$. What will be the balancing length if the radius of the wire $AB$ is doubled (in $, cm$)?