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 is set up as shown,to determine an unknown resistance '$X$' using a standard $10 \ \Omega$ resistor. The galvanometer shows a null point when the tapping key is at the $52 \ cm$ mark. The end corrections are $1 \ cm$ and $2 \ cm$ respectively for the ends $A$ and $B$. The determined value of '$X$' is (in $\Omega$)

On interchanging the resistances,the balance point of a meter bridge shifts to the left by $10 \, cm$. The resistance of their series combination is $1 \, k\Omega$. How much was the resistance on the left slot before interchanging the resistances? ................... $\Omega$

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 $\Omega$)

The resistance per centimeter of a meter bridge wire is $r$. $A$ resistance of $X \ \Omega$ is placed in the left gap and a resistance of $25 \ \Omega$ is placed in the right gap. The balancing length from the left end is $40 \ cm$. Now,the wire is replaced by another wire of $2r$ resistance per centimeter. The new balancing length for the same settings will be at: (in $cm$)

In the figure shown,for given values of $R_1$ and $R_2$,the balance point for the jockey is at $40\,cm$ from $A$. When $R_2$ is shunted by a resistance of $10\, \Omega$,the balance point shifts to $50\,cm$. Find $R_1$ and $R_2$. $(AB = 1\,m)$