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 $Omega$)?

In a meter bridge experiment,the ratio of the left gap resistance to the right gap resistance is $2:3$. The balance length from the left end is (in $\,cm$)

In the shown arrangement of the experiment of the metre bridge,if $AC$ corresponding to null deflection of the galvanometer is $x$,what would be its value if the radius of the wire $AB$ is doubled?

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 $cm$)

Two resistances are connected in the two gaps of a meter bridge. The balancing point is obtained at $20 \ cm$. When a resistance of $15 \ \Omega$ is connected in series with the smaller resistance of the two,the balancing point shifts to $40 \ cm$. The value of the smaller resistance is (in $Omega$)