In a metre-bridge experiment,the balance point is obtained when the gaps are closed by $2 \Omega$ and $3 \Omega$ resistors. $A$ shunt of $x \Omega$ is added in parallel to the $3 \Omega$ resistor to shift the balance point by $22.5 \ cm$. The value of $x$ is: (in $\Omega$)

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 $cm$)

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

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 $l_2$?

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 a meter bridge experiment,resistances are connected as shown in the figure. Initially,resistance $P = 4\,\Omega$ and the null point $N$ is at $60\,cm$ from $A$. Now,an unknown resistance $R$ is connected in series to $P$,and the new position of the null point is at $80\,cm$ from $A$. The value of unknown resistance $R$ is