Wheatstone bridge principle is used to measure the specific resistance $\left(S_1\right)$ of given wire, having length $L$, radius $r$. If $X$ is the resistance of wire, then specific resistance is: $S_1=X\left(\frac{\pi r^2}{L}\right)$. If the length of the wire gets doubled then the value of specific resistance will be :

In a metre-bridge when a resistance in the left gap is $2\  \Omega$ and unknown resistance in the right gap, the balance length is found to be $40\  \mathrm{~cm}$. On shunting the unknown resistance with $2\  \Omega$, the balance length changes by :

Consider a $72\, cm$ long wire $AB$ as shown in the figure. The galvanometer jockey is placed at $P$ on $A B$ at a distance $x cm$ from $A$. The galvanometer shows zero deflection.

The value of $x,$ to the nearest integer, is ….. $cm$

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

In the circuit shown, a meter bridge is in its balanced state. The meter bridge wire has a resistance $0.1\, ohm/cm$. The value of unknown resistance $X$ and the current drawn from the battery of negligible resistance is