Two resistances $R_1=X\Omega$ and $R_2=1\Omega$ are connected to a wire AB of uniform resistivity, as shown in the figure. The radius of the wire varies linearly along its axis from 0.2 mm at A to 1 mm at B. A galvanometer (G) connected to the center of the wire, 50 cm from each end along its axis, shows zero deflection when A and B are connected to a battery. The value of $$x is _____.

The correct answer is 5.

Step 1: Understand the setup and the condition for zero deflection (Wheatstone Bridge principle)
From the given circuit diagram, two resistances $R_1=X\Omega$ and $R_2=1\Omega$ are connected along with a non-uniform wire AB across a battery. A galvanometer G is connected to the center point C of the wire, which is at a distance of 50 cm from each end (A and B).

Let $R_{AC}$ be the resistance of the portion of the wire from A to the midpoint C, and $R_{CB}$ be the resistance of the portion of the wire from C to B.
Since the galvanometer shows zero deflection, the circuit acts as a balanced Wheatstone bridge:

$\frac{R_1}{R_2}=\frac{R_{AC}}{R_{CB}}$

$X1=\frac{R_{AC}}{R_{CB}}$

Step 2: Calculate resistance of a tapered wire
For a conductor of uniform resistivity $\rho$ whose radius varies linearly from $r_i$ to $r_f$ over a distance $L$, the resistance of an elemental cross-section of length $dx$ at position $x$ is given by:

$dR=\frac{\mathrm{\rho <!--\; \rho \; -->}dx}{\mathrm{\pi <!--\; \pi \; -->}{r}^2}$

Integrating from $x=0$ to $x=L$ where radius $r\left(x\right)=r_i+\frac{r_f-<!--\; -\; -->r_i}{L}x$, the standard result for the total resistance of a linearly tapered rod/wire is:

$R=\frac{\mathrm{\rho <!--\; \rho \; -->}L}{\mathrm{\pi <!--\; \pi \; -->}{r}_i{r}_f}$

Step 3: Determine radii at points A, C, and B
The radius of the wire varies linearly along its axis:
- At end A ($x=0$), the radius is $r_A=0.2\text{mm}$.
- At end B ($x=100\text{cm}$), the radius is $r_B=1.0\text{mm}$.
- At the midpoint C ($x=50\text{cm}$), the radius $r_C$ is the average of $r_A$ and $r_B$:

$r_C=\frac{r_A+r_B}{2}=\frac{0.2+1.0}{2}=0.6\text{mm}$

Step 4: Compute resistances $R_{AC}$ and $R_{CB}$
Both sections AC and CB have the same length $l=50\text{cm}$.
For section AC (radius varying from $r_A$ to $r_C$):

$R_{AC}=\frac{\mathrm{\rho <!--\; \rho \; -->}l}{\mathrm{\pi <!--\; \pi \; -->}{r}_A{r}_C}$

For section CB (radius varying from $r_C$ to $r_B$):

$R_{CB}=\frac{\mathrm{\rho <!--\; \rho \; -->}l}{\mathrm{\pi <!--\; \pi \; -->}{r}_C{r}_B}$

Step 5: Solve for X
Taking the ratio of $R_{AC}$ to $R_{CB}$:

$X=\frac{R_{AC}}{R_{CB}}=\frac{\frac{\mathrm{\rho <!--\; \rho \; -->}l}{\mathrm{\pi <!--\; \pi \; -->}{r}_A{r}_C}}{\frac{\mathrm{\rho <!--\; \rho \; -->}l}{\mathrm{\pi <!--\; \pi \; -->}{r}_C{r}_B}}=\frac{r_B}{r_A}$

Substitute the given values into the equation:

$X=\frac{1.0\text{mm}}{0.2\text{mm}}=5$

Thus, the value of $X$ is 5.