Balanced Bridge with Tapered Wire Resistivity Ratio

To find the value of $X$, we model the circuit as a balanced Wheatstone bridge.

1. Condition for Zero Deflection

For zero deflection in the galvanometer connected at the center $C$ of the wire (where $AC = CB = 50\text{ cm}$), the bridge must be balanced: $\frac{R_1}{R_2} = \frac{R_{AC}}{R_{CB}}$

Given $R_1 = X\ \Omega$ and $R_2 = 1\ \Omega$: $X = \frac{R_{AC}}{R_{CB}}$

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2. Resistance of a Linearly Tapered Wire Segment

Consider a wire segment of length $l$ whose radius varies linearly along its length from an initial radius $r_i$ to a final radius $r_f$.

At a distance $x$ from one end ($0 \le x \le l$), the radius $r(x)$ is given by: $r(x) = r_i + \left(\frac{r_f - r_i}{l}\right)x$

The differential resistance $dR$ of a thin elemental slice of length $dx$ is: $dR = \frac{\rho \, dx}{\pi [r(x)]^2}$

Integrating from $x = 0$ to $x = l$: $R = \int_0^l \frac{\rho \, dx}{\pi \left[r_i + \left(\frac{r_f - r_i}{l}\right)x\right]^2}$

Substituting $u = r(x)$, so $du = \frac{r_f - r_i}{l} dx$: $R = \frac{\rho l}{\pi(r_f - r_i)} \int_{r_i}^{r_f} \frac{du}{u^2} = \frac{\rho l}{\pi(r_f - r_i)} \left[ -\frac{1}{u} \right]_{r_i}^{r_f} = \frac{\rho l}{\pi r_i r_f}$

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3. Calculating $R_{AC}$ and $R_{CB}$

Let:

• $r_A = 0.2\text{ mm}$
• $r_B = 1.0\text{ mm}$
• $r_C$ be the radius at the midpoint $C$. Since the radius varies linearly, $r_C = \frac{r_A + r_B}{2} = 0.6\text{ mm}$.
• $l_1 = l_2 = 50\text{ cm} = l$

Using the derived formula for resistance: $R_{AC} = \frac{\rho l}{\pi r_A r_C}$ $R_{CB} = \frac{\rho l}{\pi r_C r_B}$

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4. Finding $X$

Taking the ratio of $R_{AC}$ to $R_{CB}$: $\frac{R_{AC}}{R_{CB}} = \frac{\frac{\rho l}{\pi r_A r_C}}{\frac{\rho l}{\pi r_C r_B}} = \frac{r_B}{r_A}$

Substituting the given values of $r_A$ and $r_B$: $X = \frac{1.0\text{ mm}}{0.2\text{ mm}} = 5$