Question

Wire AB used in the given metre bridge has uniform dimension and its resistivity at distance x from end A varies as $\rho(x) = \frac{\rho_0}{(1-kx)^2}$ where x is in metre and $\rho_0 = 5 \times 10^{-3}$ ohm-m. If $P = 2\Omega$ and $Q = 9\Omega$, then zero deflection point C is 40 cm from end A. The value of 3k is _______ $m^{-1}$.

Answer 1

2 $m^{-1}$

Answer 2

Solution:

1. For a metre bridge the balanced condition gives:

   $$\frac{R_{AC}}{R_{CB}}=\frac{P}{Q}=\frac{2}{9}.$$

2. Since the resistivity varies as

   $$\rho(x)=\frac{\rho_0}{(1-kx)^2},$$

   the resistance of a segment from $x=a$ to $x=b$ (with constant cross‐section $A$) is

   $$R = \int_{a}^{b}\frac{\rho(x)}{A}\,dx = \frac{\rho_0}{A}\int_a^b\frac{dx}{(1-kx)^2}.$$

3. Using the substitution $u=1-kx$, $du=-k\,dx$, we get

   $$\int\frac{dx}{(1-kx)^2} = \frac{1}{k}\left[\frac{1}{1-kx}\right].$$

4. Hence, the resistances are:

   • For $AC$ (from $x=0$ to $x=0.4$): $$R_{AC}=\frac{\rho_0}{A}\cdot\frac{1}{k}\left[\frac{1}{1-0.4k}-1\right].$$
   • For $CB$ (from $x=0.4$ to $x=1$): $$R_{CB}=\frac{\rho_0}{A}\cdot\frac{1}{k}\left[\frac{1}{1-k}-\frac{1}{1-0.4k}\right].$$

5. Taking the ratio:

   $$\frac{R_{AC}}{R_{CB}} = \frac{\frac{1}{1-0.4k}-1}{\frac{1}{1-k}-\frac{1}{1-0.4k}}.$$

6. Simplify the numerator:

   $$\frac{1}{1-0.4k}-1 = \frac{1-(1-0.4k)}{1-0.4k} = \frac{0.4k}{1-0.4k}.$$

7. Simplify the denominator:

   $$\frac{1}{1-k}-\frac{1}{1-0.4k} = \frac{(1-0.4k)-(1-k)}{(1-k)(1-0.4k)} = \frac{0.6k}{(1-k)(1-0.4k)}.$$

8. Therefore,

   $$\frac{R_{AC}}{R_{CB}} = \frac{\frac{0.4k}{1-0.4k}}{\frac{0.6k}{(1-k)(1-0.4k)}} = \frac{0.4\,(1-k)}{0.6} = \frac{2}{3}(1-k).$$

9. Set equal to $\frac{2}{9}$:

   $$\frac{2}{3}(1-k)=\frac{2}{9} \quad \Longrightarrow \quad (1-k)=\frac{1}{3} \quad \Longrightarrow \quad k=\frac{2}{3}.$$

10. Requested value is $3k$:

    $$3k= 3\cdot\frac{2}{3}=2\; \text{m}^{-1}.$$

Explanation (minimal):

1. Write resistance integrals for segments AC and CB.

2. Compute integrals using substitution.

3. Take ratio and simplify to get $\frac{2}{3}(1-k)=\frac{2}{9}$.

4. Solve for $k$ then calculate $3k$.