Two resistances R_1 = X Omega and R_2 = 1 Ome | vedclass

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(C) The wire $AB$ has a total length $L = 100 	ext{ cm} = 1 	ext{ m}$. The radius $r(x)$ varies linearly from $r_A = 0.2 	ext{ mm}$ to $r_B = 1 	e

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$5$

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(C) The wire $AB$ has a total length $L = 100 	ext{ cm} = 1 	ext{ m}$. The radius $r(x)$ varies linearly from $r_A = 0.2 	ext{ mm}$ to $r_B = 1 	ext{ mm}$.Let $x$ be the distance from $A$. Then $r(x) = r_A + \frac{r_B - r_A}{L} x = 0.2 + 0.8x$ (in mm).The resistance of a small element $dx$ is $dR = \frac{ho dx}{\pi r(x)^2}$.For a balanced Wheatstone bridge,the ratio of resistances must be equal: $\frac{R_1}{R_2} = \frac{R_{AC}}{R_{CB}}$,where $R_{AC}$ and $R_{CB}$ are the resistances of the two halves of the wire.$R_{AC} = \int_0^{0.5} \frac{ho dx}{\pi (0.2 + 0.8x)^2 	imes 10^{-6}}$ and $R_{CB} = \int_{0.5}^1 \frac{ho dx}{\pi (0.2 + 0.8x)^2 	imes 10^{-6}}$.Using the integral $\int \frac{dx}{(a+bx)^2} = -\frac{1}{b(a+bx)}$,we get:$R_{AC} \propto \left[ -\frac{1}{0.8(0.2 + 0.8x)} ight]_0^{0.5} = -\frac{1}{0.8} (\frac{1}{0.6} - \frac{1}{0.2}) = \frac{1}{0.8} (5 - 1.66) = \frac{3.33}{0.8} = 4.166$.$R_{CB} \propto \left[ -\frac{1}{0.8(0.2 + 0.8x)} ight]_{0.5}^1 = -\frac{1}{0.8} (\frac{1}{1} - \frac{1}{0.6}) = \frac{1}{0.8} (1.66 - 1) = \frac{0.66}{0.8} = 0.833$.Ratio $\frac{R_{AC}}{R_{CB}} = \frac{4.166}{0.833} = 5$.Therefore,$\frac{R_1}{R_2} = 5 \implies \frac{X}{1} = 5 \implies X = 5 \Omega$.

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