JEE Advanced 2022 Paper 2 · Q03 · DC Circuits

Question

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 $0.2\,\text{mm}$ at $A$ to $1\,\text{mm}$ at $B$. A galvanometer ($G$) connected to the centre of the wire, $50\,\text{cm}$ from each end along its axis, shows zero deflection when $A$ and $B$ are connected to a battery. The value of $X$ is _____ .

[Figure: A truncated-conical wire AB of length 100 cm with radius 0.2 mm at A and 1.0 mm at B. R1 = X is attached at end A, R2 = 1 ohm at end B. A galvanometer G is tapped at the midpoint O of the wire.]

SolveKar AI · Accepted Answer

For a wire with linearly varying radius $r(x)$, $dR = \rho\,dx/(\pi r(x)^2)$. Let $r$ change linearly with $x$ along the axis. From $A$ ($r_A = 0.2$ mm) to midpoint $O$ ($r_O = 0.6$ mm): integrate $\int_0^{50} dx/r(x)^2$ with $r = 0.2 + 0.008x$ (mm), giving $R_{AO} = \rho \cdot 50/(\pi \cdot r_A r_O) = \rho\cdot 50/(\pi\cdot 0.2\cdot 0.6)$ in mm$^{-1}$ units. Similarly $R_{OB} = \rho\cdot 50/(\pi\cdot 0.6\cdot 1.0)$. Ratio $R_{AO}/R_{OB} = (0.6\cdot 1.0)/(0.2\cdot 0.6) = 1.0/0.2 = 5$. Zero galvanometer deflection (Wheatstone-style balance) requires $R_1/R_{AO} = R_2/R_{OB}$, i.e. $X/1 = R_{AO}/R_{OB} = 5$. Hence $X = 5$.

JEE Advanced 2022 Paper 2 · Q03 · DC Circuits

Solution