JEE Advanced 2022 — Paper 2 — PHY-I solution

JEE Advanced2022Paper 2PHY-I

Two resistances $R_{1} = X Ω$ and $R_{2} = 1 Ω$ are connected to a wire $A B$ 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 _______.

Your answer

Solution

For a frustum of a wire with end radii $a$ and $b$ and length $l$ , the resistance is $R = \frac{ρl}{π ab}$ .

Since the radius varies linearly from 0.2 mm at $A$ to 1 mm at $B$ , the radius at the midpoint (50 cm from each end) is $r = \frac{0.2 + 1}{2} = 0.6$ mm.

Resistance of the left half (radii 0.2 and 0.6 mm, length 0.5 m): $P = \frac{ρ ( 0.5 )}{π ( 0.2 \times 1 0 ^{- 3} ) ( 0.6 \times 1 0 ^{- 3} )}$ .

Resistance of the right half (radii 0.6 and 1 mm, length 0.5 m): $Q = \frac{ρ ( 0.5 )}{π ( 0.6 \times 1 0 ^{- 3} ) ( 1 \times 1 0 ^{- 3} )}$ .

Wheatstone balance condition $\frac{R _{1}}{P} = \frac{R _{2}}{Q}$ gives $\frac{X}{P} = \frac{1}{Q}$ , hence $X = \frac{P}{Q} = \frac{0.6 \times 1}{0.2 \times 0.6} = 5$ .

$∴ X = 5$ .

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