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QJEEAdv22Pheb8d: JEE Advanced 2022 Physics - Class 12 JEEAdv22Pheb8d Solution: Two resistances and are connected to a wire of uniform…

Two resistances R1=XΩR_{1}=X \Omega and R2=1ΩR_{2}=1 \Omega are connected to a wire ABA B of uniform resistivity, as shown in the figure. The radius of the wire varies linearly along its axis from 0.2 mm0.2 \mathrm{~mm} at AA to 1 mm1 \mathrm{~mm} at BB . A galvanometer (G)(\mathrm{G}) connected to the center of the wire, 50 cm50 \mathrm{~cm} from each end along its axis, shows zero deflection when AA and BB are connected to a battery. The value of XX is ____________.

JEE Advanced 2022 Paper 2 Online Physics - Current Electricity Question 5 English

Step-by-Step Verified Solution

According to the given condition wire is having different cross-section at two junction. (Frustum shaped conducter)

JEE Advanced 2022 Paper 2 Online Physics - Current Electricity Question 5 English Explanation 1

R=ρIπab

For the shown conductor in the diagram.

JEE Advanced 2022 Paper 2 Online Physics - Current Electricity Question 5 English Explanation 2

r=a+b2=0.2+12=0.6 .........(i)

As R=ρl A ...........(ii)

Hence, Resistence of left 50 cm50 \mathrm{~cm} wire

=ρ×0.5×106π×0.2×0.6

Resistence of Right 50 cm50 \mathrm{~cm} wire

=ρ×0.5×106π×0.6×1

For wheatstone balanced condition

R1P=R2Q(R1=X)\frac{R_1}{P} =\frac{R_2}{Q} \quad\left(R_1=X\right)

⇒ (X)×π×0.2×0.6ρ×0.5×106=(1π)×0.6×1ρ×0.5×106\frac{(X) \times \pi \times 0.2 \times 0.6}{\rho \times 0.5 \times 10^6} =\frac{(1 \pi) \times 0.6 \times 1}{\rho \times 0.5 \times 10^6}

⇒ (X)×π×0.12ρ×0.5×106=π×0.6×1ρ×0.5×106\frac{(X) \times \pi \times 0.12}{\rho \times 0.5 \times 10^6} =\frac{\pi \times 0.6 \times 1}{\rho \times 0.5 \times 10^6}

⇒ X=5X =5

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