JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Current Electricity

A Copper( Cu ) rod of length 25 cm and cross sectional area

3 mm^{2}

is joined with a similar aluminium(AI)rod as shown in figure. Find the resistance of the combination between the ends

A

and B .
(Take Resistivity of Copper

= 1.7 \times 1 0^{- 8} Ω m

Resistivity of Aluminium

= 2.6 \times 1 0^{- 8} Ω m

)

Answer: (A)

Solution

The two rods run side by side between the same pair of junctions, so they are in parallel and each carries the full length

L = 0.25

m and area

A = 3 m m^{2} = 3 \times 1 0^{- 6} m^{2}

R = \frac{ρ L}{A}

R_{Cu} = \frac{1.7 \times 1 0 ^{- 8} \times 0.25}{3 \times 1 0 ^{- 6}} = 1.42 \times 1 0^{- 3} Ω, R_{Al} = \frac{2.6 \times 1 0 ^{- 8} \times 0.25}{3 \times 1 0 ^{- 6}} = 2.17 \times 1 0^{- 3} Ω

R_{A B} = \frac{R _{Cu} R _{Al}}{R _{Cu} + R _{Al}} = \frac{1.42 \times 2.17}{1.42 + 2.17} \times 1 0^{- 3} Ω

The resistance between

A

and

B

0.858 m Ω

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option