JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Current Electricity

A wire of length 10 cm and radius

7 \times 1 0^{- 4} m

connected across the right gap of a meter bridge. When a resistance of

4.5Ω

is connected on the left gap by using a resistance box, the balance length is found to be at 60 cm from the left end. If the resistivity of the wire is

R \times 1 0^{- 7} Ω m

, then value of R is :

(A)63
(B)70
(C)66✓ Correct
(D)35

Answer: (C)

Solution

At the null point the wire (right gap, length 40 cm) balances against the 4.5

Ω

box (left gap, 60 cm):

\frac{R _{w i r e}}{40} = \frac{4.5}{60} \Rightarrow R_{w i r e} = 3 Ω

The wire has length

L = 0.1

m and radius

r = 7 \times 1 0^{- 4}

m, so its cross-section is

A = π r^{2} = 7 π \times 1 0^{- 8} m^{2}

. From

R = ρ L / A

ρ = \frac{R A}{L} = \frac{3 \times 7 π \times 1 0 ^{- 8}}{0.1} = 66 \times 1 0^{- 7} Ω m

R = 66

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Single correct option

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JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Current Electricity

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