From the null point condition of a Meter Bridge, it follows that,
604.5Ω=100−60X⇒X=6040×4.5Ω=3Ω
Also,
X==Aρlπr2ρl...(1)
From equation (1), it follows that
3=ρ×π×(7×10−4)20.1⇒ρ=0.13×π×(7×10−4)2≈66×10−7Ωm
Therefore, R=66.
A wire of length 10cm and radius 7×10−4m 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 60cm from the left end. If the resistivity of the wire is R×10−7Ωm, then value of R is :
Held on 27 Jan 2024 · Verified 6 Jul 2026.
63
70
66
35
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