When a current of (2.5±0.5) A flows through a wire, it develops a potential difference

When a current of (2.5±0.5) A flows through a wire, it develops a potential difference of (20±1) V, then the resistance of wire is

Options

(a) (8±2)Ω
(b) (8±1.6)Ω
(c) (8±1.5)Ω
(d) (8±3)Ω

Correct Answer:

(8±2)Ω

Explanation:

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  • R=V/I
    R=20/2.5
    R= 8 ohm

    Now,
    ∆ R/R=∆ V/V+∆I/I
    =1/20 +0.5/2.5
    = 1/4

    Therefore,
    ∆ R÷R=1÷4
    ∆ R=1÷4×R
    ∆R=1÷4×8
    ∆R=2

    Therefore,
    Resistance with error limits=R+ - ∆R
    =(8+ - 2)ohm

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