The voltage across a lamp is (6.0 ± 0.3) volt and the current passing through it is (4.0 ± 0.1) ampere. The power consumed in watt will be (using percentage error) (a) (24.0 ± 1.8) (b) (24.0 ± 0.75) (c) (22.0 ± 0.4) (d) (18.0 ± 0.4)

Answer: (a). P = VI = 6.0 × 4.0 = 24.0 W (Δ P)/(P) = (0.3)/(6.0) + (0.1)/(4.0) = 5% + 2.5% = 7.5% Δ P = 0.075 × 24.0 = 1.8 W So the answer is (a) (24.0 ± 1.8) W.

The voltage across a lamp is (6.0 ± 0.3) volt and the current passing… | MHT CET 2026 (19 Apr, Shift 1)

Q11MHT CET 2026 · 19 APR · SHIFT 1Easy

Errors in Measurement

The voltage across a lamp is $(6.0 \pm 0.3)$ volt and the current passing through it is $(4.0 \pm 0.1)$ ampere. The power consumed in watt will be (using percentage error)

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Answer: (a)

$P = VI = 6.0 \times 4.0 = 24.0\ \text{W}$
$\frac{\Delta P}{P} = \frac{0.3}{6.0} + \frac{0.1}{4.0} = 5\% + 2.5\% = 7.5\%$
$\Delta P = 0.075 \times 24.0 = 1.8\ \text{W}$
So the answer is (a) $(24.0 \pm 1.8)\ \text{W}$ .

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