NEET 2025 — Physics solution

NEET2025Physics

A physical quantity $P$ is related to four observations $a$ , $b$ , $c$ and $d$ as follows: $P = a^{3} b^{2} / c d$ .

The percentage errors of measurement in $a$ , $b$ , $c$ and $d$ are $1%$ , $3%$ , $2%$ and $4%$ respectively. The percentage error in the quantity $P$ is:

A
10%
B
2%
C
13%
D
15%

Solution

For $P = a^{3} b^{2} c^{- 1} d^{- 1/2}$ :

$\frac{Δ P}{P} \times 100 = 3 \frac{Δ a}{a} + 2 \frac{Δ b}{b} + \frac{Δ c}{c} + \frac{1}{2} \frac{Δ d}{d}$ (all positive in error propagation).

$= 3 (1%) + 2 (3%) + 1 (2%) + \frac{1}{2} (4%) = 3 + 6 + 2 + 2 = 13%$ .

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