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

Correct answer: C. 13 \% aligned & P=a^3 b^2 c^ -1 d^ -1 / 2 & ((Δ P)/(P) × 100)=((3 Δ a)/(a)+(2 Δ b)/(b)+(1 Δ c)/(c)+(1)/(2) (3 Δ d)/(d)) × 100 & =3(1 \%)+2(3 \%)+1(2 \%)+(1)/(2)(4 \%) & =13 \% aligned

NEET 2025

Exam: NEET
Year: 2025
Subject: Physics
Chapter: Units and Dimensions

NEET 2025 · Physics · Units and Dimensions

A physical quantity P is related to four observations a, b, c and d...

A physical quantity

P

is related to four observations

a, b, c

and

d

as follows :

P=a^3 b^2 / c \sqrt{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

Options

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

Correct answer

C. $13 \%$

Step-by-step solution

\begin{aligned} & P=a^3 b^2 c^{-1} d^{-1 / 2} \\ & \left(\frac{\Delta P}{P} \times 100\right)=\left(\frac{3 \Delta a}{a}+\frac{2 \Delta b}{b}+\frac{1 \Delta c}{c}+\frac{1}{2} \frac{3 \Delta d}{d}\right) \times 100 \\ & =3(1 \%)+2(3 \%)+1(2 \%)+\frac{1}{2}(4 \%) \\ & =13 \%\end{aligned}

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