The physical quantity P= a^ 3 b^ 2 c^ 1/3 d^ 1/4 , where the percentage error in a, b, c, and d are 1\%, (3)/(2)\% , 3\% , and 2\% respectively. Then the percentage error in P is

Correct answer: D. 7.5\ \% Given P = a^ 3 b^ 2 c^ 1/3 d^ 1/4 The maximum percentage error in P is given by: (Δ P)/(P) × 100 = 3 ((Δ a)/(a) × 100) + 2 ((Δ b)/(b) × 100) + (1)/(3) ((Δ c)/(c) ×100) + (1)/(4) ((Δ d)/(d) ×100) Substituting the given percentage errors: (Δ P)/(P) ×100 = 3(1\%) + 2( (3)/(2) \%) + (1)/(3)(3\%) + (1)/(4)(2\%) (Δ P)/(P) ×100 = 3\% + 3\% + 1\% + 0.5\% = 7.5\% Answer: 7.5\ \%

MHT CET 202624 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 24 April
Shift: Morning Shift
Subject: Physics
Chapter: Mathematics in Physics

MHT CET 2026 · 24 April · Morning Shift · Physics · Mathematics in Physics

The physical quantity P= a^ 3 b^ 2 c^ 1/3 d^ 1/4 , where the...

The physical quantity

P=\dfrac{a^{3}b^{2}}{c^{1/3}d^{1/4}}

, where the percentage error in a, b, c, and d are

1\%, \dfrac{3}{2}\%

3\%

, and

2\%

respectively. Then the percentage error in P is

Options

A $14\ \%$
B $14.5\ \%$
C $7\ \%$
D $7.5\ \%$

Correct answer

D. $7.5\ \%$

Step-by-step solution

Given

P = \dfrac{a^{3}b^{2}}{c^{1/3}d^{1/4}}

The maximum percentage error in

P

is given by:

\dfrac{\Delta P}{P} \times 100 = 3 \left(\dfrac{\Delta a}{a} \times 100\right) + 2 \left(\dfrac{\Delta b}{b} \times 100\right) + \dfrac{1}{3}

\left(\dfrac{\Delta c}{c} \times100\right) + \dfrac{1}{4} \left(\dfrac{\Delta d}{d} \times100\right)

<br><br>Substituting the given percentage errors:<br><br>

\dfrac{\Delta P}{P} \times100 = 3(1\%) + 2\left(

\dfrac{3}{2} \%\right) + \dfrac{1}{3}(3\%) + \dfrac{1}{4}(2\%)

\dfrac{\Delta P}{P} \times100 = 3\% + 3\% + 1\% + 0.5\% = 7.5\%

Answer: 7.5\ \%

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