Q87 / 90 · JEE Main 2017 (09 Apr Online) · Physics

A physical quantity $P$ is described by the relation $P = a^{\frac{1}{2}} b^{2} c^{3} d^{- 4}$ . If the relative errors in the measurement of $a$ , $b$ , $c$ and $d$ respectively, are $2 %$ , $1 %$ , $3 %$ and $5 %$ . Then the relative error in $P$ will be:

A
$12 %$
B
$8 %$
C
$25 %$
Correct answer
D
$32 %$

Solution

P = a^{\frac{1}{2}} b^{2} c^{3} d^{- 4}

{(\frac{∆ p}{p})}_{m a x} = \frac{1}{2} \frac{∆ a}{a} + 2 \frac{∆ b}{b} + 3 \frac{∆ c}{c} + 4 \frac{∆ d}{d}

100 \times \frac{∆ P}{P} = (\frac{1}{2} \frac{∆ a}{a} + 2 \frac{∆ b}{b} + 3 \frac{∆ c}{c} + 4 \frac{∆ d}{d}) \times 100

% error in P = (\frac{1}{2} \times 2 + 2 \times 1 + 3 \times 3 + 4 \times 5)

= 32 %

Hence, the correct answer is option: 'D'.

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