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Mathematics in Physics

JEE Main level, single correct, medium

109 of 121

Question 109 of 121

A physical quantity $P$ is described by the relation $P={a}^{\dfrac{1}{2}}{b}^{2}{c}^{3}{d}^{-4}$ . 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:

Show answer and solution

Answer: (D) $32%$

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

{\left(\dfrac{∆p}{p}\right)}_{max}=\dfrac{1}{2}\dfrac{∆a}{a}+2\dfrac{∆b}{b}+3\dfrac{∆c}{c}+4\dfrac{∆d}{d}

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

%\text{error in}P=\left(\dfrac{1}{2}\times 2+2\times 1+3\times 3+4\times 5\right)

=32%

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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More Physics practice

JEE Main Physics practice question on Mathematics in Physics (Errors of Measurement), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.

A physical quantity PPP is described by the relation P=a12b2c3d−4P={a}^{\dfrac{1}{2}}{b}^{2}{c}^{3}{d}^{-4}P=a21​b2c3d−4. If the relative errors in the measurement of aaa, bbb, ccc and ddd respectively, are 22%2, 11%1, 33%3 and 55%5. Then the relative error in PPP will be:

JEE Main question paper PDF

More Physics practice

A physical quantity $P$ is described by the relation $P={a}^{\dfrac{1}{2}}{b}^{2}{c}^{3}{d}^{-4}$ . 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: