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JEE PYQ: Units & Measurements - Question ID ef9378df038b (JEE Main 2017)

ID: ef9378df038bJEE Main 2017Single Correct MCQ

A physical quantity P is described by the relation

P = a

^{{\raise0.5ex\hbox{

\scriptstyle 1

} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{

\scriptstyle 2

}}}

b² c³ 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 :

Select Option

Step-by-step Explanation

Core Formula & Concept:

When a physical quantity $ P $ is expressed as a product (or quotient) of other measured quantities $ a, b, c, \dots $ raised to powers, i.e.,

P = k \cdot a^{\alpha} b^{\beta} c^{\gamma} d^{\delta} \cdots

where $ k $ is a dimensionless constant, the relative error (percentage error) in $ P $ is determined by the sum of the absolute values of the relative errors in each quantity multiplied by their respective exponents. Mathematically, the relative error in $ P $ is:

\frac{\Delta P}{P} = \left| \alpha \right| \frac{\Delta a}{a} + \left| \beta \right| \frac{\Delta b}{b} + \left| \gamma \right| \frac{\Delta c}{c} + \left| \delta \right| \frac{\Delta d}{d} + \cdots

This formula arises from the logarithmic differentiation of $ P $ with respect to each variable, ensuring that errors propagate additively when powers are involved.

Step-by-Step Derivation:

Given the relation:

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

We identify the exponents:

$ \alpha = \frac{1}{2} $ for $ a $
$ \beta = 2 $ for $ b $
$ \gamma = 3 $ for $ c $
$ \delta = -4 $ for $ d $

Relative errors in measurements:

$ \frac{\Delta a}{a} = 2\% = 0.02 $
$ \frac{\Delta b}{b} = 1\% = 0.01 $
$ \frac{\Delta c}{c} = 3\% = 0.03 $
$ \frac{\Delta d}{d} = 5\% = 0.05 $

Using the error propagation formula for products of powers:

\frac{\Delta P}{P} = \left| \frac{1}{2} \right| \frac{\Delta a}{a} + \left| 2 \right| \frac{\Delta b}{b} + \left| 3 \right| \frac{\Delta c}{c} + \left| -4 \right| \frac{\Delta d}{d}

Substitute the values:

\frac{\Delta P}{P} = \frac{1}{2} \times 0.02 + 2 \times 0.01 + 3 \times 0.03 + 4 \times 0.05

Compute each term:

$ \frac{1}{2} \times 0.02 = 0.01 $
$ 2 \times 0.01 = 0.02 $
$ 3 \times 0.03 = 0.09 $
$ 4 \times 0.05 = 0.20 $

Sum the contributions:

\frac{\Delta P}{P} = 0.01 + 0.02 + 0.09 + 0.20 = 0.32

Convert to percentage:

\frac{\Delta P}{P} \times 100\% = 32\%

Thus, the relative error in $ P $ is 32%.

Common Traps & Exam Tip:

Students often make the following mistakes:

Ignoring absolute values of exponents: Even if an exponent is negative (e.g., $ d^{-4} $), the error contribution is still positive. Forgetting this leads to incorrect subtraction.
Miscounting exponents: Misreading $ a^{1/2} $ as $ a^2 $ or $ b^2 $ as $ b^{1/2} $ drastically alters the result.
Adding percentage errors directly: Simply adding 2% + 1% + 3% + 5% = 11% is wrong. The exponents must be multiplied first.
Forgetting to convert to percentage: Leaving the result as 0.32 instead of 32% may lead to selecting option A (8%) or B (12%) by mistake.

Exam Tip: Always write down the exponents clearly and apply the error propagation formula systematically. Double-check the sign of exponents and ensure absolute values are used.

JEE PYQ: Units & Measurements - Question ID ef9378df038b (JEE Main 2017)

Select Option

Step-by-step Explanation

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