JEE PYQ: Units & Measurements - Question ID 314851dc22a7 (JEE Main 2022)

ID: 314851dc22a7JEE Main 2022Single Correct MCQ

If Z=A2B3C4Z = {{{A^2}{B^3}} \over {{C^4}}}, then the relative error in Z will be :

Select Option

Step-by-step Explanation

Core Formula & Concept:

In experiments and measurements, every physical quantity is associated with some uncertainty or error. When quantities are combined through mathematical operations (like multiplication, division, or exponentiation), the errors propagate in a specific manner. The key concept here is error propagation in functions of multiple variables.

For a general function of the form: Z=f(A,B,C,… )Z = f(A, B, C, \dots) the relative error in ZZ, denoted as ΔZZ\frac{\Delta Z}{Z}, is determined using the logarithmic differentiation method. This method is particularly useful when ZZ is expressed as a product or ratio of powers of measured quantities.

The fundamental rule is: If Z=AmBnCp⋯Z = A^m B^n C^p \cdots then the relative error in ZZ is: ΔZZ=∣m∣ΔAA+∣n∣ΔBB+∣p∣ΔCC+⋯\frac{\Delta Z}{Z} = |m| \frac{\Delta A}{A} + |n| \frac{\Delta B}{B} + |p| \frac{\Delta C}{C} + \cdots where ΔA,ΔB,ΔC\Delta A, \Delta B, \Delta C are the absolute errors in A,B,CA, B, C, respectively.

Note: The exponents become multiplicative factors, and the signs of the exponents do not affect the error (since errors are always added in magnitude).

--- Step-by-Step Derivation:

We are given: Z=A2B3C4Z = \frac{A^2 B^3}{C^4}

Step 1: Take the natural logarithm of both sides

ln⁡Z=ln⁡(A2B3C4)=ln⁡(A2)+ln⁡(B3)−ln⁡(C4)\ln Z = \ln\left(\frac{A^2 B^3}{C^4}\right) = \ln(A^2) + \ln(B^3) - \ln(C^4) ⇒ln⁡Z=2ln⁡A+3ln⁡B−4ln⁡C\Rightarrow \ln Z = 2\ln A + 3\ln B - 4\ln C

Step 2: Differentiate both sides with respect to the variables

Differentiating implicitly (treating ΔA,ΔB,ΔC\Delta A, \Delta B, \Delta C as small changes): dZZ=2⋅dAA+3⋅dBB−4⋅dCC\frac{dZ}{Z} = 2 \cdot \frac{dA}{A} + 3 \cdot \frac{dB}{B} - 4 \cdot \frac{dC}{C} In the context of errors, we interpret dA→ΔAdA \rightarrow \Delta A, dB→ΔBdB \rightarrow \Delta B, dC→ΔCdC \rightarrow \Delta C, and take absolute values to ensure errors add up (since errors are always positive in magnitude): ΔZZ=2ΔAA+3ΔBB+4ΔCC\frac{\Delta Z}{Z} = 2 \frac{\Delta A}{A} + 3 \frac{\Delta B}{B} + 4 \frac{\Delta C}{C} Note: The negative sign from −4ln⁡C-4\ln C becomes positive in the error expression because we are concerned with the magnitude of the error, not its direction.

Step 3: Compare with the given options

The derived expression is: ΔZZ=2ΔAA+3ΔBB+4ΔCC\frac{\Delta Z}{Z} = \frac{2\Delta A}{A} + \frac{3\Delta B}{B} + \frac{4\Delta C}{C} This matches Option C.

--- Common Traps & Exam Tip:

Trap 1: Sign Confusion
Many students mistakenly keep the negative sign from the denominator term C4C^4. They write: ΔZZ=2ΔAA+3ΔBB−4ΔCC\frac{\Delta Z}{Z} = \frac{2\Delta A}{A} + \frac{3\Delta B}{B} - \frac{4\Delta C}{C} This is incorrect because errors are always additive in magnitude. The negative sign in the function does not translate to a negative sign in the error expression.

Trap 2: Forgetting to Multiply by Exponents
Some students add the relative errors directly: ΔAA+ΔBB+ΔCC\frac{\Delta A}{A} + \frac{\Delta B}{B} + \frac{\Delta C}{C} This ignores the fact that the exponents 2,3,42, 3, 4 amplify the errors. Always multiply each relative error by the absolute value of its exponent.

Exam Tip:
Whenever you see a function involving powers, use logarithmic differentiation. It simplifies the process and reduces the chance of sign errors. Remember: Errors add in magnitude, not direction.

Thus, the correct answer is Option C.

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