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Units and Measurements question

2022 · 25 Jun · Shift 1 · Q44

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Units and Measurements
2022 · 25 Jun · Shift 1 · Q44

Units and Measurements question

2022 · 25 Jun · Shift 1 · Q44

JEE MainPhysicsUnits and MeasurementsMCQ+4 / −1

If

Z = {{{A^2}{B^3}} \over {{C^4}}}

, then the relative error in Z will be :

A
${{\Delta A} \over A} + {{\Delta B} \over B} + {{\Delta C} \over C}$
B
${{2\Delta A} \over A} + {{3\Delta B} \over B} - {{4\Delta C} \over C}$
C
${{2\Delta A} \over A} + {{3\Delta B} \over B} + {{4\Delta C} \over C}$
D
${{\Delta A} \over A} + {{\Delta B} \over B} - {{\Delta C} \over C}$

View written solutionFree

Correct answer: C

Given

$Z = \frac{A^2 B^3}{C^4}$

We need the relative error in $Z$ .

Rule for propagation of relative errors

For a quantity of the form

$Q = \frac{X^m Y^n}{Z^p}$

the maximum relative error is

$\frac{\Delta Q}{Q} = m\frac{\Delta X}{X} + n\frac{\Delta Y}{Y} + p\frac{\Delta Z}{Z}$

Important: In error calculation, powers come as coefficients, and all contributions are added.

Apply the rule to the given expression

Here,

power of $A$ is $2$
power of $B$ is $3$
power of $C$ is $4$

Therefore,

$\frac{\Delta Z}{Z} = 2\frac{\Delta A}{A} + 3\frac{\Delta B}{B} + 4\frac{\Delta C}{C}$

Compare with options

A: Incorrect, powers are missing.
B: Incorrect, error terms are not subtracted.
C: Correct.
D: Incorrect, both powers and sign treatment are wrong.

Hence, the correct option is:

$\boxed{C}$

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