© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write down the given expression
The physical quantity is
$X = \frac{A^2 \, B^{1/2}}{C^{1/3} \, D^3}$.
Step 2: Recall the formula for maximum percentage error
If a quantity $X$ depends on variables $A$, $B$, $C$, … in the form
$X = A^p \, B^q \, C^r \dots$,
then the maximum fractional error in $X$ is given by:
$\frac{\Delta X}{X} = \lvert p \rvert \frac{\Delta A}{A}
+ \lvert q \rvert \frac{\Delta B}{B}
+ \lvert r \rvert \frac{\Delta C}{C} + \dots$
Converting fractional errors to percentages, multiply each term by 100.
Step 3: Identify exponents for each variable
From $X = \frac{A^2 \, B^{1/2}}{C^{1/3} \, D^3}$, we note:
$p = 2$ for $A^2$
$q = \tfrac{1}{2}$ for $B^{1/2}$
$r = -\tfrac{1}{3}$ for $C^{1/3}$ in the denominator (which is effectively multiplication by $C^{-1/3}$)
$s = -3$ for $D^3$ in the denominator (which is effectively multiplication by $D^{-3}$)
Step 4: Substitute percentage errors for each variable
Let the percentage errors in $A, B, C,$ and $D$ be $1\%, 2\%, 3\%,$ and $4\%$ respectively. Then:
$\frac{\Delta X}{X} \times 100
= \bigl\lvert 2 \bigr\rvert \times 1\%
+ \bigl\lvert \tfrac{1}{2} \bigr\rvert \times 2\%
+ \bigl\lvert -\tfrac{1}{3} \bigr\rvert \times 3\%
+ \bigl\lvert -3 \bigr\rvert \times 4\%.$
Step 5: Calculate each term and sum them
1) $2 \times 1\% = 2\%$
2) $\tfrac{1}{2} \times 2\% = 1\%$
3) $\tfrac{1}{3} \times 3\% = 1\%$
4) $3 \times 4\% = 12\%$
Adding these up:
$2\% + 1\% + 1\% + 12\% = 16\%$
Step 6: State the final answer
Hence, the maximum percentage error in the measurement of $X$ is $16\%$.
