A physical quantity \( Q \) is found to depend on quantities \( a \), \( b \), \( c \) by the relation\[Q = \frac{a^4 b^3}{c^2}.\]The percentage errors in \( a \), \( b \), and \( c \) are 3%, 4%, and 5% respectively. Then, the percentage error in \( Q \) is:
- 66%
- 43%
- 34%
- 14%
The Correct Option is C
Solution and Explanation
Given: \(Q = \frac{a^4 b^3}{c^2}\)
The percentage error in \( Q \) can be calculated using the rule for propagation of errors in multiplication and division.
Step 1. Calculate the fractional error in \( Q \):
\(\frac{\Delta Q}{Q} = 4 \frac{\Delta a}{a} + 3 \frac{\Delta b}{b} + 2 \frac{\Delta c}{c}\)
Step 2. Substitute the given percentage errors:
\(\frac{\Delta Q}{Q} \times 100 = 4 \left( \frac{\Delta a}{a} \times 100 \right) + 3 \left( \frac{\Delta b}{b} \times 100 \right) + 2 \left( \frac{\Delta c}{c} \times 100 \right)\)
\(= 4 \times 3\% + 3 \times 4\% + 2 \times 5\%\)
\(= 12\% + 12\% + 10\%\)
Step 3. Calculate the total percentage error in \( Q \):
\(\text{Percentage error in } Q = 12\% + 12\% + 10\% = 34\%\)
Thus, the percentage error in \( Q \) is 34%.
The Correct Answer is: 34%
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