Question

A physical quantity P is related to four observables a, b, c, and d as \( P = \frac{\sqrt{ab}.d^\alpha}{\sqrt{C}} \) (\(\alpha\) is constant). The percentage errors in a, b, c and d are 0.5% in each. If the percentage error in P is 2%, then \(\alpha\) is

• A. \( \frac{5}{2} \)
• B. \( \frac{2}{5} \)
• C. \( \frac{3}{4} \)
• D. \( \frac{3}{2} \)

Hint:

If \(P = k X^p Y^q Z^r\), then percentage error in P is \( |\pm p|(% \text{err in X}) + |\pm q|(% \text{err in Y}) + |\pm r|(% \text{err in Z}) \).
Errors are always added (maximum possible error).
Ensure powers are correctly identified from the formula \(P = a^{1/2} b^{1/2} d^\alpha C^{-1/2}\). Powers are \(1/2, 1/2, \alpha, -1/2\). Absolute values of powers are used for error sum.

Answer 1

The Correct Option is D

The given relation is \( P = \frac{\sqrt{ab} \cdot d^\alpha}{\sqrt{C}} = a^{1/2} b^{1/2} d^\alpha C^{-1/2} \). The formula for relative error (or percentage error if multiplied by 100) when quantities are multiplied or divided is: If \( P = k \cdot X^p Y^q Z^r \), then \( \frac{\Delta P}{P} = |p| \frac{\Delta X}{X} + |q| \frac{\Delta Y}{Y} + |r| \frac{\Delta Z}{Z} \). For percentage errors: \( (\frac{\Delta P}{P} \times 100%) = |1/2| (\frac{\Delta a}{a} \times 100%) + |1/2| (\frac{\Delta b}{b} \times 100%) + |\alpha| (\frac{\Delta d}{d} \times 100%) + |-1/2| (\frac{\Delta C}{C} \times 100%) \) Given percentage errors: \(\frac{\Delta a}{a} \times 100% = 0.5%\) \(\frac{\Delta b}{b} \times 100% = 0.5%\) \(\frac{\Delta C}{C} \times 100% = 0.5%\) (Note: C in formula is uppercase, observable is c. Assuming c=C) \(\frac{\Delta d}{d} \times 100% = 0.5%\) Percentage error in P = \( (\frac{\Delta P}{P} \times 100%) = 2% \). We assume \(\alpha>0\) as it's usually a power. So, \( 2% = \frac{1}{2}(0.5%) + \frac{1}{2}(0.5%) + \alpha(0.5%) + \frac{1}{2}(0.5%) \) \( 2 = 0.5 \left( \frac{1}{2} + \frac{1}{2} + \alpha + \frac{1}{2} \right) \) \( 2 = 0.5 \left( 1 + \alpha + 0.5 \right) \) \( 2 = 0.5 (1.5 + \alpha) \) Divide by 0.5 (or multiply by 2): \( 4 = 1.5 + \alpha \) \( \alpha = 4 - 1.5 = 2.5 = \frac{5}{2} \). Let's recheck the options. The checkmark is on (d) \( \frac{3}{2} \). If \(\alpha = 3/2 = 1.5\). Then \( 0.5 (1.5 + \alpha) = 0.5 (1.5 + 1.5) = 0.5 (3.0) = 1.5 \). So, \(2% = 1.5%\). This is FALSE. My calculation for \(\alpha = 5/2\) yields \(2% = 0.5(1.5+2.5) = 0.5(4) = 2%\). This is correct. So \(\alpha = 5/2\). This is option (a). The checkmark in the image is on option (d) \(3/2\). But my derivation gets \(5/2\). Let's re-read "The percentage errors in a, b, c and d are 0.5% in each". So \( \frac{1}{2}(%a) + \frac{1}{2}(%b) + |\alpha|(%d) + \frac{1}{2}(%C) = %P \) \( \frac{1}{2}(0.5) + \frac{1}{2}(0.5) + |\alpha|(0.5) + \frac{1}{2}(0.5) = 2 \) \( 0.25 + 0.25 + 0.5|\alpha| + 0.25 = 2 \) \( 0.75 + 0.5|\alpha| = 2 \) \( 0.5|\alpha| = 2 - 0.75 = 1.25 \) \( |\alpha| = \frac{1.25}{0.5} = \frac{12.5}{5} = 2.5 = \frac{5}{2} \). So \(\alpha = \pm 5/2\). Since \(\alpha\) is usually a positive constant in such formulas or if not specified, assume \(\alpha>0\). \(\alpha = 5/2\). This is option (a). The checkmark on (d) \(3/2\) appears to be incorrect based on my derivation. If \(\alpha = 3/2\), then \(0.75 + 0.5(3/2) = 0.75 + 0.5(1.5) = 0.75 + 0.75 = 1.5%\). This is not 2%. I will proceed with my derived answer \(\alpha=5/2\). Option (a) is \(5/2\). The checkmark is on (d) \(3/2\). There is a discrepancy. I'll use my derived value and note that the marked option might be different. \[ \boxed{\frac{5}{2} \text{ (This is option (a). The marked option (d) seems incorrect.)}} \]