JEE PYQ: Units & Measurements - Question ID ebffa6d3c26a (JEE Main 2018)

When a physical quantity \( A \) is expressed as a product or quotient of other measured quantities \( P, Q, R, \dots \), the relative error (or percentage error) in \( A \) is determined by the propagation of errors rule.

• If \( A = P^a Q^b R^c \dots \), then the maximum relative error in \( A \) is given by: $\frac{\Delta A}{A} = |a| \frac{\Delta P}{P} + |b| \frac{\Delta Q}{Q} + |c| \frac{\Delta R}{R} + \dots$
• Percentage errors are simply the relative errors multiplied by 100.

Given: $A = \frac{P^3 Q^2}{\sqrt{R} \, S}$

1. Rewrite the expression in exponential form: $A = P^3 Q^2 R^{-1/2} S^{-1}$
2. Identify the exponents:
   • \( P \) has exponent \( 3 \)
   • \( Q \) has exponent \( 2 \)
   • \( R \) has exponent \( -\frac{1}{2} \)
   • \( S \) has exponent \( -1 \)
3. Apply the error propagation formula: $\frac{\Delta A}{A} = 3 \frac{\Delta P}{P} + 2 \frac{\Delta Q}{Q} + \frac{1}{2} \frac{\Delta R}{R} + 1 \frac{\Delta S}{S}$
4. Convert percentage errors to relative errors:
   • \( \frac{\Delta P}{P} = 0.5\% = 0.005 \)
   • \( \frac{\Delta Q}{Q} = 1\% = 0.01 \)
   • \( \frac{\Delta R}{R} = 3\% = 0.03 \)
   • \( \frac{\Delta S}{S} = 1.5\% = 0.015 \)
5. Substitute the values: $\frac{\Delta A}{A} = 3(0.005) + 2(0.01) + \frac{1}{2}(0.03) + 1(0.015)$
6. Calculate each term:
   • \( 3 \times 0.005 = 0.015 \)
   • \( 2 \times 0.01 = 0.02 \)
   • \( \frac{1}{2} \times 0.03 = 0.015 \)
   • \( 1 \times 0.015 = 0.015 \)
7. Sum the contributions: $0.015 + 0.02 + 0.015 + 0.015 = 0.065$
8. Convert back to percentage: $\frac{\Delta A}{A} \times 100 = 0.065 \times 100 = 6.5\%$

Students often forget to take the absolute value of exponents, especially negative ones. For instance, \( R^{-1/2} \) contributes \( \frac{1}{2} \frac{\Delta R}{R} \), not \( -\frac{1}{2} \frac{\Delta R}{R} \). Errors always add up in magnitude, never subtract. Always double-check the sign of exponents when applying the error propagation rule.