JEE PYQ: Units & Measurements - Question ID 4b5086395d45 (JEE Main 2021)

In the chapter Units & Measurements, when a physical quantity \( y \) is expressed as a product of powers of other measured quantities, the relative (percentage) error in \( y \) is given by the sum of the absolute values of the percentage errors in each quantity multiplied by its respective exponent.

Mathematically, if $y = k \cdot m^{a} \cdot r^{b} \cdot g^{c} \cdot l^{d}$ where \( k \) is a dimensionless constant, then the percentage error in \( y \) is: $$\frac{\Delta y}{y} \times 100 = \left| a \right| \left(\frac{\Delta m}{m} \times 100\right) + \left| b \right| \left(\frac{\Delta r}{r} \times 100\right) + \left| c \right| \left(\frac{\Delta g}{g} \times 100\right) + \left| d \right| \left(\frac{\Delta l}{l} \times 100\right)$$

This principle is derived from the logarithmic differentiation of the expression for \( y \).

Given: $y = m^{2} \cdot r^{-4} \cdot g^{x} \cdot l^{-\frac{3}{2}}$ The percentage errors are: - \( \frac{\Delta y}{y} \times 100 = 18\% \) - \( \frac{\Delta m}{m} \times 100 = 1\% \) - \( \frac{\Delta r}{r} \times 100 = 0.5\% \) - \( \frac{\Delta l}{l} \times 100 = 4\% \) - \( \frac{\Delta g}{g} \times 100 = p\% \)

Using the error propagation formula: $\frac{\Delta y}{y} \times 100 = \left| 2 \right| \cdot 1\% + \left| -4 \right| \cdot 0.5\% + \left| x \right| \cdot p\% + \left| -\frac{3}{2} \right| \cdot 4\%$ Simplify the absolute values: $18 = 2 \cdot 1 + 4 \cdot 0.5 + \left| x \right| \cdot p + \frac{3}{2} \cdot 4$ Calculate each term: $$2 \cdot 1 = 2 \\ 4 \cdot 0.5 = 2 \\ \frac{3}{2} \cdot 4 = 6$$ Substitute back: $$18 = 2 + 2 + \left| x \right| p + 6 \\ 18 = 10 + \left| x \right| p \\ \Rightarrow \left| x \right| p = 8 \quad \text{(Equation 1)}$$

Now, we need another relation to find \( x \) and \( p \). Since the question does not provide additional constraints, we consider the dimensional consistency of the formula for \( y \).

Assume the dimensions of the quantities are: - \( m \): mass \([M]\) - \( r \): length \([L]\) - \( g \): acceleration due to gravity \([LT^{-2}]\) - \( l \): length \([L]\)

The dimension of \( y \) is: $$[y] = [m]^{2} \cdot [r]^{-4} \cdot [g]^{x} \cdot [l]^{-\frac{3}{2}} \\ = M^{2} \cdot L^{-4} \cdot (LT^{-2})^{x} \cdot L^{-\frac{3}{2}} \\ = M^{2} \cdot L^{-4 + x - \frac{3}{2}} \cdot T^{-2x} \\ = M^{2} \cdot L^{x - \frac{11}{2}} \cdot T^{-2x}$$

For \( y \) to be dimensionally consistent, its dimensions must match some physical quantity. However, since no specific quantity is mentioned, we assume \( y \) is dimensionless (a common case in such problems unless specified otherwise). Thus: $M^{2} \cdot L^{x - \frac{11}{2}} \cdot T^{-2x} = M^{0} L^{0} T^{0}$ This gives us three equations by equating exponents: 1. \( 2 = 0 \) (for mass) → Not possible.

This suggests that \( y \) is not dimensionless. Instead, let’s assume \( y \) has dimensions of some physical quantity. However, since the problem does not specify, we consider an alternative approach: the exponents of dimensions must balance for \( y \) to represent a valid physical quantity.

A more plausible assumption is that \( y \) has dimensions of \( L^{a} T^{b} \), but without loss of generality, we can consider the exponents of \( L \) and \( T \) to be zero for simplicity (as the problem likely expects us to focus on error propagation). Thus, we proceed with the earlier derived Equation 1: $\left| x \right| p = 8$

Now, we examine the options to find consistent values of \( x \) and \( p \): - Option A: \( x = 5 \), \( p = \pm 2 \) → \( 5 \cdot 2 = 10 \neq 8 \) - Option B: \( x = 4 \), \( p = \pm 3 \) → \( 4 \cdot 3 = 12 \neq 8 \) - Option C: \( x = \frac{16}{3} \), \( p = \pm \frac{3}{2} \) → \( \frac{16}{3} \cdot \frac{3}{2} = 8 \) (matches Equation 1) - Option D: \( x = 8 \), \( p = \pm 2 \) → \( 8 \cdot 2 = 16 \neq 8 \)

Thus, Option C satisfies the error propagation condition. The value of \( p \) is given as \( \pm \frac{3}{2} \), which accounts for the absolute value in the error formula.

1. Ignoring Absolute Values: Students often forget to take absolute values of exponents when calculating percentage errors, leading to incorrect signs. Always use \( \left| \text{exponent} \right| \).

2. Dimensional Analysis Misapplication: Assuming \( y \) is dimensionless without justification can lead to confusion. In such problems, focus on the error propagation formula unless dimensional consistency is explicitly required.

3. Arithmetic Errors: Simple addition or multiplication mistakes in calculating the total percentage error are common. Double-check calculations, especially when dealing with fractional exponents.

4. Overcomplicating the Problem: The question primarily tests error propagation, not dimensional analysis. While dimensional consistency can be used to verify, it is not always necessary if the error condition is sufficient to determine the answer.