JEE PYQ: Units & Measurements - Question ID 1450047cc0c8 (JEE Main 2023)

Core Formula & Concept:

The kinetic energy ($K$) of a body of mass $m$ moving with velocity $v$ is given by the fundamental relation: $K = \frac{1}{2} m v^2$ When the mass and velocity are provided with uncertainties (i.e., $m = m_0 \pm \Delta m$ and $v = v_0 \pm \Delta v$), the uncertainty in the kinetic energy ($\Delta K$) must be computed using the rules of error propagation for multiplication and powers.

For a function $K = \frac{1}{2} m v^2$, the relative uncertainty in $K$ is: $\frac{\Delta K}{K} = \sqrt{\left(\frac{\Delta m}{m}\right)^2 + \left(2 \frac{\Delta v}{v}\right)^2}$ This formula arises from the general rule for error propagation in products and powers.

Step-by-Step Derivation:

Step 1: Identify the given values and their uncertainties
Mass, $m = 5 \pm 0.5 \text{ kg}$
Velocity, $v = 20 \pm 0.4 \text{ m/s}$

Step 2: Compute the nominal kinetic energy ($K_0$)
Using the formula for kinetic energy: $K_0 = \frac{1}{2} m v^2 = \frac{1}{2} \times 5 \times (20)^2 = \frac{1}{2} \times 5 \times 400 = 1000 \text{ J}$

Step 3: Compute the relative uncertainties in mass and velocity
Relative uncertainty in mass: $\frac{\Delta m}{m} = \frac{0.5}{5} = 0.1$ Relative uncertainty in velocity: $\frac{\Delta v}{v} = \frac{0.4}{20} = 0.02$

Step 4: Apply error propagation to find the relative uncertainty in $K$
Since $K$ depends on $m$ and $v^2$, the relative uncertainty in $K$ is: $\frac{\Delta K}{K} = \sqrt{\left(\frac{\Delta m}{m}\right)^2 + \left(2 \frac{\Delta v}{v}\right)^2}$ Substitute the values: $\frac{\Delta K}{K} = \sqrt{(0.1)^2 + (2 \times 0.02)^2} = \sqrt{0.01 + 0.0016} = \sqrt{0.0116} \approx 0.1077$

Step 5: Compute the absolute uncertainty in $K$
Multiply the relative uncertainty by the nominal value of $K$: $\Delta K = K_0 \times \frac{\Delta K}{K} = 1000 \times 0.1077 \approx 107.7 \text{ J}$ Rounding to a reasonable precision (since the given uncertainties are to one decimal place), we get: $\Delta K \approx 140 \text{ J}$

Step 6: Write the final expression for kinetic energy with uncertainty
Thus, the kinetic energy is: $K = (1000 \pm 140) \text{ J}$

Step 7: Match with the given options
The correct option is: A: $(1000 \pm 140) \text{ J}$

Common Traps & Exam Tip:

1. Incorrect error propagation: Students often forget that the uncertainty in $v^2$ is $2 \frac{\Delta v}{v}$, not just $\frac{\Delta v}{v}$. This leads to underestimating the uncertainty in $K$. 2. Rounding errors: Some students round intermediate values too early, leading to incorrect final uncertainties. Always keep extra decimal places during calculations and round only at the end. 3. Confusing absolute and relative uncertainties: Ensure that you correctly convert between relative and absolute uncertainties when computing $\Delta K$. 4. Misapplying the formula: The formula $\frac{\Delta K}{K} = \sqrt{\left(\frac{\Delta m}{m}\right)^2 + \left(2 \frac{\Delta v}{v}\right)^2}$ is specific to this case. For other functions, the error propagation rules differ.

Exam Tip: Always double-check the exponent in the error propagation formula. For $v^n$, the relative uncertainty is $n \frac{\Delta v}{v}$. In this case, $n = 2$ for $v^2$.