JEE PYQ: Units & Measurements - Question ID eefe3b4e2bb8 (JEE Main 2026)

ID: eefe3b4e2bb8JEE Main 2026Single Correct MCQ

The percentage error in the calculated volume of a sphere, if there is 2%2 \% error in its diameter measurement, is ____\_\_\_\_ .

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Step-by-step Explanation

Core Formula & Concept:

In the chapter Units & Measurements, we study how errors in measured quantities propagate to derived quantities. The key concept here is error propagation in multiplication and exponentiation.

For a sphere, the volume VV is given by: V=43πr3V = \frac{4}{3} \pi r^3 where rr is the radius. Since the diameter DD is related to the radius by r=D2r = \frac{D}{2}, we can express the volume in terms of the diameter: V=43π(D2)3=π6D3V = \frac{4}{3} \pi \left(\frac{D}{2}\right)^3 = \frac{\pi}{6} D^3

When a quantity depends on another quantity raised to a power, the percentage error in the derived quantity is magnified by that power. Specifically, if VDnV \propto D^n, then: ΔVV×100=n×(ΔDD×100)\frac{\Delta V}{V} \times 100 = n \times \left(\frac{\Delta D}{D} \times 100\right) This is the fundamental rule of error propagation for powers.

Step-by-Step Derivation:

1. Express the volume in terms of diameter: V=π6D3V = \frac{\pi}{6} D^3 This shows that VD3V \propto D^3, i.e., volume depends on the cube of the diameter. 2. Given that there is a 2%2\% error in the measurement of the diameter, we write: ΔDD×100=2%\frac{\Delta D}{D} \times 100 = 2\% 3. Using the rule of error propagation for powers, the percentage error in VV is: ΔVV×100=3×(ΔDD×100)\frac{\Delta V}{V} \times 100 = 3 \times \left(\frac{\Delta D}{D} \times 100\right) Substituting the given error: ΔVV×100=3×2%=6%\frac{\Delta V}{V} \times 100 = 3 \times 2\% = 6\% 4. Therefore, the percentage error in the calculated volume of the sphere is 6%6\%.

Common Traps & Exam Tip:

  • Confusing radius and diameter: Many students mistakenly use the radius in the error propagation formula instead of the diameter. Since the question specifies error in diameter, ensure you express volume in terms of diameter before applying error rules.
  • Incorrect power application: Some students forget that the error is multiplied by the exponent. For example, they might think a 2%2\% error in diameter leads to only 2%2\% error in volume, ignoring the cubic dependence.
  • Sign errors: While percentage errors are always positive, students sometimes introduce negative signs unnecessarily. Focus on magnitudes.
Exam Tip: Always express the derived quantity in terms of the measured quantity (here, diameter) before applying error propagation rules. This ensures clarity and avoids miscalculations.

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