Maximum Fractional Error JEE Mains 2025 Physics: Question Solved (8th April Shift 2)
This post presents a detailed solution to a JEE Mains 2025 Physics question from the 8th April Shift 2 paper. The topic covered is Maximum Fractional Error and Error Propagation in derived physical quantities. This is a key concept in measurement and experimental physics, and often appears in competitive exams like JEE.
Question
The original question is:
A quantity Q is formulated as X−2Y3/2Z−2/5. X, Y and Z are independent parameters which have fractional errors of 0.1, 0.2 and 0.5, respectively, in measurement. The maximum fractional error of Q is:
- 0.1
- 0.6
- 0.7
- 0.8
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Detailed Step-by-Step Solution
Step 1: Understand the formula for fractional error in derived quantities
For a quantity expressed as Q = XaYbZc, the maximum fractional error in Q is given by:
ΔQ/Q = |a|(ΔX/X) + |b|(ΔY/Y) + |c|(ΔZ/Z)
Step 2: Identify powers and given fractional errors
- a = −2 → |a| = 2
- b = 3/2 → |b| = 1.5
- c = −2/5 → |c| = 0.4
- ΔX/X = 0.1
- ΔY/Y = 0.2
- ΔZ/Z = 0.5
Step 3: Substitute and calculate
ΔQ/Q = (2 × 0.1) + (1.5 × 0.2) + (0.4 × 0.5)
= 0.2 + 0.3 + 0.2 = 0.7
Final Answer:
Option (3) 0.7
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Conclusion
This question from JEE Mains 2025 tested core understanding of error propagation in physics. Mastering such problems helps in scoring high in both Physics sections of JEE Mains and Advanced.