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JEE Main 2026 (5 Apr, Shift 2) · Errors in Measurement

Errors in Measurement question from JEE Main 2026 (5 Apr, Shift 2)

Q33JEE Main 2026 · 5 APR · SHIFT 2Difficult

Errors in Measurement

In an experiment to determine the resistance of a given wire using Ohm's law, the voltmeter and ammeter readings are noted as 10 V and 5 A, respectively. The least counts of voltmeter and ammeter are 500 mV and 200 mA, respectively. The estimated error in the resistance measurement is ‾ Ω\underline{\qquad}\ \Omega

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Answer: (d)

  1. R=VI=105=2 ΩR = \frac{V}{I} = \frac{10}{5} = 2\ \Omega
  2. ΔRR=ΔVV+ΔII=0.510+0.25=0.05+0.04=0.09\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} = \frac{0.5}{10} + \frac{0.2}{5} = 0.05 + 0.04 = 0.09
  3. ΔR=0.09×2=0.18 Ω\Delta R = 0.09 \times 2 = 0.18\ \Omega
  4. So the answer is (d) 0.18 Ω0.18\ \Omega.
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