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JEE Main 2023 (6 Apr, Shift 1) · Errors in Measurement

Errors in Measurement question from JEE Main 2023 (6 Apr, Shift 1)

Q67JEE Main 2023 · 6 APR · SHIFT 1Easy

Errors in Measurement

Two resistances are given as R1=(10±0.5) ΩR_1 = (10 \pm 0.5)\ \Omega and R2=(15±0.5) ΩR_2 = (15 \pm 0.5)\ \Omega. The percentage error in the measurement of equivalent resistance when they are connected in parallel is -

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Answer: (c)

  1. Rp=R1R2R1+R2=15025=6 ΩR_p = \frac{R_1R_2}{R_1 + R_2} = \frac{150}{25} = 6\ \Omega
  2. From 1Rp=1R1+1R2\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}: ΔRpRp2=ΔR1R12+ΔR2R22\frac{\Delta R_p}{R_p^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}
  3. ΔRp=36(0.5100+0.5225)=0.26 Ω\Delta R_p = 36\left(\frac{0.5}{100} + \frac{0.5}{225}\right) = 0.26\ \Omega
  4. Percentage error =0.266×100≈4.33%= \frac{0.26}{6} \times 100 \approx 4.33\%
  5. So the answer is (c).
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