A student measures the current as (2.0 ± 0.1) A and resistance as (50 ± 2) Ω. The power dissipated (P = I²R) will have a percentage error of:

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Appears in NEET UG
Solution ✔ Verified
  • A7%
  • B9%
  • C10%
  • D12%
Explanation

Percentage error in I = (0.1/2.0)*100% = 5%. Percentage error in R = (2/50)*100% = 4%. For P = I²R, the percentage error is 2*(Percentage error in I) + Percentage error in R = 2*(5%) + 4% = 10% + 4% = 14%. My options are for 12%. Let me check. 2 * 5% + 4% = 14%. This means the options do not contain 14%.

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