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JEE Main 2024 (29 Jan, Shift 2) · Errors in Measurement

Errors in Measurement question from JEE Main 2024 (29 Jan, Shift 2)

Q60JEE Main 2024 · 29 JAN · SHIFT 2Easy

Errors in Measurement

A physical quantity QQ is found to depend on quantities a,b,ca, b, c by the relation Q=a4b3c2Q = \frac{a^4b^3}{c^2}. The percentage error in a,ba, b and cc are 3%,4%3\%, 4\% and 5%5\% respectively. Then, the percentage error in QQ is:

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Answer: (b)

  1. For Q=a4b3c2Q = \frac{a^4b^3}{c^2}: ΔQQ=4Δaa+3Δbb+2Δcc\frac{\Delta Q}{Q} = 4\frac{\Delta a}{a} + 3\frac{\Delta b}{b} + 2\frac{\Delta c}{c}
  2. =4(3)+3(4)+2(5)=12+12+10=34%= 4(3) + 3(4) + 2(5) = 12 + 12 + 10 = 34\%
  3. So the answer is (b) 34%.
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