A physical quantity p= √abc^2/d^3 determined by measuring a,b,c and d sepa-
d
rately with the percentage error of 2%, 3%, and 1% respectively. Minimum
amount of error is contributed by the measurement of
Answers
Answer:
error percentage is 9℅
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Given info : A physical quantity P = √(abc²)/d³ determined by measuring a , b , c and d separately with the percentage worker of 2 % , 3 %, 2 % and 1 % respectively.
To find : the percentage error in p is..
solution : here p = √(abc²)/d³
= a½ b½ c/d³
∴ % error in p = 1/2 × % error in a + 1/2 × % error in b + 1/2 × % error in c + 3 × % error in d
= 1/2 × 2 % + 1/2 × 3 % + 1/2 × 2 % + 3 × 1 %
= 1 % + 1.5 % + 1 % + 3 %
= 6.5 %
Therefore the percentage error in p is 6.5 %
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