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A physical quantity P is described by the relation P = a1/2 b2 c3 d–4. If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1%, 3% and 5%, then the relative error in P will be
- a)8%
- b)12%
- c)32%
- d)25%
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A physical quantity P is described by the relation P = a1/2 b2 c3 d–4...
Solution:
Given, P = a1/2 b2 c3 d–4
Relative error in a = 2%, relative error in b = 1%, relative error in c = 3% and relative error in d = 5%.
To find the relative error in P, we use the formula:
Relative Error in P = √[(∂P/∂a)² (Δa/a)² + (∂P/∂b)² (Δb/b)² + (∂P/∂c)² (Δc/c)² + (∂P/∂d)² (Δd/d)²]
where Δa/a, Δb/b, Δc/c and Δd/d are the relative errors in a, b, c and d respectively.
∂P/∂a = 1/2 a-1/2 b2 c3 d–4, ∂P/∂b = 2a1/2 b c3 d–4, ∂P/∂c = 3a1/2 b2 c2 d–4 and ∂P/∂d = –4a1/2 b2 c3 d–5.
Substituting the values in the formula, we get:
Relative Error in P = √[(1/2 a-1/2 b2 c3 d–4)² (0.02)² + (2a1/2 b c3 d–4)² (0.01)² + (3a1/2 b2 c2 d–4)² (0.03)² + (–4a1/2 b2 c3 d–5)² (0.05)²]
= √[(1/4 b2 c6 d–8) (0.02)² + (4a b2 c6 d–8) (0.01)² + (9a b4 c4 d–8) (0.03)² + (16a b4 c6 d–10) (0.05)²]
= √[0.0001 b2 c6 d–8 + 0.0004 a b2 c6 d–8 + 0.0027 a b4 c4 d–8 + 0.004 a b4 c6 d–10]
= √[0.0001 P² (b2/c2) (c3/d3) (d4/a4) + 0.0004 P² (b/c)² (c3/d3) (d4/a4) + 0.0027 P² (b/c)² (c/d)² (d2/a2) + 0.004 P² (b/c)² (c/d)³]
= P √[0.0001 (b2/c2) (c3/d3) (d4/a4) + 0.0004 (b/c)² (c3/d3) (d4/a4) + 0.0027 (b/c)² (c/d)² (d2/a2) + 0.004 (b/c)² (c/d)³]
= P √[0.0001 (1/1.
Given, P = a1/2 b2 c3 d–4
Relative error in a = 2%, relative error in b = 1%, relative error in c = 3% and relative error in d = 5%.
To find the relative error in P, we use the formula:
Relative Error in P = √[(∂P/∂a)² (Δa/a)² + (∂P/∂b)² (Δb/b)² + (∂P/∂c)² (Δc/c)² + (∂P/∂d)² (Δd/d)²]
where Δa/a, Δb/b, Δc/c and Δd/d are the relative errors in a, b, c and d respectively.
∂P/∂a = 1/2 a-1/2 b2 c3 d–4, ∂P/∂b = 2a1/2 b c3 d–4, ∂P/∂c = 3a1/2 b2 c2 d–4 and ∂P/∂d = –4a1/2 b2 c3 d–5.
Substituting the values in the formula, we get:
Relative Error in P = √[(1/2 a-1/2 b2 c3 d–4)² (0.02)² + (2a1/2 b c3 d–4)² (0.01)² + (3a1/2 b2 c2 d–4)² (0.03)² + (–4a1/2 b2 c3 d–5)² (0.05)²]
= √[(1/4 b2 c6 d–8) (0.02)² + (4a b2 c6 d–8) (0.01)² + (9a b4 c4 d–8) (0.03)² + (16a b4 c6 d–10) (0.05)²]
= √[0.0001 b2 c6 d–8 + 0.0004 a b2 c6 d–8 + 0.0027 a b4 c4 d–8 + 0.004 a b4 c6 d–10]
= √[0.0001 P² (b2/c2) (c3/d3) (d4/a4) + 0.0004 P² (b/c)² (c3/d3) (d4/a4) + 0.0027 P² (b/c)² (c/d)² (d2/a2) + 0.004 P² (b/c)² (c/d)³]
= P √[0.0001 (b2/c2) (c3/d3) (d4/a4) + 0.0004 (b/c)² (c3/d3) (d4/a4) + 0.0027 (b/c)² (c/d)² (d2/a2) + 0.004 (b/c)² (c/d)³]
= P √[0.0001 (1/1.
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Community Answer
A physical quantity P is described by the relation P = a1/2 b2 c3 d–4...
Given , P = a1/2 b2 c3 d–4,
Maximum relative error,
= 1/2 x 2 + 2 x 1 + 3 x 3 + 4 x 5
= 32%
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A physical quantity P is described by the relation P = a1/2 b2 c3 d–4. If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1%, 3% and 5%, then the relative error in P will bea)8%b)12%c)32%d)25%Correct answer is option 'C'. Can you explain this answer?
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