Please do both Questions. 31 The electrical resistance R of a wire is directly p

Question

Please do both Questions. 31 The electrical resistance R of a wire is directly propor- tional to its length and inversely proportional to the square of its diameter. If the length is measured with a possible error of t 1% and the diameter is measured with a possible error of 13%, what is the maximum percent age error in the calculated value of R? 32 The flow of blood through an arteriole is given by PR4 where l is the length of the arteriole, R is /(8vl), the radius, P is the pressure difference between the two ends, and v is the viscosity of the blood (see Section 6.8) Suppose that v and are constant. Use differentials to approximate the percentage change in the blood flow if the radius decreases by 2% and the pressure increases by 3%.

Explanation / Answer

Question 31)

Answer :-

Electrical resistance R of wire is directly proportional to the length of the wire and inversely proportional to the square of the diameter.

So we can write it like R = k*l/d^2    where k is constant , l = length and d = diameter.

Possible error in lenght = 1% and possible error in diameter = 3%

then new length = 1.01*l and new diameter = 1.03*d

So the R = k*(1.01)*l /(1.03d)^2 = k*1.01l /(1.0609d^2)

So R = (0.9520)kl/d^2

So the percentage error in R = [New resistance - true resistance ]*100 = [0.9520 - 1] *100 = [0.47978]*100 = - 4.8

So the percentage error in R is -4.8%

Possible error in lenght = -1% and possible error in diameter = -3%

then new length = 0.99*l and new diameter = 0.97*d

So the R = k*(0.99)*l /(0.97d)^2 = k*(0.99)l /(0.9409d^2)

So R = (1.0521)kl/d^2

So the percentage error in R = [New resistance - true resistance ]*100 = [1.0521 - 1] *100 = [0.0521]*100 = 5.21%

So the percentage error in R is 5.21%

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