The electrical resistance R of a uniform wire of length L and diameter d is given by R = \… — JAMB UTME

The electrical resistance $R$ of a uniform wire of length $L$ and diameter $d$ is given by $R = \frac{4\rho L}{\pi d^2}$ , where $\rho$ is the constant resistivity of the material. If the length $L$ is measured with a percentage error of $2.0\%$ and the diameter $d$ is measured with a percentage error of $1.5\%$ , what is the maximum percentage error in the calculated value of $R$ ?

A
$3.5\%$
$5.0\%$ Cevap
C
$0.5\%$
D
$4.25\%$

Cevap

The maximum percentage error in the calculated resistance

R

5.0\%

For a physical quantity defined by

R = \frac{4\rho L}{\pi d^2}

, the maximum percentage error is determined by adding the percentage error of

L

to twice the percentage error of

d

. Calculating

2.0\% + 2(1.5\%) = 5.0\%

yields the correct maximum percentage error.

Adım Adım Çözüm

Identify the relation for maximum percentage error propagation in a physical formula.

For

R = \frac{4\rho L}{\pi d^2}

, the fractional error equation is

\frac{\Delta R}{R} = \frac{\Delta L}{L} + 2\left(\frac{\Delta d}{d}\right)

When quantities are multiplied or divided, their relative errors add, and any exponent becomes a multiplier for that quantity's relative error.

Express the relation in terms of percentage errors.

\% \text{ Error in } R = (\% \text{ Error in } L) + 2 \times (\% \text{ Error in } d)

Multiplying the fractional error expression by

100\%

converts all relative errors into percentage errors.

Substitute the given values into the formula.

\% \text{ Error in } R = 2.0\% + 2 \times (1.5\%) = 2.0\% + 3.0\% = 5.0\%

Substituting

2.0\%

for length error and

1.5\%

for diameter error yields the combined maximum percentage error.

Anahtar Kavram

Propagation of errors in physical quantities involving exponents

Soru

Cevap

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