Problem 6 on Units & Dimensions

Description:

A physical quantity P is related to four observables a,b,c and d as follows −

P = a²b²/d √c

The percentage errors of measurement in a, b, c and ð‘‘ are 1%, 3%, 4% and 2%, respectively.

What is the percentage error in the quantity P?

If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result?

The concept required to solve this problem is,

When we have two numbers, a and b, which have Δa and Δb errors associated with them respectively.

a = a₀ ± Δa

b = b₀ ± Δb

Then, for R = a × b or R = a/b,

R = R₀ ± ΔR

R = a₀ × b₀ or R₀ = a₀/b₀

ΔR/R = Δa/a₀ + Δb/b₀

Relative Errors in the original quantities get added up in the final result of product.

Also, if, R = aⁿ,

R = R₀ ± ΔR

R₀ = a₀ⁿ

ΔR/R = n(Δa/a₀)

Relative Error in the original quantity gets multiplied to the exponent in the final result of product.

Same concept of relative error will apply to percentage error as well because,

Percentage Error = Relative Error × 100

Therefore, in the given problem we find out percentage error associated with each a³, b², √c and d first,

Now, taking the original expression,

P = a²b²/d √c

Percentage Error in P −

ΔP/P × 100 = (ΔR₁/R₁ + ΔR₂/R₂ + ΔR₃/R₃ + ΔR₄/R₄) × 100

ΔP/P × 100 = 3 + 6 + 2 + 2 = 13%

In the second part of the problem,

given the value of P = 3.763.

Therefore, Error in P is, ΔP = 0.13 × 3.763 = 0.48919 (13% of the value of P)

The error, ΔP, contains a non-zero digit right after the decimal point.

Therefore, we round off after the first decimal place in P.

Therefore, final result,

P = 3.8

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