Units and Measurement

Question : 13

Total: 33

A physical quantity P is related to four observables a, b, c and d as follows :
P=a3b2∕(√cd)
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P? If the valueof P calculated using the above relation turns out to be 3.763, to what valueshould you round off the result?

Solution:

a3b2

√cd

Percentage error in P is given by

ΔP

×100=3(

Δa

×100 )+2(

Δb

×100 )+

(

Δc

×100 )+ (

Δd

×100)

Given,

Δa

×100=1%,

Δb

×100=3%,

Δc

×100=4%,

Δd

×100=2%
Therefore,

ΔP

×100
=3×1%+2×3%+

×4%+2%
=13%
Thus, the result has two significant figures, therefore, if P turns out to be3.763, the result would be rounded off to 3.8.

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