Question
Question: A body has weight \(22.42\) gm and has a measured volume of \(4.7\) cc. The possible error in the me...
A body has weight 22.42 gm and has a measured volume of 4.7 cc. The possible error in the measurement of mass and volume is 0.01 gm and 0.1 cc. The maximum error in the density will be
A. 22%
B. 2.2%
C. 0.22%
D. 0.022%
Solution
We can find the maximum error in the density of the body by using error analysis of quantities and errors in a product and division. Also, we use the concept of relative or fractional error and percentage error.
Formulae used:
Relative / fractional error =aΔa
Percentage error % =aΔa×100
Maximum error in division =(aΔa+bΔb)
Where, Δa,Δb - are the errors in quantities a,brespectively.
Complete step by step answer:
We are given that a body has weight 22.42 gm and has a measured volume of 4.7 cc and the possible error in the measurement of mass and volume is 0.01 gm and 0.1 cc, then,
m=22.42 gm , Δm=0.01 gm
And v=4.7 cc, Δv=0.1 cc
Then, the relative error in measurement of mass and volume will be
mΔm=22.420.01=22421 and vΔv=4.70.1=471
We know that, Density of the body is mass per volume of the body.
d=vm
The error in the density will be
dΔd=mΔm−vΔv
But, the maximum error in the density will be as
(dΔd)max=mΔm+vΔv
Percentage maximum error in measurement of density is
(dΔd)max%=(mΔm+vΔv)×100%
Substituting the values, we get
(dΔd)max%=(22421+471)×100%
⇒(dΔd)max%=0.0217×100%
∴(dΔd)max%=2.17%≈2.2%
Hence, option B is correct.
Note: We should identify the original quantity and the error of the measurement.The maximum value of fractional error in the division or product of two or more quantities is equal to the sum of the fractional errors in the individual quantities.