wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

PARAGRAPH
If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z=x/y. If the errors in x,y and z are Δx,Δy and Δz, respectively, then
z±Δz=x±Δxy±Δy=xy(1±Δxx)(1±Δyy)1.
The series expansion for (1±Δyy)1, to first power in Δy/y. is 1(Δy/y). The relative errors in independent variables are always added. So the error in z will be
Δz=z(Δxx+Δyy) .
The above derivation makes the assumption that Δx/x1,Δy/y1. Therefore, the higher powers of these quantities are neglected.

Consider the ratio r=(1a)(1+a) to be determined by measuring a dimensionless quantity a. If the error in the measurement of a is Δa(Δaa<<1), then what is the error Δr in determining r?

A
Δa(1+a)2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2Δa(1+a)2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2Δa(1a2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2aΔa(1a2)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 2Δa(1+a)2
r=(1a1+a)
Δrr=Δ(1a)(1a)+Δ(1+a)(1+a)
Δrr=Δa(1a)+Δa(1+a)
Δrr=Δa(1+a+1a)(1a)(1+a)
Δr=2Δa(1a)(1+a)(1a)(1+a)=2Δa(1+a)2

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Total Surface Area of a Cuboid
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon