Advertisement Remove all ads

What is the Percentage Error in the Quantity P and 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 - Physics

A physical quantity is related to four observables a, b, c and as follows:

`P=(a^3b^2)/((sqrtcd))`

The percentage errors of measurement in aband d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P? If the value of calculated using the above relation turns out to be 3.763, to what value should you round off the result?

Advertisement Remove all ads

Solution 1

`P = (a^3b^2)/(sqrtcd)`

`(triangleP)/P = (3trianglea)/a + (2triangleb)/b + 1/2 (trianglec)/c + (triangled)/d`

`((triangleP)/P xx100)%=(3xx(trianglea)/axx10+2xx(triangleb)/bxx100+1/2xx(trianglec)/c xx 100 + (triangled)/d xx 100)%`

= `3xx1+2xx3+1/2xx4+2`

=3+6+2+2 = 13%

Percentage error in P = 13 %

Value of is given as 3.763.

By rounding off the given value to the first decimal place, we get P = 3.8.

Solution 2

As `P = (a^3b^2)/(sqrtcd) = a^3b^2 c^((-1)/2) d^(-1)`

∴Maximum fractional error in the measurement

`(triangleP)/P = 3 (trianglea)/a + 2 (triangleb)/b + 1/2 (trianglec)/c + (triangled)/d`

As `(trianglea)/a` = 1%, `(triangleb)/b` = 3%, `(trianglec)/c` = 4% and `(triangled)/d` = 2%

∴Maximum fractional error in the measurement

`(triangleP)/P = 3xx1%+2xx3%+ 1/2xx4% + 2%`

= 3% + 6% + 2% + 2% = 13%

if P = 3.763, then traingleP = 13% of P

=`(13P)/100 = (13xx3.763)/100 = 0.489`

As the error lies in first decimal place, the answer should be rounded off to first decimal place. Hence, we shall express the value of P after rounding it off as P = 3.8.

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 2 Units and Measurements
Q 13 | Page 36
Advertisement Remove all ads

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×