NCERT Solutions Class 11 Physics - Solved Exercise Question 2.13

Question 2.13:

A physical quantity P is related to four observables a, b, c and d as follows :
P = a3b/ (√c )d
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 value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?

Solution:

P = a3b2 / (√c)d
∆P/P = 3(∆a/a) +2(∆b/b) + (1/2)(∆c/c) + (∆d/d)
((∆P/P) X 100) % = (3(∆a/a) X 100 + 2(∆b/b) X 100 + (1/2)(∆c/c) X 100 + (∆d/d) X 100) %
= 3 X 1 + 2 X 3 + (1/2) X 4 + 2
= 13%
Percentage error in P = 13 %
Value of P is given as 3.763.

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

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