JEE MAIN - Physics (2023 - 10th April Morning Shift - No. 2)
A physical quantity P is given as
$$P = {{{a^2}{b^3}} \over {c\sqrt d }}$$
The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4% respectively. The percentage error in the measurement of quantity P will be
توضیح
The percentage error in a quantity that is a product or quotient of other quantities is given by the sum of the percentage errors in those quantities, each multiplied by the power to which it is raised in the expression for the quantity.
Given the physical quantity P as
$$P = \frac{a^2b^3}{c\sqrt{d}}$$
The percentage error in P, denoted as $\Delta P/P$, is given by:
$$\frac{\Delta P}{P} = 2 \left(\frac{\Delta a}{a}\right) + 3 \left(\frac{\Delta b}{b}\right) + \left(\frac{\Delta c}{c}\right) + \frac{1}{2} \left(\frac{\Delta d}{d}\right)$$
Substituting the given percentage errors for a, b, c, and d:
$$\frac{\Delta P}{P} = 2(0.01) + 3(0.02) + 0.03 + \frac{1}{2}(0.04) = 0.02 + 0.06 + 0.03 + 0.02 = 0.13$$
Therefore, the percentage error in P is 13%.