Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
A physical quantity $$P$$ is given as $$P = \frac{a^2 b^3}{c\sqrt{d}}$$. 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
$$P = \frac{a^2 b^3}{c \sqrt{d}} = a^2 b^3 c^{-1} d^{-1/2}$$
$$\frac{\Delta P}{P}\% = 2\left(\frac{\Delta a}{a}\right)\% + 3\left(\frac{\Delta b}{b}\right)\% + 1\left(\frac{\Delta c}{c}\right)\% + \frac{1}{2}\left(\frac{\Delta d}{d}\right)\%$$
$$\frac{\Delta P}{P}\% = 2(1) + 3(2) + 1(3) + \frac{1}{2}(4) = 2 + 6 + 3 + 2 = 13\%$$
Click on the Email ☝️ to Watch the Video Solution
Create a FREE account and get:
Educational materials for JEE preparation