Back to Directory
NEET PHYSICSMedium

In an experiment, four quantities aa, bb, cc and dd are measured with percentage error 1%1\%, 2%2\%, 3%3\% and 4%4\% respectively. Quantity PP is calculated as follows: P=a3b2cdP=\frac{a^3b^2}{cd}. Percentage error in PP is:

A

10%10\%

B

7%7\%

C

4%4\%

D

14%14\%

Step-by-Step Solution

  1. Identify the Given Formula: The quantity PP is given by P=a3b2cdP = \frac{a^3b^2}{cd}.
  2. Apply Error Propagation Rules: For a quantity that is a product or quotient of other quantities, the maximum relative error is the sum of the relative errors of the individual quantities multiplied by their respective powers. Thus, ΔPP=3Δaa+2Δbb+Δcc+Δdd\frac{\Delta P}{P} = 3\frac{\Delta a}{a} + 2\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d}.
  3. Calculate Percentage Error: Multiplying by 100100 converts this to percentage error: % error in P=3(% error in a)+2(% error in b)+(% error in c)+(% error in d)\% \text{ error in } P = 3(\% \text{ error in } a) + 2(\% \text{ error in } b) + (\% \text{ error in } c) + (\% \text{ error in } d).
  4. Substitute Given Values: % error in P=3(1%)+2(2%)+1(3%)+1(4%)\% \text{ error in } P = 3(1\%) + 2(2\%) + 1(3\%) + 1(4\%) % error in P=3%+4%+3%+4%=14%\% \text{ error in } P = 3\% + 4\% + 3\% + 4\% = 14\%.
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started
Solved: PHYSICS Question for NEET | Sushrut