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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^3 b^2}{cd}. The percentage error in PP is

A

14%14\%

B

10%10\%

C

7%7\%

D

4%4\%

Step-by-Step Solution

Given the physical quantity P=a3b2cdP = \frac{a^3 b^2}{cd}. The maximum percentage error in PP is determined by the sum of the percentage errors of the individual quantities multiplied by their respective powers. The formula for percentage error is: ΔPP×100=3(Δaa×100)+2(Δbb×100)+(Δcc×100)+(Δdd×100)\frac{\Delta P}{P} \times 100 = 3\left(\frac{\Delta a}{a} \times 100\right) + 2\left(\frac{\Delta b}{b} \times 100\right) + \left(\frac{\Delta c}{c} \times 100\right) + \left(\frac{\Delta d}{d} \times 100\right) Given the percentage errors are: Δaa×100=1%\frac{\Delta a}{a} \times 100 = 1\% Δbb×100=2%\frac{\Delta b}{b} \times 100 = 2\% Δcc×100=3%\frac{\Delta c}{c} \times 100 = 3\% Δdd×100=4%\frac{\Delta d}{d} \times 100 = 4\% Substituting these values into the error equation: % 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\%.

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