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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Mathematics in Physics

JEE Main · Physics · Mathematics in Physics

In the measurement of a physical quantity X= A^2 B C^ 1 / 3 D^3 ....

In the measurement of a physical quantity X=A2BC1/3D3X=\frac{A^2 B}{C^{1 / 3} D^3}. The percentage errors introduced in the measurements of the quantities A,B,CA, B, C, and DD are 2%,2%,4%2 \%, 2 \%, 4 \% and 5%5 \%, respectively. Then, the minimum amount of percentage error in the measurement of XX is contributed by

Options

  1. AAA
  2. BBB
  3. CCC
  4. DDD

Correct answer

C. CC

Step-by-step solution

x=A2BC1/3D3% error in x,Δxx=2ΔAA+ΔBB+13ΔCC+3ΔDD2(ΔAA×100%)+(ΔBB×100%)+13(ΔCC×100%)+3(ΔDD×100%)\begin{aligned} & x=\frac{A^2 B}{C^{1 / 3} D^3} \\ & \% \text { error in } x, \\ & \frac{\Delta x}{x}=2 \frac{\Delta A}{A}+\frac{\Delta B}{B}+\frac{1}{3} \frac{\Delta C}{C}+\frac{3 \Delta D}{D} \\ & 2\left(\frac{\Delta A}{A} \times 100 \%\right)+\left(\frac{\Delta B}{B} \times 100 \%\right) \\ & +\frac{1}{3}\left(\frac{\Delta C}{C} \times 100 \%\right)+3\left(\frac{\Delta D}{D} \times 100 \%\right) \end{aligned} Minimum percentage error =13(ΔCC×100%)=13×4=1.1%=\frac{1}{3}\left(\frac{\Delta C}{C} \times 100 \%\right)=\frac{1}{3} \times 4=1.1 \%
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