Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

In an experiment, the percentage errors that occurred in the measurement of physical quantities $A$ , $B$ , $C$ , and $D$ are $1\%$ , $2\%$ , $3\%$ , and $4\%$ respectively. Then, the maximum percentage of error in the measurement of $X$ , where $X = \frac{A^2 B^{1/2}}{C^{1/3} D^3}$ , will be:

$10\%$

$\left(\frac{3}{13}\right)\%$

$16\%$

$-10\%$

Step-by-Step Solution

Identify the formula for the physical quantity: The quantity $X$ is given by $X = \frac{A^2 B^{1/2}}{C^{1/3} D^3}$ .
Apply the rule for maximum relative error: For a quantity expressed as a product or quotient of other quantities, the maximum relative error is the sum of the individual relative errors multiplied by their respective powers. Thus, the percentage error is: $\frac{\Delta X}{X} \times 100 = 2\left(\frac{\Delta A}{A} \times 100\right) + \frac{1}{2}\left(\frac{\Delta B}{B} \times 100\right) + \frac{1}{3}\left(\frac{\Delta C}{C} \times 100\right) + 3\left(\frac{\Delta D}{D} \times 100\right)$
Substitute the given percentage errors: $\% \text{ error in } X = 2(1\%) + \frac{1}{2}(2\%) + \frac{1}{3}(3\%) + 3(4\%)$ $\% \text{ error in } X = 2\% + 1\% + 1\% + 12\% = 16\%$ .

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