JEE MAIN - Physics (2025 - 3rd April Evening Shift - No. 25)
A physical quantity C is related to four other quantities $\mathrm{p}, \mathrm{q}, \mathrm{r}$ and s as follows
$$ C=\frac{p q^2}{r^3 \sqrt{s}} $$
The percentage errors in the measurement of $\mathrm{p}, \mathrm{q}, \mathrm{r}$ and s are $1 \%, 2 \%, 3 \%$ and $2 \%$, respectively. The percentage error in the measurement of $C$ will be__________%
Explanation
To determine the percentage error in the measurement of $ C $, which is related to $ p $, $ q $, $ r $, and $ s $ as:
$ C = \frac{p q^2}{r^3 \sqrt{s}} $
we first express it in terms of powers:
$ C = p^1 q^2 r^{-3} s^{-1/2} $
The percentage error in $ C $ can be calculated using the formula for the propagation of error, which is:
$ \left(\frac{\Delta C}{C}\right)_{\max} = \left|\frac{\Delta p}{p}\right| + 2\left|\frac{\Delta q}{q}\right| + 3\left|\frac{\Delta r}{r}\right| + \frac{1}{2}\left|\frac{\Delta s}{s}\right| $
Given the percentage errors for $ p $, $ q $, $ r $, and $ s $ are $ 1\% $, $ 2\% $, $ 3\% $, and $ 2\% $ respectively, we substitute these values into the formula:
$ \left(\frac{\Delta C}{C}\right)_{\max} = 1\% + 2 \times 2\% + 3 \times 3\% + \frac{1}{2} \times 2\% $
Calculating each part:
Contribution from $ p $: $ 1\% $
Contribution from $ q $: $ 4\% $ (since $ 2 \times 2\% = 4\% $)
Contribution from $ r $: $ 9\% $ (since $ 3 \times 3\% = 9\% $)
Contribution from $ s $: $ 1\% $ (since $ \frac{1}{2} \times 2\% = 1\% $)
Adding these contributions together:
$ 1\% + 4\% + 9\% + 1\% = 15\% $
Thus, the maximum percentage error in the measurement of $ C $ is $ 15\% $.
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