JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 3)
A physical quantity $$Q$$ is found to depend on quantities $$a, b, c$$ by the relation $$Q=\frac{a^4 b^3}{c^2}$$. The percentage error in $$a, b$$ and $$c$$ are $$3 \%, 4 \%$$ and $$5 \%$$ respectively. Then, the percentage error in $$Q$$ is :
43%
34%
66%
14%
Explanation
$$\begin{aligned} & \mathrm{Q}=\frac{\mathrm{a}^4 \mathrm{~b}^3}{\mathrm{c}^2} \\ & \frac{\Delta \mathrm{Q}}{\mathrm{Q}}=4 \frac{\Delta \mathrm{a}}{\mathrm{a}}+3 \frac{\Delta \mathrm{b}}{\mathrm{b}}+2 \frac{\Delta \mathrm{c}}{\mathrm{c}} \end{aligned}$$
$$\frac{\Delta \mathrm{Q}}{\mathrm{Q}} \times 100=4\left(\frac{\Delta \mathrm{a}}{\mathrm{a}} \times 100\right)+3\left(\frac{\Delta \mathrm{b}}{\mathrm{b}} \times 100\right)+2\left(\frac{\Delta \mathrm{c}}{\mathrm{c}} \times 100\right)$$
$$\begin{aligned} \% \text { error in } \mathrm{Q} & =4 \times 3 \%+3 \times 4 \%+2 \times 5 \% \\ & =12 \%+12 \%+10 \% \\ & =34 \% \end{aligned}$$
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