JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 24)
The surface of water in a water tank of cross section area $$750 \mathrm{~cm}^{2}$$ on the top of a house is $$h \mathrm{~m}$$ above the tap level. The speed of water coming out through the tap of cross section area $$500 \mathrm{~mm}^{2}$$ is $$30 \mathrm{~cm} / \mathrm{s}$$. At that instant, $$\frac{d h}{d t}$$ is $$x \times 10^{-3} \mathrm{~m} / \mathrm{s}$$. The value of $$x$$ will be ____________.
Answer
2
Explanation
$\begin{aligned} & \mathrm{AV}=\mathrm{av} \\\\ & 750 \times 10^{-4} \times\left(\frac{d h}{d t}\right)=\left(500 \times 10^{-6}\right)\left(30 \times 10^{-2}\right) \\\\ & \frac{d h}{d t}=\frac{15 \times 10^{-5}}{75 \times 10^{-3}} \\\\ & =\frac{1}{5} \times 10^{-2} \\\\ & =2 \times 10^{-3} \mathrm{~m} / \mathrm{s} \\\\ & \therefore x=2\end{aligned}$
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