The surface of water in a water tank of cross section area $750 \mathrm{~cm}^2$ on the top of a house is '…

The surface of water in a water tank of cross section area $750 \mathrm{~cm}^2$ on the top of a house is ' $h$ ' $m$ 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 $\frac{\mathrm{dh}}{\mathrm{dt}}$ is $\mathrm{x}=10^{-3} \mathrm{~m} / \mathrm{s}$. The value of ' x ' will be
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Solution

$\begin{aligned} & \text { From equation of continuity, } \mathrm{A}_1 \mathrm{v}_1=\mathrm{A}_2 \mathrm{v}_2 \\ & \left(750 \times 10^{-4}\right) \cdot v_1=\left(500 \times 10^{-6}\right) \times\left(30 \times 10^{-2}\right) \\ & \mathrm{v}_1=\frac{500 \times 10^{-6} \times 30 \times 10^{-2}}{750 \times 10^{-4}}=2 \times 10^{-3} \mathrm{~m} / \mathrm{s} \\ & \text { But, } \mathrm{v}_1=\frac{\mathrm{dh}}{\mathrm{dt}}=\mathrm{x} \times 10^{-3} \mathrm{~m} / \mathrm{s} \\ & \therefore \quad \mathrm{x}=2 \end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 1)

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