MHT CET · Physics · Mechanical Properties of Fluids
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
- A 2
- B 3
- C 4
- D 6
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text { From equation of continuity, } \mathrm{A}_1 \mathrm{v}_1=\mathrm{A}_2 \mathrm{v}_2 \\
& \left(750 \times 10^{-4}\right) \cdot v_1=\left(500 \times 10^{-6}\right) \times\left(30 \times 10^{-2}\right) \\
& \mathrm{v}_1=\frac{500 \times 10^{-6} \times 30 \times 10^{-2}}{750 \times 10^{-4}}=2 \times 10^{-3} \mathrm{~m} / \mathrm{s} \\
& \text { But, } \mathrm{v}_1=\frac{\mathrm{dh}}{\mathrm{dt}}=\mathrm{x} \times 10^{-3} \mathrm{~m} / \mathrm{s} \\
& \therefore \quad \mathrm{x}=2
\end{aligned}\)
& \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}\)
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