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  • Water from a tap emerges vertically downwards with an initial speed of <img alt="1.0 \mathrm{~ms}^{-1}" src="https://entrancecorner.oncodecogs.com/gif.latex?1.0%20%5Cmathrm%7B%7Ems%7D%5E%7B-1%7D" /

Water from a tap emerges vertically downwards with an initial speed of 1.0 \mathrm{~ms}^{-1}. The cross-sectional area of the tap is 10^{-4} \mathrm{~m}^2. Assume that the pressure is constant throughout the stream of water and that the flow is steady. The cross-sectional area of the stream\mathrm{0.15 \mathrm{~m} \text { below the tap is (take } g=10 \mathrm{~ms}^{-2} \text { ) }}

Option: 1

\mathrm{5.0 \times 10^{-4} \mathrm{~m}^2}


Option: 2

\mathrm{1.0 \times 10^{-5} \mathrm{~m}^2}


Option: 3

\mathrm{5.0 \times 10^{-5} \mathrm{~m}^2}


Option: 4

\mathrm{2.0 \times 10^{-5} \mathrm{~m}^2}


Answers (1)

The equation of continuity of flow is \mathrm{v_1 a_1=v_2 a_2}  where \mathrm{v_1=1.0 \mathrm{~ms}^{-1}, a_1=10^{-4} \mathrm{~m}^2, v_2=} velocity of stream at \mathrm{h=0.15 \mathrm{~m}}  below the tap. The value of \mathrm{v_2}  is given by

                    \mathrm{ \begin{aligned} v_2^2 & =v_1^2+2 g h \\ & =1.0+2 \times 10 \times 0.15=4.0 \end{aligned} }
or

                                     \mathrm{ v_2=2.0 \mathrm{~ms}^{-1} \text {. } }
Now

                       \mathrm{ a_2=\frac{v_1 a_1}{v_3}=\frac{1.0 \times 10^{-4}}{2.0}=5.0 \times 10^{-5} \mathrm{~m}^2 }

Posted by

Ramraj Saini

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