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Physics Question on mechanical properties of fluid

Water is flowing with velocity 4 m s−14 \,m\,s^{-1} in a cylinder of diameter 8 cm8\,cm, it is connected to a pipe with it end tip of diameter 2 cm2\,cm, calculate the velocity of water at this free end.

A

4 m s−14 \,m\,s^{-1}

B

8 m s−18 \,m\,s^{-1}

C

32 m s−132 \,m\,s^{-1}

D

64 m s−164 \,m\,s^{-1}

Answer

64 m s−164 \,m\,s^{-1}

Explanation

Solution

Here, v1=4 m s−1,A1=πr12=16π×10−4 m2v_{1}=4\, m \,s^{-1}, A_{1}= \pi r^{2}_{1}=16\pi \times 10^{-4}\,m^{2} A2=πr22=π×10−4m2A_{2}=\pi r^{2}_{2}=\pi \times 10^{-4} m^{2} Using, A1v1=A2v2A_{1}v_{1} = A_{2}v_{2} ∴v2=A1v1A2\therefore v_{2}=\frac{A_{1}v_{1}}{A_{2}} =16π×10−4×4π×10−4=\frac{16 \pi \times 10^{-4} \times4}{\pi\times10^{-4}} =64 m s−1= 64 \,m\, s^{ -1} Velocity of water at free end, v2=64 m s−1v_{2} = 64 \,m \,s^{-1}