Water is flowing continuously from a tap having an internal diameter 8 x 10^-3 m. The water velocity as it leaves the tap is 0.4 ms^-1. The diameter of the water stream at a distance 2 x 10^-1 m below the tap is close to (a) 5.0 x 10^-3 m (b) 7.5 x 10^-3 m (c) 9.6 x 10^-3 m (d) 3.6 x 10^-3 m

Given

d1 = 8 x 10-3 m

v1 = 0.4 ms-1

h = 0.2 m

According to equation of motion,

[latex]v_{2}=\sqrt{v_{1}^{2}+2gh}=\sqrt{(0.4)^{2}+2+10\times 0.2}[/latex]

We get,

= 2 ms-1

According to equation of continuity

a1v1 = a2v2

[latex]\pi \times \left ( 8\times 10^{-3} \right )^{2}\times 0.4=\pi \times d_{2}^{2}\times 2\\d_{2}^{2}=(8\times 10^{-3})^{2}\times 0.4\times \frac{1}{2}\\d_{2}^{2} = 12.8 \times 10^{-6}[/latex]

We get,

d2 = 3.6 x 10-3 m

Therefore, the correct option is (d)

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