Water is flowing continuously from a tap having an internal diameter $8 \times 10^{-3}\ m $ The water velocity as it leaves the tap is $0.4$ $ms^{-1}$ છે. The diameter of the water stream at a distance $2 \times 10^{-1}$ $m$ below the tap is close to .......$\times 10^{-3}\;m$
AIEEE 2011Diffcult
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From Bernoulli's theorem,
${P_0} + \frac{1}{2}\rho v_1^2\rho gh = {P_0} + \frac{1}{2}\rho v_2^2 + 0$
${v_2} = \sqrt {v_1^2 + 2gh} = \sqrt {0.16 + 2 \times 10 \times 0.2} = 2.03m/s$
From equation of continuity
${A_2}{v_2} = {A_1}{v_1}$
$\pi \frac{{D_2^2}}{4} \times {v_2} = \pi \frac{{D_1^2}}{4}{v_1}$
$ \Rightarrow {D_1} = {D_2}\sqrt {\frac{{{v_1}}}{{{v_2}}}} = 3.55 \times {10^{ - 3}}m$
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