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(C) Let the velocity at the tap be $v_1 = 0.4 \ m/s$ and the diameter be $D_1 = 8 	imes 10^{-3} \ m$.Let the velocity at a distance $h = 0.2 \ m$ bel

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$3.6$

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(C) Let the velocity at the tap be $v_1 = 0.4 \ m/s$ and the diameter be $D_1 = 8 	imes 10^{-3} \ m$.Let the velocity at a distance $h = 0.2 \ m$ below the tap be $v_2$ and the diameter be $D_2$.Using the equation of motion $v_2^2 = v_1^2 + 2gh$,we get:$v_2 = \sqrt{(0.4)^2 + 2 	imes 10 	imes 0.2} = \sqrt{0.16 + 4} = \sqrt{4.16} \approx 2.04 \ m/s$.From the equation of continuity,$A_1 v_1 = A_2 v_2$,where $A = \frac{\pi D^2}{4}$.Thus,$D_1^2 v_1 = D_2^2 v_2$.$D_2 = D_1 \sqrt{\frac{v_1}{v_2}} = (8 	imes 10^{-3}) \sqrt{\frac{0.4}{2.04}} \approx (8 	imes 10^{-3}) \sqrt{0.196} \approx (8 	imes 10^{-3}) 	imes 0.443 \approx 3.54 	imes 10^{-3} \ m$.Rounding to the nearest provided option,the value is close to $3.6 	imes 10^{-3} \ m$.

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 $0.4 \ m/s$. The diameter of the water stream at a distance $2 \times 10^{-1} \ m$ below the tap is close to .......$\times 10^{-3} \ m$.

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