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Question: Water is flowing continuously from a tap having an internal diameter \(8 \times {10^{ - 3}}\) \(m\)....

Water is flowing continuously from a tap having an internal diameter 8×10−38 \times {10^{ - 3}} mm.The water velocity as it leaves the tap is 0⋅4ms−10 \cdot 4m{s^{ - 1}}The diameter of the water steam at a distance 2×10−12 \times {10^{ - 1}} mmbelow the tap is close to:
(A) 9.6×10−3m9.6 \times {10^{ - 3}}m
(B) 3.6×10−3m3.6 \times {10^{ - 3}}m
(C) 5.0×10−3m5.0 \times {10^{ - 3}}m
(D) 7.5×10−3m7.5 \times {10^{ - 3}}m

Explanation

Solution

Hint
Firstly we will find the velocity of stream at a distance of 2×10−1m2 \times {10^{ - 1}}m below the tap.Then we will apply the continuity equation .And finally we get the diameter of the water stream. Continuity equation is the conservation of volume of the liquid in the top and bottom pipe i.e. volume of liquid at top and bottom will always be the same.

Complete step by step process
We have given:
Diameter of tap is equal to (d1{d_1}) = 8×10−3m8 \times {10^{ - 3}}m
Velocity of stream as it leaves tap (v1{v_1}) = 0⋅4ms−10 \cdot 4m{s^{ - 1}}
Now let velocity of stream at bottom = v2{v_2},
and let the diameter of the stream at bottom = d2{d_2}.
Now we apply third equation of motion-
v22=v12+2gh{v_2}^2 = {v_1}^2 + 2gh
Where, gg is acceleration due to gravity,
and hh is the height between top and bottom of pipes.
After putting the values of known variables-
v22=(0⋅4)2+2×10×2×10−1{v_2}^2 = {\left( {0 \cdot 4} \right)^2} + 2 \times 10 \times 2 \times {10^{ - 1}}
v22=1⋅6+4{v_2}^2 = 1 \cdot 6 + 4
v22=5⋅6{v_2}^2 = 5 \cdot 6
v2=2{v_2} = 2(approximately)
Now apply equation of continuity:
A1V1=A2V2{A_1}{V_1} = {A_2}{V_2}
Where, A1{A_1} is area of top,
A2{A_2}is area of bottom,
V1{V_1} is velocity of top,
V2{V_2} is the velocity of the bottom.
So putting the value of all known variables.
π(8×10−3)2×0⋅4=π×d22×2\pi {\left( {8 \times {{10}^{ - 3}}} \right)^2} \times 0 \cdot 4 = \pi \times {d_2}^2 \times 2
d22=(8×10−3)2×0⋅4×12{d_2}^2 = {\left( {8 \times {{10}^{ - 3}}} \right)^2} \times 0 \cdot 4 \times \dfrac{1}{2}
d22=12⋅8×10−6{d_2}^2 = 12 \cdot 8 \times {10^{ - 6}}
d2=3⋅57×10−3{d_2} = 3 \cdot 57 \times {10^{ - 3}}
Hence, diameter at distance 2×10−1m2 \times {10^{ - 1}}m below the tap will close to 3⋅6×10−3m3 \cdot 6 \times {10^{ - 3}}m
Option (B) will give the correct answer.

Note
Use the correct dimension of the measurement. Correctly apply the equation of continuity. Use the third equation of motion properly. The continuity equation in fluid dynamics describes that for any steady state process, the rate at which mass leaves the system is equal to the rate which mass enters a system.