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Step-by-Step Solution
Step 1: Identify the Known Quantities
• Initial diameter of the tap, $D_{1} = 8 \times 10^{-3}\,\text{m}$
• Initial velocity of water as it leaves the tap, $v_{1} = 0.4\,\text{m/s}$
• Vertical drop from the tap, $h = 0.2\,\text{m}$
• Acceleration due to gravity, $g \approx 10\,\text{m/s}^2$ (as used in the given solution references).
Step 2: Apply Bernoulli’s Theorem to Find the Velocity After Falling
When the water drops a height $h$, its velocity increases due to gravity. Ignoring losses, Bernoulli’s equation between the tap outlet (point 1) and the position below (point 2) gives:
$P_0 + \tfrac{1}{2}\rho v_{1}^{2} + \rho g h = P_0 + \tfrac{1}{2}\rho v_{2}^{2}$
Here, $P_0$ is the atmospheric pressure at both points, so it cancels out. Solving for $v_{2}$:
$v_{2} = \sqrt{v_{1}^{2} + 2gh}
Substitute the values:
$v_{2} = \sqrt{(0.4)^{2} + 2 \times 10 \times 0.2} = \sqrt{0.16 + 4} = \sqrt{4.16} \approx 2.03\,\text{m/s}$
Step 3: Use the Equation of Continuity
Next, we apply the continuity equation, which states that the volume flow rate remains constant:
$A_{1} v_{1} = A_{2} v_{2},$
where $A_{1}$ and $A_{2}$ are the cross-sectional areas of the stream at points 1 and 2 respectively. Since the cross-sectional area of a circular stream is given by $A = \tfrac{\pi D^{2}}{4}$, we have:
$\tfrac{\pi D_{1}^{2}}{4} \, v_{1} = \tfrac{\pi D_{2}^{2}}{4} \, v_{2}$
Canceling common factors:
$D_{1}^{2} \, v_{1} = D_{2}^{2} \, v_{2}$
Solving for $D_{2}$:
$D_{2} = D_{1} \sqrt{\tfrac{v_{1}}{v_{2}}}$
Substitute $D_{1} = 8 \times 10^{-3}\,\text{m}$, $v_{1} = 0.4\,\text{m/s}$, and $v_{2} = 2.03\,\text{m/s}$:
$D_{2} = 8 \times 10^{-3} \,\text{m} \times \sqrt{\tfrac{0.4}{2.03}} \approx 3.55 \times 10^{-3} \,\text{m}$
This is very close to $3.6 \times 10^{-3}\,\text{m}$.
Step 4: Conclude the Diameter
Hence, the diameter of the water stream after falling $0.2\,\text{m}$ is approximately $3.6 \times 10^{-3}\,\text{m}$. This matches the correct answer provided in the choices.
