Liquid drops are falling slowly one by one from vertical glass tube. | PGMN Solutions

Liquid drops are falling slowly one by one from vertical glass tube. The relation between the weight of a drop ' w ', the surface tension ' T ' and the radius ' r ' of the bore of the tube is (Angle of contact is zero)
(A) w=πr^2 " " T
(B) w=2π^2 rT
(C) w=(4/3) π^2 rT
(D) w=2πrT


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🚀 Step-by-Step Solution to Solve This Question 🚀
📌 Chapter: Mechanical Properties of Fluids
📌 Topic: Weight of Liquid Drops and Surface Tension

🔹 Step 1: When a liquid drop forms at the opening of a vertical tube, the surface tension of the liquid creates a force that holds the drop at the opening. The weight of the drop is balanced by the force due to the surface tension.

🔹 Step 2: The surface tension acts along the circumference of the tube's opening. Since the tube has a circular cross-section, the force due to surface tension is equal to the surface tension multiplied by the circumference of the bore (which is proportional to the radius of the tube).

🔹 Step 3: The weight of the drop is the force due to gravity, and it is related to the mass of the drop, which in turn depends on the volume of the drop. The volume of the drop is related to the radius of the bore of the tube.

🔹 Step 4: After balancing the forces, the correct relationship between the weight of the drop and the surface tension is:

✅ Final Answer: (D)
w=2πrT 🚀
🔥 Pro Tip: The weight of the drop is determined by the balance between the surface tension at the tube's opening and the gravitational force acting on the drop. The surface tension force depends on the circumference of the tube's bore.

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