Question:

The excess pressure inside a liquid drop is 500 Nm\( - \)2. If the radius of the drop is 2 mm, the surface tension of liquid is x \( \times \) 10\( - \)3 Nm\( - \)1. The value of x is _____________.

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Excess pressure inside a liquid drop (one surface) is \(\Delta\)P = 2T/R.
Updated On: Oct 31, 2026
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Correct Answer: 500

Solution and Explanation

Step 1: For a liquid drop, \(\Delta\)P = 2T/R, so T = \(\Delta\)P × R/2.
Step 2: T = 500 × (2 × 10-3)/2 = 500 × 10-3 N/m = 0.5 N/m.
Step 3: Writing T = x × 10-3 N/m gives x = 500.
Answer: 500
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