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Q.

The excess pressure inside a liquid drop is 500 . 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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answer is 500.

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Detailed Solution

$P = P_{0} + \frac{2 T}{R} \Rightarrow P - P_{0} = \frac{2 T}{R}$

$\Rightarrow 500 = \frac{2 \times T}{2 \times 10^{- 3}}$

$T = 500 \times 10^{- 3}$

$So, x = 500$

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