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Q.

Surface tension of mercury is $0.465 N m^{- 1} .$ The excess pressure inside a mercury drop of diameter 6mm is

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a

310 Pa

b

410 Pa

c

510 Pa

d

610 Pa

answer is A.

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Detailed Solution

$D = 6 m m, r = 3 \times 10^{- 3} m$
$P = \frac{2 s}{r} = \frac{2 \times 0.465}{3 \times 10^{- 3}} = 310 P a$

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