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A cube with a mass = 20g, wettable with, water floats on the surface of water .Each face of the cube is $α = 3 cm$ long. Surface tension of water is 70dyne/cm. The distance of the lower face of the cube from the surface of water is $\frac{x}{y}$ cm. Find $(x + y)$ where $x$ and $y$ are co-primes. $(g = 980 cm s^{- 2})$

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Detailed Solution

If y is the required distance ,then $a^{2} yρg = mg + 4 aσ$
$or y = \frac{mg + 4 aσ}{a^{2} ρg} = \frac{20 \times 980 + 4 \times 3 \times 70}{3 \times 3 \times 1 \times 980} cm = \frac{1022}{441} cm$

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