A cube with a mass - 20 g wettable water floats on the surface of water- Each face of the cube is - - 3 cm long- Surface tension of water is 70 dyn-cm- The distance of the lower face of the cube from the surface of water is g - 980 cm-s2-

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Standard IX

Physics

Pressure

A cube with a...

Question

A cube with a mass $= 20 g$ wettable water floats on the surface of water. Each face of the cube is $α = 3 c m$ long. Surface tension of water is $70 d y n / c m$ . The distance of the lower face of the cube from the surface of water is $(g = 980 c m / s^{2})$ .

2.3 c m

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4.6 c m

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9.7 c m

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12.7 c m

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Solution

The correct option is A $2.3 c m$
Since the angle of contact for cube and water is $0^{o}$ , Surface tension provides a downward force to the cube. And for the equilibrium of the cube.
$Downward force=upward force \Rightarrow weight + surface tension force=buoyant force \Rightarrow m g + 4 α T = l α^{2} ρ g Here l = distance of the lower face of the cube from the surface of water \Rightarrow l = \frac{m g + 4 α T}{α^{2} ρ g} \Rightarrow l = \frac{20 \times 980 + 4 \times 3 \times 70}{3^{2} \times 1 \times 980} \Rightarrow l = 2.3175 c m \approx 2.3 c m$

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A cube with a mass =20 g wettable water floats on the surface of water. Each face of the cube is α=3 cm long. Surface tension of water is 70 dyn/cm. The distance of the lower face of the cube from the surface of water is (g=980 cm/s2).

A cube with a mass $= 20 g$ wettable water floats on the surface of water. Each face of the cube is $α = 3 c m$ long. Surface tension of water is $70 d y n / c m$ . The distance of the lower face of the cube from the surface of water is $(g = 980 c m / s^{2})$ .