A cube of mass `m =800 g` floats on the suface of water. Water wets it completely. The cube is `10 cm` on each edge. By what additional distance is it buoyed up or down by surface tension? Surface tension of water `=0.07Nm^(-1)`
Text Solution
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If surface tension is neglected the condition for floating gives `800xx10^(-3)g=(0.1^(2)xrho)g` or `x=0.08m` `(rho=1000kgm^(-3)` for water) Since water wets the cube, the angle of contact is zero and force of surface tension acts vertically downwards. So it is buoyed down by surface tension. `:. 800xx10^(-3)g+4xx0.1xx0.07=(0.1^(2)x'rho)g` or `x'=0.08+0.028/98=0.08+2.8xx10^(-4)` Therefore the additional distance `=2.8xx10^(-4)m`
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CENGAGE PHYSICS-PROPERTIES OF SOLIDS AND FLUIDS-INTEGER_TYPE
