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A liquid drop of density rho floats half-submerged in another liquid of density sigma. The surface tension between them is 7.5 * 10⁻⁴ N/cm. Find the radius of the drop in cm. (Take g = 10 m/s².)

1. 15/sqrt(2*rho - sigma)
2. 15/sqrt(rho - sigma)
3. 3/(2*sqrt(rho - sigma))
4. 20/sqrt(2*rho - sigma)

Correct answer: 15/sqrt(2*rho - sigma)

Solution

For a drop half immersed: weight = upthrust from submerged half + surface tension force. Weight = rho*(4/3*pi*r³)*g. Upthrust = sigma*(1/2)*(4/3*pi*r³)*g. Surface tension supports the rest: T*2*pi*r. Balance: rho*(4/3*pi*r³)*g = sigma*(2/3*pi*r³)*g + T*2*pi*r. Simplify and solve for r leading to r = sqrt(3T/((2*rho - sigma)*g)) form. With T = 7.5*10⁻⁴ N/cm = 7.5*10⁻² N/m and g = 10, the constant works out to 15 (in cm units), giving r = 15/sqrt(2*rho - sigma) cm.

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