Multiple choice

A capillary tube of radius 0.25 mm is dipped vertically in a liquid of density $800 kg m^{-3}$ and of surface tension $3\times 10^{-2} Nm^{-2}$.The angle of contact of liquid-glass is given by $cos\theta = 0.3$. If $g=10ms^{-2}$, the rise of liquid in the capillary tube is______ $( Cm)$:

  1. $9$
  2. $0.9$
  3. $9\ x\ 10^ {-3}$
  4. $0.09$
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B Correct answer
Explanation

The formula for capillary rise is h = (2 * T * cos(theta)) / (r * rho * g). T = 3 * 10^-2, cos(theta) = 0.3, r = 0.25 * 10^-3 m, rho = 800, g = 10. h = (2 * 3 * 10^-2 * 0.3) / (0.25 * 10^-3 * 800 * 10) = (1.8 * 10^-2) / (2) = 0.9 * 10^-2 m = 0.9 cm.

AI explanation

The capillary rise formula is h = 2*T*cos(theta) / (r*d*g). Plugging in the given values, h = 2 multiplied by 3*10^-2 multiplied by 0.3, divided by the product of 0.00025, 800, and 10. The numerator evaluates to 0.018 and the denominator is 2, which results in a height of 0.009 meters. Converting this to centimeters by multiplying by 100 gives a liquid rise of 0.9 cm.

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