Multiple choice

A vessel has a small hole at its bottom. If water can be poured into it upto a height of $7 cm$ without leakage ($g=10 ms^{-2}$), the radius of the hole is (surface tension of water is $0.07 Nm^{-1}$).

  1. $2\ mm$
  2. $0.2\ mm$
  3. $0.1\ mm$
  4. $0.4\ mm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The pressure difference due to surface tension at the hole is 2T/r. This must balance the hydrostatic pressure rho*g*h. 2T/r = rho*g*h. r = 2T / (rho*g*h). Using T=0.07, rho=1000, g=10, h=0.07m: r = 2*0.07 / (1000*10*0.07) = 2 / 10000 = 0.0002 m = 0.2 mm.

AI explanation

The excess pressure provided by the water surface tension to prevent leakage is given by the formula h*rho*g = 2*T*cos(theta). Assuming the angle of contact is 0 degrees, we solve for the radius of the hole using r = 2*T / (h*rho*g). Plugging in the given values gives r = (2 * 0.07) / (0.07 * 1000 * 10), which simplifies to 0.0002 meters. Converting this radius to millimeters by multiplying by 1000 gives 0.2 mm. The result is 0.2 mm.

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