JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Fluid Mechanics

When liquid medicine of density

ρ

is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension

T

when the radius of the drop is

R

. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

If the radius of the opening of the dropper is

r

, the vertical force due to the surface tension on the drop of radius

R

(assuming

r ≪ R

) is

(A) $2 π r T$
(B) $2 π R T$
(C) $\frac{2 π r ^{2} T}{R}$ ✓ Correct
(D) $\frac{2 π R ^{2} T}{r}$

Answer: (C)

Solution

The surface tension acts along the perimeter of the opening, which is

2 π r

. However, the drop is spherical with radius

R

, and the contact line makes an angle with the vertical.

The vertical component of the surface tension force is:

F = 2 π r T sin θ

where

sin θ \approx \frac{r}{R}

for small angles (since

r ≪ R

).

Therefore:

F = 2 π r T \times \frac{r}{R} = \frac{2 π r ^{2} T}{R}

The answer is **(C)

\frac{2 π r ^{2} T}{R}

**.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Single correct option

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JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Fluid Mechanics

Solution

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