If the radius of the drop when it detaches from the dropper is approximately [2010]

Question

When liquid medicine of density $ρ$ is to 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.

Q. If $r = 5 \times 10^{- 4} m, ρ = 10^{3} {kgm}^{- 3}, g = 10 {ms}^{- 2}, T = 0.11 {Nm}^{- 1},$ the radius of the drop when it detaches from the dropper is approximately [2010]

Smart Achievers · Accepted Answer

(1)

When the drop is about to detach from the dropper

$Weight = vertical force due to surface tension (m g = F_{v})$

$∴$ $\frac{4}{3} π R^{3} ρ g = \frac{2 π r^{2} T}{R}$

$\Rightarrow R^{4} = (\frac{3}{2} \frac{r^{2} T}{ρ g}) = \frac{3}{2} \times \frac{{(5 \times 10^{- 4})}^{2} \times 0.11}{1000 \times 10} = 4.12 \times 10^{- 12}$

$or, R = 1.42 \times 10^{- 3} m$

Solution

1 $1.4 \times 10^{- 3} m$

2 $3.3 \times 10^{- 3} m$

3 $2.0 \times 10^{- 3} m$

4 $4.1 \times 10^{- 3} m$

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Solution

1 1.4×10-3 m

2 3.3×10-3 m

3 2.0×10-3 m

4 4.1×10-3 m

Ans. (1) When the drop is about to detach from the dropper Weight=vertical force due to surface tension (mg=Fv) ∴ 43πR3ρg=2πr2TR ⇒R4=(32r2Tρg)=32×(5×10-4)2×0.111000×10=4.12×10-12 or, R=1.42×10-3 m

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1 $1.4 \times 10^{- 3} m$

2 $3.3 \times 10^{- 3} m$

3 $2.0 \times 10^{- 3} m$

4 $4.1 \times 10^{- 3} m$