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When liquid medicine of density rho is t...

When liquid medicine of density `rho` is to be put in the eye, it is done with the help of a dropper. As the bylb on the top of the dropper is pressed. A drop frons 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 become smaller than the weight of the drop the drop gets detached from the dropper.
If `r=5xx10^(-4)m,rho=10^(3) kg m^(-3), g=10ms^(-2), T=0.11Nm^(-1),` The radius of the drop when it detaches from the dropper is approximately,

A

`1.4 xx 10^(-3) m`

B

`3.3 xx 10^(-3) m`

C

`2.0 xx 10^(-3) m`

D

`4.1xx 10^(-3) m`

Text Solution

Verified by Experts

The correct Answer is:
A
When drop detaches from the dropper then net vertical force dur to surface tension is equal to weight of drop `(2pir^(2)T)/(R)=mg`
`rArr (2pir^2T)/(R)=p 4/3pi R^(3)g`
On solving and substituting given values we get `R=1.4 xx 10^(-3) m`.
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