JEE Advance - Physics (2010 - Paper 2 Offline - No. 3)
If r = 5 $$ \times $$ 10−4 m, $$\rho $$ = 103 kg m−3 , g = 10 m/s2 , T = 0.11 Nm−1 , the radius of the drop when it detaches from
the dropper is approximately
1.4 $$ \times $$ 10−3 m
3.3 $$ \times $$ 10−3 m
2.0 $$ \times $$ 10−3 m
4.1 $$ \times $$ 10−3 m
Explanation
The drop falls down when the weight of the drop becomes larger than the surface tension. Hence,
$$mg = {{2\pi {r^2}T} \over R}$$
That is,
$${4 \over 3}\pi {R^3}\rho g = {{2\pi {r^2}T} \over R}$$
where
r = 5 $$\times$$ 10$$-$$4 m ; $$\rho$$ = 103 kg m$$-$$3 ; g = 10 m/s2 ; T = 0.11 Nm$$-$$1
Therefore,
$${R^4} = {{3{r^2}t} \over {2\rho g}} = {{3 \times {{(5 \times {{10}^{ - 4}})}^2} \times 0.11} \over {2 \times {{10}^3} \times 10}} = 4.125 \times {10^{ - 12}}$$ m4
$$ \Rightarrow R = 1.425 \times {10^{ - 3}}$$ m
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