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A drop of water falling in air reaches the ground with terminal momentum P. Another drop of water of twice the radius reaches the ground with terminal velocity will have momentum.

$(a)\;2\;P \\(b)\;8\;P \\(c)\;16\;P \\(d)\;32\;P $
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Terminal velocity of a given spherical body in a medium of iscosity $\eta$ is given by
$v= \large\frac{2}{9} $$g (\rho - \sigma)$$ \large\frac{r^2}{\eta}$
Where $ \rho$ density of body
$\sigma$ density of liquid
$r$ radius of spherical body
$v\; \alpha\; r^2$
also mass $\alpha \;r^3$
Momentum = mass $\times$ velocity
Momentum $\alpha\; r^5$
Momentum of drop whose radius is twice is $2^5 P$
$\qquad=32\;P$
Hence d is the correct answer.
answered Mar 18, 2014 by meena.p
 

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