Two equal drops are falling through air with a steady velocity of 5, cm/second . If two drops coales

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9-1.Fluid Mechanics

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Two equal drops are falling through air with a steady velocity of $5\, cm/second$. If two drops coalesce to form one drop then new terminal velocity will be

A

$5 \times {\left( 4 \right)^{1/3}}\,cm/s$

B

$5\sqrt 2 \,cm/s$

C

$\frac{5}{{\sqrt 2 }}\,cm/s$

D

$10\, cm/s$

Solution

$2 \times \frac{4}{3}\pi {r^3} = \frac{4}{3}\pi {R^3}$

$R = {2^{1/3}}r$

Now, ${V_T} \propto {r^2}$

${V_T} = {2^{2/3}} \times 5 = {4^{1/3}} \times 5$ $cm/s$

Standard 11

Physics

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