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9-1.Fluid Mechanics
hard
સમાન ત્રિજયાના બે ટીપાં $5cm/sec$ ના વેગથી ગતિ કરતાં ભેગા થઇ જાય,તો મોટાં ટીપાંનો ટર્મિનલ વેગ કેટલો થાય?
A
$10 cm \;per sec$
B
$2.5 cm\; per sec$
C
$5 \times {(4)^{1/3}}cm\; per sec$
D
$5 \times \sqrt 2 \,cm\; per sec$
Solution
(c)If two drops of same radius r coalesce then radius of new drop is given by R
$\frac{4}{3}\pi {R^3} = \frac{4}{3}\pi {r^3} + \frac{4}{3}\pi {r^3}$==> ${R^3} = 2{r^3} \Rightarrow R = {2^{1/3}}r$
If drop of radius r is falling in viscous medium then it acquire a critical velocity v and $v \propto {r^2}$
$\frac{{{v_2}}}{{{v_1}}} = {\left( {\frac{R}{r}} \right)^2} = {\left( {\frac{{{2^{1/3}}r}}{r}} \right)^2}$
==> ${v_2} = {2^{2/3}} \times {v_1} = {2^{2/3}} \times (5) = 5 \times {(4)^{1/3}}m/s$
Standard 11
Physics
