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9-1.Fluid Mechanics
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સમાન ત્રિજયાના બે ટીપાં $5cm/sec$ ના વેગથી ગતિ કરતાં ભેગા થઇ જાય,તો મોટાં ટીપાંનો ટર્મિનલ વેગ 

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

$\frac{4}{3}\pi {R^3} = \frac{4}{3}\pi {r^3} + \frac{4}{3}\pi {r^3}$ ==> ${R^3} = 2{r^3} \Rightarrow R = {2^{1/3}}r$ $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

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