A spherical ball of density - and radius 0-003 m is dropped into a tube containing a viscous fluid filled up to the 0 cm mark as shown in the figure- Viscosity of the fluid -1-2160 N-m -2 and its density - L -2-1260 kg- m -3- Assume the ball reaches a terminal speed by the 10 cm mark- Find the time taken in sec by the ball to traverse the distance between the 10 cm and 20 cm mark- g-acceleration due to gravity - 10 m s -2

We know that terminal velocity v=2(ρ_metal ρ_fluid)gR^29μ Here, R= 3× 10^ 3 m Putting the value we get , ρ_metal ρ_fluid=ρ_metal/2 We know that ter ...

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Standard XII

Physics

Terminal Velocity

A spherical b...

Question

A spherical ball of density $ρ$ and radius $0.003 m$ is dropped into a tube containing a viscous fluid filled up to the $0 c m$ mark as shown in the figure. Viscosity of the fluid $= 1.2160 {N . m}^{- 2}$ and its density $ρ_{L} = ρ / 2 = 1260 k g . m^{- 3}$ . Assume the ball reaches a terminal speed by the $10 c m$ mark. Find the time taken (in sec) by the ball to traverse the distance between the $10 c m$ and $20 c m$ mark. (g=acceleration due to gravity $= 10 m s^{- 2}$

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Solution

We know that terminal velocity $v = \frac{2 (ρ_{m e t a l} - ρ_{f l u i d}) g R^{2}}{9 μ}$
Here, R= $3 \times 10^{- 3} m$
Putting the value we get , $ρ_{m e t a l} - ρ_{f l u i d} = ρ_{m e t a l} / 2$
We know that terminal velocity $v = \frac{21260 \times g \times 9 \times 10^{- 6}}{18 \times 1.2160}$
$v = 2 \times 10^{- 2} m / s$
Distance traveled =10 cm =0.1m
$t = \frac{0.1}{v} = 5 s e c$

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We know that terminal velocity v=2(ρmetal−ρfluid)gR29μHere, R=3×10−3mPutting the value we get ,ρmetal−ρfluid=ρmetal/2We know that terminal velocity v=21260×g×9×10−618×1.2160v=2×10−2m/sDistance traveled =10 cm =0.1m t=0.1v=5 sec