Question

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question==>>$<b><u>A spherical ball of radius [tex]1\times 10^{-4}$ and density $10^{4}\frac{kg}{m^{3}}$ falls freely under the gravity through a distance 'h' before entering a tank l of water. If after entering water the velocity of ball doesn't change , find 'h'. The coeffecient of VISCOSITY of water is $9.8\times 10^{-6}$>>[IIT 90's]

Finally, is VISCOSITY varies for a single material..???


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Answer 1

Heya...!!!

.... Solution...✍️

A spherical ball of radius 1×10-4 m and density 104 kg/m3 falls freely under gravity through a distance h before entering a tank of water. If after entering the water the velocity of the ball does not change, find h. The viscosity of water is 9.8 × 10-6N -m2

Solution : After falling through a height h, the velocity of the ball becomes

v = √(2gh)

After entering water, this velocity does not change, this velocity is equal to the terminal velocity,

√(2gh)=2/gr^2 [ p - O ] / n g

2gh 2/9 × [ (10 - 4)^2 ( 10^4 - 10^3/9.8×10^-6 ]^-2

h = 20×20
----------
2×9.8


or h = 20.41m



Solving this

It gives h = 20m...✍️

Answer 2

$\huge\mathfrak{Answer:-}$

The velocity of the ball becomes,

$v = \sqrt{2gh}$ (After falling through a certain height)

After entering the water,

This velocity $\bold{does\: not}$ change.

$\bold{This\:velocity\:=\:Terminal\:velocity}$

$\sqrt{2gh} = \frac{2}{9} r^{2} ( \frac{p -δ }{η} ) g$

$2gh = [ \frac{2}{9} \times (10^{ - 1} )^{2} \frac{(10^{1} - 10^{3} )}{9.8 \times 10^{ - 6} }] ^{2}$

$h = \frac{20 \times 20}{2 \times 9.8}$

$h = \frac{400}{19.6}$

$h = 20.41m$

$\text{Be\:Brainly}$❤️