A small spherical ball of radius 0.1 mm and density 10⁴ kg m^-3 falls freely under gravity through a distance h before entering a tank of water. If, after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of h will be________m
(Given, $g = 10 {ms}^{- 2}$ , viscosity of water $= 1.0 \times 10^{- 5} N - {sm}^{- 2}$ )

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 20.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Speed after falling through height h Should be equal to terminal velocity
$\sqrt{2 gh} = \frac{2}{9} \frac{r^{2} (d - ρ) g}{η} \Rightarrow \sqrt{2 gh} = \frac{2}{9} \frac{10^{- 8} (10000 - 1000) \times 10}{10^{- 5}} \begin{array}{l} \Rightarrow \sqrt{2 gh} = \frac{2}{9} \times 10^{- 8} \frac{9 \times 10^{4}}{10^{- 5}} = 20 \\ \Rightarrow 2 \times 10 \times h = 400 \\ \Rightarrow h = 20 m \end{array}$