JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Fluid Mechanics

A spherical ball of radius

1 \times 1 0^{- 4} m

and density

1 0^{5} kg / m^{3}

falls freely under gravity through a distance

h

before entering a tank of water, If after entering in water the velocity of the ball does not change, then the value of

h

is approximately:
(The coefficient of viscosity of water is

9.8 \times 1 0^{- 6} Ns / m^{2}

)

Answer: (C)

Solution

Since the ball's velocity is unchanged on entering water, its speed at entry equals the terminal velocity in water:

v_{T} = \frac{2 r ^{2} ( ρ - σ ) g}{9 η}

With

r = 1 0^{- 4}

m, ball density

ρ = 1 0^{5}

kg/m

^{3}

, water density

σ = 1 0^{3}

kg/m

^{3}

η = 9.8 \times 1 0^{- 6}

Pa·s,

g = 10

m/s

^{2}

v_{T} = \frac{2 ( 1 0 ^{- 4} ) ^{2} ( 1 0 ^{5} - 1 0 ^{3} ) ( 10 )}{9 ( 9.8 \times 1 0 ^{- 6} )} = \frac{2 \times 1 0 ^{- 8} \times 99000 \times 10}{8.82 \times 1 0 ^{- 5}} \approx 224.5 m/s

Since the ball fell freely through height

h

before entering:

v_{T} = 2 g h

, so

h = \frac{v _{T}^{2}}{2 g} = \frac{( 224.5 ) ^{2}}{20} \approx 2518 m

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option