The velocity of a small ball of mass 0.3 g and density \(8 \mathrm{~g} / \mathrm{cc}\) when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is \(1.3 \mathrm{~g} / \mathrm{cc}\), then the value of viscous force acting on the ball will be \(\mathrm{x} \times 10^{-4} \mathrm{~N}\), the value of x is \_\_\_\_. \(\left[\right.\) use \(\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]\)

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Fluid Mechanics

The velocity of a small ball of mass 0.3 g and density

8 g / cc

when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is

1.3 g / cc

, then the value of viscous force acting on the ball will be

x \times 1 0^{- 4} N

, the value of x is ____.

[

use

g = 10 m / s^{2}]

Answer: 25

When terminal velocity is reached, the net force is zero:

F_{V} + F_{B} = m g

where

F_{V}

is viscous force and

F_{B}

is buoyant force.

The viscous force is:

F_{V} = m g - F_{B} = (ρ_{B} - ρ_{L}) V g

The volume of the ball is:

V = \frac{m}{ρ _{B}} = \frac{0.3 \times 1 0 ^{- 3}}{8 \times 1 0 ^{3}} = \frac{0.3}{8} \times 1 0^{- 6} m^{3}

Substituting:

F_{V} = (8 - 1.3) \times 1 0^{3} \times \frac{0.3}{8} \times 1 0^{- 6} \times 10

= 6.7 \times 1 0^{3} \times \frac{0.3}{8} \times 1 0^{- 5} = \frac{6.7 \times 0.3}{8} \times 1 0^{- 2}

= \frac{2.01}{8} \times 1 0^{- 2} \approx 0.251 \times 1 0^{- 2} = 25.1 \times 1 0^{- 4} N

The answer is 25.

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Integer answer