Question

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 × 10^–4 N. The value of x is _______. [use g = 10 m/s²]

Answer 1

Correct Answer: 25

To solve this problem, we need to find the viscous force acting on a small ball when it reaches terminal velocity after being dropped in glycerine. The ball's mass is 0.3 g and its density is 8 g/cc, while the density of glycerine is 1.3 g/cc. We'll use the gravitational acceleration \(g = 10 \, \text{m/s}^2\).

1. Convert units: 

The mass of the ball is \(0.3 \, \text{g} = 0.0003 \, \text{kg}\).

The density of the ball is \(8 \, \text{g/cc} = 8000 \, \text{kg/m}^3\).

The density of glycerine is \(1.3 \, \text{g/cc} = 1300 \, \text{kg/m}^3\).

2. Calculate the volume of the ball:

Using \( \text{density} = \frac{\text{mass}}{\text{volume}} \), we have:

\(\text{Volume of the ball} = \frac{0.0003 \, \text{kg}}{8000 \, \text{kg/m}^3} = 3.75 \times 10^{-8} \, \text{m}^3\).

3. Calculate the buoyant force:

Buoyant force \(F_b = \text{Density of glycerine} \times g \times \text{Volume of the ball}\):

\(F_b = 1300 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 \times 3.75 \times 10^{-8} \, \text{m}^3\).

\(F_b = 4.875 \times 10^{-4} \, \text{N}\).

4. Calculate gravitational force:

The gravitational force acting on the ball is:

\(F_g = \text{mass} \times g = 0.0003 \, \text{kg} \times 10 \, \text{m/s}^2 = 3 \times 10^{-3} \, \text{N}\).

5. Calculate the viscous force at terminal velocity:

At terminal velocity, the net force is zero, so:

\(F_g - F_b - F_v = 0\) or \(F_v = F_g - F_b\).

\(F_v = 3 \times 10^{-3} \, \text{N} - 4.875 \times 10^{-4} \, \text{N} = 2.5125 \times 10^{-3} \, \text{N}\).

6. Determine the value of \(x\):

We are asked for the viscous force in the form \(x \times 10^{-4} \, \text{N}\), so:

\(2.5125 \times 10^{-3} \, \text{N} = 25.125 \times 10^{-4} \, \text{N}\).

Thus, \(x = 25.125\).

Conclusion:

The value of \(x\) is 25.125. This falls within the specified range of 25,25.

Answer 2

Fv=6πηrvT
F_v+F_v=mg
⇒F_v=mg–F_B
=10×(8–1.3)×\(\frac{0.3}{8}×10^{−3}\)
\(=2.5125×10^{–3} N\)
\(≃25×10^{−4} N\)
So, the answer is 25.