A sperical object of mass 1kg and radius 1 m is falling vertically downward inside a viscous liquid in a gravity free space. At a certain instant the velocity of the sphere is 2 m/s. if the coefficient of viscosity of the liquid is `(1)/(18pi)N-S//m^(2)` then velocity of ball will become 0.5 m/s after a tie

Options

A. In 4 s
B. 2 In 4 s
C. 3 In s
D. 2 In 2 s

Explanation

To solve the problem, we need to determine the time it takes for a spherical object to decrease its velocity from 2 m/s to 0.5 m/s while falling through a viscous liquid. We will use the equation for viscous drag and integrate to find the time.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Sphere:**
The only force acting on the sphere in the viscous liquid is the viscous drag force, which can be expressed as:
\[ F_{\text{drag}} = -6 \pi \eta r v \]
where:
- \( \eta \) is the coefficient of viscosity,
- \( r \) is the radius of the sphere,
- \( v \) is the velocity of the sphere.

2. **Set Up the Equation of Motion:**
According to Newton's second law, the net force acting on the sphere is equal to the mass times the acceleration:
\[ m \frac{dv}{dt} = -6 \pi \eta r v \]
Given that the mass \( m = 1 \, \text{kg} \), we can rewrite the equation as:
\[ \frac{dv}{dt} = -\frac{6 \pi \eta r}{m} v \]

3. **Substitute the Given Values:**
We know:
- \( \eta = \frac{1}{18\pi} \, \text{N-s/m}^2 \)
- \( r = 1 \, \text{m} \)
- \( m = 1 \, \text{kg} \)

Substituting these values into the equation gives:
\[ \frac{dv}{dt} = -\frac{6 \pi \left(\frac{1}{18\pi}\right) (1)}{1} v = -\frac{6}{18} v = -\frac{1}{3} v \]

4. **Separate Variables and Integrate:**
We can separate the variables:
\[ \frac{dv}{v} = -\frac{1}{3} dt \]
Integrating both sides from the initial velocity \( v_0 = 2 \, \text{m/s} \) to the final velocity \( v = 0.5 \, \text{m/s} \) and from \( t = 0 \) to \( t = T \):
\[ \int_{2}^{0.5} \frac{1}{v} dv = -\frac{1}{3} \int_{0}^{T} dt \]

5. **Calculate the Integrals:**
The left side becomes:
\[ \ln(v) \bigg|_{2}^{0.5} = \ln(0.5) - \ln(2) = \ln\left(\frac{0.5}{2}\right) = \ln\left(\frac{1}{4}\right) = -\ln(4) \]
The right side becomes:
\[ -\frac{1}{3} T \]

6. **Set the Equations Equal:**
Setting the two integrals equal gives:
\[ -\ln(4) = -\frac{1}{3} T \]

7. **Solve for Time \( T \):**
Rearranging gives:
\[ T = 3 \ln(4) \]

### Final Answer:
The time taken for the sphere to reduce its velocity from 2 m/s to 0.5 m/s is:
\[ T = 3 \ln(4) \, \text{seconds} \]