A sperical ball of mass m, radiua r is connected to string of length `(l gt gt r)`. This string is then connected to fixed point and this arrangement allowed to oscillate in a medium whose density is `rho` and viscosity is `eta`. If amplitude of oscillation becomes half of the initial value after n seconds then viscosity of liquid is given by (Assume that resistance force on ball follows stoke's law)

Options

A. `(m)/(pirn)`
B. `(mln(2^(1//3)))/(pirn)`
C. `(mln8)/(pirn)`
D. `(mln2)/(pirn)`

Explanation

To solve the problem, we need to find the viscosity of the liquid (η) given that the amplitude of oscillation of a spherical ball becomes half of its initial value after n seconds. We will use the principles of damped oscillations and Stokes' law.

### Step-by-Step Solution:

1. **Understanding Damped Oscillation**:
In damped oscillation, the amplitude \( A(t) \) of the oscillation decreases over time. The relationship can be expressed as:
\[ A(t) = A_0 e^{-\frac{b}{2m} t} \]
where \( A_0 \) is the initial amplitude, \( b \) is the damping coefficient, and \( m \) is the mass of the ball.

2. **Setting Up the Condition**:
According to the problem, after \( n \) seconds, the amplitude becomes half of its initial value:
\[ A(n) = \frac{A_0}{2} \]
Substituting this into the equation gives:
\[ \frac{A_0}{2} = A_0 e^{-\frac{b}{2m} n} \]

3. **Simplifying the Equation**:
Dividing both sides by \( A_0 \) (assuming \( A_0 \neq 0 \)):
\[ \frac{1}{2} = e^{-\frac{b}{2m} n} \]

4. **Taking the Natural Logarithm**:
To solve for \( b \), take the natural logarithm of both sides:
\[ \ln\left(\frac{1}{2}\right) = -\frac{b}{2m} n \]
This simplifies to:
\[ -\ln(2) = -\frac{b}{2m} n \]
or
\[ \ln(2) = \frac{b}{2m} n \]

5. **Solving for the Damping Coefficient \( b \)**:
Rearranging gives:
\[ b = \frac{2m \ln(2)}{n} \]

6. **Relating Damping Coefficient to Viscosity**:
From Stokes' law, the damping coefficient \( b \) for a spherical object in a viscous medium is given by:
\[ b = 6 \pi \eta r \]
where \( \eta \) is the viscosity and \( r \) is the radius of the sphere.

7. **Equating the Two Expressions for \( b \)**:
Set the two expressions for \( b \) equal to each other:
\[ 6 \pi \eta r = \frac{2m \ln(2)}{n} \]

8. **Solving for Viscosity \( \eta \)**:
Rearranging gives:
\[ \eta = \frac{2m \ln(2)}{6 \pi r n} \]
Simplifying this, we get:
\[ \eta = \frac{m \ln(2)}{3 \pi r n} \]

### Final Answer:
The viscosity of the liquid is given by:
\[ \eta = \frac{m \ln(2)}{3 \pi r n} \]