A wire of `9.8xx10^(-3) kg/m` passes over a frictionless light pulley fixed on the top of a frictionless inclined plane which makes an angle of `30^(@)` with the horizontal. Masses m and M are tied at the two ends of wire such that m rests on the plane and M hangs freely vertically downwards. the entire system is in equilibrium and a transverse wave propagates along the wire with a velocities of `100 m//s`.

Options

A. `m=20 kg`
B. `M=5 kg`
C. `(m)/(M)=(1)/(2)`
D. `(m)/(M)=2`

Explanation

To solve the problem step by step, we will analyze the forces acting on the masses and use the information given about the wave velocity and linear mass density.

### Step 1: Understanding the Setup
We have a wire with a linear mass density of \( \mu = 9.8 \times 10^{-3} \, \text{kg/m} \) passing over a frictionless pulley. One mass \( m \) is on an inclined plane at an angle of \( 30^\circ \), and the other mass \( M \) is hanging vertically. The system is in equilibrium, and a transverse wave propagates along the wire with a velocity of \( v = 100 \, \text{m/s} \).

### Step 2: Free Body Diagram (FBD)
For mass \( m \) on the incline:
- The gravitational force acting on it can be resolved into two components:
- Parallel to the incline: \( m g \sin(30^\circ) \)
- Perpendicular to the incline: \( m g \cos(30^\circ) \)

For mass \( M \) hanging vertically:
- The gravitational force acting on it is simply \( M g \).

### Step 3: Writing the Equations of Equilibrium
For mass \( m \) on the incline, the tension \( T \) in the wire must balance the component of the gravitational force acting down the incline:
\[ T = m g \sin(30^\circ) \]

For mass \( M \) hanging vertically, the tension \( T \) must balance the weight of \( M \):
\[ T = M g \]

### Step 4: Relating the Two Equations
Since both expressions represent the same tension \( T \), we can set them equal to each other:
\[ m g \sin(30^\circ) = M g \]

### Step 5: Simplifying the Equation
We can cancel \( g \) from both sides (as long as \( g \neq 0 \)):
\[ m \sin(30^\circ) = M \]

Using \( \sin(30^\circ) = \frac{1}{2} \):
\[ m \cdot \frac{1}{2} = M \]
\[ m = 2M \]

### Step 6: Finding the Tension in the Wire
The velocity of the wave in the wire is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
We can rearrange this to find the tension \( T \):
\[ T = v^2 \cdot \mu \]
Substituting the known values:
\[ T = (100)^2 \cdot (9.8 \times 10^{-3}) \]
\[ T = 10000 \cdot 9.8 \times 10^{-3} \]
\[ T = 98 \, \text{N} \]

### Step 7: Finding the Mass \( M \)
From the equation \( T = M g \):
\[ 98 = M \cdot 9.8 \]
\[ M = \frac{98}{9.8} = 10 \, \text{kg} \]

### Step 8: Finding the Mass \( m \)
Using the relationship \( m = 2M \):
\[ m = 2 \cdot 10 = 20 \, \text{kg} \]

### Final Results
- The mass \( M = 10 \, \text{kg} \)
- The mass \( m = 20 \, \text{kg} \)

### Conclusion
The values of \( m \) and \( M \) are \( 20 \, \text{kg} \) and \( 10 \, \text{kg} \) respectively.