A wire of `9.8 xx 10^(-3) kg` per meter mass passes over a fricationless pulley fixed on the top of an inclined frictionless plane which makes an angle of `30^(@)` with the horizontal. Masses `M_(1)` & `M_(2)` are tied at the two ends of the wire. The mass `M_(1)` rests on the plane and the mass `M_(2)` hangs freely vertically downwards. the whole system is in equilibrium. Now a transverse wave propagates along the wire with a velocity of 100m/sec. Find the ratio of masses M 1 ​ to M 2 ​ .

Answer

To solve the problem, we need to find the ratio of the masses \( M_1 \) and \( M_2 \) when the system is in equilibrium. Here’s how we can approach the problem step by step:

### Step 1: Identify the Forces Acting on the Masses
- For mass \( M_1 \) on the inclined plane:
- The gravitational force acting downwards is \( M_1 g \).
- This force can be resolved into two components:
- Perpendicular to the incline: \( M_1 g \cos(30^\circ) \)
- Parallel to the incline: \( M_1 g \sin(30^\circ) \)

- For mass \( M_2 \) hanging vertically:
- The gravitational force acting downwards is \( M_2 g \).

### Step 2: Write the Equilibrium Conditions
Since the system is in equilibrium, the tension \( T \) in the wire must balance the forces acting on both masses.

- For \( M_2 \):
\[ T = M_2 g \quad \text{(1)} \]

- For \( M_1 \):
The tension must balance the component of the weight acting down the incline:
\[ T = M_1 g \sin(30^\circ) \quad \text{(2)} \]

### Step 3: Equate the Two Expressions for Tension
From equations (1) and (2), we can set them equal to each other:
\[ M_2 g = M_1 g \sin(30^\circ) \]

### Step 4: Simplify the Equation
We can cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ M_2 = M_1 \sin(30^\circ) \]

### Step 5: Substitute the Value of \( \sin(30^\circ) \)
We know that \( \sin(30^\circ) = \frac{1}{2} \):
\[ M_2 = M_1 \cdot \frac{1}{2} \]

### Step 6: Find the Ratio of \( M_1 \) to \( M_2 \)
To find the ratio \( \frac{M_1}{M_2} \):
\[ \frac{M_1}{M_2} = \frac{M_1}{\frac{1}{2} M_1} = 2 \]

### Final Answer
The ratio of the masses \( M_1 \) to \( M_2 \) is:
\[ \frac{M_1}{M_2} = 2 \]

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