A mass of M kg is suspended by a weightless string. The horizontal force required to displace it until string makes an angle of `45^@` with the initial vertical direction is:

Options

A. (a)`(Mg)/(sqrt2)`
B. (b)`Mg(sqrt(2)- 1)`
C. (c)`Mg (sqrt(2) + 1)`
D. (d)`Mg sqrt(2)`

Explanation

To solve the problem, we need to find the horizontal force required to displace a mass \( M \) suspended by a weightless string until the string makes an angle of \( 45^\circ \) with the vertical. Let's break down the solution step by step.

### Step 1: Understand the Forces Acting on the Mass
When the mass is displaced to make an angle of \( 45^\circ \) with the vertical, two main forces act on the mass:
1. The gravitational force \( F_g = Mg \) acting downward.
2. The tension \( T \) in the string acting along the string.

### Step 2: Draw a Free Body Diagram
Draw a diagram showing the mass at an angle of \( 45^\circ \). Label the forces:
- The weight \( Mg \) acting downward.
- The tension \( T \) acting along the string.
- The horizontal force \( F \) that we need to find.

### Step 3: Resolve the Forces
When the string makes an angle of \( 45^\circ \):
- The vertical component of the tension \( T \) must balance the weight:
\[ T \cos(45^\circ) = Mg \]
- The horizontal component of the tension provides the centripetal force:
\[ T \sin(45^\circ) = F \]

### Step 4: Use Trigonometric Values
Since \( \cos(45^\circ) = \sin(45^\circ) = \frac{1}{\sqrt{2}} \), we can rewrite the equations:
1. From the vertical balance:
\[ T \cdot \frac{1}{\sqrt{2}} = Mg \quad \Rightarrow \quad T = Mg \sqrt{2} \]
2. Substitute \( T \) into the horizontal force equation:
\[ F = T \cdot \frac{1}{\sqrt{2}} = (Mg \sqrt{2}) \cdot \frac{1}{\sqrt{2}} = Mg \]

### Step 5: Conclusion
Thus, the horizontal force required to displace the mass until the string makes an angle of \( 45^\circ \) with the vertical is:
\[ F = Mg \]

### Final Answer
The required horizontal force is \( F = Mg \).

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