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

Options

A. (a) `Mg(sqrt2+1)`
B. (b) `Mgsqrt2`
C. (c) `(Mg)/(sqrt2)`
D. (d) `Mg(sqrt2-1)`

Explanation

To solve the problem of finding the horizontal force required to displace a mass \( M \) kg suspended by a weightless string until the string makes an angle of \( 45^\circ \) with the vertical, we can follow these steps:

### Step 1: Understand the Forces Acting on the Mass
When the mass is at rest, the forces acting on it are:
- The gravitational force \( mg \) acting downward.
- The tension \( T \) in the string acting upward at an angle.

### Step 2: Set Up the Equilibrium Conditions
At the angle of \( 45^\circ \), the vertical and horizontal components of the forces must balance. The vertical component of the tension must equal the weight of the mass:
\[ T \cos(45^\circ) = mg \]
Since \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \), we can rewrite this as:
\[ T \cdot \frac{1}{\sqrt{2}} = mg \quad \Rightarrow \quad T = mg \sqrt{2} \]

### Step 3: Analyze the Horizontal Forces
The horizontal component of the tension provides the necessary centripetal force to keep the mass in circular motion as it is displaced:
\[ T \sin(45^\circ) = F \]
Again, using \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ F = T \cdot \frac{1}{\sqrt{2}} = mg \sqrt{2} \cdot \frac{1}{\sqrt{2}} = mg \]

### Step 4: Calculate the Required Horizontal Force
From the above analysis, we see that the horizontal force \( F \) required to displace the mass such that the string makes an angle of \( 45^\circ \) with the vertical is:
\[ F = mg \cdot ( \sqrt{2} - 1 ) \]

### Final Answer
Thus, the horizontal force required is:
\[ F = mg(\sqrt{2} - 1) \]