1. Understand the situation

A mass $M$ is hanging at rest by a light string. A horizontal force $F$ is applied so that the string is displaced and finally makes an angle $\theta = 45^\circ$ with the initial vertical.

We need the horizontal force required to hold it in this position.

2. Draw forces on the mass

At the final equilibrium position, three forces act on the mass:

• Weight: $Mg$ downward
• Tension: $T$ along the string
• Horizontal applied force: $F$ toward the side

Since the mass is in equilibrium at $45^\circ$, resolve forces along horizontal and vertical directions.

3. Vertical equilibrium

The vertical component of tension balances the weight:

$T\cos\theta = Mg$

With $\theta = 45^\circ$,

$T\cos 45^\circ = Mg$

$T \cdot \frac{1}{\sqrt{2}} = Mg$

$T = Mg\sqrt{2}$

4. Horizontal equilibrium

The horizontal component of tension balances the applied force:

$T\sin\theta = F$

So,

$F = T\sin 45^\circ$

Substitute $T = Mg\sqrt{2}$:

$F = Mg\sqrt{2} \cdot \frac{1}{\sqrt{2}} = Mg$

5. Compare with options

The required horizontal force is

$\boxed{Mg}$

Now check the options:

• A: $Mg(\sqrt{2}+1)$
• B: $Mg\sqrt{2}$
• C: $\dfrac{Mg}{\sqrt{2}}$
• D: $Mg(\sqrt{2}-1)$

None of these equals $Mg$.

6. Possible source of confusion

If the question had asked for the work done by the horizontal force in slowly displacing the mass to $45^\circ$, then one may get an expression involving $Mg(\sqrt{2}-1)$ after integration. But for the force required at the final position, the answer is clearly

$\boxed{Mg}$

Therefore, the stored correct answer appears inconsistent with the stated question.