A mass $ {\text{M kg}} $ is suspended by a weightless string. What is the horizontal force required to hold the mass $ {60^ \circ } $ with the vertical?
(A) $ Mg $
(B) $ Mg\sqrt 3 $
(C) $ Mg\left( {\sqrt 3 + 1} \right) $
(D) $ \dfrac{{Mg}}{{\sqrt 3 }} $

Hint : We need to balance all the forces in the opposite directions for the mass to be stable. The force acting on the string can be divided into several components.

Complete step by step answer
Contact forces are exerted when two or more physical objects that are in contact. These forces are classified based on the objects in contact.
The force is tension if one of these objects in contact happens to be a string, rope, cable or spring.
It has been given in the question that a mass $ {\text{M kg}} $ is suspended by a weightless string.

The force acting on the string can be divided into several components.
Let the force acting on the mass in the horizontal direction be $ F $ .The horizontal component of force is given by-
 $\Rightarrow {F_{horizontal}} = T\sin \theta $ where $ T $ is the tension in the string and $ \theta $ represents the direction in which the component is resolved.
Thus, $ F = T\sin \theta $ in the horizontal direction- $ \left( 1 \right) $
For the weight of mass $ {\text{M kg}} $ suspended on the string, weight acts vertically downwards.
The weight $ W = Mg $ where mass is $ M $ and $ g $ is the acceleration due to gravity.
The vertical component of the force that balances the weight of the mass $ {\text{M kg}} $ is given by-
 $\Rightarrow {F_{vertical}} = T\cos \theta $ where $ T $ is the tension in the string and $ \theta $ represents the direction in which the component is resolved.
Thus, $ Mg = T\cos \theta $ in the vertical direction - $ \left( 2 \right) $
Dividing equation $ \left( 1 \right) $ by $ \left( 2 \right) $ ,
 $\Rightarrow \dfrac{F}{{Mg}} = \dfrac{{T\sin \theta }}{{T\cos \theta }} $
We are required to find the horizontal force required to hold the mass $ {60^ \circ } $ with the vertical.
 $ \therefore \theta = {60^ \circ } $
Assigning the value of $ \theta = {60^ \circ } $ , and simplifying the equation further,
 $\Rightarrow F = Mg\dfrac{{\sin {{60}^ \circ }}}{{\cos {{60}^ \circ }}} $
We know from our prior mathematical knowledge that, $ \sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2} $ and $ \cos {60^ \circ } = \dfrac{1}{2} $ .
Substituting these values,
 $\Rightarrow F = Mg\dfrac{{2\sqrt 3 }}{2} $
 $ \Rightarrow F = Mg\sqrt 3 $
 $ \therefore $ The horizontal force required to hold the mass $ {60^ \circ } $ with the vertical is $ Mg\sqrt 3 $ .
The correct answer is Option B.

Note
Tension is a force along the length of a medium, especially a force carried by a flexible medium, such as a rope or cable. While considering a rope, the tension force is felt by every section of the rope in both the directions, apart from the endpoints. The endpoints experience tension on one side and the force from the weight attached. Throughout the string, the tension varies in some circumstances.