A mass M is hung with a light inextensible string as shown in figure. Find the tension of the horizontal string.

Let \(T_1\) be the tension in the horizontal string, \(T_2\) be the tension in the slanted string, and \(T_3\) be the tension in the vertical string.

For the mass \(M\) in equilibrium, the tension in the vertical string is:
\(T_3 = Mg\)

Considering the equilibrium of the knot O, we can resolve the forces in the horizontal and vertical directions.

Balancing the forces in the vertical direction:
\(T_2 \sin 45^{\circ} = T_3 = Mg\)

Balancing the forces in the horizontal direction:
\(T_1 = T_2 \cos 45^{\circ}\)

Dividing the vertical force equation by the horizontal force equation, we get:
\(\dfrac{T_2 \sin 45^{\circ}}{T_2 \cos 45^{\circ}} = \dfrac{Mg}{T_1}\)

\(\tan 45^{\circ} = \dfrac{Mg}{T_1}\)

Since \(\tan 45^{\circ} = 1\), we have:
\(1 = \dfrac{Mg}{T_1}\)

\(T_1 = Mg\)

Thus, the tension in the horizontal string is \(Mg\).

Answer: \(Mg\)