What is the tension in the string in the following case:A small smooth ring R of mass 0.1 kg is threaded on a light string. The ends of the string...

In the problem the ring that is threaded in the string has
a mass of 0.1 kg. The string is attached at the ends A and B with the bead hanging in
the middle at point R

The tension is the same throughout
the string. The end AR is inclined at an angle 40 degree to the horizontal. We can
divide the tension T in this portion into a horizontal and a vertical component given by
T*cos 40 and T*sin 40 respectively. As the ring is stationary there is no net force
acting on it. The tension in the other end RB is also equal to T but acts in the
opposite direction. Let the part RB form an angle A with the
horizontal.

So we have T*cos A = T* cos
40

=> A = 40
degrees.

The part RB of the string is also inclined at 40
degrees to the horizontal.

The vertical components of the
two parts add up the weight of the ring.

2*T*sin 40 =
0.1*9.8

=> T = 0.1*9.8 / 2*sin
40

=> T = 0.7623
N

The tension in the string is 0.7623 N, and
the part RB is also inclined at 40 degrees to the
horizontal.