As it is given that 2 kg mass is suspended by 12 cm long thread and then a horizontal force is applied on it so that it remains in equilibrium at 30 degree angle
So here we can use force balance in X and Y directions
now for X direction or horizontal direction we can use
[tex]F = Tsin30[/tex]
for vertical direction similarly we can say
[tex]mg = T cos30[/tex]
so here we first divide the two equations
[tex]\frac{F}{mg} = \frac{sin 30}{cos 30}[/tex]
[tex]\frac{F}{mg} = tan 30[/tex]
[tex]F = mg tan30[/tex]
now plug in all values in the above equation
[tex]F = 2 * 9.8 * tan30[/tex]
[tex]F = 11.3 N[/tex]
Part b)
now in order to find the tension in the thread we can use any above equation
[tex]F = T sin30[/tex]
[tex]11.3 = T sin30[/tex]
[tex]T = \frac{11.3}{sin30}[/tex]
[tex]T = 22.6 N[/tex]
so tension in the thread will be 22.6 N