In the figure given below, O is the centre of an equilateral triangle ABC and $\overrightarrow{{\mathrm{F}}_1},$ $$ $\overrightarrow{{\mathrm{F}}_2},$ $$ $\overrightarrow{{\mathrm{F}}_3}$ are three forces acting along the sides AB, BC and AC. What should be the magnitude of $\overrightarrow{F_3}$ $$ so that total torque about O is zero?
        
1.  $|$ ${\overrightarrow{F}}_3|$ $=$ $\left|\overrightarrow{F_1}\right|$ $+$ $|\overrightarrow{F_2|}$

2. $|$ $\overrightarrow{F_3}|$ $=$ $\left|\overrightarrow{F_1}\right|$ $-$ $|\overrightarrow{F_2|}$

3. $|$ $\overrightarrow{F_3}|$ $=$ $\overrightarrow{F_1}$ $+$ $2\overrightarrow{F_2}$

4. Not possible

A wheel having a moment of inertia of \(2\) kg–m² about its vertical axis rotates at the rate of \(60\) rpm about the axis. The torque which can stop the wheel's rotation in one minute would be:
1. \(\frac{\pi }{12}\) N-m

2.  \(\frac{\pi }{15}\) N-m

3.  \(\frac{\pi }{18}\) N-m

4.  \(\frac{2\pi }{15}\) N-m