A particle is moving on a circular path of a 10 m radius. At any instant of time, its speed is 5 m s^-1 and the speed is increasing at a rate of 2 m s^-2. The magnitude of net acceleration at this instant is 3.2 m s^-2. The force acting on the particle is

Given: $\stackrel{\rightharpoonup }{r}=A \cos \omega t \hat{i} +B \sin \omega t \hat{j}$

Velocity , 

         $$\begin{gathered} \overrightarrow{v}=\frac{d\overrightarrow{r}}{dt} \\ =\frac{d}{dt}\left(A \cos \omega t \hat{i}+ B \sin \omega t \hat{j}\right) \\ =-A\omega \sin \omega t \hat{i} +B\omega \cos \omega t \hat{j} \end{gathered}$$

Acceleration,

$$\begin{gathered} \overrightarrow{a}=\frac{d\overrightarrow{v}}{dt} \\ =-A{\omega }^2\cos \omega t \hat{i}-B{\omega }^2\sin \omega t \hat{j} \\ \overrightarrow{a}=-{\omega }^2\left[A \cos \omega t \hat{i}+B \sin \omega t \hat{j}\right] \\ =-{\omega }^2\overrightarrow{r} \end{gathered}$$

The force acting on the particle is 

$$\begin{gathered} \overrightarrow{F}=m\overrightarrow{a} \\ =m(-{\omega }^2\overrightarrow{r}) \\ =-m{\omega }^2\overrightarrow{r} \end{gathered}$$

Therefore, option 2 is correct.