A particle of mass 1 kg is moving on a circular path of radius 1 m. Its kinetic energy is K=bt⁴, where b= 1 J/s. The force acting on the particle at t = 1s is (2x)^½N. The value of x is____________.

$\text{ Here, }\frac{1}{2}m{v}^2=b{t}^4\text{ or }m{v}^2=2b{t}^4$

The centripetal force on the body is $F_c=\frac{m{v}^2}{R}=\frac{2b{t}^4}{R}$

$\Rightarrow v=\left(\sqrt{\frac{2b}{m}}\right)t^2$

$\text{ and }{a}_t=\frac{dv}{dt}=\sqrt{\frac{2b}{m}}\frac{d(t{)}^2}{dt}=\sqrt{\frac{8b}{m}}t$

$\therefore \text{ Tangential force is }{F}_t=m{a}_t=m\sqrt{\frac{8b}{m}}t$

The force on the particle is

$\begin{array}{l}F=\sqrt{{\left(\frac{2b{t}^4}{R}\right)}^2+8mb{t}^2}=\sqrt{4+8}=\sqrt{12}=2\sqrt{3}\mathrm{N}\\ \therefore 2\sqrt{3}=(2x{)}^{1/2}\Rightarrow 4\times 3=2x\\ \Rightarrow x=6\end{array}$