JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Kinematics

The stone moves in a horizontal circle of radius $R=1.8$ m while completing $28$ revolutions each minute, so its angular speed is $\omega =\frac{28\times 2\pi }{60}=\frac{28\times 2\times \frac{22}{7}}{60}=\frac{44}{15}$ rad s${}^{-1}$. For uniform circular motion the acceleration points to the centre and has magnitude $a={\omega }^{2}R$. Substituting, $a={\left(\frac{44}{15}\right)}^2\times 1.8=\frac{1936}{225}\times 1.8=\frac{1936}{125}$ ms${}^{-2}$. Comparing with $\frac{1936}{x}$, the value of $x$ is $125$.