A ball of mass $(m)$ $0.5\ \text{kg}$ is attached to the end of a string having length $(L)$ $0.5\ \text{m}$. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is $324\ \text{N}$. The maximum possible value of angular velocity of ball (in $\text{radian/s}$) is

Step 1: Identify the given parameters.

The mass of the ball, the length of the string, and the maximum tension are given.

$$m = 0.5\text{ kg}$$ $$L = 0.5\text{ m}$$ $$T_{\max} = 324\text{ N}$$

Step 2: Set up the equations of motion for the conical pendulum.

Let $\theta$ be the angle that the string makes with the vertical axis, and let $r$ be the radius of the horizontal circular path.

$$r = L\sin\theta$$

The vertical component of the tension balances the weight of the ball.

$$T\cos\theta = mg$$

The horizontal component of the tension provides the necessary centripetal force.

$$T\sin\theta = m\omega^2 r$$

Step 3: Simplify the tension equation.

Substitute $r = L\sin\theta$ into the centripetal force equation.

$$T\sin\theta = m\omega^2 (L\sin\theta)$$

Divide both sides by $\sin\theta$.

$$T = m\omega^2 L$$

Step 4: Calculate the maximum angular velocity.

The maximum angular velocity $\omega_{\max}$ occurs when the tension reaches its maximum limit $T_{\max}$.

$$T_{\max} = m\omega_{\max}^2 L$$ $$\omega_{\max} = \sqrt{\frac{T_{\max}}{mL}}$$

Substitute the given values into the equation.

$$\omega_{\max} = \sqrt{\frac{324}{0.5 \times 0.5}}$$ $$\omega_{\max} = \sqrt{\frac{324}{0.25}}$$ $$\omega_{\max} = \sqrt{1296}$$ $$\omega_{\max} = 36\text{ rad/s}$$

Answer: $36\text{ rad/s}$