A ball of mass $0.5 \mathrm{~kg}$ is attached to a string of length $50 \mathrm{~cm}$. The ball is rotated on a horizontal circular path about its vertical axis. The maximum tension that the string can bear is $400 \mathrm{~N}$. The maximum possible value of angular velocity of the ball in $\mathrm{rad} / \mathrm{s}$ is, :

<p>To find the maximum possible angular velocity (ω) of the ball, we need to consider the maximum tension the string can bear without breaking. This tension provides the centripetal force needed to keep the ball in a circular path.</p> <p>The centripetal force (F_c) required for circular motion is given by the formula:</p> <p>$ F_{c} = m r \omega^2 $</p> <p>Where:</p> <p>m = mass of the ball (0.5 kg)</p> <p>r = radius of the circle (0.5 m, since 50 cm = 0.5 m)</p> <p>ω = angular velocity in rad/s</p> <p>The maximum tension the string can bear is also the maximum centripetal force (F_c,max) that can be provided by the string, which is 400 N.</p> <p>Now we can set up the equation with the given values:</p> <p>$ 400 = 0.5 \times 0.5 \times \omega^2 $</p> <p>$\Rightarrow$ $ 400 = 0.25 \times \omega^2 $</p> <p>$\Rightarrow$ $ \omega^2 = \frac{400}{0.25} $</p> <p>$\Rightarrow$ $ \omega^2 = 1600 $</p> <p>$\Rightarrow$ $ \omega = \sqrt{1600} $</p> <p>$\Rightarrow$ $ \omega = 40 \, \mathrm{rad/s} $</p> <p>Therefore, the maximum possible angular velocity of the ball is 40 rad/s, which corresponds to Option C.</p>