Time Taken by Slipping Cylinder to Start Rolling

To find the time $t$ when the solid cylinder starts rolling without slipping, we analyze the effect of kinetic friction on both linear and angular motions.

The linear acceleration (deceleration) caused by kinetic friction $f_k = \mu_K m g$ is given by: $a = \mu_K g$

The angular acceleration due to the torque about the center of mass is: $\alpha = \frac{f_k R}{I} = \frac{\mu_K m g R}{\frac{1}{2}m R^2} = \frac{2\mu_K g}{R}$

The velocity $v(t)$ and angular velocity $\omega(t)$ as functions of time $t$ are: $v(t) = v_0 - a t = v_0 - \mu_K g t$ $\omega(t) = \omega_0 + \alpha t = \frac{v_0}{4R} + \frac{2\mu_K g}{R} t$

Pure rolling begins when the condition $v(t) = R \omega(t)$ is satisfied: $v_0 - \mu_K g t = R \left(\frac{v_0}{4R} + \frac{2\mu_K g}{R} t\right)$

Solving for $t$: $\frac{3}{4}v_0 = 3\mu_K g t \implies t = \frac{v_0}{4\mu_K g}$

Substituting the given values ($v_0 = 49\text{ m/s}$, $\mu_K = 0.25$, and $g = 9.8\text{ m/s}^2$): $t = \frac{49}{4 \times 0.25 \times 9.8} = 5\text{ s}$

Thus, the correct option is B.