To solve this problem, we need to find the maximum safe velocity of a car negotiating a banked curve.$\\$Given:$\\$• Radius of the curve $R \\$• Banking angle $\theta \\$• Coefficient of static friction $\mu_s \\$The forces acting on the car are:$\\$• Gravitational force $mg$ acting downwards.$\\$• Normal force $N$ acting perpendicular to the surface.$\\$• Frictional force $f_s$ acting parallel to the surface.$\\$For a car to move safely on a banked road, the net force towards the center of the curve provides the necessary centripetal force.$\\$The centripetal force $F_c$ is given by:$\\ F_c = \frac{mv^2}{R} \\$The components of forces along the incline are:$\\$• Normal force component: $N \cos\theta \\$• Frictional force component: $f_s = \mu_s N \\$The components of forces perpendicular to the incline are:$\\$• Normal force component: $N \sin\theta \\$• Gravitational force component: $mg \cos\theta \\$Balancing the forces perpendicular to the incline:$\\ N \cos\theta = mg + \mu_s N \sin\theta \\$Solving for $N: \\ N = \frac{mg}{\cos\theta - \mu_s \sin\theta} \\$Balancing the forces along the incline for centripetal force:$\\ \frac{mv^2}{R} = N \sin\theta + \mu_s N \cos\theta \\$Substitute $N$ from the previous equation:$\\ \frac{mv^2}{R} = \left(\frac{mg}{\cos\theta - \mu_s \sin\theta}\right) (\sin\theta + \mu_s \cos\theta) \\$Simplify:$\\ v^2 = \frac{gR (\mu_s + \tan\theta)}{1 - \mu_s \tan\theta} \\$Therefore, the maximum safe velocity $v$ is:$\\ v = \sqrt{gR \left(\frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}\right)} \\$This corresponds to Option 4.