In this problem, a car round sun, a flat circular track of radius 225 m. So let's say this is the circular track on which the car is rounding and radius of this track are is 225 m coefficient of static friction is given, which is 0.65 So we have to draw first force diagram of the car. So this is this is suing, let us a card is going in forward direction or in a poor direction. So this is suing the situation when we are viewing the situation from the, from the top. So what will be the uh what will be the its view when we view the situation from back side? So basically, now I'm going to draw the situation when we are seeing from on backside. So there will be the car and these are the tires which is on the road or the circular track. So this circular track will apply a normal reaction forcing vertically upward direction and there will be weight of the car which will be acting in downward direction. Now due to the centrifugal force, this car will have a tendency to move in outside the track. So to avoid the slipping, a friction force will act in the towards the center direction or in other sense, if we can see that these forces are acting in vertical in vertical direction, so but the circular motion is performing in the horizontal horizontal circle, so there must be a horizontal force which will be providing the centripetal acceleration. So the force should be friction force. Therefore the friction force must act towards the center. So the fourth diagram after the car will be normal reaction force in a poor direction, EMC in downward direction and freaks and force towards the center. No, in part we we have to calculate what is the maximum speed of the car so that it so that it is to stay on the track or it doesn't slip from the track. So for that, what we can see that this friction force is going to provide the centripetal acceleration and normal forces going to balance the MG component. Therefore, in vertical direction, what we can write and must be close to MG. Let us say this is equation when and freaks and force is going to provide centripetal acceleration, therefore freaks. And force will be close to M V squared divided by our. Now we can replace freaks and force with the value of us into and we know that static friction force is always less than equals two mules into. And so here we can replace Fs with the M V. Squared by our term. So what will be the equation M V squared by our will be less than equals two mules into. And so and in the uh equation one, we know that energy calls to MZ Therefore mp square by our should be less than it calls to us into MG. Now, mass can be cancelled from both sides of the expression. So what will be the value of the square? So we square should be less than equals two new S into our into G? Therefore velocity should be less than equals to square root over um us into our end to G. So what is the value of us here? In the problem, it has given that message 0.65 radius of the track is 225 and acceleration due to gravity is 9.8. So velocity has to be less than equals to this quantity, which if you calculate so this terms gives 37.85 m for a second. so velocity should be less than the value of less than the 37.85 m per second. Therefore, maximum permissible velocity. Therefore maximum velocity is 37.85 m per second. So this will be the answer for this problem.