 Hello everyone, this is Shila Ratna Banshwade from Alchan Institute of Technology, Solapur. Today we are going to see the topic sections of solids and in that we will particularly see the solid cone. Let us move further. The learning outcome of this video will be the students will be able to draw the sectional view of cone. At this moment of the video, I suggest you to pause the video and recollect or remember the structure of cone. As we all know, cones are solids of revolution. That is, when we rotate a right angle triangle about the side adjacent to 90 degrees, we get a cone. Learning outcomes at the end of this session, the students will be able to draw the sectional view of cone. At this point of the video, I suggest you to pause the video and recollect the structure of a cone. As we all know, cones have a circular base and a point at the top called as the apex or the vertex. Thus, we can say that points on the circular form join to a common point called as apex forming the cone. Also, cones are called as solids of revolution. That is, to obtain a cone, if we rotate a right angle triangle about the side adjacent to 90 degrees, we get a cone. Thus, cones are called as solids of revolution. Now moving further, in the previous video, we have seen an example on cone. So in this video, we will see the second example. A cone base 70 mm diameter and axis 90 mm is lying on one of its generators in HP with the axis parallel to VP. It is cut by a section plane inclined at an angle of 30 degrees to HP and perpendicular to VP. And the same plane passes at a distance of 30 mm above the base along the axis so that the apex is retained, draw the front view and sectional top view. Let us look the question into detail. So we have a cone of 70 mm base diameter that is base circle will be of 70 mm and axis height 90 mm. So the vertical axis, axis is the point joining the center of the base. Axis is a line joining center of the base and apex. It is a straight line. So that axis or the length of axis is 90 mm. Now this cone is lying on generators. As we all know that cone is a solid of revolution. It does not have specific edges or faces. It has smooth surface and for that we consider generators which are imaginary. So there are no edges in cone. So that cone is lying on one of the generators in HP. In HP means the cone should be tilted totally on the ground with the upper axis parallel to VP. You have to tilt the cone in such a way that the axis is parallel to VP. That is you have you can see the axis length that is of 90 degrees in the front view. Now this cone is cut by a section plane inclined at an angle of 30 degrees to HP. Here we have a core of 70 mm diameter and axis 90 mm. That is the base circle should be of 70 mm and the axis is the line joining the center of the base to the common point that is apex. That is of 90 mm. As we know the cone does not have specified edges or corners. It has generators. Generators are imaginary lines joining this base and the apex. Now this cone is resting on HP on that generator. In such a way that the axis is parallel to VP. And the axis is parallel to VP. That means you have to you can see the true length of the axis that is true length true height of the cone. Now this cone is cut by a section plane inclined at an angle of 30 degrees to HP and perpendicular to VP. Now when the cone is resting on generators in that position it is cut by a section plane inclined at an angle of 30 degrees to HP. Inclination with HP is seen in front view plus the axis line is perpendicular to VP. This cutting plane also passes through a point at a distance of 30 mm from the base. So for cutting plane we have three conditions. The cutting plane is inclined at 30 degrees to HP. It is perpendicular to VP and it passes through a point at a distance of 30 mm from the base above the base and you have to cut or we have to cut the cone in such a way that the apex part or the top most part of the cone should be retained whereas the remaining part can be removed. For this we have to draw the front view and sectional top view. Let us move further. This is the XY line or the reference line. Now to draw the cone resting on one of its generators we will consider that or we assume that initially the cone is resting on HP on its base. So when we consider that the cone is resting on base in the top view we see the circle as you can see on the screen. We know the dimensions of the circle that is 70 mm. So we draw a circle of 70 mm further we divide the circle in equal number of parts. Now it is up to the student whether to divide the circle into 8 number of parts, 12 number of parts, 16 parts or more. But minimum the circle has to be divided into 8 number of parts for better drawing conditions. We name the parts as 1, 2, 3, 4 up to 8. Now let us move to complete the front view. Now as I said the cone is resting on base vertically. So in the front view you can see the triangular shape of the cone. Let us project the points, point 1 projected vertically upwards, point 2 and 3, point 2 and 8, 3 and 7, 4 and 6 and 5. We draw the cone in the desired fashion. Now we join the generators and we name the points. So this is the condition that we have assumed before the given condition that the cone is resting on HP on its base. Now the generator should be in contact with HP that is the generator's O1 dash or O5 dash should be coinciding with the XY line. The XY line here represents HP. So in the second condition we redraw this first image as shown on the screen. So here we take O5 dash coinciding with XY line and we redraw this figure over here. We name the parts, we name the points as shown on the screen. So this is the required front view or the front view of the cone as per the given condition that is it is resting on one of its generators. Now moving further to complete the top view we require the points from this front view as well as the previous top view. So we project O from the second front view and corresponding point O from the previous top view to get this point. Similarly we project O from the front view, second front view and O from the previous top view. So we get this point. Similarly we obtain all the points. Let us name the points. So these are the points in the second top view or the required top view. We will join these points and we will join the remaining points. So this is the condition given in question. That is the cone is lying on one of its generators. Now in this condition we need to cut the cone with an axis or with a cutting plane perpendicular to VP and 30 degrees to HP in such a way that the apex part, this part should be retained. So one can take 30 degrees with HP in this direction as well as in this direction. But as we need to retain this part we cut the cone in this direction. So this is the cutting plane. We cut the cone in this direction. Now to get the cutting points we consider that this is the cutting plane passing through the cone. Wherever this cutting plane intersects with the generators we consider those points. So these are the points. The cutting plane cuts O1 dash at point P1 similarly O2 dash at P2 O3 dash at P3 O4 dash at P4 and so on. Now we have to transfer these cutting plane points on the second top view or the desired top view. So we project these points in the second top view that is point P1 will be projected on the generator O1 in the top view. So we get this section plane as seen on the screen. When we join all these points so this is the required sectional part of the cone when it is resting on its generators. Now let us dark the remaining parts. So this is the part that is being retained that is the apex part is being retained whereas the other part is being eradicated. So here you can see the direction of observation is this. This part is being inclined and you can see it over here and this part is the right hand side part which is in the dark form. So while drawing on sheet practically you need to draw this with light shade or a thin line and then you can dark the lines or thick the lines. So this is the complete projection of the cone when it is cut by the thank you.