 Hello and welcome to the session. Let's discuss the following question. It says, A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. So we are given a capsule which is in the shape of a cylinder with two hemispheres stuck to each of its ends. And we are given that the diameter of the capsule is 5 mm and the length of the entire capsule is 14 mm. And we have to find the surface area of this capsule. So let's now move on to the solution. Now we are given that diameter of the capsule is 5 mm. Now since diameter of the capsule is 5 mm, therefore the diameter of the hemispherical ends will also be 5 mm. And the radius would be half of it. Therefore radius of hemispherical ends would be 5 by 2 mm. Let's denote it by r and also the radius of cylindrical part would be r that is 5 by 2 mm. Now to find the surface area of the capsule, what we need to do is we need to add the surface area of the cylindrical part and the surface areas of two hemispherical parts. And to find the surface area of the cylindrical part, we need to know the height of the cylindrical part and here in this case it would be the length and the length of the cylindrical part would be the total length minus the radii of two hemispherical ends. Right, the length of the cylindrical part minus radii two hemispheres. Now the total length 14 mm and radii of each hemispherical part is 5 by 2 mm. So it would be 5 by 2 plus 5 by 2. So this is equal to 14 minus 5 by 2 plus 5 by 2 is 5. So this is equal to 9 mm. So the length of the cylindrical part that is this length is 9 mm. Now the surface area of the capsule is equal to the surface area of cylindrical portion plus the surface area of two hemispheres. Right, now let this surface area be A. Therefore A is equal to surface area of cylindrical portion that is cylinder is 2 pi rh is the length. Let us denote this by h plus surface area of two hemispheres that is 2 pi r square plus 2 pi r square. So this implies A is equal to 2 pi rh plus 4 pi r square. So this implies A is equal to taking 2 pi r common we have h plus r into 2 pi r. Right, let us now substitute the values of pi r and h. Pi is 22 by 7 r is 5 by 2 mm radius is 5 by 2 mm into h is 9 plus 5 by 2. So this implies A is equal to 22 by 7 into 5 into taking LCM we have 23 upon 2. Since here we have taken 2 pi r common. So we will be having 2 r here. So it is 2 into 5 by 2. So here 2 gets cancelled with 2 and we have 9 plus 5 that is 14 not 23 by 2. So we have 14 here. It gets cancelled by 7 to give 2. So this implies A is equal to 22 into 5 into 2. And this implies A is equal to 220 cm mm square. So required answer is 220 mm square. The surface area of the capsule is 220 mm square. So this completes the question on the session. Bye for now. Take care. Have a good day.