 Hello and welcome to the session. I am Deepika here. Let's discuss the question which says an engineering student was asked to make a model shape like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that resonates, assume the outer and inner dimensions of the model to be nearly the same. Now we know that volume of cylinder is equal to pi r square h and volume of cone is equal to 1 by 3 pi r square h where is the radius h is the height. So this is a key idea behind our question. We will take the help of this key idea to solve the above question. So let's start the solution. The model is made of a cylinder and two cones. We have to find the volume of the air inside the model. Volume of the air inside the model is equal to volume of air inside the cone plus volume of air inside the cylinder plus volume of this cone are given of model is equal to 3 cm. Therefore radius of model is equal to 3 by 2 cm that is this is 3 by 2 cm that is r is equal to 3 by 2 cm. It is also given that the length of the model is 12 cm that is the total height of the model is equal to 12 cm. Now according to our question each cone has a height of 2 cm now this is a height of cone so this is 2 cm and this is also 2 cm. Let h1 is the height of the cone therefore h1 is equal to 2 cm. Let 2 is the height of this cylinder then h2 is equal to total height that is 12 cm minus height of two cones and this is equal to 8 cm. So height of the cylinder is equal to 8 cm. Now volume of air the model volume of air according to our key idea volume of cone is equal to 1 by 3 pi r square h so this is equal to 1 by 3 pi r square h1 as h1 is the height of the cone plus pi r square h2 as h2 is the height of this cylinder plus 1 by 3 pi r square h1. Let us take 1 by 3 pi r square common so we have volume of air inside the model is equal to 1 by 3 pi r square into h1 plus 3 h2 plus h1 and this is equal to 1 by 3 take pi is equal to 22 by 7 now radius of the model is 3 by 2 cm h1 is the height of the cone which is 2 cm plus 3 into h2 is the height of the cylinder which is 8 cm h1 is the height of the cone so this is plus 2 cm q is equal to 1 by 3 into 22 upon 7 into 3 by 2 into 3 by 2 into 24 plus 2 cm q and this is equal to 1 by 3 into 22 upon 7 into 3 by 2 into 3 by 2 into 28 cm q on cancellation we have this is equal to 22 into 3 cm q and this is equal to 66 cm q therefore volume of the air contained in the model that Rachel made is equal to 66 cm q hence the answer for the above question is 66 cm q i hope the solution is clear to you buy and take care