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H Harsh Kankaria
Given, a cylinder with a base. The radius of the cylinder =  Height of the cylinder =  We know,  The lateral surface area of a cylinder of radius  and height   =    Area of the cylindrical penholder = Lateral areal + Base area Area of 35 penholders  =  Therefore, the area of carboard required is
Given, a cylindrical lampshade Daimeter of the base =  Height of the cylinder  =  Total height of lampshade=   We know,  Curved surface area of a cylinder of radius  and height   =  Now, Cloth required for covering the lampshade = Curved surface area of the cylinder Therefore,  cloth will be required for covering the lampshade.
Given, a closed cylindrical petrol tank. The diameter of the tank =  Height of the tank =  Now, Total surface area of the tank = Now, let  of steel sheet be actually used in making the tank Since  of steel was wasted, the left  of the total steel sheet was used to made the tank.  The total surface area of the tank =  Therefore,  of steel was actually used in making the tank.

H Harsh Kankaria
Given, a closed cylindrical petrol tank. The diameter of the tank =  Height of the tank =  We know,  The lateral surface area of a cylinder of radius  and height  =   The lateral surface area of a cylindrical tank =  Therefore, the lateral or curved surface area of a closed cylindrical petrol storage tank is

H Harsh Kankaria
Given, Length of the cylindrical pipe =  Diameter =   The total radiating surface will be the curved surface of this pipe. We know, The curved surface area of a cylindrical pipe of radius  and length  =   CSA of this pipe =  Therefore, the total radiating surface of the system is

H Harsh Kankaria
Given, Inner diameter of the circular well =   Depth of the well =   The inner curved surface area of the circular well is Now, the cost of plastering the curved surface per  = Rs. 40  Cost of plastering the curved surface of   = Therefore, the cost of plastering the well is
Given, The inner diameter of the circular well =   Depth of the well =  We know, The curved surface area of a cylinder =   The curved surface area of the well =  Therefore, the inner curved surface area of the circular well is
Given, a right circular cylinder Curved surface area of the cylinder =  Radius of the base =  Let the height of the cylinder be   We know,  Curved surface area of a cylinder of radius  and height   =     Therefore, the required height of the cylinder is

H Harsh Kankaria
Given, Radius of the cylindrical pillar, r =  Height of the cylinder, h  =  We know, Curved surface area of a cylinder =   Curved surface area of the pillar =  Now, Cost of painting  of the pillar =  Cost of painting the curved surface area of the pillar =  Therefore, the cost of painting curved surface area of the pillar is

H Harsh Kankaria
Given, The diameter of the cylindrical roller =  Length of the cylindrical roller =  The curved surface area of the roller =   Area of the playground =  Therefore, the required area of the playground =

H Harsh Kankaria
Note: There are two surfaces, inner and outer. Given, Height of the cylinder,  Outer diameter  =  Inner diameter  =  Outer curved surface area =  Inner curved surface area =  Area of the circular rings on top and bottom =   The total surface area of the pipe =  Therefore, the total surface area of the cylindrical pipe is
Note: There are two surfaces, inner and outer. Given, Height of the cylinder,  Outer diameter  =  Inner diameter  =  Outer curved surface area =  Therefore, the outer curved surface area of the cylindrical pipe is
Note: There are two surfaces, inner and outer. Given, Height of the cylinder,  Outer diameter  =  Inner diameter  =   Inner curved surface area =  Therefore, the inner curved surface area of the cylindrical pipe is
Given, Height of the cylindrical tank =  Base diameter =   We know,  The total surface area of a cylindrical tank =  Therefore, square metres of the sheet is