Q&A - Ask Doubts and Get Answers

Sort by :
Clear All
Q

Q : 11     The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius \small 3\hspace{1mm}cm and height  \small 10.5\hspace{1mm}cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?
 

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   

Q : 10    In Fig. \small 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of \small 20\hspace{1mm}cm and height of \small 30\hspace{1mm}cm. A margin of \small 2.5\hspace{1mm}cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

                

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.

Q : 9    Find (ii)  how much steel was actually used, if \small \frac{1}{12}  of the steel actually used was wasted in making the tank.
 

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.

Q : 9     Find

             (i)  the lateral or curved surface area of a closed cylindrical petrol storage tank that is \small 4.2\hspace{1mm}m in diameter and \small 4.5\hspace{1mm}m high.
 

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 

Q : 8    In a hot water heating system, there is a cylindrical pipe of length \small 28\hspace{1mm}m and diameter \small 5\hspace{1mm}cm. Find the total radiating surface in the system.
 

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 

Q : 7    The inner diameter of a circular well is \small 3.5\hspace{1mm}m. It is \small 10\hspace{1mm}m deep. Find

            (ii) the cost of plastering this curved surface at the rate of Rs 40 per \small m^2.

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

Q : 7    The inner diameter of a circular well is \small 3.5\hspace{1mm}m. It is \small 10\hspace{1mm}m deep. Find

            (i) its inner curved surface area.

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

Q : 6     Curved surface area of a right circular cylinder is \small 4.4\hspace{1mm}m^2. If the radius of the base of the cylinder is \small 0.7\hspace{1mm}m, find its height.
 

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 

Q : 5    A cylindrical pillar is  \small 50\hspace{1mm}cm in diameter and 3.5\ m in height. Find the cost of painting the curved surface of the pillar at the rate of \small Rs \hspace{1mm} 12.50  per \small m^2

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 

Q: 4    The diameter of a roller is \small 84\hspace{1mm}cm and its length is \small 120\hspace{1mm}cm. It takes \small 500 complete revolutions to move once over to level a playground. Find the area of the playground in \small m^2.
 

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 = 

Q : 3    A metal pipe is \small 77\hspace{1mm}cm long. The inner diameter of a cross section is 4 cm, the outer diameter being \small 4.4 \hspace{1mm}cm (see Fig. \small 13.11). Find its

          (iii) total surface area.

          

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

Q : 3    A metal pipe is \small 77\hspace{1mm}cm long. The inner diameter of a cross section is \small 4\hspace{1mm}cm, the outerdiameter being  \small 4.4\hspace{1mm}cm (see Fig. \small 13.11). Find its

              (ii) Outer surface area.

                

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

Q : 3    A metal pipe is \small 77\hspace{1mm}cm long. The inner diameter of a cross section is \small 4\hspace{1mm}cm, the outer diameter being \small 4.4\hspace{1mm}cm (see Fig.  \small 13.11). Find its

           (i) inner curved surface area,

            

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

Q : 2    It is required to make a closed cylindrical tank of height \small 1\hspace{1mm}m and base diameter \small 140\hspace{1mm}cm from a metal sheet. How many square metres of the sheet are required for the same?
 

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   

Q: 1    The curved surface area of a right circular cylinder of height  \small 14\hspace{1mm}cm is  \small 88\hspace{1mm}cm^2. Find the diameter of the base of the cylinder.
 

Given, The curved surface area of the cylinder =  And, the height of the cylinder,  We know, Curved surface area of a right circular cylinder =    Therefore, the diameter of the cylinder =   
Exams
Articles
Questions