In this post we will discuss surface area and volume questions of right circular cone. All the questions are solved step by step for your understanding so that you do not get stuck with any doubt.
Questions about cone are asked repeatedly in competition exams, so it is important for you to understand the basic concept and formulas. Some formulas covered in this post is:
1. Volume of cone
2. Curved Surface Area of cone
3. Lateral Height of cone
4. Total Surface area of cone
Remember that all the formulas are extremely important to solve volume and surface area related question. Please invest some of your time to remember each of them
Questions on Cone
(01) A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the sides of 3 cm to form a cone. The volume of cone so formed is?
Here
r = radius = 3 cm
h= height = 4 cm
Putting the values in the above formula we get
\\ volume\quad of\quad cone\quad =\quad \frac { 1 }{ 3 } \pi { \quad 3 }^{ 2 }\quad *\quad 4\ \ \\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \ =\quad 3\quad *\quad 4*\quad \pi \\\quad \\ \quad \quad \quad \quad \quad \quad \quad =\quad 12\quad \pi \quad { cm }^{ 3 }Hence we got the volume of cone from the above calculation
(02) If a right circular cone of height 24 cm has a volume of 1232 cu cm, then the area of its curved surface is?
Given
Height = h = 24 cm
Volume = 1232 cu cm
We know that volume of cube is = \frac { 1 }{ 3 } \pi r^{ 2 }h
1232=\frac { 1 }{ 3 } \pi r^{ 2 }h\\ \ \\ 1232\quad =\frac { 1\quad *\quad 22*\quad { r }^{ 2 }\quad *\quad 24 }{ 3\quad *\quad 7 } \\\ \\ r^{ 2 }\quad =\quad 7*7\\\ \\ r=7From the above calculation we got radius of cone = 7 cm
Curved surface area of cone =πrl
where
r =radius
l = slant height
\\ In\quad the\quad cone\quad applying\quad pythagors\quad theorem\\\ \\ { l }^{ 2 }={ h }^{ 2 }+{ r }^{ 2 }\\\ \\ l=\sqrt { { h }^{ 2 }+{ r }^{ 2 } } \\\ \\ l=\sqrt { 24^{ 2 }+{ 7 }^{ 2 } } \\\ \\ l=\sqrt { 625 } \\\ \\ l=\quad 25\quad cm\quad \\ \quad Hence the slant height of the cone is 25 cm
Now putting the value of L in the curved surface area formula
curved\quad surface\quad area\quad of\quad cone=\pi rl\\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad=\frac { 22*7*25 }{ 7 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\\ \\=22*25\quad =>\quad 550{ cm }^{ 2 }hence 550 sq cm is the area of curved surface of the cone
(03)The length of canvas 1.1 m wide required to build a conical tent of height 14 m and the floor area 346.5 sq m is ?
Given
Height = h = 14 meter
floor area = 346.5 sq m
Area\quad of\quad floor\quad =\pi { r }^{ 2 }=\quad 346.5\\\ \\ { r }^{ 2 }=\quad \frac { 346.5 }{ \pi } \\\ \\ { r }^{ 2 }=\quad \frac { 346.5\quad *\quad 7 }{ 22 } \\\ \\ { r }^{ 2 }=\quad 110.25\\ r=10.5\quad m\\\ \\ For\quad slant\quad height\quad we\quad will\quad apply\quad Pythagoras\quad formula\quad in\quad cone\\\ \\ { l }^{ 2 }={ h }^{ 2 }+{ r }^{ 2 }\\ l=\sqrt { { h }^{ 2 }+{ r }^{ 2 } } \\ l=\quad \sqrt { { 14 }^{ 2 }+110.25 } \\ l=\quad \sqrt { 196+110.25 } \\ l=\quad 17.5\quad meter\\\ \\ \\ we\quad know\quad curved\quad surface\quad area=\quad \pi rl\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 22 }{ 7 } *10.5*17.5\quad =>577.5\quad { m }^{ 2 }\\ \\ \\ \\ Its given that,
width of the canvas = 1.1 meter
length of canvas required =577.5/1.1 =525 meter
Hence the length of the canvas is 525 meter
(04) A cone of height 7 cm and base radius 3 cm is carved from a rectangular block of wood of dimensions 10 cm * 5 cm * 2 cm . Find the percentage of wood wasted?
Volume of block= Length * Breadth * Height
= 10 * 5 * 2
= 100 cu cm
\\ Volume\quad of\quad cone\quad carved\quad by\quad wood\quad =\quad \frac { 1 }{ 3 } \pi { r }^{ 2 }h\\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { 22*3*3*7 }{ 3 } \quad =>\quad 66{ cm }^{ 3 }Volume of wood wasted = (100 – 66) = 34 cu cm
%of wood wastage = (100 – 66)/100 *100 = 34%
Hence 34% of wood wastage
(05) A right circular cone and right circular cylinder have equal base and equal height. If the radius of the base and the height are in the ratio 5 : 12, then find the ratio of the total surface area of cylinder to that of cone.
Given
Radius of Base / Height = 5/12
Surface\quad area\quad of\quad cylinder=\quad 2\pi { r }^{ 2 }+2\pi { r }h\\\ \\putting\quad values\quad of\quad r=5\quad and\quad h=12\\\ \\ Surface\quad area\quad of\quad cylinder\quad =2*\pi *5*5\quad +2*\pi *5*12\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =2*\pi \quad (25+60)\quad =>\quad 170\pi \\\ \\ Surface\quad area\quad of\quad cone\quad =\quad \pi rl\quad +\quad \pi r^{ 2 }\\ \\ Here\quad l=\sqrt { { h }^{ 2 }+{ r }^{ 2 } } \\ =>\quad l\quad =\sqrt { { 12 }^{ 2 }+{ 5 }^{ 2 }\quad } \\ =>\quad l\quad =\quad 13\\\ \\ Now\quad the\quad surface\quad area\quad of\quad cone\quad =\quad \pi *5*13\quad +\pi { 5 }^{ 2 }\quad =>\quad 90\pi \\\ \\ According\quad to\quad question\\ \frac { Surface\quad area\quad of\quad cylinder }{ Surface\quad area\quad of\quad cone } =\frac { 170\pi }{ 90\pi } \quad =>\frac { 17 }{ 9 } \\ \\ Hence the surface area of cylinder to cone is 17 : 9
(06) A right cylindrical vessel is full of water. How many cones with the same radius and height as those of right circular cylinder will be needed to store that water?
volume\quad of\quad cylinder\quad =\pi { r }^{ 2 }h\\ volume\quad of\quad cone\quad =\quad \frac { 1 }{ 3 } \pi { r }^{ 2 }h\\\ \\ Let\quad x\quad cones\quad be\quad needed\quad to\quad store\quad all\quad the\quad water\\\ \\ According\quad to\quad question\\ number\quad of\quad cones\quad =\quad \frac { volume\quad of\quad cylinder }{ volume\quad of\quad cones } \quad =\quad \frac { \pi { r }^{ 2 }h }{ \frac { 1 }{ 3 } \pi { r }^{ 2 }h } \quad =>\quad 3\\\ \\ Hence\quad 3\quad cone\quad is\quad needed\quad to\quad fill\quad up\quad the\quad whole\quad water\\ (07) A solid metallic cylinder of base radius 3 cm and height 5 cm is melted to form cones, each of height 1 cm and base radius 1 mm. Find the number of cone that can be made with this process?
Radius of cylinder = 3 cm
Height of cylinder = 5 cm
Radius of cone = 1 mm = 1/10 cm = 0.1 cm { as 1 cm = 10 mm}
Height of cone = 1 cm
volume\quad of\quad cylinder\quad =\quad number\quad of\quad cone\quad *\quad Volume\quad of\quad cone\\\ \\ \pi { r }^{ 2 }H\quad =\quad number\quad of\quad cones\quad *\quad \frac { 1 }{ 3 } \pi { r }^{ 2 }h\\\ \\ Cancelling\quad \pi \quad from\quad each\quad sides,\quad we\quad get\quad \\\ \\ 3*3*5\quad =\quad number\quad of\quad cones\quad *\quad \frac { 0.\quad 1*\quad 0.1\quad *\quad 1 }{ 3 } \\\ \\ number\quad of\quad cones\quad =\quad \frac { 45\quad *\quad 3 }{ 0.01 } ==>13500Hence we will get 13500 cones from the process
(08) Water flows at the rate of 10 meters per minute from a cylindrical pipe 5mm in diameter. How long will it take to fill up a conical vessel whose diameter at the base is 40 cm and the depth is 24 cm?
Diameter of cylindrical pipe = 5 mm = 5/10 cm = 0.5 cm
Radius of cylindrical pipe => R = diameter/2 => 0.5/2 = 0.25 cm
Water flows in per minute = 10 meters = 10 * 100 =1000 cm
Volume\quad of\quad water\quad that\quad flows\quad in\quad 1\quad minute=\pi { R }^{ 2 }H\\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \pi \quad *\quad 0.25\quad *\quad 0.25\quad *1000\\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =62.5\pi \quad { cm }^{ 3 }\\ Diameter of conical vessel =40 cm
Radius of conical vessel, r = diameter /2 = 40/2 => 20 cm
Height of conical vessel, h = 24 cm
According\quad to\quad question\\ Volume\quad flown\quad in\quad conical\quad vessel\quad =\frac { \pi { r }^{ 2 }h }{ 3 } \\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { \pi *20*20*24 }{ 3 } =3200\pi \quad cu\quad cm\quad \\\ \\ Time\quad required\quad to\quad fill\quad the\quad vessel\quad =\frac { volume\quad flown\quad in\quad conical\quad vessel }{ volume\quad of\quad water\quad flowing\quad per\quad minute } \\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { 3200\pi }{ 62.5\pi } \quad =>\quad 51\quad minute\quad 12\quad seconds(09) A solid cylindrical block of radius 12 cm and height 18 cm is mounted with a conical block of radius 12 cm and height 5 cm. Find the total lateral surface of the solid formed.
Given
Radius of cylinder, r = 12 cm
Height of cylinder, h =18 cm
Radius of cone, R= 12 cm
Height of cone, H = 5 cm
Slant height of cone l=\sqrt { { 12 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 169 } =13
Lateral surface of solid = curved surface of cone + surface area of cylinder + Surface area of bottom
Lateral surface of solid= \pi Rl\quad +\quad 2\pi rh\quad +\pi { r }^{ 2 }
=\pi \quad (Rl\quad +\quad 2rh\quad +{ r }^{ 2 })\\ =\pi \quad (156\quad +\quad 432\quad +\quad 144)\\ =\pi \quad 732\\ =2300.57