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10th Standard
Maths
A cylindrical drum has a height of 20 cm and base radius of 14 cm. Find its curved surface area and the total surface area.
The curved surface area of a right circular cylinder of height 14 cm is 88 cm2 . Find the diameter of the cylinder.
A garden roller whose length is 3 m long and whose diameter is 2.8 m is rolled to level a garden. How much area will it cover in 8 revolutions?
If one litre of paint covers 10 m2, how many litres of paint is required to paint the internal and external surface areas of a cylindrical tunnel whose thickness is 2 m, internal radius is 6 m and height is 25 m.
The radius of a conical tent is 7 m and the height is 24 m. Calculate the length of the canvas used to make the tent if the width of the rectangular canvas is 4 m?
If the total surface area of a cone of radius 7cm is 704 cm2, then find its slant height.
Find the diameter of a sphere whose surface area is 154 m2.
The radius of a spherical balloon increases from 12 cm to 16 cm as air being pumped into it. Find the ratio of the surface area of the balloons in the two cases.
If the base area of a hemispherical solid is 1386 sq. metres, then find its total surface area?
A sphere, a cylinder and a cone are of the same radius, where as cone and cylinder are of same height. Find the ratio of their curved surface areas.
The slant height of a frustum of a cone is 5 cm and the radii of its ends are 4 cm and 1 cm. Find its curved surface area.
The external radius and the length of a hollow wooden log are 16 cm and 13 cm respectively. If its thickness is 4 cm then find its T.S.A.
4 persons live in a conical tent whose slant height is 19 cm. If each person require 22 cm2 of the floor area, then find the height of the tent.
The ratio of the radii of two right circular cones of same height is 1 : 3. Find the ratio of their curved surface area when the height of each cone is 3 times the radius of the smaller cone.
The radius of a sphere increases by 25%. Find the percentage increase in its surface area.
Find the volume of a cylinder whose height is 2 m and whose base area is 250 m2.
The volume of a solid right circular cone is 11088 cm3. If its height is 24 cm then find the radius of the cone.
The ratio of the volumes of two cones is 2 : 3. Find the ratio of their radii if the height of second cone is double the height of the first.
A 14 m deep well with inner diameter 10 m is dug and the earth taken out is evenly spread all around the well to form an embankment of width 5 m. Find the height of the embankment.
If the circumference of a conical wooden piece is 484 cm then find its volume when its height is 105 cm.
If the ratio of radii of two spheres is 4 : 7, find the ratio of their volumes.
Water is flowing at the rate of 15 km per hour through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Find the time in which the level of water in the tanks will rise by 21 cm.
A conical flask is full of water. The flask has base radius r units and height h units, the water poured into a cylindrical flask of base radius xr units. Find the height of water in the cylindrical flask.
The barrel of a fountain-pen cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used for writing 330 words on an average. How many words can be written using a bottle of ink containing one fifth of a litre?
A hemi-spherical tank of radius 1.75 m is full of water. It is connected with a pipe which empties the tank at the rate of 7 litre per second. How much time will it take to empty the tank completely?
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r units.
When ‘h’ coins each of radius ‘r’ units and thickness 1 unit is stacked one upon the other, what would be the solid object you get ? Also find its C.S.A.
When the radius of a cylinder is double its height, find the relation between its C.S.A. and base area.
Two circular cylinders are formed by rolling two rectangular aluminum sheets each of dimensions 12 m length and 5 m breadth, one by rolling along its length and the other along its width. Find the ratio of their curved surface areas.
Give practical example of solid cone.
Find surface area of a cone in terms of its radius when height is equal to radius
Compare the above surface area with the area of the base of the cone
When a sector of a circle is transformed to form a cone, then match the conversions taking place between the sector and the cone.
| Sector | Cone |
| Radius | Circumference of the base |
| Area | Slant height |
| Arc length | Curved surface area |
Find the value of the radius of a sphere whose surface area is 36\(\pi\) sq. units
How many great circles can a sphere have?
Find the surface area of the earth whose diameter is 12756 kms.
Shall we get a hemisphere when a sphere is cut along the small circle?
T.S.A of a hemisphere is equal to how many times the area of its base?
How many hemispheres can be obtained from a given sphere?
Give two real life examples for a frustum of a cone.
Can a hemisphere be considered as a frustum of a sphere
If the height is inversely proportional to the square of its radius, the volume of the cylinder is ____________.
What happens to the volume of the cylinder with radius r and height h, when its
(a) radius is halved (b) height is halved
Is it possible to find a right circular cone with equal
(a) height and slant height
(b) radius and slant height
(c) height and radius.
There are two cones with equal volumes. What will be the ratio of their radius and height?
Consider the cones given in Fig
(i) Without doing any calculation, find out whose volume is greater?
(ii) Verify whether the cone with greater volume has greater surface area.
(iii) Volume of cone A : Volume of cone B = ?
A cone, a hemisphere and a cylinder have equal bases. The heights of the cone and cylinder are equal and are same as the common radius. Are they equal in volume?
Give any two real life examples of sphere and hemisphere
Is it possible to obtain the volume of the full cone when the volume of the frustum is known?
Answers
Given that, height of the cylinder h = 20 cm ; radius r =14 cm
Now, C.S.A. of the cylinder = 2p\(\pi\)h sq. units
C.S.A. of the cylinder = \(2\times \frac { 22 }{ 7 } \times 14\times 20=2\times 22\times 2\times 20\)
T.S.A. of the cylinder \(=2\pi r(h+r)\)sq.units
\(=2\times \frac { 22 }{ 7 } \times 14\times (20+14)=2\times \frac { 22 }{ 7 } \times 14\times 34\)
= 2992 cm2
Therefore, C.S.A. = 1760 cm2 and T.S.A. = 2992 cm2
Given that, C.S.A. of the cylinder = 88 sq. cm
2\(\pi\)rh = 88
\(2\times \frac { 22 }{ 7 } \times 14=88\) (given h = 14cm)
2r = \(\frac { 88\times 7 }{ 22\times 14 } =2\)
Therefore, diameter = 2 cm
Given that, diameter d = 2.8 m and height = 3 m
radius r = 1.4 m
Area covered in one revolution = curved surface area of the cylinder
= 2\(\pi\)rh sq. units
\(2\times \frac { 22 }{ 7 } \times 1.4\times 3=26.4\)
Area covered in 1 revolution = 26.4 m2
Area covered in 8 revolutions = 8 x 26.4 = 211.2
Therefore, area covered is 211.2 m2
Given that, height h = 25 Given that, height h = 25 m; thickness = 2 m.
internal radius r = 6 m
Now, external radius R = 6 + 2 = 8m
C.S.A. of the cylindrical tunnel = C.S.A. of the hollow cylinder
C.S.A. of the hollow cylinder = 2\(\pi\)(R + r)h sq.units
\(2\times \frac { 22 }{ 7 } (8+6)\times 25\)
Hence, C.S.A. of the cylindrical tunnel = 2200 m2
Area covered by one litre of paint = 10 m2
Number of litres required to paint the tunnel \(=\frac { 2200 }{ 10 } =220\)
Therefore, 220 litres of paint is needed to paint the tunnel.
Let r and h be the radius and height of the cone respectively.
Given that, radius r = 7 m and height h = 24 m
Hence, l = \(\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(=\sqrt { 49+576 } \)
\(l=\sqrt { 625 } =25m\)
C.S.A. of the conical tent = \(\pi\)rl sq. units
Area of the canvas \(=\frac { 22 }{ 7 } \times 7\times 25={ 550 }m^{ 2 }\)
Now, length of the canvas \(\frac{Area\ of\ the\ canvas}{width}=\frac{550}{4}=137.5m\)
Therefore, the length of the canvas is 137.5 m
Given that, radius r = 7 cm
Now, total surface area of the cone = \(\pi\)r(l + r)sq. units
T.S.A = 704 cm2
704 = \(\frac{22}{7}\times7(l+7)\)
32 = l + 7 implies l = 25 cm
Therefore, slant height of the cone is 25 cm.
Let r be the radius of the sphere. Given that, surface area of sphere = 154 m2
4\(\pi\)r2 = 154
\(4\times \frac { 22 }{ 7 } \times { r }^{ 2 }=154\)
gives \({ r }^{ 2 }=154\times \frac { 1 }{ 4 } \times \frac { 7 }{ 22 } \)
hence, \({ r }^{ 2 }=\frac { 49 }{ 4 } \)We get r = \(\frac{7}{2}\)
Therefore, diameter is 7 m
Let r1 and r2 be the radii of the balloons.
Given that, \(\frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 12 }{ 16 } =\frac { 3 }{ 4 } \)
Now, ratio of C.S.A. of balloons \(=\frac { 4\pi { r }_{ 1 }^{ 2 } }{ 4\pi { r }_{ 2 }^{ 2 } } =\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } ={ \left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) }^{ 2 }={ \left( \frac { 3 }{ 4 } \right) }^{ 2 }=\frac { 9 }{ 16 } \)
Therefore, ratio of C.S.A. of balloons is 9:16.
Let r be the radius of the hemisphere.
Given that, base area = \(\pi\)r2 = 1386 sq. m
T.S.A. = 3 \(\pi\)r2 sq.m
= 3 x 1386 = 4158
Therefore, T.S.A. of the hemispherical solid is 4158 m2.
Required Ratio = C.S.A. of the sphere: C.S.A. of the cylinder : C.S.A. of the cone
\(4\pi { r }^{ 2 }:2\pi rh:\pi rl,\ (l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =\sqrt { { 2r }^{ 2 } } =\sqrt { 2r } units)\)
\(=4:2:\sqrt { 2 } =2\sqrt { 2 } :\sqrt { 2 } :1\).
Let l, R and r be the slant height, top radius and bottom radius of the frustum.
Given that, l = 5 cm, R = 4 cm, r = 1 cm
Now, C.S.A. of the frustum \(\pi\)(R + r)l sq.units
\(\frac { 22 }{ 7 } \times (4+1)\times 5\)
\(=\frac { 550 }{ 7 } \)
Therefore, C.S.A. = 78.57 cm2
External radius of hollow cylinder R = 16 cm
length h = 13 cm
Thickness R - r = 4
16 - r = 4
r = 12 cm
Total surface area of hollow cylinder \(=2 \pi(\mathrm{R}+\mathrm{r})(\mathrm{R}-\mathrm{r}+\mathrm{h}) \text { sq. units }\)
\(=2 \times \frac{22}{7} \times(16+12)(4+13) \)
\(=2 \times \frac{22}{7} \times 28 \times 17 \)
= 2992 sq. cm
Each person requires 22 m2 of floor area.
Required base area = 22 x 4 = 88 m2
\(\pi r^{2} =88 \)
\(r^{2} =\frac{88 \times 7}{22}=4 \times 7 \)
\(r =2 \sqrt{7} \mathrm{~m} \)
slant height = 19 m
height of the tent, h \(=\sqrt{l^{2}-r^{2}}\)
\(=\sqrt{(191)^{2}-(2 \sqrt{7})^{2}} \)
\(=\sqrt{361-28} =\sqrt{330}=18.25 \mathrm{~m} \)
Height of the tent = 18.25 m
Let the radii of two cones be r1 and r2 and heights be h1 and h2
Given ratio of their radii = \(\frac{r_{1}}{r_{2}}=\frac{1}{3}\)
\(r_{1}=\frac{r_{2}}{3}
\)
\(h_{1}=3 r_{1}, h_{2}=3 r_{1}
\)
[ r1 is the radius of smaller cone]
Slant heights \(l_{1} =\sqrt{h_{1}^{2}+r_{1}^{2}}
\)
\(=\sqrt{9 r_{1}^{2}+r_{1}^{2}}=\sqrt{10} r_{1}
\)
\(l_{2} =\sqrt{h_{2}^{2}+r_{2}^{2}}
\)
\(=\sqrt{9 r_{1}^{2}+9 r_{1}^{2}}=\sqrt{18 r_{1}^{2}}=3 \sqrt{2} r_{1}\)
Ratio of curved surface areas
\(=\frac{\text { CSA of I cone }}{\text { CSA of II cone }}
\)
\(=\frac{\pi r_{1} l_{1}}{\pi r_{2} l_{2}} =\frac{r_{1}\left(\sqrt{10} r_{1}\right)}{\left(3 r_{1}\right)\left(3 \sqrt{2} r_{1}\right)}
\)
\(=\frac{\sqrt{10}}{9 \sqrt{2}}= \frac{\sqrt{5} \sqrt{2}}{9 \sqrt{2}}=\frac{\sqrt{5}}{9}
\)
Ratio of C.S.A = \(\sqrt{5}: 9\)
Let the radius of the sphere be 'r' cm
Surface area = \(4 \pi r^{2}\)
when radius is increased by 25% ,then new diameter = r + 25% + r
\(=r+\frac{25 r}{100}=\frac{5 r}{4}\)
Surface area of new sphere
\(=4 \pi\left(\frac{5 r}{4}\right)^{2} \)
\(=4 \pi\left(\frac{25 r^{2}}{16}\right) \)
\(=\frac{25 \pi r^{2}}{4} \)
Increase in surface area = \(\frac{25 \pi r^{2}}{4}-4 \pi r^{2}\)
\(=\frac{25 \pi r^{2}-16 \pi r^{2}}{4} \)
\(=\frac{9 \pi r^{2}}{4} \)
Percentage increase in surface area
\(=\frac{9 \pi r^{2} / 4}{4 \pi r^{2}} \times 100 \% \)
\(=\frac{900}{16} \%=56.25 \% \)
Let r and h be the radius and height of the cylinder respectively.
Given that, height h = 2 m, base area = 250 m2
Now, volume of a cylinder = \(\pi\)r h 2 cu. units
= base area x h
= 250 x 2 = 500 m3
Therefore, volume of the cylinder = 500 m3
Let r and h be the radius and height of the cone respectively.
Given that, volume of the cone = 11088 cm3
\(\frac { 1 }{ 3 } { \pi r }^{ 2 }h=11088\)
\(\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times { r }^{ 2 }\times 24=11088\)
\({ r }^{ 2 }=441\)
Therefore, radius of the cone r = 21 cm.
Let r1 and h1 be the radius and height of the cone - I and let r2 and h2 be the radius and height of the cone-II.
Given h2 = 2h1 = 2 and \(\frac { Volume\ of\ the\ cone\ I }{ Volume\ of\ the\ cone\ II } =\frac { 2 }{ 3 } \)
\(\frac { \frac { 1 }{ 3 } { \pi r }_{ 1 }^{ 2 }{ h }_{ 1 } }{ \frac { 1 }{ 3 } { \pi r }_{ 2 }^{ 2 }{ h }_{ 2 } } =\frac { 2 }{ 3 } \)
\(\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } \times \frac { { h }_{ 1 } }{ 2{ h }_{ 2 } } =\frac { 2 }{ 3 } \)
\(\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } =\frac { 4 }{ 3 } \text {gives} \frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 2 }{ \sqrt { 3 } } \)
Therefore, ratio of their radii = 2 : \(\sqrt3\)
Radius of well = 5 m
Depth of well = 14 m
Volume of earth taken out \(=\pi r^{2} h \)
\(=\frac{22}{7} \times(5)^{2} \times 14 \)
= 1100 m3
Now, it is spread to form an embankment, which is in the form of hollow cylinder
Inner radius = 5m
Width of embankment = 5 m
Outer radius = 5 + 5 = 10 m
height = h
Volume of hollow cylinder = \(\pi h\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)\)
\(\therefore \pi h\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)=1100 \)
\(\frac{22}{7} \times h\left(10^{2}-5^{2}\right)=1100 \)
height of the embankment
\(h=\frac{1100 \times 7}{22 \times 75}=4.67 \mathrm{~m}\)
Given circumference = 484 cm
\(2 \pi r =484
\)
\(2 \times \frac{22}{7} \times r =484
\)
\(r =\frac{484 \times 7}{44}=77 \mathrm{~cm}
\)
height h = 105 cm
Volume of cone \(=\frac{1}{3} \pi r^{2} h \text { cu. units }
\)
\(=\frac{1}{3} \times \frac{22}{7} \times 77 \times 77 \times 105
\)
= 652190 cm3
Let r1, r2 be the radii of two spheres
\(\text { Given } \frac{r_{1}}{r_{2}}=\frac{4}{7} \Rightarrow \mathrm{r}_{1}=\frac{4 r_{2}}{7}\)
Ratio of the volumes = \(\frac{V_{1}}{V_{2}}=\frac{\frac{4}{3} \pi r_{1}^{3}}{\frac{4}{3} \pi r_{2}^{3}}\)
\(=\frac{\left(\frac{4 r_{2}}{7}\right)^{3}}{r_{2}^{3}}=\frac{4^{3}}{7^{3}}\)
Ratio of volumes \(V_{1}: V_{2}=64: 343=\frac{64}{343}\)
Diameter of cylindrical pipe = 14 cm
Radius = 7 cm
Length of the pipe = Speed of the water
= 15 km = 15000 m
Length of the water tank = 50 m
Width of the water tank = 44 m
Height of the water tank = Water level
= 21 cm
= 0.21 cm
volume of water tank = l x b x h cu. units
= 50 x 44 x 0.21 = 462 m3
Volume of cylindrical Pipe = Volume of Rectangular tank
\(\frac{\pi r^{2} h}{} h =462
\)
\(\frac{22}{7} \times 0.07 \times 0.07 \times h =462
\)
\(\mathrm{h} =\frac{462 \times 7}{22 \times 0.07 \times 0.07}
\)
\(=\frac{3234}{0.1078}=30000
\)
Time required \(=\frac{30000}{15000}=2 \text { hrs. }
\)
Radius of conical flask = 'r' units
Height of conical flask = 'h' units
Volume of conical flask = Volume of water
\(=\frac{1}{3} \pi r^{2} h \text { cu. units }\)
Since, water is poured into the cylindrical flask
Volume of cylinder = Volume of water
\(\pi(\mathrm{xr})^{2} H=\frac{1}{3} \pi r^{2} h\)
[xr - radius of cylinder, H - height]
\(\mathrm{X}^{2} \mathrm{r}^{2} \mathrm{H}=\frac{r^{2}}{3} h\)
Height of the water in cylinder flask
\(\mathrm{H}=\frac{h}{3 x^{2}}\)
Height of the barrel = h = 7 cm
Diameter = 5 mm
radius r = \(\frac{5}{2}=2.5 \mathrm{~mm}=0.25 \mathrm{~cm}\)
Volume of cylindrical barrel = \(\pi r^{2} h\)
\(=\frac{22}{7} \times 0.25 \times 0.25 \times 7\)
= 1.375 cm2
Given 1.375 cm3 of ink is used for writing 330 words
Number of words that can be written with one - fifth of a litre
\({\left[\because 1000 \mathrm{~cm}^{3}=1 \mathrm{ltr} ; \frac{1}{5} \times 1000 \mathrm{~cm}^{3} ; 200 \mathrm{~cm}^{3}\right]}
\)
\(=\frac{330}{1.375} \times 200=48000 \text { words }
\)
Radius of hemispherical tank 'r' = 1.75 m
Volume of hemispherical tank \(=\frac{2}{3} \pi r^{3} \text { cu. units }\)
\(=\frac{2}{3} \times \frac{22}{7} \times(1.75)^{3}\)
= 11.225 m3
= 11225 litre
Given that cylindrical pipe empties the tank at the rate of 7 litre per second.
Time Required to empty the tank completely
\(=\frac{\text { Volume }}{\text { Rate }}
\)
\(=\frac{11225}{7}=1604 \sec (\mathrm{app})
\)
= 27 min (app)
Radius of hemisphere = r units
Radius of cone = Radius of hemisphere
Height of cone = Radius of hemisphere
Maximum volume of cone = \(\frac{1}{3} \pi r^{2} h \text { cu. units }\)
\(=\frac{1}{3} \pi\left(r^{2}\right) r=\frac{1}{3} \pi r^{3} \text { cu. units }\)
The solid is a cylinder
radius = r
height = h x 1 = h
C.S.A = 2\(\pi\)rh sq. units
Radius is double the height r = 2h
\( \text { C.S.A }=4 \pi h^{2} \text { as } 2 \pi(2 \mathrm{~h}) \mathrm{h} \)
\(\text{ Base area} =4 \pi h^{2}\ as \ \pi(2 h)^{2}\)
C.S.A = Base area
\(\left.\begin{array}{l} \text { ratio of curved } \\ \text { surface areas } \end{array}\right\}=\frac{2 \pi r(12)}{2 \pi r(5)}=\frac{12}{5}=12: 5\)
Ice cream Cone
h = r,
\(
l =\sqrt{h^{2}+r^{2}}=\sqrt{r^{2}+r^{2}}=\sqrt{2} r
\)
\(\text { Surface Area } =\pi r(l+r)
\)
\(=\pi r(\sqrt{2} r+r)
\)
\(=\pi r^{2}(\sqrt{2}+1)
\)
Area of base of cone = \(\pi\)r2
Surface area of the cone obtained in (5)
\(=(\sqrt{2}+1) \pi r^{2}=(\sqrt{2}+1) \text { times more }\)
| Sector | Cone |
| Radius | Slant height |
| Area | Curved surface area |
| Arc length | Circumference of the base |
surface area = 36 \(\pi\)
\(4 \pi r^{2}=36 \pi \Rightarrow r^{2}=9 \Rightarrow r=3\)
Two circles
Diameter of earth = 12756 kms
Radius \(r=\frac{12756}{2}=6378 \mathrm{kms}\)
Surface Area \(
=4 \pi r^{2}=4 \times \frac{22}{7} \times(6378)^{2}
\)
\(=\frac{3579741792}{7}\)
= 511391584.571 sq. km
No, it is not possible to get the hemisphere, when a sphere is cut along the small circle.
3 times.
2 hemispheres
Bucket, Table lamp.
No
\(
\text { Height } \alpha \frac{1}{(\text { radius })^{2}}
\)
\(\text { Height } =k\left(\frac{1}{(\text { radius })^{2}}\right)[\because k \text { - constant }]
\)
\(\text { Volume } =\pi r^{2} h
\)
\(=\pi r^{2}\left(\frac{k}{r^{2}}\right)=k \pi\)
(a) radius is halved then radius \(\frac{r}{2}\) height is h
\(
\text { Volume } =\pi r^{2} h=\pi\left(\frac{r}{2}\right)^{2} \mathrm{~h}=\frac{\pi r^{2} h}{4}
\)
\(=\frac{1}{4} \text { (Volume of original cylinder). }
\)
(b) When height is halved., then height = \(\frac{h}{2}\)
\(
\text { Volume } =\pi r^{2}\left(\frac{h}{2}\right)=\frac{\pi r^{2} h}{2}
\)
\(=\frac{1}{2} \text { (Volume of original cylinder). }\)
(i) height = Slant height
i.e., h = l = cone is not possible
(b) r = l = cone is not possible
(c) h = r = cone is possible.
Ratios of radii and heights are same
(i) Volume of cone B is greater as the radius is greater
(iii) Surface area of cone B > Surface area of cone A
\(\frac{\text { Volume of cone } A}{\text { Volume of cone B }}=\frac{12 \pi}{16 \pi}=3: 4\)
No
Sphere - Globe, Ball
Hemisphere - Left side of the brain
Not possible
