Not possible

Sphere - Globe, Ball
Hemisphere - Left side of the brain

No

(i) Volume of cone B is greater as the radius is greater

(iii) Surface area of cone B > Surface area of cone A
\(\frac{\text { Volume of cone } A}{\text { Volume of cone B }}=\frac{12 \pi}{16 \pi}=3: 4\)

Ratios of radii and heights are same

(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.

(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 = \(\frac{h}{2}\)
\( \text { Volume } =\pi r^{2}\left(\frac{h}{2}\right)=\frac{\pi r^{2} h}{2} \)
\(=\frac{1}{2} \text { (Volume of original cylinder). }\)

\( \text { Height } \alpha \frac{1}{(\text { radius })^{2}} \)
\(\text { Height } =k\left(\frac{1}{(\text { radius })^{2}}\right)[\because k \text { - constant }] \)
\(\text { Volume } =\pi r^{2} h \)
\(=\pi r^{2}\left(\frac{k}{r^{2}}\right)=k \pi\)

No

Bucket, Table lamp.

2 hemispheres

3 times.

No, it is not possible to get the hemisphere, when a sphere is cut along the small circle.

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

Two circles

surface area = 36 \(\pi\)
\(4 \pi r^{2}=36 \pi \Rightarrow r^{2}=9 \Rightarrow r=3\)

Sector	Cone
Radius	Slant height
Area	Curved surface area
Arc length	Circumference of the base

Area of base of cone = \(\pi\)r²
Surface area of the cone obtained in (5)
\(=(\sqrt{2}+1) \pi r^{2}=(\sqrt{2}+1) \text { times more }\)

h = r,
\( l =\sqrt{h^{2}+r^{2}}=\sqrt{r^{2}+r^{2}}=\sqrt{2} r \)
\(\text { Surface Area } =\pi r(l+r) \)
\(=\pi r(\sqrt{2} r+r) \)
\(=\pi r^{2}(\sqrt{2}+1) \)

Ice cream Cone

\(\left.\begin{array}{l} \text { ratio of curved } \\ \text { surface areas } \end{array}\right\}=\frac{2 \pi r(12)}{2 \pi r(5)}=\frac{12}{5}=12: 5\)

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

The solid is a cylinder
radius = r
height = h x 1 = h
C.S.A = 2\(\pi\)rh sq. units

Radius of hemisphere = r units
Radius of cone = Radius of hemisphere
Height of cone = Radius of hemisphere
Maximum volume of cone = \(\frac{1}{3} \pi r^{2} h \text { cu. units }\)
\(=\frac{1}{3} \pi\left(r^{2}\right) r=\frac{1}{3} \pi r^{3} \text { cu. units }\)

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)

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 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
\(\mathrm{H}=\frac{h}{3 x^{2}}\)

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 \(=\frac{30000}{15000}=2 \text { hrs. } \)

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 \(V_{1}: V_{2}=64: 343=\frac{64}{343}\)

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

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
\(h=\frac{1100 \times 7}{22 \times 75}=4.67 \mathrm{~m}\)

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 : \(\sqrt3\)

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 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 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 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 = \(\sqrt{5}: 9\)

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

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

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

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 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.

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 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

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 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, 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.

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, 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, 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