The volume of water the bowl can contain
\(=\frac{2}{3} \pi r^{3}\)
\(=\frac{2}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5 \mathrm{~cm}^{3}=89.8 \mathrm{~cm}^{3}\)

100

(i) 968 cm²    
(ii) 1210 cm²

66 m²

21.6 m

28 cm

6.615 cm²

(i) Surface area of the sphere = \(4\pi { r }^{ 2 }\)
(ii) For cylinder
Radius of the base = r
Height  = 2r
\(\therefore\) Curved surface area of the cylinder
\(=2\pi \left( r \right) \left( 2r \right) =4\pi { r }^{ 2 }\)
(iii) Ratio of the areas obtained in (i) and (ii)
 \(=\frac { Surface\ area\ of\ the\ sphere }{ Curved\ surface\ area\ of\ the\ cylinder } \)
\(\\ =\frac { 4\pi { r }^{ 2 } }{ 4\pi { r }^{ 2 } } =\frac { 1 }{ 1 } =1:1.\)

Base radius (r) = 7 cm
Height (h)  = 24 cm
\(\therefore \)  Slant height (l) = \(\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
 \(=\sqrt { { \left( 7 \right) }^{ 2 }+{ \left( 24 \right) }^{ 2 } } =\sqrt { 49+576 } \)
\(\\ =\sqrt { 625 } =25cm\)
\(\therefore \)  Curved surface area of a cap = \(\pi rl\)
\(=\frac { 22 }{ 7 } \times 7\times 25=550{ cm }^{ 2 }\)
\(\therefore \)  Curved surface area of 10 caps
= 550 \(\times\) 10 = 5500 cm²
 Hence, the area of the sheet required to make 10 such caps is 5500 cm² .

r = 3 cm
h = 10.5 cm
\(\therefore \)  Cardboard required for 1 competitor
 \(=2\pi rh+\pi { r }^{ 2 }\)
\(\\ =2\times \frac { 22 }{ 7 } \times 3\times 10.5+\frac { 22 }{ 7 } { \left( 3 \right) }^{ 2 }\)
\(\\ =198+\frac { 198 }{ 7 } =198\left( 1+\frac { 1 }{ 7 } \right) \)
\(\\ =\frac { 198\times 8 }{ 7 } { cm }^{ 2 }\)
Cardboard required for 35 competitors
\(=\frac { 198\times 8 }{ 7 } \times 35{ cm }^{ 2 }=7920{ cm }^{ 2 }\)
Hence, 7920 cm² of cardboard was required to be bought for the competition.

h = 77 cm
2r = 4 cm
r = 2 cm
2R = 4.4 cm
R = 2.2 cm
(i) Inner curved surface area \(=2\pi rh\)
 \(=2\times \frac { 22 }{ 7 } \times 2\times 77=968{ cm }^{ 2 }\)
(ii) Outer curved surface area =
 \(=2\times \frac { 22 }{ 7 } \times 2.2\times 77=1064.8{ cm }^{ 2 }\)
(iii) Total surface area
 \(=2\pi Rh+2\pi rh+2\pi \left( { R }^{ 2 }-{ r }^{ 2 } \right) \)
\(\\ =1064.8+2\times \frac { 22 }{ 7 } \times 2\times 77+2\times \frac { 22 }{ 7 } \times \left\{ { \left( 2.2 \right) }^{ 2 }-{ \left( 2 \right) }^{ 2 } \right\} \)
\(\\ =1064.8+968+2\times \frac { 22 }{ 7 } \times \left( 4.84-4 \right) \)
\(\\ =1064.8+968+2\times \frac { 22 }{ 7 } \times 0.84\)
\(\\ =1064.8+968+5.28=2038.08{ cm }^{ 2 }.\)

(i) For herbarium 
l = 30 cm, b = 25 cm,
h = 25 cm
\(\therefore \)  Area of the glass = 2 (lb + bh + hl)
= 2[(30)(25) + (25)(25) + (25)(30)]
= 2[750 + 625 + 750] = 4250 cm².
(ii) The tape needed for all the 12 edges
= 4 (l + b + h)
= 4(30 + 25 + 25) = 320 cm.

Required number \(=\frac { 60\times 30\times 30 }{ 15\times 6\times 4 } =150\)

Number of cubes = \(\frac { { \left( 20 \right) }^{ 3 } }{ { \left( 5 \right) }^{ 3 } } =64\)

\(\frac { 2 }{ 3 } \times \left( 6\times 5\times 4 \right) 80{ m }^{ 3 }\)

v = 5 \(\times\) (6 \(\times\) 2 \(\times\) 1.5)

Volume = 15 \(\times\) 10 \(\times\) 8 = 1200 cm³

Length of the rod \(=\sqrt { { \left( 10 \right) }^{ 2 }+{ \left( 10 \right) }^{ 2 }+{ \left( 5 \right) }^{ 2 } } \)

Radius of the wall (r) =  \(\frac{10}{2}\) m = 5 m
Depth of the wall (h )= 10 m
ஃ  Volume of the earth dug out
= Volume of the well
=\(\pi r^2h=\pi (5)^2(10)=250\pi m^3\)
Radius of the well with embankment (R)
= 5 + 5 = 10 m
ஃ  Area of the embankment
= Area of the well with embankment-Area of the well without embankment
\(=\pi R^2-\pi r^2=\pi (R+r)(R-r)\)
\(= \pi(10+5)(10-5)=75 \pi m^2\)
ஃ Height of the embankment  =\(\frac{Volume\ of \ the\ earth\ dug\ out }{Area \ of \ the \ embankment}\)
\(=\frac{250 \pi}{75\pi}=\frac{10}{3}m\)