Height of cylinder = 20 cm
radius of cylinder = 3.5 cm = radius of each hemisphere
Total surface area of the article = 2X C.S.A of a hemisphere
= \(2\times 2\pi r^{ 2 }+2\pi rh\rightleftharpoons 2\pi r(2r+h)\)
= \(2\times \frac { 22 }{ 7 } \times 3.5\left[ 2\times 3.5+20 \right] \)
= 44 x 0.5[7+20] = 44 x 0.5 x 27 cm² = 594.0 cm²

Here, the decorative block is a combination of a cube and a hemisphere.
For cubical portion,
Each edge = 6 cm
For hemispherical portion,
Diameter = 2.1 cm
\(\therefore \quad \text { Radius, } r=\frac{2.1}{2} \mathrm{~cm}\)
Now, total surface area of the cube = 6 x (£dge)2= 6 x6 x6= 216cm²
Here, the part of the cube where the hemisphere is attached, is not included in the surface area.
So, the total surface area of the decorative block = Total surface area of cube - Area of base of hemisphere + Curved surface area of hemisphere
\(=216-\pi r^{2}+2 \pi r^{2}=216+\pi r^{2}=216+\frac{22}{7} \times \frac{2.1}{2} \times \frac{2.1}{2}\)
=216+3.465 = 219.465cm²
Clearly, the total area to be painted = Total surface area of the decorative block - Area of base of cube
= 219.465 _62 = 219.465 -36 = 183.465 cm²

Cone is cut by plane PB and PQDB is a frustum.
OA = AC = 10 cm, AB = r₁, CD = r₂
Vertical angle = 60°, so Semi vertical angle = 30°

In right-angled triangle OAB,
\({{AB}\over{OA}}\) = tan 30°
\(\Rightarrow\)   r₁ = \({{10}\over{\sqrt{3}}}\)cm
In right-angled triangle OCD,
\({{CD}\over{OC}}\) = tan 30°
\(\Rightarrow\)     r₂ = \({{20}\over{\sqrt{3}}}\)cm.
Frustum is drawn into wire of diameter \({{1}\over{12}}\) cm and length x cm ( say )
\(\therefore\)  Volume of wire = volume of frustum
\(\Rightarrow\)  \({\pi r}^{2}\)x = \({{1}\over{3}}{\pi h}({ r }_{ 1 }^{ 2 } + { r }_{ 2 }^{ 2 }+{ r }_{ 1 }.{ r }_{ 2 })\)
\(\Rightarrow\)  \({\pi}\left( {{1}\over{24}} \right)^{2}\times x = {{\pi \times10}\over{3}}\left[ { \left( \frac { 10 }{ \sqrt { 3 } } \right) }^{ 2 }+{ \left( \frac { 20 }{ \sqrt { 3 } } \right) }^{ 2 } +\frac { 10 }{ \sqrt { 3 } } .\frac { 20 }{ \sqrt { 3 } } \right]\)
\(\Rightarrow\)  \({{1}\over{576}}\times{x}={{10}\over{3}}\left( {{100}\over{3}}+{{400}\over{3}}+{{200}\over{3}} \right)\)
= \({{10}\over{3}}\times{{700}\over{3}}\)
\(\Rightarrow\)  x = \({{79000}\over{9}}\times 576\)
= 7000 X 64 cm
= 448000 cm
\(\Rightarrow\) length of wire = \({{448000}\over{100000}}\) km = 4.48 km

Surface area of remaining block =
surface area of cuboid + curved surface area of cylinder - surface area of two circular end of cylinder hole
= 2[lb + bh + lh] + \(2\pi rh-2\pi r^2\)
= 2[15 x 10 + 10 x 5 + 15 x 5] + 2 x \(\frac{22}{7}\times\frac{7}{2}\) x 5 - 2 x \(\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\)
= 2(150 + 50 + 75) + 110 - 77 cm²
= 583 cm² 

Since the mould is in the shape of a frustum of a cone, the quantity (volume) of molasses that can be poured into it \(=\frac{\pi}{3} h\left(r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right)\)
where r₁ is the radius of the larger base and r₂ is the radius of the smaller base.
\(=\frac{1}{3} \times \frac{22}{7} \times 14\left[\left(\frac{35}{2}\right)^{2}+\left(\frac{30}{2}\right)^{2}+\left(\frac{35}{2} \times \frac{30}{2}\right)\right] \mathrm{cm}^{3}=11641.7 \mathrm{~cm}^{3} .\)
It is given that 1 cm3 of molasses has mass 1.2g. So, the mass of the molasses that can be poured into each mould = (11641.7 × 1.2) g
= 13970.04 g = 13.97 kg
= 14 kg (approx.)

Given, internally dimensions of cistern
=150 cm x 120 cm x 110 cm
ஃ Volume of cistern = 150 x 120 x 110
=1980000 cm³
Given, volume of water in cistern = 129600 cm³
ஃ Remaining volume of cistern to be filled = 1980000 -129600 = 1850400 cm³
Let required number of bricks be n.
ஃ    Volume of each brick = 22.5 x 7.5 x 6.5
= 1096.875 cm³
Water absorbed by 1 brick\(=\frac { 1096.875 }{ 17 } { cm }^{ 3 }\)
Then, water absorbed by n bricks=\(=n\left( \frac { 1096.875 }{ 17 } \right) { cm }^{ 3 }\)
Now, volume of remaining water + Water absorbed by n bricks 
= n X Volume of each brick
\(\Rightarrow \ 1850400+n\left( \frac { 1096.875 }{ 17 } \right) =n(1096.875)\\ \Rightarrow \ n\times 1096.875\times \frac { 16 }{ 17 } =1850400\\ \Rightarrow \ n=\frac { 1850400\times 17 }{ 1096.875\times 16 } =1792.4102\approx 1792(approx)\)
Hence, 1792 bricks can be put in a cistern.

Given, radius and height of cylinder, r = 3.5 cm and h = 14cm.
Also, radius of sphere, R=\(\frac { 7 }{ 12 } cm\)
Now, volume of cylinder \(=\pi { r }_{ 1 }^{ 2 }h=\frac { 22 }{ 7 } \times { (3.5) }^{ 2 }\times 14\\ \quad \quad \quad =22\times 12.25\times 2=539\quad { cm }^{ 3 }\)
Now, volume of solid metallic balls \(=\frac { 4 }{ 3 } \pi { r }_{ 2 }^{ 3 }\)
\(=\frac { 4 }{ 3 } \pi { \left( \frac { 7 }{ 12 } \right) }^{ 3 }=\frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times \frac { 7\times 49 }{ 12\times 12\times 12 } =\frac { 11\times 49 }{ 3\times 3\times 6\times 12 } { cm }^{ 3 }\\ \therefore \quad Number\quad of\quad balls=\frac { Volume\quad of\quad cylinder }{ Volume\quad of\quad solid\quad metallic\quad balls } \\ =\frac { \frac { 539 }{ 11\times 49 } }{ 3\times 3\times 6\times 12 } =\frac { 539\times 3\times 3\times 6\times 12 }{ 11\times 49 } =648\)
Hence, the required number of balls so formed are 648.

Capacity of the glass = \(\frac { 1 }{ 3 } \pi { r }h\left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right) \)
=\(\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 14\left( { 6 }^{ 2 }+{ 2 }^{ 2 }+6\times 2 \right) \)
=\(\frac { 44 }{ 3 } \times 52\)
= 762.67 cm³

Since 14 cm, 12 cm and 10 cm are inner dimensions of the closed box.
Capacity of the box = 14 x 12 x 10
= 1680 cm³

\(2100 \ \pi { m }^{ 2 }\)

373.41 cm²

Here, h=40 cm, circumference = 22 cm
2\(\pi\)r=22
\(\Rightarrow \ r=\frac { 22\times 7 }{ 2\times 22 } \)
\(\Rightarrow \ r=\frac { 7 }{ 2 } \)
= 3.5cm

(i) (a)
(ii) (d): Slant height of conical cavity \(l=\sqrt{h^{2}+r^{2}}\)
\(=\sqrt{(24)^{2}+(7)^{2}}=\sqrt{576+49}=\sqrt{625}=25 \mathrm{~cm}\)
(iii) (b): Curved surface area of conical cavity = \(\pi r l\)
\(=\frac{22}{7} \times 7 \times 25=550 \mathrm{~cm}^{2}\)
(iv) (c) : External curved surface area of cylinder 
\(=2 \pi r h=2 \times \frac{22}{7} \times 7 \times 24=1056 \mathrm{~cm}^{2}\)
(v) (a): Volume of conical cavity \(=\frac{1}{3} \pi r^{2} h\)
\(=\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24=1232 \mathrm{~cm}^{3}\)

We have, radius of each coin = 3.5 cm
\(=\frac{35}{10} \mathrm{~cm}=\frac{7}{2} \mathrm{~cm}\)
Thickness of each coin \(=0.5 \mathrm{~cm}=\frac{1}{2} \mathrm{~cm}\)
So,height of cylinder made by Meera (h1) \(=12 \times \frac{1}{2}=6 \mathrm{~cm}\)
and height of cylinder made by Dhara (h2)
\(=8 \times \frac{1}{2}=4 \mathrm{~cm}\)
(i) (b): Curved surface area of cylinder made by Meera \(=2 \times \frac{22}{7} \times \frac{7}{2} \times 6=132 \mathrm{~cm}^{2}\)
(ii) (b): Required ratio
\(=\frac{\text { Curved surface area of cylinder made by Meera }}{\text { Curved surface area of cylinder made by Dhara }}\)
\(=\frac{2 \pi r h_{1}}{2 \pi r h_{2}}=\frac{h_{1}}{h_{2}}=\frac{6}{4}=\frac{3}{2} \text { i.e., } 3: 2\)
(iii) (a): Volume of cylinder made by Dhara \(=\pi r^{2} h_{2}\)
\(=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 4=154 \mathrm{~cm}^{3}\)
(iv) (c): Required ratio
\(=\frac{\text { Volume of cylinder made by Meera }}{\text { Volume of cylinder made by Dhara }} \)
\(=\frac{\pi r^{2} h_{1}}{\pi r^{2} h_{2}}=\frac{h_{1}}{h_{2}}=\frac{6}{4}=\frac{3}{2} \text { i.e., } 3: 2\)
(v) (a): When two coins are shifted from Meera's cylinder to Dhara's cylinder, then length of both cylinders become equal. So, volume of both cylinders become equal