Given that, tent is a combination of a cylinder and a cone.

For conical portion,
Slant height, l = 2.8 m
Radius, r = Radius of cylinder  \(=\frac{\text { Diameter }}{2}=\frac{4}{2}=2 \mathrm{~m}\)
For cylindrical portion,
Radius, r = \(\frac{4}{2}=2 \mathrm{~cm}\)
Height, h = 2.1 m
\(\therefore\) Required surface area of the tent = CSA of cone + CSA of cylinder
\(\begin{aligned} & =\pi r l+2 \pi r h=\pi r(l+2 h) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times 2 \times(2.8+2 \times 2.1) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{44}{7}(2.8+4.2)=\frac{44}{7} \times 7=44 \mathrm{~m}^2 \end{aligned}\)
Now, cost of the canvas of the tent at the rate of Rs 500 per m² = Surface area \(\times\) Cost per m² = 44 \(\times\) 500 = Rs 22000

Given, a cubical block is surmounted by a hemisphere. Therefore, diameter of hemisphere must be equal to the side of cubical block and it is the greatest diameter of hemisphere.

For cubical portion,
Edge = 7 cm
For hemispherical portion,
Diameter = 7 cm
\(\therefore\)   Radius, r = \(\frac{7}{2}\)cm
Now, required surface area of solid = TSA of the cube + CSA of hemisphere - Area of circular base of hemisphere.
\(\begin{aligned} & =6 \times(\text { Edge })^2+2 \pi r^2-\pi r^2 \\ \end{aligned}\)
\(\begin{aligned} & =6 \times(7)^2+2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}-\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \\ \end{aligned}\)
\(\begin{aligned} & =294+11 \times 7-\frac{11 \times 7}{2} \\ \end{aligned}\)
\(\begin{aligned} & =294+77-\frac{77}{2} \end{aligned}\)
= 371 - 38.5 = 332.5 cm²

Given that,
Height (h) of the conical part = Height (h) of the cylindrical part = 2.4 cm
Diameter of the cylindrical part = 1.4 cm
Therefore, radius (r) of the cylindrical part = 0.7 cm

Slant height of the conical activity l=\(\sqrt{\left(Radius \ of \ base\right)^{2}+\left(Height \ of \ cylinder\right)^{2}}\)
=\(\sqrt{\left(0.7\right)^{2}+\left(2.4\right)^{2}}\)=\(\sqrt{0.49+5.76}\)
=\(\sqrt{6.25}\)=2.5 cm

Total surface area of the remaining solid will be = CSA of cylindrical part + CSA of conical part + Area of cylindrical base
= 2πrh + πrl + πr²
\(=2 \times \frac{22}{7} \times 0.7 \times 2.4+\frac{22}{7} \times 0.7 \times 2.5+\frac{22}{7} \times 0.7 \times 0.7\)
= 4.4 x 2.4 + 2.2 x 2.5 + 2.2 x 0.7
=10.56+5.50+1.54 = 17.60 cm²
The total surface area of the remaining solid to the nearest cm² is 18 cm².

Radius of hemisphere and cone = 3.5 cm
Height of conical portion = 9 cm
\(\therefore\)   Volume of the ice-cream in the cone
= Volume of hemisphere + Volume of cone
\(={{2}\over{3}}{\pi r}^{3}+{{1}\over{3}}{\pi r}^{2}h={{1}\over{3}}{\pi r}^{2}(2r+h)\)
\(={{1}\over{3}}\times{{22}\over{7}}\) x 3.5 x 3.5 x [ 2x 3.5 + 9 ]
\(={{1}\over{3}}\times{{22}\over{7}}\) x 3.5 x 3.5 x 16
= 205.33 cm³

Here, radius of cylindrical portion and conical portion = \(\frac{105}{2}\) m.
Slant height of conical portion = 53 m
Total canvas used in making the tent = C.S.A. of cylindrical portion + C.S.A. of conical portion 
=2πrh+πrl
=2x\(\frac{22}{7}\times{\frac{105}{2}}\times{3}+\frac{22}{7}\times{\frac{105}{2}}\times{53}\)
=990+8745=9735 m²

Volume of first cube =3x3x3
= 27 cm³ 
Volume of second cube = 4x4x4
= 64 cm³ 
Volume of third cube = 5x5x5
= 125 cm³ 
Volume of single cube = 27 + 64 + 125
= 216 cm³ 
Edge Of single cube = 6 cm
Hence, the edge of single cube so formed is 6 cm. 

Let ADC is a cone with vertical angle 60⁰
Now, cone is cut into two parts, parallel to its base at height 10cm.
Radius of larger end of the frustum=R₁
In ΔAO'C, tan 30⁰=\(\frac{O'C}{O'A}\)
\(\frac{1}{\sqrt{3}}\)=\(\frac{O'C}{20}\)=\(\frac{R_{1}}{20}\)⇒ R₁=\(\frac{20}{\sqrt{3}}\)cm

In ΔAOB, tan 30⁰=\(\frac{r_{1}}{OA}\)
⇒ \(\frac{1}{\sqrt{3}}\)=\(\frac{r_{1}}{10}\)⇒ r₁=\(\frac{10}{\sqrt{3}}\)cm
Volume of frustum=\(\frac{\pi{h}}{3}\)(R₁²+r₁²+R₁r₁)
=\(\frac{22}{7}\times{\frac{10}{3}}\left[\left(\frac{20}{\sqrt{3}}\right)^{2}+\left(\frac{10}{\sqrt{3}}\right)^{2}+\frac{20\times{10}}{\sqrt{3}\times{\sqrt{3}}}\right]\)
=\(\frac{22}{7}\times{\frac{10}{3}}\left[\frac{400}{3}+\frac{100}{3}+\frac{200}{3}\right]=\frac{22}{7}\times\frac{10}{3}\times{\frac{700}{3}}=\frac{22\times{10}\times{100}}{9}\)
A wire be formed having diameter \(\frac{1}{16}\)cm and length be h cm
Volume of wire so obtained=πr²h
=\(\frac{22}{7}\times{\left(\frac{1}{16\times{2}}\right)}^{2}\times{h}=\frac{22}{7}\times{\frac{1}{32}}\times\frac{1}{32}\times{h}\)
According to the question,
\(\frac{22\times{10}\times{100}}{9}=\frac{22}{7}\times{\frac{1}{32}}\times{\frac{1}{32}}\times{h}\)
⇒ \(\frac{22\times{10}\times{100}\times{32}\times{32}\times{7}}{22\times{9}}\)=h⇒ 796444.44 cm ⇒ h=7964.4 m

Let the depth of drainlet be h m
Volume of the earth taken out
=260 x 200 x h -240 x 180 x h
= 100 x h[26 x 20-24 x 18] 
Level is raised by 25 cm=\(\frac{25}{100}\)=\(\frac{1}{4}\)m
Volume of earth taken out = volume of the platform 
⇒ 100 x h[520-24x18]=240x180x\(\frac{1}{4}\)
⇒ (88)h x 100=60x180
⇒ h=1.227m=1.23m

Let the number of cubes be n
Volume of a cuboid=l x b x h
=30 x 30 x 42.6cm²
Volume of cube=3³
Volume of cuboid=nx volume of one cube
30 x 30 x 42.6=3 x 3 x 3 x n \(\Rightarrow\ n={30\times30\times42.6\over3\times3\times3}=1420\)

\(616\sqrt {5} \ cm^{2}\)

\({125\over3} \pi \ cm^{3}\)

Rs 2250

Let the height of the hall be h m.
Then, sum of areas of four walls=2(l+b)h m²
=2(20+16)h m²=72 h m²
Sum of the areas of the floor and the flat roof=(20x16+20x16)m²=640 m²
It is given that the sum of areas of four walls is equal to the sum of the areas of the floor and roof.
72h=640⇒h=\(\frac{640}{72}\)m=\(\frac{80}{9}\)m=8.88m
So, height of the hall=8.88m
Volume of the hall=(20x16x\(\frac{80}{9}\))m³=\(\frac{25600}{9}\)m³=2844.4 m³ 

Let radius of the base be r
r+h=37 cm
h=(37-r)cm
Total surface area=2πr(r+h)⇒1628=2x\(\frac{22}{7}\)xrx37
r=1628x\(\frac{7}{22\times{2}\times{37}}\)=7cm
h=(37-7)cm=30 cm
Volume of cylinder=πr²h=\(\frac{22}{7}\)x(7)²x30 cm³=22x7x30 cm³
=4620 cm³

Given, model is a combination of a cylinder and two cones. clearly, volume of the air will be equal to the sum of the volumes of two cones and one cylinder.
We have, diameter of the model, BC = ED = 3 cm
\(\therefore\) Radius of cone = radius of cylinder, \(r=\frac{3}{2}=1.5 \mathrm{~cm}\)
Length of cone, h₁ = 2 cm
Total length of the model, AF = 12 cm

\(\therefore\) Length of the cylinder, OO' = AF - (AO + O' F)
= 12 - (2 + 2) = 8 cm = h₂
Now, volume of the air inside the model = Volume of air inside (cone + cylinder + cone)
\(\begin{aligned} & =\left(\frac{1}{3} \pi r^2 h_1+\pi r^2 h_2+\frac{1}{3} \pi r^2 h_1\right) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{3} \pi r^2\left(h_1+3 h_2+h_1\right) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{3} \times \frac{22}{7} \times 1.5 \times 1.5(2+3 \times 8+2) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times \frac{2.25}{3} \times(2+24+2) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times 0.75 \times 28=22 \times 3=66 \mathrm{~cm}^3 \end{aligned}\)