Radius of hemisphere portion = Radius of cylindrical portion = \(\frac{7}{2}m\)
Total length of the tanker = 20 m
\(\therefore\) Length of cylindrical portion = 20 - 7 = 13 m

Now, Capacity (volume) of the oil tanker
= Volume of cylinder + 2 x Volume of hemisphere
= \(\pi r^2h+2\times\frac{2}{3}\pi r^3\)
= \(\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\times13+2\times\frac{2}{3}\times\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\times\frac{7}{2}\)
= 500.5 + 179.67 = 680.17 m³

Total height of the haystack = 31 cm
Height of conical portion = 15 cm
Radius of conical portion = 20 cm
\(\therefore\) Height of cone frustum = 31 - 15 = 16 cm
Slant height of conical portion = \(\sqrt{(20)^2+(15)^2}\)
= \(\sqrt{400+225}\)
= \(\sqrt{625}\) = 25 cm
Slant heigt of the cone frustum
= \(\sqrt{h^2+(R-r)^2}=\sqrt{16^2+(20-8)^2}\)
= \(\sqrt{256+144}=\sqrt{400}\) = 20 cm
Now, the lateral surface area of the haystack
= \(\pi rl+\pi l(R+r)\)
= 3.14 x 20 x 25 + 3.14 x 20 x 28
= 1570 + 1758.4 = 3328.4 cm² 

Let height of the cone be x cm
Radius of the cone = 3.5 cm = radius of hemisphere

Volume of the toy = \({{1}\over{3}}{\pi r}^{2}h+{{2}\over{3}}{\pi r}^{3}\)
\(166{{5}\over{6}}={{1}\over{3}}\times{{22}\over{7}}{(3.5)}^{2}\times x+{{2}\over{3}}\times{{22}\over{7}}\times{(3.5)}^{3}\)
\({{1001}\over{6}}={{269.5}\over{21}}x+{{1886.5}\over{21}}\)
\(\Rightarrow\)   \({{1001}\over{6}}-{{1886.5}\over{21}}={{269.5}\over{21}}x\)
\(\Rightarrow\)    \({{3234\times21}\over{42\times269.5}}=x\Rightarrow 6\) cm²
\(\therefore\)   Height of toy = 6 cm + 3.5 cm = 9.5 cm
Surface area of hemicspherical part = \({2\pi r}^{2}\)
= \(2\times{{22}\over{7}}\times3.5\times3.5=77\) cm²
Cost to paint at a rate of Rs 10 per cm²
= 77 x 10 = Rs 770

Radius of hemispherical bowl r = 9cm
\(\therefore \) Volume of water in the bowl = volume of hemispherical bowl
= \(\frac { 2 }{ 3 } \pi r^{ 3 }=\frac { 2 }{ 3 } \pi \times (9)^{ 3 }=486\pi cm^{ 3 }\)
Radius of culindrical vessel = R= 6 cm
Let height of water in the vessel be h cm
\(\therefore \) Volume of water = \(\pi r^{ 2 }h=\pi (6)^{ 2 }h=36\pi hcm^{ 3 }\)
\(\Rightarrow 36\pi h=480\pi \Rightarrow h=13.5cm\)

Given, side of the cube = Diameter of the hemisphere = l units
\(\therefore\) Radius of the hemisphere, \(r=\frac{l}{2}\) units

Now, required surface area of the remaining solid = TSA of the cube + CSA of hemisphere - Area of circular base of hemisphere
\(\begin{aligned} & =6 \times(\text { Edgc })^2+2 \pi r^2-\pi r^2 \\ \end{aligned}\)
\(\begin{aligned} & =6 \times l^2+2 \pi \times\left(\frac{l}{2}\right)^2-\pi\left(\frac{l}{2}\right)^2 \\ \end{aligned}\)
\(\begin{aligned} & =6 l^2+2 \pi \times \frac{l^2}{4}-\pi \frac{l^2}{4}=6 l^2+\pi \frac{l^2}{4} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{l^2}{4}(\pi+24) \text { sq units } \end{aligned}\)

Diameter d=9m ⇒r=\(\frac{9}{2}\)m
Volume of wheat=volume of cone
=\(\frac{1}{3}\times{\frac{22}{7}}\times{\frac{9}{2}}\times{\frac{9}{2}}\times{3.5}\)m³
=74.25 m³
Area of canvas required=C.S.A. =πrl
l=\(\sqrt{r^{2}+h^{2}}=\sqrt{\left(\frac{9}{2}\right)^{2}+\left(3.5\right)^{2}}\)
=570m
Area of canvas=3.14x4.5x5.7 m²
80.541 m²

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

Radius of the cylinder=radius of the sphere=r
and height of the cylinder=diameter of the sphere
=2xradius=2xr=2r

Curved surface area of the cylinder
=2πrh=2πrx2r=4πr² sq.units

Surface area=154 cm²
\(\Rightarrow\ \ 4\pi r^2=154\ \ \Rightarrow r^2={154\times7\over4\times22}\)
\(\Rightarrow\ \ r=\sqrt{7\times7\over2\times2}={7\over2}=3.5\)

Diameter of tent = 24m radius = 12m
Slant height=13m
Curved surface area =\(\pi r l=\frac{22}{7} \times 12 \times=\frac{3432}{7} m^2\)

Volume of container = \(10459\frac { 3 }{ 7 } cm^{ 3 }=\frac { 73216 }{ 7 } cm^{ 3 }\)
\(\Rightarrow \frac { 1 }{ 3 } \pi h(r^{ 2 }_{ 1 }+r^{ 2 }_{ 2 }+r_{ 1 }+r_{ 2 })=\frac { 73216 }{ 7 } \)
\(\Rightarrow \frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times h(20^{ 2 }+8^{ 2 }+20\times 8)=\frac { 73216 }{ 7 } \)
\(\Rightarrow \frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times h[400+64+160]=\frac { 73216 }{ 7 } \)
\(\Rightarrow \frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times h(624)=\frac { 73216 }{ 7 } \)
\(\Rightarrow h=\frac { 73216\times 3\times 7 }{ 7\times 22\times 624 } =16cm\)
l= \(\sqrt { h^{ 2 }+(r_{ 1 }+r_{ 2 }) } ^{ 2 }\)
= \(\sqrt { (16)^{ 2 }+(20-8)^{ 2 } } =\sqrt { 256+144 } =\sqrt { 400 } =20\)
Area of metal sheet used = \(\pi (r_{ 1 }+r_{ 2 })+\pi r^{ 2 }\)
= \(\frac { 22 }{ 7 } \times 20(20+8)+\frac { 22 }{ 7 } \times (8)^{ 2 }\)

\(\frac { 22 }{ 7 } \times 20\times 28+\frac { 22 }{ 7 } \times 64=\frac { 13728 }{ 7 } cm^{ 2 }\)
Cost of metal sheet used = Rs \(\frac { 13728 }{ 7 } \) X 1.40 = Rs 2745.60

Radii and height of the frustum i.e., bucket are
R= 20 cm ,r = 8cm , h= 16 cm 
Volume of bucket = \(\frac { 1 }{ 3 } \pi h(R^{ 2 }+r^{ 2 }+Rr)\)
= \(\frac { 1 }{ 3 } \times 27\times 16(8^{ 2 }+20^{ 2 }+8\times 20)\)

\(\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 16(64+100+160)\)
\(\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 16\times 624cm^{ 3 }=10459.42cm^{ 3 }\)
Area of sheet required = area of the bucket 
\(=\pi (R+r)+\pi r^{ 2 }\)
Where l =  r\(\sqrt { h^{ 2 }+(R-r)^{ 2 } } \)
\(l=\sqrt { (16)^{ 2 }+(20-8)^{ 2 } } =\sqrt { 256+(12)^{ 2 } } =\sqrt { 256+144 } =\sqrt { 400 } =20\)
Area of sheet required =  \(\pi (R+r)+\pi r^{ 2 }\)
= \(\frac { 22 }{ 7 } \left[ 20(20+8)+64 \right] =\frac { 22 }{ 7 } (20\times 28+64)\)
\(\frac { 22 }{ 7 } \left( 560+64 \right) =\frac { 22 }{ 7 } \times 624=1961.142cm^{ 2 }\)
Cost of 100 cm² of metal sheet = Rs 15
Cost of 1 cm² of metal sheet = Rs \(\frac { 15 }{ 100 } \)
Cost of 1961.142 cm² = Rs \(\frac { 15 }{ 100 } \times 1961.142=Rs\quad 294.17\)​​​​​​​

Internal radius of pipe = 1cm
Rate = 80 cm/s
Quality of liquid passed in 1 sec = \(\pi r^{ 2 }h\)
\(=\frac { 22 }{ 7 } \times 1\times 1\times 80cm^{ 3 }=\frac { 1760 }{ 7 } cm^{ 3 }\)
Quantity of liquid passed in \(\frac { 1 }{ 2 } \) hr i.e. 30 min = \(\left( \frac { 1760 }{ 7 } \times 1800 \right) cm^{ 3 }\)
Volume of tank filled in \(\frac { 1 }{ 2 } \) hour = \(\pi \times (40)^{ 2 }\times h\)
Volume of tank filled in \(\frac { 1 }{ 2 } \) hour = quantity of water passed through pipe in  \(\frac { 1 }{ 2 } \) hour
\(\Rightarrow \frac { 22 }{ 7 } \times (40)^{ 2 }\times h=\frac { 1760 }{ 7 } \times 1800\Rightarrow h=90cm\)

Given, one gulabjamun is a combination of a cylinder and two hemispheres. Here, total length of one gulabjamun = 5 cm and diameter = 2.8 cm

\(\therefore\) Radius of cylindrical part = Radius of hemispherical part
\(=r=\frac{2.8}{2}=1.4 \mathrm{~cm}\)
Height of cylindrical part,
h = PQ - (PR + SQ) = 5 - (1.4 + 1.4)
= 5 - 2.8 = 2.2 cm
\(\therefore\) Volume of one gulabjamun = 2 \(\times\) Volume of hemispherical part + Volume of cylindrical part
\(\begin{aligned} & =2 \times\left[\frac{2}{3} \pi r^{3}\right]+\pi r^2 h=\frac{4}{3} \pi r^{3}+\pi r^2 h \\ \end{aligned}\)
\(\begin{aligned} & =\pi r^2\left[\frac{4 r}{3}+h\right]=\frac{22}{7} \times 1.4 \times 1.4\left[\frac{4}{3} \times 1.4+2.2\right] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times \frac{14}{10} \times \frac{14}{10}\left[\frac{4}{3} \times \frac{14}{10}+\frac{22}{10}\right] \\ \end{aligned}\)
\(\begin{aligned} & =22 \times \frac{1}{5} \times \frac{7}{5}\left[\frac{28}{15}+\frac{11}{5}\right]=\frac{154}{25} \times \frac{61}{15}=\frac{9394}{375} \mathrm{~cm}^3 \end{aligned}\)
Now, volume of 45 gulabjamuns
\(=45 \times \frac{9394}{375}=\frac{3 \times 9394}{25}=\frac{28182}{25}=1127.28 \mathrm{~cm}^3\)
Since, one gulabjamun contains sugar syrup upto about 30% of its volume.
Hence, quantity of syrup found in 45 gulabjamuns
\(=1127.28 \times \frac{30}{100}=1127.28 \times \frac{3}{10}=338.184 \approx 338 \mathrm{~cm}^3\)

Radius = 14 m
Height of cylinder = 3m
Height of cone = (13.5-  3) = 10.5m 
Slant height of cone = \(\sqrt { r^{ 2 }+h^{ 2 } } \)
\(=\sqrt { (14)^{ 2 }+(10.5)^{ 2 } } =\sqrt { 306.25 } =17.5m\)

Area to be painted = \(2\pi rh+\pi rl=\pi r(2h+l)\)
= \(\frac { 22 }{ 7 } \times 14(2\times 3+17.5)\)
= 44(6+17.5) = 1034 m²
Cost of painting 1 m² =  Rs 2
Cost of painting 1034 m² = Rs 2 x 1034 = Rs 2068