Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere (see Figure). The radius BO of the hemisphere (as well as of the cone) \(=\frac{1}{2} \times 4 \mathrm{~cm}=2 \mathrm{~cm} .\)
So, volume of the toy = \(\begin{aligned} & =\frac{2}{3} \pi r^3+\frac{1}{3} \pi r^2 h \end{aligned}\)
\(\begin{aligned} & =\left[\frac{2}{3} \times 3.14 \times(2)^3+\frac{1}{3} \times 3.14 \times(2)^2 \times 2\right] \mathrm{cm}^3=25.12 \mathrm{~cm}^3 \end{aligned}\)
Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of the base of the right circular cylinder = HP = BO = 2 cm, and its height is EH = AO + OP = (2 + 2) cm = 4 cm
So, the colume required = volume of the right circular cylinder - volume of the toy
= 3.14 \(\times\)2² \(\times\)4 - 25.12) cm³
= 25.12 cm³
hence, the required difference of the two volumes = 25.12 cm³.

The volume of air inside the shed (when there are no people or machinery) is given by the volume of air inside the cuboid and inside the half cylinder, taken together.
Now, the length, breadth and height of the cuboid are 15 m, 7 m and 8 m, respectively. Also, the diameter of the half cylinder is 7 m and its height is 15 m.
So, the required volume = volume of the cuboid \(+\frac{1}{2}\) volume of the cylinder 
\(=\left[15 \times 7 \times 8+\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 15\right] \mathrm{m}^{3}=1128.75 \mathrm{~m}^{3}\)
Next, the total space occupied by the machinery = 300 m³
And the total space occupied by the workers = 20 × 0.08 m³ = 1.6 m³
Therefore, the volume of the air, when there are machinery and workers
= 1128.75 – (300.00 + 1.60) = 827.15 m³

The total surface area of the cube = 6 × (edge)² = 6 × 5 × 5 cm² = 150 cm².
Note that the part of the cube where the hemisphere is attached is not included in the surface area.
So, the surface area of the block = TSA of cube – base area of hemisphere + CSA of hemisphere
\(=150-\pi r^{2}+2 \pi r^{2}=\left(150+\pi r^{2}\right) \mathrm{cm}^{2} \)
\(=150 \mathrm{~cm}^{2}+\left(\frac{22}{7} \times \frac{4.2}{2} \times \frac{4.2}{2}\right) \mathrm{cm}^{2} \)
\(=(150+13.86) \mathrm{cm}^{2}=163.86 \mathrm{~cm}^{2}\)

Since the inner diameter of the glass = 5 cm and height = 10 cm,
the apparent capacity of the glass \(=\pi r^{2} h\)
\(=3.14 \times 2.5 \times 2.5 \times 10 \mathrm{~cm}^{3}=196.25 \mathrm{~cm}^{3}\)
But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass.
i.e., it is less by \(\frac{2}{3} \pi r^{3}=\frac{2}{3} \times 3.14 \times 2.5 \times 2.5 \times 2.5 \mathrm{~cm}^{3}=32.71 \mathrm{~cm}^{3}\)
So, the actual capacity of the glass = apparent capacity of glass – volume of the hemisphere
= (196.25 – 32.71) cm³
=163.54 cm³

Let r₁ and r₂ denote the radius of the base of the cyclinder and cone respectively.
Then r₁ = r₂ = 8 cm. Let h₁ and h₂ cm be the heights of the cylinder and the cone respectively.

Then h₁ =240 cm and h₂ = 36 cm
Now volume of the cylinder = \(\pi r^{ 2 }_{ 1 }h_{ 1 }cm^{ 3 }\)
= \((\pi \times 8\times 8\times 240)cm^{ 3 }\)
\((\pi \times 64\times 240)cm^{ 3 }\)
Volume of the cone = \(\frac { 1 }{ 3 } \pi r^{ 2 }_{ 2 }h_{ 2 }cm^{ 3 }\)
=\(\left( \frac { 1 }{ 3 } \pi \times 8\times 8\times 36 \right) cm^{ 3 }\)
\(=\left( \frac { 1 }{ 3 } \pi \times 64\times 36 \right) cm^{ 3 }\)
\(\therefore \) Total volume of the iron 
= volume of cylinder + volume of cone 
= \(\left( \pi \times 64\times 240+\frac { 1 }{ 3 } \pi \times 64\times 36 \right) cm^{ 3 }\)
= \(\pi +64\times (240+12)cm^{ 3 }\)
= 22X64X36 cm³
Hence, total weight of the pillar 
= volume X weight per cm³
= (22 x 64 x 36) 10 g = 506880 g.
= 506.88 kg. 

Let the level of water raise in the tank in x hours = 15000x metres
Length of the water flow in x hours = 15000x metres
 Diameter of the pipe = 14 cm
radius = \(\frac { 14 }{ 2 } =7cm=\frac { 7 }{ 100 } cm\)
Volume of water = \(\pi r^{ 2 }h\)
Volume of water flow in x hours in the pond  =  \(\frac { 22 }{ 7 } \times \frac { 7 }{ 100 } \times \frac { 7 }{ 100 } \times 15000x\)
According to the question 
\(50\times 44\times \frac { 21 }{ 100 } =\frac { 22 }{ 7 } \times \frac { 7 }{ 100 } \times \frac { 7 }{ 100 } \times 15000x\)
\(\Rightarrow 22\times 21=\frac { 154\times 15 }{ 10 } x\Rightarrow x=\frac { 22\times 21\times 10 }{ 154\times 15 } =2\)
Hence, the level of water in the pond will rise by 21 cm in 2 hours. 

Volume of hollow sphere = volume of cone
 \(\frac{4}{3}\pi (R^3 - r^3)=\frac{1}{3}\pi r^2h\)
\(\Rightarrow \ \frac{4}{3}(5^3 - r^3)=\frac{1}{3}\times(7)^2 \times8\)
\(\Rightarrow \ \frac{4}{3}(125 - r^3)=\frac{1}{3}\times49 \times8\)
\(​​\Rightarrow \ 500 -4r^3 = 49\times 8\)
\(\Rightarrow \ -4r^3 = 392 - 500\)
\(\Rightarrow \ -4r^3 = -108\)
\(\Rightarrow \ r^3 =27\)
\(\Rightarrow \ r =3\) cm
Diameter, d = 6 cm

Vertical angle of conical part 90⁰
\(\Rightarrow \) Semi-vertical angle of conical part = 45⁰
In \(\triangle \) AOB \(\angle 1=45^{ 0 }\Rightarrow OA=OB=\frac { 4.3 }{ 2 } m\)
Slant height of cone = \(\sqrt { r^{ 2 }+h^{ 2 } } =\sqrt { \left( \frac { 4.3 }{ 2 } \right) ^{ 2 } } =\frac { 4.3 }{ 2 } \sqrt { 2 } m\)
= \(\sqrt { 2\times \frac { (4.3)^{ 2 } }{ 2 } } =\frac { 4.3 }{ 2 } \sqrt { 2 } m\)
Surface area of solid = S.A. Of cone + S.A. of cylinder 
= \(\pi rl+2\pi rH=\pi r(l+2H)\)
= \(\frac { 22 }{ 7 } \times \frac { 4.3 }{ 2 } \left( \frac { 4.3 }{ 2 } \sqrt { 2 } +2\times 3.8 \right) =71.896m^{ 2 }\)
Volume of the building = volume of the cylinder + volume of the cone 
= \(\pi r^{ h }+\frac { 1 }{ 3 } \pi r^{ 2 }H=\pi \times r^{ 2 }\left( 3.8\times \frac { 1 }{ 3 } \times \frac { 4.3 }{ 2 } \right) \)
\(=\frac { 22 }{ 7 } \times \frac { 4.3 }{ 2 } \times \frac { 4.3 }{ 2 } \left( 3.8+\frac { 4.3 }{ 6 } \right) =65.61m^{ 2 }\)

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
\(=1127.28 \times \frac{30}{100}=1127.28 \times \frac{3}{10}=338.184 \approx 338 \mathrm{~cm}^3\)

Total cost of covering the walls with wallpaper =  Rs 362.50
Cost of paper per m = Rs 1.25
\(\Rightarrow \) Length of paper = \(\frac { Rs\quad 362.50 }{ Rs\quad 1.25 } =290m\)
Area of paper required = area of 4 walls =area Of 2 doors =area of I almirah 
\(\Rightarrow 290\times \frac { 40 }{ 100 } =\left[ 2h(8+6)-2\times 2\times 1-3\times 2 \right] m^{ 2 }\)
\(\Rightarrow 29\times 4=(28h-4-6)m^{ 2 }\)
\(\Rightarrow \) 116=28h-10\(\Rightarrow \) 126\(\Rightarrow \) h= \(\frac { 126 }{ 28 } =4.5m\)