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 = \(\pi r^{ 2 }h=\pi (6)^{ 2 }h=36\pi hcm^{ 3 }\)
\(\Rightarrow 36\pi h=480\pi \Rightarrow h=13.5cm\)

Diameter of roller = 84 cm \(\Rightarrow \) radius of roller = 42 cm
Area levelled in one revolutions = \(2\times \frac { 22 }{ 7 } \times 120\times 42cm^{ 2 }=\frac { 221760 }{ 7\times 10000 } m^{ 2 }\)
Area levelled in 500 revolutions = \(500\times \frac { 221760 }{ 7 } cm^{ 2 }=500\times \frac { 221760 }{ 7\times 10000 } m^{ 2 }\)
Cost of levelling 1 m² = 30 paise
Total cost of levelling = Rs \(\frac { 30 }{ 100 } \times 500\times \frac { 221760 }{ 7\times 10000 } \) = Rs 475.20

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

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, height of the cone=8 cm and radius of the cone = 5 cm

Volume of cone=\(\frac{1}{3}\)πr²h
=\(\frac{1}{3}\)π(5)²8 cm³=\(\frac{200}{3}\)π cm³
Radius of one spherical lead shot=0.5 cm
Volume of one spherical lead shot=\(\frac{4}{3}\)πr ³=\(\frac{4}{3}\)π(0.5)³ cm³
=\(\frac{4\times{0.125}}{3}\)π cm³=\(\frac{0.5}{3}\)π cm³
When spherical lead are dropped in the vessel, one fourth of water flows out
Let number of lead shots be n
Volume of n spherical shots = \(\frac{1}{4}\) volume of conical vessel
n\(\left(\frac{0.5}{3}\pi\right)\)=\(\frac{1}{4}\left(\frac{200}{3}\pi\right)\)⇒n(0.5)=50⇒ n=\(\frac{50\times{10}}{5}\)=100 

Given, length of cuboid (l) = 15 cm, breadth of cuboid (b) = 10 cm and height of cuboid (h) = 3.5 cm
\(\therefore\) Volume of cuboid = l \(\times\) b \(\times\) h
= 15 \(\times\) 10 \(\times\)3.5 = 525 cm³
Also, radius of conical depression,r = 0.5 cm
and height of conical depression, h = 1.4 cm
\(\therefore\) Volume of one conical depression \(=\frac{1}{3} \pi \times r^2 \times h\)
\(\begin{aligned} & =\frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 1.4 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{3} \times \frac{1}{2} \times \frac{1}{2} \times \frac{2}{10}=\frac{11}{30} \mathrm{~cm}^3 \end{aligned}\)
Now, volume of 4 conical depressions = 4 \(\times\)Volume of one conical depression

\(=4 \times \frac{11}{30}=\frac{22}{15} \mathrm{~cm}^3\)
Hence, the volume of wood in the entire stand = Volume of cuboid - Volume of 4 conical depressions
\(=525-\frac{22}{15}=525-1.47=523.53 \mathrm{~cm}^3\)

Given, solid is a combination of a cone and a hemisphere.
Also, radius of the cone, r = radius of the hemisphere
= 1 cm

Height of the cone, h = 1 cm
\(\therefore\) Required volume of the solid = Volume of the cone + Volume of the hemisphere
\(\begin{aligned} & =\frac{1}{3} \pi r^2 h+\frac{2}{3} \pi r^3=\frac{1}{3} \pi(1)^2(1)+\frac{2}{3} \pi(1)^3 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{\pi}{3}+\frac{2}{3} \pi=\frac{(\pi+2 \pi)}{3}=\frac{3 \pi}{3}=\pi \mathrm{cm}^3 \end{aligned}\)

Let radius be r and the height be h.
Curved surface area of cylinder = 2πrh
Curved surface area of cone =πr\(\sqrt{r^{2}+h^{2}}\)
\(\frac{2\pi{rh}}{\pi{r}\sqrt{r^{2}+h^{2}}}\)=\(\frac{8}{5}\Rightarrow \frac{h}{\sqrt{r^{2}+h^{2}}}=\frac{4}{5}\)
⇒ 25h²=16(r²+h²)
⇒ 25h²=16r²+16h²
⇒ 9h²=16r²
⇒ \(\frac{9}{16}=\frac{r^{2}}{h^{2}}\Rightarrow \frac{r}{h}=\frac{3}{4}\)
=r:h=3:4

Let radius of the base be r

Height of conical part=h
Slant height of conical pat=l
l=\(\sqrt{h^{2}+r^{2}}\)...(i)
ATQ= 2πr²=πrl⇒ l=2r
Equation (i) becomes
2r=\(\sqrt{h^{2}+r^{2}}\)
4r²=h²+r²
h²=3r²⇒\(\frac{r^{2}}{h^{2}}\)=\(\frac{1}{3}\)
\(\frac{r}{h}\)=\(\frac{1}{\sqrt{3}}\)
r:h=1:\(\sqrt{3}\)

Lateral surface area = \(94.2 \ cm^{2}\) , h = 5 cm
\(2\pi rh = 94.2\) 
\(\Rightarrow \) 2 x 3.14 x r x 5 = 94.2
\(\Rightarrow \) \(r = {94.2 \over 2 \times 3.14 \times 5}\) = 3 cm

Given: height of cylinder=10 cm
base radius=3.5 cm
Curved surface area of cylinder =2πrh=\(\frac{2\times22\times35\times10}{7\times10}\)cm²=220 cm²
Inner surface area of hemispherical cavity=2πr²=2x\(\frac{22}{7}\times\frac{35\times35}{10\times10}\)cm²=77 cm²
Inner surface area of both hemispherical cavity=77 cm²+77 cm²=154 cm²
Total surface area of solid = curved surface area of solid + inner surface area of both hemispherical ends
= 220 cm² + 154 cm² = 374 cm²

Denote radius of cone by r, slant height of cone by l, height of cone by h, radius of cylinder by r' and height of cylinder by h'. Then r = 2.5 cm, h = 6 cm, r' = 1.5 cm, h' = 26 – 6 = 20 cm and 
\(l=\sqrt{r^{2}+h^{2}}=\sqrt{2.5^{2}+6^{2}} \mathrm{~cm}=6.5 \mathrm{~cm}\)
Here, the conical portion has its circular base resting on the base of the cylinder, but the base of the cone is larger than the base of the cylinder. So, a part of the base of the cone (a ring) is to be painted.
So, the area to be painted orange = CSA of the cone + base area of the cone – base area of the cylinder
\(=\pi r l+\pi r^{2}-\pi\left(r^{\prime}\right)^{2} \)
\(=\pi\left[(2.5 \times 6.5)+(2.5)^{2}-(1.5)^{2}\right] \mathrm{cm}^{2} \)
\(=\pi[20.25] \mathrm{cm}^{2}=3.14 \times 20.25 \mathrm{~cm}^{2} \)
\(=63.585 \mathrm{~cm}^{2}\)
Now, the area to be painted yellow = CSA of the cylinder + area of one base of the cylinder
\(=2 \pi r^{\prime} h^{\prime}+\pi\left(r^{\prime}\right)^{2} \)
\(=\pi r^{\prime}\left(2 h^{\prime}+r^{\prime}\right) \)
\(=(3.14 \times 1.5)(2 \times 20+1.5) \mathrm{cm}^{2} \)
\(=4.71 \times 41.5 \mathrm{~cm}^{2} \)
\(=195.465 \mathrm{~cm}^{2}\)

This top is exactly like the object we have discussed. So, we can conveniently use the result we have arrived at there. That is :
TSA of the toy = CSA of hemisphere + CSA of cone
Now, the curved surface area of the hemisphere = \(\frac{1}{2}\left(4 \pi r^{2}\right)=2 \pi r^{2} \)
\(=\left(2 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2}\right) \mathrm{cm}^{2}\)
Also, the height of the cone = height of the top – height (radius) of the hemispherical part
\(=\left(5-\frac{3.5}{2}\right) \mathrm{cm}=3.25 \mathrm{~cm}\)
So, the slant height of the cone \((l)=\sqrt{r^{2}+h^{2}}=\sqrt{\left(\frac{3.5}{2}\right)^{2}+(3.25)^{2}} \mathrm{~cm}=3.7 \mathrm{~cm}(\text { approx. })\)
Therefore, CSA of cone \(=\pi r l=\left(\frac{22}{7} \times \frac{3.5}{2} \times 3.7\right) \mathrm{cm}^{2}\)
This gives the surface area of the top as
\(=\left(2 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2}\right) \mathrm{cm}^{2}+\left(\frac{22}{7} \times \frac{3.5}{2} \times 3.7\right) \mathrm{cm}^{2}\)
\(=\frac{22}{7} \times \frac{3.5}{2}(3.5+3.7) \mathrm{cm}^{2}=\frac{11}{2} \times(3.5+3.7) \mathrm{cm}^{2}=39.6 \mathrm{~cm}^{2}(\text { approx. })\)
You may note that ‘total surface area of the top’ is not the sum of the total surface areas of the cone and hemisphere

Let radius of original ball be 4x \(\Rightarrow \) radius of each small ball be x  
Number of balls = \(\frac { Volume \ of \ original \ ball }{ Volume \ of \ 1 \ small \ ball } \)
\(=\frac { \frac { 4 }{ 3 } \pi \times (4x)^{ 3 } }{ \frac { 4 }{ 3 } \pi x^{ 3 } } =\frac { 4^{ 3 }x^{ 3 } }{ x^{ 3 } } =4^{ 3 }=64\)
S.A. of 1 small balls = 64X4 \(\pi x^{ 2 }\)
S.A of 64 small balls = 64X \(4\pi x^{ 2 }\)
S.A of original ball = \(4\pi (4x^{ 2 })=4\times \pi \times 16x^{ 2 }\)
\(=\frac { S.A \ of \ small \ balls \ combined \ together }{ S.A \ of \ original \ ball } \ =\frac { 64\times 4\pi x^{ 2 } }{ 4\pi \times 16x^{ 2 } } =4:1\)

Height (h₁) of larger cylinder = 220 cm
Radius (r₁) of larger cylinder = \(\frac{24}{2}\)= 12 cm
Height (h₂) of smaller cylinder = 60 cm
Radius (r₂) of smaller cylinder = 8 cm
Total volume of pole = Volume of larger cylinder + Volume of smaller cylinder
\(\begin{aligned} & =\pi r_1^2 h_1+\pi r_2^2 h_2 \\ \end{aligned}\)
\(\begin{aligned} & =\left(\pi(12)^2 \times 220\right)+\left(\pi(8)^2 \times 60\right) \\ \end{aligned}\)
\(\begin{aligned} & =\pi[144 \times 220+64 \times 60] \\ \end{aligned}\)
\(\begin{aligned} & =3.14[31,680+3,840] \\ \end{aligned}\)
\(\begin{aligned} & =3.14 \times 35520=111,532.8 \mathrm{~cm}^3 \end{aligned}\)
Mass of 1 cm³ iron = 8 g
Mass of 111532.8 cm³ iron = 11532.8 \(\times\)8 = 892262.4 g