Volume of hemisphere = \(\frac { 2 }{ 3 } { \pi r }^{ 3 }=\frac { 2 }{ 3 } \pi \left( 9\times 9\times 9 \right) \)
Let height of water in cylindrical vessel be h cm.
\(\therefore\)   Volume of cylindrical vessel = Volume of hemisphere
\(\pi\) (6 x 6) x h = \(\frac{2}{3}\pi\) (9 x 9 x 9)
h = \(\frac { 2 }{ 3 } \times \frac { 9\times 9\times 9 }{ 6\times 6 } \) = 13.5
\(\therefore\)  Height of water in cylindrical vessel = 13.5 cm.

Volume of kerosene in case
= 20 x 20 x 45 = 25 x 15 x h (Volume in second cane)
\(\Rightarrow\)  h = \(\frac{20\times20\times45}{25\times15}\) = 48 cm

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
\(\begin{aligned} & =6 \times(\text { Edgc })^2+2 \pi r^2-\pi r^2 \\ \end{aligned}\)
\(\begin{aligned} & =6 \times l^2+2 \pi \times\left(\frac{l}{2}\right)^2-\pi\left(\frac{l}{2}\right)^2 \\ \end{aligned}\)
\(\begin{aligned} & =6 l^2+2 \pi \times \frac{l^2}{4}-\pi \frac{l^2}{4}=6 l^2+\pi \frac{l^2}{4} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{l^2}{4}(\pi+24) \text { sq units } \end{aligned}\)

Volume of upper cone\(=\frac { 1 }{ 3 } \pi { \left( \frac { r }{ 2 } \right) }^{ 2 }\times \frac { h }{ 2 } \)
\(=\frac { 1 }{ 3 } \pi \frac { { r }^{ 2 } }{ 2 } \times \frac { h }{ 2 } \)
\(=\frac { 1 }{ 3 } \pi \frac { { r }^{ 2 }h }{ 2 } \)
Volume of cone\(=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
\(\frac { Volume \ of \ upper \ part \ of \ cone }{ Volume \ of \ cone } =\frac { \frac { 1 }{ 3 } \pi \frac { { r }^{ 2 }h }{ 2 } }{ \frac { 1 }{ 3 } \pi { r }^{ 2 }h } \)
\(=\frac { 1 }{ 8 } \)
= 1:8

\(\frac { Volume \ of \ { 1 }^{ st } \ sphere }{ Volume \ of \ { 11 }^{ nd } \ sphere } =\frac { \frac { 4 }{ 3 } \pi { r }_{ 1 }^{ 3 } }{ \frac { 4 }{ 3 } \pi { r }_{ 2 }^{ 3 } } =\frac { 64 }{ 27 } \)
\(\therefore \ =\frac { { r }_{ 1 }^{ 3 } }{ { r }_{ 2 }^{ 3 } } =\frac { 64 }{ 27 } \)
\(=\frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 4 }{ 3 } \)
Ratio of their surface areas=\(\frac { Surface \ are \ of \ { I }^{ st } \ sphere }{ Surface \ are \ of \ { II }^{ nd } \ sphere } \)
\(=\frac { 4\pi { r }_{ 1 }^{ 2 } }{ 4\pi { r }_{ 2 }^{ 2 } } ={ \left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) }^{ 2 }\)
\(={ \left( \frac { 4 }{ 3 } \right) }^{ 2 }=\frac { 16 }{ 9 } \)
=16:9

\(\because\) Whole surface of each part=\(=2\pi { r }^{ 2 }+\pi { r }^{ 2 }=3\pi { r }^{ 2 }\)
\(\therefore\) Total surface of two parts=\(=3\pi { r }^{ 2 }+3\pi { r }^{ 2 }=6\pi { r }^{ 2 }\)

The largest sphere of 1 m can be cut from cylindrical log of wood of base 1 m and height 4 m 
\(\therefore Volume \ of \ sphere=\frac { 4 }{ 3 } \pi { \left( 1 \right) }^{ 3 }{ m }^{ 3 }=\frac { 4 }{ 3 } \pi { m }^{ 3 }\)

Length of the longest stick that can be placed inside the box = length of its diagonal
Diagonal of the cuboid=\(\sqrt{l^{2}+b^{2}+h^{2}}\)
=\(\sqrt{(12)^{2}+(4)^{2}+(3)^{2}}\)
=\(\sqrt{144+16+9}=\sqrt{169}\)
= 13 cm

Radius of lower end (r₁)= 8 cm
Radius of upper end (r₂) = 20 cm
Height of frustum = 24 cm 

Volume of the container=\(\frac{\pi{h}}{3}\)(r₁²+r₂²+r₁r₂)
=\(\frac{22}{7}\times{\frac{24}{3}}\)[(8)²+(20)²+8x20]
=\(\frac{22}{7}\)x8(64+400+160)
=\(\frac{22}{7}\)x8x624 cm³=15869.14 cm³
=15.68914 l
Cost of milk which can completely fill the container=Rs.21 x 15.68914
=Rs.329.47

r= 1 cm,h = ? , H = 16 cm, R₁ = 10 cm, Thickness = 2.5 mm = 0.25 cm
Internal radius ofcylinder = (10 — 0.25) cm = 9.75 cm = r₁ 
Volume of hollow cylinder = πH(R₁²-r₁²)...(i)
Volume of solid cylinder=Volume of hollow cylinder
Obtain volume of hollow cylinder from (i)
⇒ πr²h=πH(R₁²-r₁²)​​​​​​​...(ii)
Volume of hollow cylinder=πx16x[(10)²-(9.75)²]=79π cm³
Volume of solid cylinder = volume of hollow cylinder ​​​​​​​=79π
=πr²h=79π
⇒ πx(1)²xh=79π⇒ h=79 cm

Radius of cylinder=radius of cone=radius of hemisphere

Total height of the toy = 30 cm 
height of cone =30 cm-(height of cylinder + r of hemisphere) 
= 30 cm - (13+5) cm =12 cm
l of cone=\(\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =\sqrt { { 12 }^{ 2 }+{ 5 }^{ 2 } } \)
\(=\sqrt { 144+25 } =\sqrt { 169 } =13cm\)
C.S.A. of toy = C.S.A of cone + C.S.A of cylinder + C.S.A. of hemisphere
\(=\pi rl+2\pi rh+{ 2\pi r }^{ 2 }=\pi r(l+2h+2r)\)
\(=\frac { 22 }{ 7 } \times 5(13+2\times 13+2\times 5)\)
\(=\frac { 22 }{ 7 } \times 5(13+26+10)=770{ cm }^{ 2 }\)

Let surface area of sphere be x cm²
⇒ 4πr²=x cm² ⇒ r=\(\sqrt{\frac{x}{4\pi}}\)cm
Volume of sphere=\(\frac{4}{3}\pi \times \left(\frac{x}{4\pi}\right)^{3/2}\)cm³
Also surface area of cube be x cm²
⇒ 6 x side²=x cm² ⇒ side=\(\sqrt{\frac{x}{6}}\)cm 
Volume of cube=\(\left(\frac{x}{6}\right)^{3/2}\)cm³
Volume of sphere : Volume of cube
=\(\frac{4}{3}\pi \times \left(\frac{x}{4\pi}\right)^{3/2}\):\(\left(\frac{x}{6}\right)^{3/2}\)=1:\(\sqrt{\frac{\pi}{6}}\)

False.

Total surface area of the figure = \(\frac { 2\pi { r }h+\pi { r }^{ 2 } }{ upper\quad cylinder } +\frac { \pi { r }^{ 2 }+2\pi { r }h }{ lower\quad cylinder } =4\pi { r }h+2\pi { r }^{ 2 }\) 

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

(i) 57.750 L
(ii) Social cohension 

Radius of hemisphere = 4.2 cm
Volume of hemisphere  =\(\frac { 2 }{ 3 } \pi \) r³
=\(\frac { 2 }{ 3 } \pi \times (4.2)^{ 3 }{ cm }^{ 3 }\)   =49.329 \(\pi \) cm³
∴  volume of 2 hemispheres
=2 X  =98.392\(\pi \) cm³ = 98.784\(\pi \) cm³​​​​​​​
Height of cylinder =10 cm
Radius =4.2 cm
∴  volume of cylinder =\(\pi \) r²h
=\(\pi \) X (4.2)²X 10=176.4 \(\pi \)
∴  volume of metal left 
=176.4\(\pi \) -98.3784\(\pi \) =77.616\(\pi \) cm³
Radius of wire =0.7 cm
let length of wire be x
Volume of wire =\(\pi \) X 0.7 X 0.7 X x
  =0.49\(\pi \) x cm³
⇒   0.49\(\pi \)x =77.616 \(\pi \)
⇒ x= 158.4 cm
∴ length of wire =158.4 cm

Radius (r₁) of 1^st sphere = 6 cm
Radius (r₂) of 2^nd sphere = 8 cm
Radius (r₃) of 3^rd sphere = 10 cm
Let the radius of the resulting sphere be r.
The object formed by recasting these spheres will be same in volume as the sum of the volumes of these spheres.
Volume of 3 spheres = Volume of resulting sphere
\(\begin{array}{l} \frac{4}{3} \pi\left[r_{1}^{3}+r_{2}^{3}+r_{3}^{3}\right]=\frac{4}{3} \pi r^{3} \\ \frac{4}{3} \pi\left[6^{3}+8^{3}+10^{3}\right]=\frac{4}{3} \pi r^{3} \end{array}\)
r³ = 216 + 512 + 1000 = 1728
r = 12 cm
Therefore, the radius of the sphere so formed will be 12 cm.

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