Coin is in the form of a cylinder
Diameter of the coin = 1.5 cm
Radius of the coin = \(\frac{1.5}{2}\)
Thickness = height = 2 mm = \(\frac{2}{10}=0.2 \mathrm{~cm}\)
Volume of coin (cylinder) = \(\pi r^{2} h\)
\(=\pi\left(\frac{1.5}{2}\right)^{2}(0.2) \)
\(=0.1125 \pi \mathrm{cm}^{3} \)
Diameter of cylinder = 4.5 cm
radius = \(\frac{4.5}{2}=2.25 \mathrm{~cm}\)
height = 10 cm
volume = \(\pi r^{2} h\ sq. units \)
= \(\pi(2.25)^{2}(10) \)
= \(50.625 \pi \)
No.of coins \(=\frac{\text { Volume of cylinder }}{\text { Volume of Coin }} \)
\(=\frac{50.625 \pi}{0.1125 \pi}=450 \text { coins. } \)

Solid sphere
radius = 6 cm
Volume \(=\frac{4}{3} \pi r^{3} \text { cu. units } \)
\(=\frac{4}{3} \pi(6)^{3} \)
\(=\frac{4}{3} \pi(216)=288 \pi \mathrm{cm}^{3} \)
Hollow cylinder
Internal radius = 'r'
External radius = 'R' = 5 cm
Height h = 32 cm
Volume of Hollow Cylinder
\(=\pi h\left(\mathrm{R}^{2}-r^{2}\right)\ cu. units \)
\(=\pi(32)\left(25-r^{2}\right) \mathrm{cm}^{3} \)
Given that solid sphere is melted to form a hollow cylinder.
Volume of Hollow Cylinder = Volume of Sphere
\(32 \pi\left(25-r^{2}\right) =288 \pi \)
\(25-r^{2} =\frac{288}{32}=9 \)
r² = 25 - 9 = 16
Internal radius r = 4 cm
Thickness = External radius - Internal radius
= R - r = 5 - 4 = 1 cm.

Edge of cube = 7 cm
surface area of a cube = 6a² sq. units
= 6(7)²
= 294 cm²
radius of hemisphere = \(\frac{7}{2} \mathrm{~cm}\)
[Only C.S.A is considered as the hemisphere surmounted]
C.S.A of hemisphere \(=2 \pi r^{2} \text { sq. units } \)
\(=2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \)
= 77 Cm²
Surface area of = T.S.A of cube + C.S.A the solid of hemisphere area of circular region (bottom of hemisphere)

\(=294+77-\left(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\right)\)
= 371 - 38.5
= 332.5 cm²

Let r be the radius of the hemisphere.
Given that, diameter of the hemisphere = side of the cube = l
Radius of the hemisphere = \(\frac{l}{2}\)
TSA of the remaining solid = Surface area of the cubical part + C.S.A. of the hemispherical part − Area of the base of the hemispherical part
= 6 x (Edge)² + 2\(\pi\)r²−\(\pi\)r²
= 6 x (Edge)² +  \(\pi\)r²
\(=6{ \times (l) }^{ 2 }+\pi { \left( \frac { l }{ 2 } \right) }^{ 2 }=\frac { 1 }{ 4 } (24+\pi ){ l }^{ 2 }\)
Total surface area of the remaining solid \(=\frac { 1 }{ 4 } (24+\pi ){ l }^{ 2 }\)sq. units

Let l, b and h₁ be the length, breadth and height of the cuboid. Also let us take r and h₂ be the radius and height of the cylinder.
Now, Volume of the box = Volume of the cuboid + \(\frac{1}{2}\) (Volume of cylinder)
\((l\times b\times { h }_{ 1 })+\frac { 1 }{ 2 } ({ \pi r }^{ 2 }{ h }_{ 2 })cu.units\)
\(=\left( 30\times 15\times 10 \right) +\frac { 1 }{ 2 } \left( \frac { 22 }{ 7 } \times \frac { 15 }{ 2 } \times 30 \right) \)
= 4500 + 2651.79 = 7151. 79
Therefore, Volume of the box = 7151.79 cm³

Let r and h be the radius and height of the cylinder respectively.
Given that, diameter d = 12 cm, radius r = 6 cm
Total height of the toy is 25 cm
Therefore, height of the cylindrical portion = 25 - 6 = 19 cm
T.S.A. of the toy = C.S.A. of the cylinder + C.S.A. of the hemisphere + Base Area of the cylinder
\(2\pi rh+2\pi { r }^{ 2 }+\pi { r }^{ 2 }\)
\(=\pi r(2h+3r)\quad sq.units\)
\(\frac { 22 }{ 7 } \times 6\times 56=1056\)
Therefore, T.S.A. of the toy is 1056 cm²

Let r₁ and r₂ be the radii of sphere and hemisphere respectively.
Given TS.A of sphere = T.S.A of hemisphere
\(4 \pi r_{1}^{2} =3 \pi r_{2}^{2} \)
\(\frac{r_{1}^{2}}{r_{2}^{2}} =\frac{3}{4} \Rightarrow \frac{r_{1}}{r_{2}}=\frac{\sqrt{3}}{2} \)
Ratio of their volumes : \(\frac{V_{1}}{V_{2}}=\frac{\frac{4}{3} \pi r_{1}^{3}}{\frac{2}{3} \pi r_{2}^{3}}=2\left(\frac{r_{1}}{r_{2}}\right)^{3}\)
\(=2\left(\frac{\sqrt{3}}{2}\right)^{3} \)
\(=\frac{3 \sqrt{3}}{4}=3 \sqrt{3}: 4 \)
Ratio of their volumes = \(3 \sqrt{3}: 4\)

Let r₁, r₂ be the radii of two cones,
Given r₁ = r₁ and let h₁, h₂ be the heights of two
V₁ = Volumes of I cone = \(\frac{1}{3} \pi r_{1}^{2} h_{1}=\frac{\pi}{3} r_{1}^{2} h\)
= 3600 cm³
V₂ = Volume of II cone = \(\frac{1}{3} \pi r_{2}^{2} h_{2}=5040 \mathrm{~cm}^{3}\)
\(Now, \frac{V_{1}}{V_{2}}=\frac{3600}{5040} \Rightarrow \frac{\frac{\pi}{3} r_{1}^{2} h_{1}}{\frac{\pi}{3} r_{2}^{2} h_{2}}=\frac{3600}{5040}\left[\because r_{1}=r_{2}\right]. \)
\(\frac{h_{1}}{h_{2}}=\frac{5}{7}=5: 7\)

Diameter of Glass = 20 cm
radius = 10 cm
water upto height = 9 cm
radius of cylindrical metal = 5 cm
height of cylindrical metal = 4 cm
Volume of water displaced = Volume of cylindrical metal
\(\pi r_{1}^{2} h_{1}=\pi r_{2}^{2} h_{2} \)
\((10)^{2} h_{1}=(5)^{2}(4) \)
\(h_{1}=\frac{100}{100}=1 \mathrm{~cm} \)
Hence, the increase in water level is 1 cm.

Let r be the radius of the hemisphere.
Given that, volume of the hemisphere = 29106 cm³
Now, volume of new hemisphere = \(\frac{2}{3}\)(Volume of original sphere)
= \(\frac{2}{3}\) x 29106
Volume of new hemisphere = 19404 cm³
\(\frac { 2 }{ 3 } \pi { r }^{ 3 }=19404\)
\({ r }^{ 3 }=\frac { 19404\times 3\times 7 }{ 2\times 22 } =9261\)
\(r=\sqrt [ 3 ]{ 9261 } =21cm\)
Therefore, r = 21 cm