(a) Volume of cylinder, V _cylinder =450cm³

Since, the cone has the same radius and height as the cylinder, so its volume is

\(V_{\text {cone }}=\frac{1}{3} \pi r^2 h\)

\(=\frac{1}{3} \times V_{\text {cylinder }}\)

\(=\frac{1}{3} \times 450=150 \mathrm{~cm}^3\)

\(V_{\text {empty }}=V_{\text {cylinder }}-V_{\text {cone }}\)

= 450- 150 = 300 cm³

Given, radius of base of right circular cone = 5 cm and radius of a sphere = 5 cm

It is given that volume of cone is equal to the volume of

\(\text { sphere i.e. } \frac{1}{3} \pi r^2 h=\frac{4}{3} \pi r^3\)

⇒ \(\frac{1}{3} \pi(5)^2 h=\frac{4}{3} \pi(5)^3\)

∴ h= 20 cm

Let r be the radius of hemisphere. Given, volume of a solid hemisphere is \(\frac{396}{7} \mathrm{~cm}^3\)

\(\therefore \frac{2}{3} \pi r^3=\frac{396}{7}\)

⇒ \(r^3=\frac{396 \times 3 \times 7}{2 \times 22 \times 7}=27\)

⇒ r=3cm

Total surface area of the solid hemisphere

\(=3 \pi r^2=3 \times \frac{22}{7} \times 3^2=\frac{594}{7} \mathrm{~cm}^2\)

Hint Height of the cone = Edge of the cube= 4.2 cm

and diameter of the cone = Edge of cube = 4.2 cm

Ans. 19.404 cm³

Hint Volume of remaining solid = Volume of cylinder volume of a conical cavity

Ans. 628 cm³

Hint Required volume = Volume of the cuboid \(+\frac{1}{2} \times \text { Volume of the cylinder }\)

\(=15 \times 7 \times 8+\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 15\)

=1128.75 m³

Total space occupied by the machinery = 400 m³

and total space occupied by the workers

= 15 x 0.08

= 1.2 m³

∴ Total volume of the air, when there are machinery and workers = 1128.75 – 400 - 1.2

Ans. 727.55 m³

Given, the side of a cube = 14cm

∴ The volume of the cube = (Side)³ = (14)³ =2744 cm³

∵ The largest possible cone is carved out from one face of cube.

∴ The radius of cone is half the side of the cube

i.e. \(r=\frac{14}{2}=7 \mathrm{~cm}\)∴

and height of the cone is equal to the side of the cube

i.e. h=14 cm

∴ The volume of the cone =\(\frac{1}{3} \times \pi r^2 h .\)

\(=\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 14\)

\(=\frac{2156}{3} \mathrm{~cm}^3\)

So, the volume of remaining solid

=Volume of cube - Volume of cone

\(=2744-\frac{2156}{3}\)

\(=\frac{8232-2156}{3}=\frac{6076}{3} \mathrm{~cm}^3\)

Now, surface area of cube = 6x (Side)²

=6x (14) 1176 cm²

and the curved surface area of the cone

=πrl

\(=\pi r \sqrt{h^2+r^2}\left[\because l^2=h^2+r^2\right]\)

\(=\frac{22}{7} \times 7 \times \sqrt{14^2+7^2}\)

\(=\frac{22}{7} \times 7 \times 7 \times \sqrt{5}\)

=22 x7x 2.23

=343.42

The area of the base of the cone =πr²

\(=\frac{22}{7} \times 7^2=154 \mathrm{~cm}^2\)

So, the surface area of the remaining solid = Surface area of cube +Curved surface area of cone - Base area of cone

= 1176 -154 +343.42

= 1365.42 cm²

Given that tent is a combination of a cylinder and a Cone.

Height of the cone, h₁=Total height- Height of cylinder = 13.5-3= 10.5m

Radius of cone and radius of cylinder r= 14 m

Slant height l of cone \(=\sqrt{h_1^2+r^2}=\sqrt{(10.5)^2+(14)^2}\)

=17.5 m

Curved surface area of cone= πrl

=π x 14 x 17.5= 245 πm²

∴ Area of canvas required =245π + 84π + 26

\(=245 \times \frac{22}{7}+84 \times \frac{22}{7}+26\)

=770+ 264 + 26= 1060 m²

∴ Cost of canvas = ₹(500 × 1060) = ₹530000