NCERT Solution for Surface Areas and Volumes

∴ Total volume of the two cubes = 2 × 64 cm^{3}

= 128 cm^{3}

Let the edge of each cube = x

∴ x^{3} = 64 = 4^{3}

∴ x = 4 cm

Now, Length of the resulting cuboid l = 2x cm

Breadth of the resulting cuboid b = x cm

Height of the resulting cuboid h = x cm

∴ Surface area of the cuboid = 2 (lb + bh + hl) = 2[(2x . x) + (x . x) + (x . 2x)]

= 2[(2 × 4 × 4) + (4 × 4) + (4 × 2 × 4)] cm^{2}

= 2 [32 + 16 + 32] cm^{2} = 2[80] cm^{2} = 160 cm^{2}.

Radius (r) = 7 cm

Height (h) = 6 cm

∴ Curved surface area

= 2πrh

For hemispherical part:

Radius (r) = 7 cm

∴ Surface area = 2πr^{2}

∴ Total surface area

= (264 + 308) cm^{2} = 572 cm^{2}.

∴ h = (15.5 – 3.5) cm = 12.0 cm

Surface area of the conical part

= πrl

Surface area of the hemispherical part

= 2πr^{2}

∴ Total surface area of the toy

= πrl + 2πr^{2} = πr (l + 2r) cm^{2}

∵ l^{2} = (12)^{2} + (3.5)^{2}

⇒ The greatest diameter of the hemisphere = 7 cm

Surface area of the solid

= [Total S.A. of the cubical block] + [S.A. of the hemisphere] – [Base area of the hemisphere]

= (6 × l2) + 2πr^{2} – πr^{2}

∴ The greatest diameter of the curved hemisphere = l

∴ Surface area of one hemispherical part = 2πr^{2}

Height = 2.4 cm

Diameter = 1.4 cm

⇒ Radius (r) = 0.7 cm

⇒ Total surface area of the cylindrical part

Height of the cylinder (h) = 10 cm

∴ Total surface area = 2πrh + 2πr^{2} = 2πr(h +r)

Total height = 12 cm

Height of a cone (h_{1}) = 2 cm

∴ Height of both cones = 2 × 2 = 4 cm

⇒ Height of the cylinder (h_{2}) = (12 – 4) cm = 8 cm.

Now, volume of the cylindrical part = πr^{2}h_{2}

Total height of the gulab jamun = 5 cm.

Diameter = 2.8 cm

⇒ Radius = 1.4 cm

∴ Length (height) of the cylindrical part = 5 cm – (1.4 + 1.4) cm

= 5 cm – 2.8 cm =2.2 cm

= 15 × 35 cm^{3}

= 525 cm^{3}

Since each depression is conical with base radius (r) = 0.5 cm and depth (h) = 1.4 cm,

∴ Volume of each depression (cone)

Base radius (r) = 5 cm

∴ Volume of the big cylinder = pr^{2}h = π(12)2 × 220 cm^{3}

Also, height of smaller cylinder (h1) = 60 cm

Base radius (r^{1}) = 8 cm

∴ Volume of the smaller cylinder πr_{1}^{2}h_{1} = π(8)^{2} × 60 cm^{3}

∴ Volume of iron

= [Volume of big cylinder] + [Volume of the smaller cylinder]

= π × 220 × 12^{2} + π × 60 × 82 cm^{3}

Base radius of the conical part = 60 cm.

= πr^{2}h = 3.14 × 12 × 8 cm^{3}

⇒ Volume of water in the vessel = 346.51 cm^{3}

Since, the child finds the volume as 345 cm^{3}

∴ The child’s answer is not correct

⇒ The correct answer is 346.51 cm^{3}.

r^{1} = 6 cm

r^{2} = 8 cm

r^{3} = 10 cm

⇒ Volume of the given spheres are:

⇒ Volume of the earth taken out = 22 × 7 × 5 m^{3}

Now this earth is spread out to form a cuboial platform having

length = 22 m

breadth = 14 m

Let ‘h’ be the height of the platform.

∴ Volume of the platform = 22 × 14 × h m^{3}

∴ 22 × 14 × h = 22 × 7 × 5

Thus, the required height of the platform is 2.5 m.

Let the height of the embankment = ‘H’ metre.

Internal radius of the embankment (r) = 1.5 m.

External radius of the embankment R = (4 + 1.5)m = 5.5 m.

Radius (r) = 18 cm

Height (h) = 32 cm

Volume = πr^{2}h

Depth of the canal = 1.5 m

Since the above amount (volume) of water is spread in the form of a cuboid of height as 8 cm

Since, the water flows through the pipe at 3 km/hr.

∴ Length of water column per hour

(h) = 3 km = 3 × 1000 m = 3000 × 100 cm = 300000 cm.

∴ Volume of water = πr^{2}h = π × 102 × 300000 cm^{3} = π × 30000000 cm^{3}

Now, for the cylindrical tank,

Diameter = 10 m

Slant height (l) = 4 cm

The radius of the upper base (r^{2}) = 4 cm

Slant height (l) = 15 cm

∴ Area of the material required

= [Curved surface area of the frustum] + [Area of the top end]

and h = 16 cm

∴ Volume of the frustum

Let l be the length and D be diameter of the wire drawn from the frustum. Since the wire is in the form of a cylinder,

∴ Volume of the wire = πr^{2}l

∴ Length of wire required to cover the whole surface = Length of wire required to complete 40 rounds

= 40 × 31.4 cm = 1256 cm

∴ Volume of the cistern = 150 × 120 × 110 cm^{3} = 1980000 cm^{3}

Volume of water contained in the cistern = 129600 cm^{3}

∴ Free space (volume) which is not filled with water

= 1980000 – 129600 cm^{3} = 1850400 cm^{3}

Now, Volume of one brick = 22.5 × 7.5 × 6.5 cm^{3}

Diameter = 8 cm

⇒ Radius (r) = 4 cm

Height = 10 cm

Curved surface area of the frustum PQRS

= [curved surface area of the rt circular cone OPQ] – [curved surface area of the rt circular cone ORS]

[Volume of the frustum RPQS]

= [Volume of right circular cone OPQ] – [Volume of right circular cone ORS]