## RS Aggarwal Class 9 Ex 13B Chapter 13

**Q.1:** **Find the volume, the curved surface area, the total surface area of a cone having base radius 35 am and height 84 cm.**

**Solution:**

Here, r = 35 cm and h = 84 cm

∴ Volume of cone =

= ^{3}

∴ Curved surface area =

=

= ^{2}

∴ Total surface area = 𝜋r(l + r)

Now, l =

=

∴ Total surface area =

=

**Q.2:** **Find the volume, the curved surface area and the total surface area of a cone whose height and slant height are 6 cm and 10 cm respectively. (take 𝜋 = 3.14)**

**Solution:**

Here, height (h) = 6 cm and slant height (l) = 10 cm

∴ radius (r) =

∴ Volume of cone =

= ^{3}

∴ Curved surface area = 𝜋rl = (3.14 × 8 × 10) cm^{2} = 251.2 cm^{2}

∴ Total surface area = 𝜋r(l + r) = 𝜋r(10 + 8) = (3.14 × 8 × 18) cm^{2} = 452.16 cm^{2}

**Q.3:** **The volume of a right circular cone is (100𝜋) cm ^{3} and its height is 12 cm. Find its slant height and its curved surface area.**

**Solution:**

Here, Volume = (100𝜋) cm^{3}, height (h) = 12 cm

Volume of the cone =

Slant height (l) =

l =

∴ Slant height (l) = 13 cm

∴ Curved surface area = 𝜋rl = (5)(13) cm^{2 } = 65𝜋 cm^{2}

**Q.4:** **A cone of slant height 25 cm has a curved surface area 550 cm ^{2}. Find the height and volume of the cone.**

**Solution:**

Here, curved surface area = 550 cm^{2} and slant height (l) = 25 cm

∴ Curved surface area = 𝜋rl

Now, height (h) =

∴ Height of the cone = 24 cm.

Volume of the cone = ^{3 }

∴ Volume of the cone = 1232 cm^{3}

**Q.5:** **Find the volume of a cone having radius of the base 35 cm and slant height 37 cm.**

**Solution:**

Here, radius (r) = 35 cm and slant height (l) = 37 cm

∴ Height (h) = 12 cm.

Volume of the cone =

= ^{3 }

∴ Volume of the cone = 15400 cm^{3}

**Q.6:** **The curved surface area of a cone is 4070 cm ^{2} and its diameter is 70 cm. Find its slant height.**

**Solution:**

Here, curved surface area = 4070 cm^{2 }

Diameter = 70 cm

∴ Curved surface area = 𝜋rl

∴ Slant height = 37 cm.

**Q.7:** **How many meters of cloth, 2.5 m wide, will be required to make a conical tent whose base radius is 7 m and height 24 m?**

**Solution:**

Here, radius = 7 m and height = 24 m

∴ slant height (l) =

=

Now, area of the cloth = 𝜋rl

∴ Length of cloth =

∴ Length of cloth required to make a conical tent = 220 m.

**Q.8:** **A right circular cone is 3.6 cm high and the radius of its base is 1.6 cm. It is melted and re-casted into a right circular cone having a base radius 1.2 cm. Find its height.**

**Solution:**

Here, height of cone = 3.6 cm and radius = 1.6 cm

After melting, its radius = 1.2 cm

Volume of original cone = Volume of cone after melting

∴

∴ Height of new cone = 6.4 cm

**Q.9:** **Two cones have their heights in the ratio 1 : 3 and their radii of their bases in the ratio 3 :1. Show that their volumes are in the ratio 3 : 1.**

**Solution:**

Let their heights be ‘h’ and ‘3h’

And, their radii be ‘3r’ and ‘r’

Then,

And,

∴ V_{1}:V_{2}= 3 : 1

**Q.10:** **A circus tent is in cylindrical to a height of 3 metres and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.**

**Solution:**

Radius of the cylinder, R =

Slant height (l) = 53 m

∴ Area of canvas = (2𝜋RH + 𝜋Rl)

= ^{2} = 9735 m^{2}

∴ Length canvas =

**Q.11:** **A conical tent is to accommodate 11 persons. Each person must have 4 m ^{2} of the space on the ground and 20 m^{3} of air to breathe. Find the height of the cone.**

**Solution:**

Let the radius be ‘r’ meters and height be ‘h’ meters.

Area of the base = (11 × 4) m^{2} = 44 m^{2}

Therefore, 𝜋r^{2} = 44

Volume of the cone =

∴ Volume of the cone = (11 × 20) m^{3} = 220 m^{3}

∴ The height of the cone = 15 m.

**Q.12:** **A cylindrical bucket, 32 cm high and 18 cm of the radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and the slant height of the heap.**

**Solution:**

Here, the height of the cylindrical bucket = 32 m and radius = 18 cm.

Now, let the radius of the heap be R cm and its slant height be ‘l’ cm.

Then,

∴ Radius of the heap = 36 cm

Slant height (l) =

∴ Slant height of the heap = 43.27 cm.

**Q.13:** **A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, show that the radius and height of each have the ratio 3 :4.**

**Solution:**

Let the curved surface area of the cylinder and cone b 8x and 5x.

Then, 2𝜋rh = 8x . . . . . . . (i)

And,

Squaring both sides of equation (i), we have

From (ii), we have,

Squaring both sides,

∴ The ratio of radius and height = 3:4

**Q.14:** **An iron pillar consists of a cylindrical portion 2.8 m high and 20 cm in diameter and a cone 42 cm high is surrounding it. Find the weight of the pillar, given that 1 cm ^{3 }of iron weighs 7.5 g.**

**Solution:**

Here, height (h) of cylinder = 2.8 m = 280 cm and diameter = 20 cm

Height of the cone (H) = 42 cm

∴ Volume of the pillar =

=

=

= ^{3}

∴ Weight of pillar =

**Q.15:** **The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be (1/27) of the volume of the given cone, at what height above the base, the section has been made?**

**Solution:**

Let the smaller cone have radius = ‘r’ cm and height = ‘h’ cm

And, let the radius of the given original cone be R cm

Since, the two triangles,

Given that the volume of the small cone is

∴

From the figure,

AC = (OA – OC) = (30 – 10) cm = 20 cm

∴ The required height = 20 cm

**Q.16:** **From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and base is removed. Find the volume of the remaining solid (take 𝜋 = 3.14).**

**Solution:**

Here, height (h) =10 cm and radius = 6 cm

∴ Volume of the remaining solid =

=

=

=

∴ Volume of the remaining solid = 753.6 cm^{3}

**Q.17:** **Water flows at a rate of 10 meters per minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the surface 40 cm and depth 24 cm?**

**Solution:**

Diameter of the pipe = 5 mm = 0.5 cm

Radius of the pipe = 0.25 cm

Length of the pipe = 10 m = 1000 cm

Volume that flows in 1 minute = [𝜋 × (0.25)^{2} × 24] cm^{3}

∴ Required time =