The document Ex 13.6 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by VP Classes.

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Note: For the following questions, assume Ï€ = (22/7), unless stated otherwise

**Question 1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm ^{3} = 1 l) Solution: **Let the base radius of the cylindrical vessel be â€˜râ€™ cm.

âˆ´ Circumference = 2Ï€r

â‡’ 2Ï€r = 132 [âˆµ Circumference = 132 cm]

âˆµ Height of the vessel = 25 cm

âˆ´ Volume = Ï€r^{2} x h [âˆµ Volume of a cylinder = Ï€r^{2}h]

= (22/7) (21)^{2} x 25 cm^{3}

= (22/7) x 21 x 21 x 25 cm^{3}

= 22 x 3 x 21 x 25 cm^{3}

= 34650 cm^{3 }

âˆµ Capacity of the vessel = Volume of the vessel

âˆ´ Capacity of cylindrical vessel = 34650 cm^{3 }

Since 1000 cm^{3 }= 1 litre

â‡’ 34650 cm^{3} = (34650/1000) litres = 34.65 l

**Question 2. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm ^{3} of wood has a mass of 0.6 g. Solution: **Here, Inner diameter of the cylindrical pipe = 24 cm

â‡’ Inner radius of the pipe (r) = (24/2) cm = 12 cm

Outer diameter of the pipe = 28 cm

â‡’ Outer radius of the pipe (R) = (28/2)cm = 14 cm

Length of the pipe (h) = 35 cm

âˆµ Inner volume of the pipe = Ï€r

Outer volume of the pipe = Ï€r

âˆ´ Amount of wood (volume) in the pipe = Outer volume â€“ Inner volume

= Ï€R^{2}h - Ï€r^{2}h

= Ï€h(R^{2}-r^{2})

= Ï€h(R+r)(R-r) [ âˆµ a^{2} - b^{2} = (a+b)(a-b)]

= 22/7 x 35 x (14+12) x (14-12) cm^{3}

= 22 x 5 x 26 x 2 cm^{3}

Mass of the wood in the pipe = [Mass of wood in 1 m^{3} of wood] x [Volume of wood in the pipe]

= [0.6g] x [22 x 5 x 26 x 2] cm^{3}

= (6/10)x 22 x 10 x 26 g = 6 x 22 x 26 g

= 3432 g = (3432/1000)= 3.432 kg [âˆµ 1000 g = 1 kg]

Thus, the required mass of the pipe is 3.432 kg.

**Question 3. A soft drink is available in two packs (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much? Solution:** For rectangular pack: Length (l) = 5 cm

Breadth (b) = 4 cm Height (h) = 15 cm

âˆ´ Volume = l x b x h = 5 x 4 x 15 cm

= 300 cm

â‡’ Capacity of the rectangular pack = 300 cm

For cylindrical pack: Base diameter = 7 cm

â‡’ Radius of the base (r) = (7/2)cm

Height (h) = 10 cm

âˆ´ Volume = Ï€r^{2}h = (22/7) x (7/2)^{2} x 10 cm^{3}

= (22/7) x (7/2) x (7/2) x 10 cm^{3}

= 11 x 7 x 5 cm^{3} = 385 cm^{3}

â‡’ Volume of the cylindrical pack = 385 cm^{3 }...(2)

From (1) and (2),

we have 385 cm^{3} â€“ 300 cm^{3} = 85 cm^{3}

â‡’ The cylindrical pack has the greater capacity by 85 cm^{3}.

**Question 4. If the lateral surface of a cylinder is 94.2 cm ^{2 }and its height is 5 cm, then find: (i) radius of its base (ii) its volume. (Use Ï€ = 3.14) Solution:** Height of the cylinder (h) = 5 cm Let the base radius of the cylinder be â€˜râ€™.

(i) Since lateral surface of the cylinder = 2 Ï€rh

But lateral surface of the cylinder = 94.2 cm

Thus, the radius of the cylinder = 3 cm

(ii) Volume of a cylinder = Ï€r^{2}h

â‡’ Volume of the given cylinder = 3.14 x (3)^{2} x 5 cm^{3}

Thus, the required volume = 141.3 cm^{3}

**Question 5. It costs **â‚¹** 2200 to paint the inner curved surface of cylindrical vessel 10 m deep. If the cost of painting is at the rate of **â‚¹** 20 per m ^{2}; find: (i) inner curved surface of the vessel (ii) radius of the base (iii) capacity of the vessel. Solution: **(i) To find inner curved surface

Total cost of painting = â‚¹ 2200

Rate of painting = â‚¹ 20 per m^{2}

âˆ´ Area =

â‡’ Inner curved surface of the vessel = 110 m^{2}

(ii) To find radius of the base Let the base radius of the cylindrical vessel.

âˆµ Curved surface of a cylinder = 2 Ï€rh

âˆ´ 2Ï€rh = 110

â‡’ The required radius of the base = 1.75 m

(iii) To find the capacity of the vessel

Since, volume of a cylinder = Ï€r^{2}h

Since, 1 m^{3} = 1000000 cm^{3} = 1000 l = 1 kl

âˆ´ 96.5 m^{3} = 96.5 kl

Thus, the required volume = 96.25 kl

**Question 6. The capacity of closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it? Solution:** Capacity of the cylindrical vessel = 15.4 l

â‡’ Volume of the vessel = (15.4/1000)m^{3}

Height of the vessel = 1m Let â€˜râ€™ metres be the radius of the base of the vessel

âˆ´ Volume = Ï€r^{2}h

Now, total surface area of the cylindrical vessel

Thus, the required sheet = 0.4708 m^{2}

**Question 7. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite. Solution:** Since, 10 mm = 1cm

âˆ´ 1 mm = (1/10) cm

For graphite cylinder

Thus, the required volume of the graphite = 0.11 cm^{3 }

For the pencil Diameter of the pencil = 7 mm = (7/10)cm

âˆ´ Radius of the pencil (R) = (7/20) cm

Height of the pencil (h) = 14 cm

Volume of the pencil =

Volume of the wood Volume of the wood = [Volume of the pencil] â€“ [Volume of the graphite]

= 5.39 cm^{3} â€“ 0.11 cm^{3} = 5.28 cm^{3 }

Thus, the required volume of the wood is 5.28 cm^{3}.

**Question 8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients? Solution:** The bowl is cylindrical.

Diameter of the base = 7 cm

â‡’ Radius of the base (r) = (7/3) cm

Height (h) = 4 cm

= 11 x 7 x 2 cm^{3} = 154 cm^{3}

i.e. Volume of soup in a bowl = 154 cm^{3}

â‡’ Volume of soup in 250 bowls = 250 x 154 cm^{3}

= 38500 cm^{3}

^{= 38500 / 100 liters }

Thus, the hospital needs to prepare 38.5 litres of soup daily for 250 patients.

**VOLUME OF A RIGHT CIRCULAR CONE**

**REMEMBER**

Volume of a cone = (1/3) Ï€r^{2}h, where â€˜râ€™ is its base radius and â€˜hâ€™ is its height.

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