Jharkhand Board JAC Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.6 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.6

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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)

[Assume, π = \(\frac{22}{7}\)]

Answer:

Let the radius of the cylindrical vessel be r.

Height (h) of vessel = 25 cm

Circumference of vessel =132 cm

2πr = 132 cm

r = \(\frac{132 × 7}{2 × 22}\) cm = 21 cm

Volume of cylindrical vessel = πr^{2}h = × (21)^{2} × 25 cm^{3} = 34650 cm^{3}

= \(\frac{34650}{1000}\) liters = 34.65 litres

[∵ 1 litre = 1000 cm^{3}]

Therefore, such vessel can hold 34.65 litres of water.

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 cm3 of wood has a mass of 0.6 g.

Assume, n= \(\frac{22}{7}\)

Answer:

Inner radius (r ) of cylindrical pipe

= \(\frac{24}{2}\) cm = 12 cm

Outer radius (r2) of cylindrical pipe = \(\frac{28}{2}\) cm = 14 cm

Height (h) of pipe = Length of pipe = 35 cm

Volume of pipe = π\(\left(r_1^2-r_2^2\right)\)h

= [\(\frac{22}{7}\) × (14^{2} – 12^{2}) × 35] cm^{3}

= 110 × 52 cm^{3} = 5720 cm^{3}

Mass of 1 cm3 wood = 0.6 g

Mass of 5720 cm3 wood = (5720 × 0.6) g

= 3432 g = 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?

Assume, π = \(\frac{22}{7}\)

Answer:

(i) The tin can will be cuboidal in shape while the plastic cylinder will be cylindrical in shape.

Length (l) of tin can = 5 cm

Breadth (b) of tin can = 4 cm

Height (h) of tin can = 15 cm

Capacity of tin can = l × b × h = (5 × 4 × 15) cm^{3} = 300 cm^{3}

(ii)

Radius (r) of circular end of plastic cylinder = \(\frac{7}{2}\) cm = 3.5 cm

Height (H) of plastic cylinder = 10 cm

Capacity of plastic cylinder = πr^{2}H

= \(\frac{22}{7}\) × (3.5)^{2} × 10 cm^{3}

= 11 × 35 cm^{3} = 385 cm^{3}

Therefore, plastic cylinder has the greater capacity.

Difference in capacities = (385 – 300) cm^{3} = 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 π= 3.14]

Answer:

(i) Height (h) of cylinder = 5 cm

Let radius of cylinder be r.

CSA of cylinder = 94.2 cm^{2}

2πrh = 94.2 cm^{2}

(2 × 3.14 × r × 5) cm = 94.2 cm^{2}

r = 3 cm

(ii) Volume of cylinder = πr^{2}h

= (3.14 x (3)2 x 5) cm^{3} = 141.3 cm^{3}

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Question 5.

It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m^{2}, find:

(i) Inner curved surface area of the vessel

(ii) Radius of the base

(iii) Capacity of the vessel

Assume π = \(\frac{22}{7}\)

Answer:

(i) ₹ 20 is the cost of painting 1 m^{2} area.

₹ 2200 is the cost of painting (\(\frac{1}{20}\) × 20) m^{2} area = 110 m^{2} area

Therefore, the inner surface area of the vessel is 110 m^{2}.

(ii) Let the radius of the base of the vessel be r.

Height (h)of the vessel = 10 m

Surface area = 2πrh = 110 m^{2}.

⇒ (2 × \(\frac{22}{7}\) × r × 10) = 110 m^{2}

⇒ r = \(\frac{7}{4}\) m = 1.75 m

(iii) Volume of vessel = πr^{2}h

= [\(\frac{22}{7}\) × (1.75)^{2} × 10] m^{3}

= 96.25 m^{3}

Therefore, the capacity of the vessel is 96.25 m3 or 96250 litres.

Question 6.

The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

Assume, π = \(\frac{22}{7}\)

Answer:

Let the radius of the cylindrical vessel be r.

Height (h) of cylindrical vessel = 1 m

Volume of cylindrical vessel = 15.4 litres = 0.0154 m^{3}

πr^{2}h = 0.0154

\(\frac{22}{7}\) × r^{2} × 1 m = 0.0154^{3}

⇒ r = 0.07 m

Total surface area of vessel = 2πr(r + h)

= [2 × \(\frac{22}{7}\) × 0.07(0.07 + 1)] m^{2}

= 0.44 ×1.07 m^{2}

= 0.4708 m^{2}

Therefore 0.4708 m^{2} of the metal sheet would be required to make the cylindrical vessel.

Question 7.

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

Assume, π = \(\frac{22}{7}\)

Answer:

Radius (r) of pencil = \(\frac{7}{2}\) mm = \(\frac{0.7}{2}\) cm = 0.35 cm

Radius (r ) of graphite = \(\frac{1}{2}\) mm = \(\frac{0.1}{2}\) cm = 0.05 cm

Height (h) of pencil =14 cm

Volume of wood in pencil = π\(\left(r_1^2-r_2^2\right)\)h

= [\(\frac{22}{7}\) {(0.35)^{2} – (0.05)^{2}} × 14] cm^{3}

= [\(\frac{22}{7}\) (0.1225 – 0.0025) × 14] m^{3}

= 44 × 0.12 cm^{3} = 5.28 cm^{3}

Volume of graphite

= π\({r_{1}^{2}}\)h

= \(\frac{22}{7}\) × (0.05)^{2} × 14] cm^{3}

= 44 × 0.0025 cm^{3} = 0.11 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?

Assume, π = \(\frac{22}{7}\)

Answer:

Radius (r) of cylindrical bowl = \(\frac{7}{2}\) cm = 3.5 cm

Height (h) of bowl, up to which bowl is filled with soup = 4 cm

Volume of soup in 1 bowl = πr^{2}h

= [\(\frac{22}{7}\) × (3.5)^{2} × 4] cm^{3}

= (11 × 3.5 × 4) cm^{3}= 154 cm^{3}

Volume of soup given to 250 patients = (250 × 154) cm^{3} = 38500 cm^{3} = 38.5 litres.