These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.6

Question 1.

The circumference of the base of a cylinderical vessel is 132 cm and its height is 25 cm. How many liters of water can it hold? (1000 cm^{3} = 1L)

Solution:

We have given that

Circumference of base of cylinder = 132 cm

2πr = 132

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

and height of cylinderical vessel = 25 cm

Volume of cylindrical vessel = πr^{2}h

= \(\frac {22}{7}\) × 21 × 21 × 25

= 34650 cm^{3}

Now, we know that 1000 cm^{3} = 1L

34650 cm^{3} = 34.65 L

Therefore, the cylindrical vessel can hold 34.65 liters 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 cm^{3} is wood has a mass of 0.6 g.

Solution:

Volume of required wood = volume of outer cylinder – volume of inner cylinder.

= π(14)^{2} × 35 – π(12)^{2} × 35

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

= \(\frac {22}{7}\) × 35 × (196 – 144)

= \(\frac {22}{7}\) × 35 × 52

= 5720 cm^{3}

Now, we have given that, mass of 1 cm^{3} of wood = 0.6

Therefore, mass of 5720 cm^{3} of wood = 0.6 × 5720 g.

= 3432g

= 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 wide 4 cm having a height of 15 cm and (ii) a plastic cylinder with a circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much.

Solution:

In the first case,

The tin can with a rectangular base of length 5 cm and wide 4 cm having a height of 15 cm.

∴ The volume of this can = l × b × h

= 15 × 4 × 5

= 300 cm^{3} …….(i)

In the second case,

The circular plastic cylinder has a base diameter is 7 cm and the height of the cylinder is 10 cm.

∴ Volume of plastic cylinder = πr^{2}h

= \(\frac {22}{7}\) × 3.5 × 3.5 × 10

= 385 cm^{3} ……(ii)

From (i) and (ii), it is clear that

The volume of the circular cylinder has a greater capacity of 85 cm^{3}.

Question 4.

If the lateral surface of a cylinder is 94.2 cm^{3} and its height is 5 cm, then find

(i) radius of its base

(ii) volume of the cylinder.

Solution:

(i) We have given that lateral surface of cylinder = 94.2 cm^{3} and height is 5 cm.

Now, we know that, lateral surface area of cylinder = 2πrh

94.2 = 2 × 3.14 × r × 5

r = \(\frac{94.2}{2 \times 3.14 \times 5}\) = 3 cm

(ii) We know that

Volume of cylinder = 2πrh

= 3.14 × 3 × 3 × 5

= 141.3 cm^{3}.

Question 5.

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

(i) Inner curved surface area of the vessel.

(ii) radius of the base.

(iii) capacity of the vessel.

Solution:

(i) We have given that

Rate of painting = Rs. 20 per m^{2}

and total cost of curved surface area of cylindrical vessel = Rs. 2200

It means that inner curved surface area of cylindrical vessel = \(\frac{2200}{20}\) = 110 m^{2}

(ii) we know that curved surface ara of cylinder = 2πrh

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

⇒ r = \(\frac{110 \times 7}{2 \times 22 \times 10}\) = 1.75 m

(iii) We know that volume of cylinder = πr^{2}h

= \(\frac {22}{7}\) × 1.75 × 1.75 × 10

= 96.25 m^{3}

Therefore, Capacity of the vessel = 96.25 m^{3} = 96.25 kl

(1 m^{3} = 1000 l =1 kl)

Question 6.

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

Solution:

We have given that

Capacity of closed cylindrical vessel = 15.41 = \(\frac{15.4}{1000}\) m^{3}

and height of the cylindrical vessel = 1 m.

We know that,

volume of cylinder = πr^{2}h

⇒ \(\frac{15.4}{1000}\) = \(\frac {22}{7}\) × r^{2} × 1

⇒ r^{2} = 0.0049

⇒ r = 0.07m

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

= 2 × \(\frac {22}{7}\) × 0.07 (1 + 0.07)

= 0.4708 m^{2}

Therefore, a metal sheet would be needed to make the required cylinder is 0.4708 m^{2}.

Question 7.

A lead pencil consists of a cylinder of wood will a 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.

Solution:

We have given that

Diameter of pencil = 7 mm

Radius of pencil = 3.5 mm

and height of pencil = 14 cm = 140 mm

Volume of pencil = πr^{2}h

= \(\frac {22}{7}\) × 3.5 × 3.5 × 140

= 5390 mm^{3}

Agian, we have

Diamter of lead = 1 mm

Radius of lead = 0.5 mm

and height of lead = 14 cm = 140 mm

Volume of lead = πr^{2}h

= \(\frac {22}{7}\) × 0.5 × 0.5 × 140

= 110 mm^{3}

Volume of graphite = 110 mm^{3} = 0.11 cm^{3}

and volume of wood = 5390 – 110

= 5280 mm^{3}

= 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:

We have given that

Diameter of pencil = 7 mm

Radius of cylindrical bowl = 3.5 cm

and height of soup to serve = 4 cm.

Volume of soup to serve one patient = πr^{2}h

= \(\frac {22}{7}\) × 3.5 × 3.5 × 4

= 154 cm^{3}

Therefore, volume of soup to serve 250 patients are = 250 × 154

= 38500 cm^{3}.

= 38.5 litre.

Hence, Hospital has to prepare daily 38.5 litres of soup to serve 250 patients.