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

Assume = \(\frac{22}{7}\) 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} = 1l)

Solution:

Circumference of base, C =132 cm ,

h = 25 cm

C = 2πr

132 = 2 x \(\frac{22}{7}\) x r

r = \(\frac{132×7}{2×22}\) = 21 cm

Volume of cylinder =πr^{2}h

= \(\frac{22}{7}\) x 21 x 21 x 25 = 34650 cm^{3}

= \(\frac{34650}{1000}\)

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.

Solution:

Inner diameter

d_{1} = 24 cm ⇒ r_{1} = 12 cm

Outer diameter, d_{2} = 28 cm ⇒ r_{2} = 14 cm

Volume of wood in the piple = πr_{2}^{2}h – πr_{1}^{2}h

πh(r_{2}^{2} – r_{1}^{2}) = \(\frac{22}{7}\) x 35(14^{2} – 12^{2})

= \(\frac{22}{7}\) x 35 x 52 = 5720 cm^{3}

Mass of pipe = 0.6 x 5720 = 3432 gm

= 3.432 kg.

Question 3.

A soft drink is available in two packs –

- a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and
- a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Solution:

1. Volume of cuboidal can = (5 x 4 x 15) = 300 cm^{3}

2. Volume of cylindrical can = πr^{2}h = \(\frac{22}{7}\) x 3.5 x 3.5 x 10

= 385 cm^{2}

Capacity of cylindrical can is more than the cuboidal can by 85 cm2.

Question 4.

If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then find

- radius of its base
- its volume. (Use π = 3.14)

Solution:

1. CSA of cylinder = 94.2 cm^{2}

2πrh = 94.2 cm^{2}

2 x 3.14 x 5 = 94.2 cm^{2}

r = \(\frac{94.2}{2×3.14×5}\) = 3 cm

Volume of the cylinder = πr^{2}h

= 3.14 x 3 x 3 x 5

= 3.14 x 45 = 141.3 cm^{3}

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

- inner curved surface area of the vessel
- radius of the base
- Capacity of the Vessel.

1. Inner curved surface area =

2. Let r be the radius of the base ICSA = 2πrh

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

= 1.75 m

3. Capacity of vessel = πr^{2}h = \(\frac{22}{7}\) x 1.75 x 1.75 x 10

= 96.25 m^{3}.

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?

Solution:

h = 1 m

V = 15.4l

= \(\frac{15.4}{1000}\)m^{3} = 0.0154 m^{3}

Volume of the vessel = πr^{2}h

0. 0154 = \(\frac{22}{7}\) x r^{2} x 1

∴ r^{2} = \(\frac{0.0154×7}{22}\)

⇒ r^{2} = 0.0049

∴ r = 0.07 m

Area of metal sheet required = 2nr (h + r)

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

= 0.4708 m^{2}.

Question 7.

A lead pencil consists of a cylinder ofvyood with 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:

Volume of the pencil = πr^{2}h

= \(\frac{22}{7}\) x 3.5 x 3.5 x 140 = 5390 mm^{3}

Volume of graphite = πr^{2}h

= \(\frac{22}{7}\) x 0.5 x 0.5 x 140 = 110 mm^{3}

Volume of wood = Volume of pencil – Volume of graphite

= 5390 – 110

= 5280 mm^{3}

=5.28 cm^{3}.

Question 8.

A patient in a hospital is given soup daily in a cylindrical biftvl 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:

d = 7cm

r = 3.5 cm

h = 4cm

Volume of the bowl = πr^{2}h

= \(\frac{22}{7}\) x 3.5 x 3.5 x 4 = 154 cm^{3}

Volume of soup needed for 250 patients = 154 x 250

= 38500 cm^{3}

= 38.5l.