Surface Areas & Volume

Mathematics Class Tenth

10th-Math-home

RD Sharma exercise 16.1 q6 to 10

Question (6) A cylindrical vessel having diameter equal to its height is full of water which is poured into two identical cylindrical vessels with diameter 42 cm and height 21 cm which are filled completely. Find the diameter of the cylindrical vessel.

Solution:

[ Strategy to solve the RD Sharma tenth math exercise 16.1 Question (6) (a) Find the volume of identical vessel. (b) Sum of the volume of both of the identical vessels will be equal to the volume of cylindrical vessel. (c) Using volume of cylindrical vessel find the diameter. ]

Given, Height of identical small vessel = 21 cm

Diameter of small vessel = 42 cm

Therefore, radius = 42/2 = 21 cm

Diameter of big vessel = height of big vessel.

Let radius of big vessel = r

Therefore, diameter = 2r

Therefore, height of the big vessel = 2r (Since according to question, diameter and height are equal)

Therefore, diameter of big vessel = ?

We know that, volume of a cylinder `=4/3\ pi\ r^2\ h`

Thus, volume of given small vessel = `4/3\ pi\ (21\ cm)^2\ xx\ 21\ cm`

`=4/3 \ pi\ (21\ cm)^3`

Now, volume of two vessel = volume of one vessel × 2

`=2xx4/3 \ pi\ (21\ cm)^3`

Now, since water is completely filled in small identical vessels, therefore,

Sum of the volume of identical vessels = Volume of big vessel

Thus, Volume of big vessel `=2xx4/3 \ pi\ (21\ cm)^3`

`=>4/3\ pi\ r^2 xx h` `=2xx4/3 \ pi\ (21\ cm)^3`

`=>4/3\ pi\ r^2 xx 2\ r` `=2xx4/3 \ pi\ (21\ cm)^3`

[since height = 2r]

`=>2xx4/3\ pi\ r^3` `=2xx4/3 \ pi\ (21\ cm)^3`

`=>r^3 = (2xx4/3 \ pi\ (21\ cm)^3)/( 2xx4/3\ pi)`

`=>r^3 = (21\ cm)^3`

Thus, radius (r) = 21 cm

Thus, diameter = 2 × r = 21 cm × 2 = 42 cm

Thus, diameter of big vessel = 42 cm Answer

Question (7) 50 circular plates each of diameter 14 cm and thickness 0.5 cm are placed one above the other to form a right circular cylinder. Find its total surface area.

Solution:

[Strategy to solve the RD Sharma tenth math exercise 16.1 Question (7) (a) Find the height of right circular cylinder by multiplying thickness of one plate to 50 plates. (b) Radius of circular plate will be equal to radius of cylinder. (c) Calculate the surface area of cylinder by taking radius and height.]

Given, Diameter of circular plate = 14 cm

Therefore, radius of circular plate= 14/2 = 7 cm

Now, number of circular plate placed one above other = 50

And, thickness of one plate = 0.5 cm

Thus, total thickness of plates, i.e. height of cylinder made = 0.5 cm × 50

= 25 cm

Thus, height of cylinder = 25 cm

And radius of cylinder = 7 cm

Thus, total surface area of cylinder so made = ?

We know that, surface area of cylinder `=2\ pi\ r(h+r)`

Thus, surface of cylinder formed in question `=2xx22/7xx7cm(25cm+7cm)`

`=2xx22xx31cm^2 = 1408\ cm^2`

Thus, total surface area of cylinder formed by placing all the plates one above other = 1408 cm2 Answer

Question (8) 25 circular plates, each of radius 10.5 cm and thickness 1.6 cm, are placed one above the other to form a solid circular cylinder. Find the curved surface area and the volume of the cylinder so formed.

Solution:

[Strategy to solve the RD Sharma tenth math exercise 16.1 Question (8) (a) Find the height of right circular cylinder by multiplying thickness of one plate to 25 plates. (b) Radius of circular plate will be equal to radius of cylinder. (c) Calculate the curved surface area and volume of cylinder by taking radius and height.]

Given, radius of circular plate = 10.5 cm

Thickness of one circular plate = 1.6 cm

And number of plates = 25

Thus, thickness of total plates, i.e. height of cylinder formed by placing all plates one above other

= Thickness of one plate × Total number of plates

= 1.6 cm × 25

Thus, height of cylinder formed = 40 cm

Now, we know that, curved surface area of cylinder `=2\ pi\ r\ h`

Thus, curved surface area of cylinder formed in question `=2xx22/7xx10.5\ cmxx40\ cm`

`=(18480\ cm^2)/7 = 2640\ cm^2`

Thus, curved surface area of cylinder = 2640 cm2

Now, we know that, volume of cylinder `=pi\ r^2\ h`

Thus, volume of cylinder formed in question `=22/7xx(10.5\ cm)^2xx40cm`

`=(97020\ cm^3)/7=13860\ cm^3`

Thus, volume of cylinder = 13860 cm3

Thus, curved surface = 2640 cm2 and volume = 13860 cm3 Answer

Question (9) A path 2 m wide surrounds a circular pond of diameter 40 m. How many cubic meters of gravel are required to grave the path to a depth of 20 cm?

Solution:

[Strategy to solve the RD Sharma tenth math exercise 16.1 Question (9) (a) Find the area of circular pond. (b) Find the area of pond along with path. (c) Find the area of path after subtracting area of pond from the area of pond along with path. (d) Now find the volume of path by multiplying its depth which will be equal to the volume of gravel required.]

Given, Diameter of circular pond = 40 m

Therefore, radius of circular pond = 40/2 = 20 m

Width of circular path surrounds the pond = 2 m

Now, radius of circular pond along with path = radius of pond + width of path

= 20 m + 2 m = 22 m

Depth of path to be graved = 20 cm = 20/100 = 0.2 m

Now, we know that area of circle `=pi\ r^2`

Thus, area of circular path = Area of circular pond along with path – Area of circular pond

`=pi\ (22\ m)^2\ –\ pi\ (20\ m)^2`

`=pi(484–400) m^2`

`=22/7xx84\ m^2`

= 22 × 12 m2

= 264 m2

Thus, area of path = 264 m2

Now, depth of the path to be graved = 0.2 m

Thus, volume of gravel required = Area of path × depth

= 264 m2 × 0.2 m

= 52.8 m3

Thus, required gravel to grave the path = 52.8 cubic meter Answer

Question (10) A 16 m deep well with diameter 3.5m is dug up and the earth from it is spread evenly to form a platform 27.5 m by 7 m. Find the height of the platform.

Solution:

[Strategy to solve the RD Sharma tenth math exercise 16.1 Question (9) (a) Find the volume of well dug (b) The volume of well will be equal to the volume of platform. (c) By using volume find the height of platform using width and length given in question.]

Given, diameter of well = 3.5 m

Thus, radius of well = 3.5/2 = 1.75 m

Again given, depth of well = 16 m = height of the well.

Now, earth taken from well is to be spread evenly to form a platform.

Length of platform = 27.5 m

And, width of platform = 7 m

Then height of platform = ?

We know that, volume of cylinder `=pi\ r^2\h`

Since well is in the shape of cylinder,

Thus, volume of earth dug from well = Volume of well

`=22/7xx(1.75\ m)^2xx16\ m`

`=(1078\ cm^3)/7=154\ m^3`

Thus, volume of earth dug from well = 154 m3

Now, since earth is spread to form rectangular platform, thus

Volume of platform = Volume of earth dug from well

Now, we know that volume of a rectangular platform = Length (l) × Width (w) × Height (h)

⇒ 154 m3 = 27.5 m × 7 m × h

`=> h = (154\ m^3)/(27.5\ m xx7m)`

`=> h = (154\ m^3)/(192.5\ m^2)`

Or, h = 0.8 m

Thus, height of platform = 0.8 m or 80 cm Answer

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