Surface Areas & Volume

Mathematics Class Tenth

10th-Math-home

RD Sharma Exercise 16.1 q16 to 20

Question (16) A rectangular tank 15 m long and 11 m broad is required to receive entire liquid contents from a full cylindrical tank of internal diameter 21 m and length 5 m. Find the least height of the tank that will serve the purpose.

Solution:

[Strategy to solve the RD Sharma tenth math exercise 16.1 Question (16) (a) Find the volume of cylindrical tank which will be equal to the volume of liquid content in that. (b) Using the volume of liquid content find the height of the rectangular tank.]

Given, Internal diameter of cylindrical tank = 21 m

Thus, radius = 21/2 = 10.5 m

And length, i.e. height of the cylindrical tank = 5 m

Length of rectangular tank = 15 m

Width of the rectangular tank = 11 m

Thus, height of the tank to contain the liquid = ?

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

Thus, volume of the given cylindrical tank `=22/7xx(10.5\ m)^2xx5\ m`

`=(22xx110.25xx5)/7\ m^3`

= 1732.5 m3

Thus, volume of cylindrical tank = volume of liquid content = 1732.5 m3

Now, since whole liquid content is to be poured into rectangular tank, thus volume of rectangular tank should be equal to the volume of liquid content.

Now, we know that, volume of a rectangle = length × width × height

⇒ 1732.5 m3 = 15 m × 11 m × h

⇒ 165 m2 × h = 1732.5 m3

`=>h = (1732.5\ m^3)/(165\ m^2)`

⇒ 10.5 m

Thus, height of rectangular tank should be = 10.5 m Answer

Question (17) A hemispherical bowl of internal radius 9 cm is full of liquid. This liquid is to be filled into cylindrical shaped small bottles each of diameter 3 cm and height 4 cm. How many bottles are necessary to empty the bowl?

Solution:

[Strategy to solve the RD Sharma tenth math exercise 16.1 Question (17) (a) Find the volume of hemispherical bowl, which is equal to the volume of liquid in it. (b) Find the volume of one cylindrical bottle. (c) Divide the volume of hemispherical bowl by volume of cylindrical bottle. This will give the number of bottles required to empty the bowl]

Given, Radius of hemispherical bowl = 9 cm

Height of cylindrical shaped bottle = 4 cm

And, diameter of cylindrical shaped bottle = 3 cm

Thus, radius = 3/2 = 1.5 cm

The hemispherical bowl is full of liquid, thus no. of cylindrical bottles required to empty the bowl = ?

Calculation of volume of liquid in hemispherical bowl

We know that volume of a hemisphere `=2/3\ pi\ r^3`

Thus, volume of given hemispherical bowl `=2/3\ pi\ (9\ cm)^3`

`=2xx3xx9xx9\ pi\ cm^3`

`=486\ pi\ cm^3`

Since, this hemisphere bowl is full of liquid,

Thus volume of liquid = volume of hemispherical bowl

Thus, volume of liquid `=486\ pi\ cm^3`

Calculation of volume of cylindrical bottle

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

Thus, volume of given cylindrical bottle `=pi\ (1.5\ cm)^2xx4\ cm`

`=2.25 xx4\ pi\ cm^3`

Or, Volume of cylindrical bottle `=9\ pi\ cm^3`

Calculation of number of bottles required

Now, number of bottles required to empty the liquid in bowl = Volume of liquid/Volume of one bottle

`=(486\ pi\ cm^3)/( 9\ pi\ cm^3)` = 54

Thus, number of bottles required to empty the liquid in bowl = 54 Answer

Question (18) The diameters of the internal and external surfaces of a hollow spherical shell are 6 cm and 10 cm respectively. If it is melted and recast into solid cylinder of diameter 14 cm, find the height of the cylinder.

Solution:

[Strategy to solve the RD Sharma class ten math exercise 16.1 Question (18) (a) Calculate the volume of material into spherical shell. (b) The volume of the material of spherical shell will be equal to the volume of solid cylinder. (c) Using this volume of cylinder, find the height of cylinder using formula for volume of cylinder.]

Given, External diameter of hollow spherical shell = 10 cm

Thus, external radius of hollow spherical shell (R) = 10/2 =5 cm

Internal diameter of hollow spherical shell = 6 cm

Thus, internal radius (r) = 6/2=3 cm

Diameter of cylinder = 14 cm

Thus, radius of cylinder = 14/2 = 7 cm

Thus, when given hollow sphere is melted and recast as cylinder,

Then height of cylinder = ?

Calculation of volume of hollow spherical shell

We know that, volume of material of hollow spherical shell `=4/3\ pi\ (R^3–r^3)`

Thus, volume of material of given hollow spherical shell `=4/3\ pi\ [(5\ cm)^3–(3\ cm)^3]`

`= 4/3\ pi\ (125–27)\ cm^3`

`=392/3\ pi\ cm^3`

Thus, volume of material of hollow sphere `=392/3\ pi\ cm^3`

Calculation of height of cylinder

Since, cylinder is to be cast from the material of hollow sphere, thus volume of cylinder = volume of material of hollow sphere

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

Thus, volume of given cylinder `=pi\ (7\ cm)^2xxh`

`=>392/3\ pi\ cm^3 = pi\ 7xx7\ cm^2\ h`

Thus, `h = (392\ pi\ cm^3)/(3xx7xx7xxpi\ cm^2)`

⇒ h = 2.66 cm

Thus, height of cylinder = 2.66 cm Answer

Question (19) A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Calculate the height of the cone.

Solution:

[Strategy to solve the RD Sharma class ten math exercise 16.1 Question (19) (a) Find the volume of material of hollow sphere. (b) The volume of material is to be cast into cone, thus volume of cone will be equal to the volume of material of hollow sphere. (c) Using the volume of material, i.e. volume of cone and diameter given in question find the height of cone.]

Given, External diameter of hollow sphere = 8 cm

Thus, external radius (R) of hollow sphere = 8/2= 4 cm

The Internal diameter of hollow sphere = 4 cm

Thus, internal radius (r) of hollow sphere = 4/2 = 2 cm

Diameter of base of the cone = 8 cm

Thus, base radius of cone = 8/2=4 cm

Now, hollow sphere is melted into the cone, thus height of cone = ?

Calculation of volume of material in hollow sphere

We know that, Volume of material in hollow sphere = External volume – Internal volume

`=4/3\ pi\ R^2 – 4/3\ pi\ r^2`

`=4/3\ pi\ (R^3–r^3)`

Thus, volume of material in the given hollow sphere `=4/3\ pi((4\ cm)^3–(2\ cm)^3)`

`=4/3\ pi\ (64–8)cm^3`

Or, Volume of material in the given hollow sphere `=4/3\ pi\ xx56\ cm^3`

[For ease of calculation do not calculate the value of `4/3\ pi` rather left them as it is. This will help in less calculation further.]

Calculation of height of cone

Since hollow sphere is melted into cone, thus volume of cone will be equal to the volume of material of hollow sphere.

We know that, volume of cone `=1/3\ pi\ r^2\ h`

`=>4/3\ pi\ xx56\ cm^3=1/3\ pi\ (4\ cm)^2\ h`

`=>1/3\ pi xx 16\ cm^2\ h =4/3\ pi\ xx56\ cm^3`

`:. h = (4xx3xxpixx56\ cm^3)/(3xxpi\ xx16\ cm^2)`

`=> h = 14\ cm`

Thus, height of cone = 14 cm Answer

Question (20) A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical ball is dropped into the tub and the level of the water is raised by 6.75 cm. Find the radius of the ball.

Solution :

[Strategy to solve the RD Sharma class ten math exercise 16.1 Question (20) (a) After drop of ball into tub the increased level of water is due the volume of ball. This increased volume of water is equal to the volume of ball dropped into water in cylinder. Thus, using the height of water raised and radius of cylinder, find the volume of water of cylinder. This volume will be equal to the volume of ball. (b) Now using the volume of ball find the radius of ball using formula to calculate the volume of spherical ball.]

Given, the radius of cylindrical tub = 12 cm

And, depth of water, = 20 cm.

The height of water raised when ball is dropped into cylinder = 6.75 cm

Thus, radius of ball = ?

Calculation of raised volume of water because of ball dropped into cylinder

Here since water level increased because of ball dropped into water of cylinder. The raised volume will be equal to the volume of ball. Thus, by taken the height of raised water and radius of cylinder the increased volume of water can be calculated. This increased volume of water will be equal to the volume of ball.

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

Thus, increased volume of water in cylinder `= pi\ (12\ cm)^2xx6.75\ cm`

`=pi\ 144xx6.75\ cm^3`

`=pi\ 972\ cm^3`

Thus, increased volume of water `=pi\ 972\ cm^3`

Calculation of radius of spherical ball

We know that, volume of spherical ball `=4/3\ pi\ r^3`

Since increased volume of water = volume of spherical ball

Thus, ` pi\ 972\ cm^3 =4/3\ pi\ r^3`

`=>r^3=(pi\ 972xx3\ cm^3)/(4xxpi)`

`=>r^3 = 729\ cm^3`

Thus, `r = 3sqrt(729\ cm^3)`

Or, r = 9 cm

Thus, radius of ball = 9 cm Answer

 

Alternate method

[Strategy to solve the RD Sharma class ten math exercise 16.1 Question (20) (a) Calculate the volume of water in cylinder. (b) Calculate the volume of water along with volume of ball taking raised water level into consideration. (c) Difference in volume will give the volume of ball dipped in water (d) Calculate the radius of ball using volume of ball.]

Given, radius of cylinder = 12 cm

Height of water in cylinder = 20 cm

Height of water raised after ball is dropped = 6.75 cm

Thus, radius of ball = ?

Calculation of volume of water in cylinder

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

Thus, volume of water in cylinder `=pi\ (12\ cm)^2\ 20\ cm`

`= pi\ 144\ cm^2\ 20\ cm`

Thus, volume of water in cylinder `=pi\ 2880\ cm^3`

[pi is left as it is for ease of calculation]

Calculation of volume of water along with volume of ball in cylinder

The height raised when ball is dropped in cylinder = 6.75 cm

Thus, total height of water = initial height of water in cylinder + height raised

= 20 cm + 6.75 cm = 26.75 cm

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

Thus, volume of water along with volume of ball in cylinder `=pi\ (12\ cm)^2xx26.75\ cm`

`=pi\ 144\ cm xx 26.75\ cm`

`= pi\ 3852\ cm^3`

Thus, volume of water along with volume of ball in cylinder `= pi\ 3852\ cm^3`

[pi is left as it is for ease of calculation]

Calculation of volume of ball

Now, volume of ball = volume of water along with volume of ball in cylinder – volume of water in cylinder

`= pi\ 3852\ cm^3 – pi\ 2880\ cm^3`

`=pi\ (3852–2880)\ cm^3`

Thus, volume of ball `=pi\ 972\ cm^3`

Calculation of radius of ball

We know that, volume of a spherical ball `=4/3\ pi\ r^3`

`=> pi\ 972\ cm^3 =4/3\ pi\ r^3`

`=>r^3=(pi\ 972xx3\ cm^3)/(4xxpi)`

`=>r^3 = 729\ cm^3`

Thus, `r = 3sqrt(729\ cm^3)`

Or, r = 9 cm

Thus, radius of ball = 9 cm Answer

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