Surface Areas & Volume

Mathematics Class Tenth

10th-Math-home

RD Sharma Exercise 16.1 q 1 to 5

Question (1) How many balls, each of radius 1 cm, can be made from a solid sphere of lead of radius 8 cm?

Solution:

Given, radius of solid sphere = 8 cm

Radius of balls = 1 cm

Therefore, number of balls can be made from solid sphere = ?

[Strategy to solve RD Sharma tenth math exercise 16.1 question (1): (1) Calculate the volume of given sphere. (2) Calculate the volume of one ball. (3) Divide the volume of sphere by volume of one ball. This will given the number of given balls made by given sphere.]

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

Thus, volume of given sphere `=4/3\ pi\ (8\ cm)^3`

`=4/3\ pi\ 512\ cm^3`

Thus Volume of given sphere `=4/3\ pi\ 512\ cm^3`

Now, volume of given ball `=4/3\ pi\ r^3`

`=4/3\ pi\ xx\ 1\ cm^3`

Thus, volume of given ball `=4/3\ pi\ 1\ cm^3`

Now, number of balls to be made from given sphere `={text(Volume of sphere)}/{text(Volume of sphere)}`

`=(4/3\ pi\ 512\ cm^3)/ (4/3\ pi\ 1\ cm^3)`

= 512

Thus, number of balls to be made by given sphere = 512 Answer

Question (2) How many spherical bullets each of 5 cm in diameter can be cast from a rectangular block of metal 11 dm × 1 m × 5 dm ?

Solution:

[Strategy to solve  RD Sharma tenth math exercise 16.1 question (2):  (1) Find the volume of given metal block. (2) Find the volume of given bullet. (3) Divide the volume of metal block by the volume of bulltet. This will given the number of bullets can be cast.]

Given, Dimension of rectangular block = 11 dm × 1 m × 5 dm

= (11 × 10 cm) × (1 × 100 cm) × (5 × 10 cm)

= 110 cm × 100 cm × 50 cm

We know that volume of a rectangular block = length × height × width

Thus, volume of given rectangular block of metal = 110 cm × 100 cm × 50 cm

Thus, volume of given rectangular block of metal = 550000 cm3

Now, given, diameter of spherical bullet = 5 cm

Thus, radius of given bullet = 5/2 = 2.5 cm

Now, we know that, volume of sphere `=4/3\ pi\ r^3`

Thus, volume of given bullet `= 4/3\ xx\ 22/7\ xx\ (2.5\ cm)^3`

`=88/21\ xx 15.625 cm^3`

= 65.476 cm3

Thus, volume of given bullet = 65.476 cm3

Now, number of spherical bullet to be cast from given metal block `={text(volume of rectangular block)}/{text(volume of spherical bullet)}`

`=(550000\ cm^3)/( 65.476 cm3)`

= 8400.02

= 8400 bullets

Thus, number of spherical bullets can cast from given rectangular block of metal = 8400 Answer

Question (3) A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of the two of the balls are 1.5 cm and 2 cm respectively. Determine the diameter of the thirds ball.

Solution:

[Strategy to solve  RD Sharma tenth math exercise 16.1 question (3): (1) Find the volume of given spherical ball. (2) Find the volume of both of the recasted small balls and add their volume. (3) Now subtract the volume of given spherical ball by the sum of the volume of both of the recasted balls. (4) The result will give the volume of third ball. (4) Using volume find the diameter.]

Given, radius of spherical ball = 3 cm

Radius of recasted two small balls = 1.5 cm and 2 cm respectively

Thus, diameter of third ball = ?

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

Thus, volume of given spherical ball `= 4/3\ pi\ 3\ cm^3`

Thus, volume of given spherical ball `= 4/3\ pi\ 27\ cm^3`

Now, volume of recasted ball (1) `=4/3\ pi\ (1.5\ cm)^3`

And volume of second recasted small ball (2) `=4/3\ pi\ (2\ cm)^3`

Now, Total volume of given recsted balls,

i.e. volume of ball (1) + volume of ball (2) `=4/3\ pi\ (1.5\ cm)^3` `+ 4/3\ pi\ (2\ cm)^3 `

`=4/3\ pi(3.375\ cm^3 + 8\ cm^3)`

`=4/3\ pixx11.375\ cm^3`

Now, volume of third recasted ball = Volume of given spherical ball – Sum of the volume of both of the recasted balls

`=4/3\ pi\ (27-11.375) cm^3`

Or, Volume of third recasted ball `=4/3\ pixx15.625\ cm^3`

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

`=> 4/3\ pixx15.625\ cm^3 =4/3\ pi\ r^3`

`=>r^3 = (4/3\ pixx15.625\ cm^3)/( 4/3\ pi)`

⇒ r3 = 15.625 cm3

`=>r = 3sqrt(15.625\ cm^3)`

⇒ r = 2.5 cm

Thus, radius of given third small ball = 2.5 cm

Now, we know that diameter = radius × 2

Thus, diameter of given small third ball = 2.5 cm × 2 = 5 cm

Thus, diameter of third recasted ball = 5 cm Answer

Qeustion (4) 2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire.

Solution:

[Strategy to solve RD Sharma tenth math exercise 16.1 question (4): (1) volume of brass is given (2) The volume of cubic brass will be equal to the volume of cylindrical wire. (3) Using volume and diameter of cylindrical wire find the length of the cylindrical wire by applying the formula for the volume of cylinder]

Given, Volume of brass = 2.2 cubic dm

= 2.2 dm3

= 2.2 ×10 × 10 × 10× cm3

= 2200 cm3

Since this brass is drawn into cylindrical wire, thus volume of cylindrical wire = volume of given brass

Thus, volume of cylindrical wire = = 2200 cm3

Now, as given diameter of cylindrical wire = 0.25 cm

Or, radius of given cylindrical wire = 0.25/2 = 0.125 cm

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

`=>2200\ cm^3= 22/7xx(0.125\ cm)^2xxh`

Thus, `h = (2200\ cm^3xx7)/(22xx0.125xx0.125\ cm^2)`

`=>h = 700/(0.125xx0.125)` cm

Or, h = 44800 cm

Or, h = 44800/100 m = 448 m

Thus, length of the cylindrical wire drawn = 44800 cm or 448 mAnswer

Question (5) What length of a solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of length 16 cm, external diameter 20 cm and thickness 2.5 mm?

Solution:

[Strategy to solve RD Sharma tenth math exercise 16.1 question (5) : (1) calculate the volume of material given hollow cylinder to be recast. (2) This will be the volume of given solid cylinder. (3) Now using the volume and given diameter find the length of the given solid cylinder using the formula for the volume of cylinder.]

Given, length of hollow cylinder = 16 cm

External diameter of hollow cylinder = 20 cm

Thus, external radius of hollow cylinder (R) = 20/2 = 10 cm

And, thickness of hollow cylinder = 2.5 mm = 2.5/10 cm = 0.25 cm

Now, internal radius of hollow cylinder (r) = External radius (R) – thickness

= 10 cm – 0.25 cm = 9.75 cm

Thus, internal radius of hollow cylinder (r) = 9.75 cm

Now, we know that Volume of hollow cylinder `=pi\ h(R^2-r^2)`

Where, R = External radius and `r` = Internal radius

Thus, volume of given hollow cylinder `= pixx16cm(10\ cm)^2–(9.75\ cm)^2`

`= pixx\ 16\ cm xx (100–95.0625) cm^2`

`=pixx\ 16xx4.9735 cm^3 = pi\ 79\ cm^3`

Thus, volume of materials in hollow cylinder `= 79\ pi\ cm^3`

This will be equal to the volume of solid cylinder to be taken to recast the given hollow cylinder.

Thus, volume of solid cylinder to be taken to recast the hollow cylinder `= 79\ pi\ cm^3`

Now, as given in question, diameter of solid cylinder to be taken = 2 cm

Thus, radius of this solid cylinder = 2 cm/2 = 1 cm

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

`=>79\ pi\ cm^3 = pi\ (1\ cm^2)\ h`

Thus, `h = (79\ pi\ cm^3)/(pi\ 1\ cm^2)`

Or, `h = 79\ cm`

Thus, length of the solid cylinder to be taken = 79 cm Answer

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