From the solid cylinder whose height is $2.4cm$ and diameter $1.4cm$, a conical cavity of the same height and same diameter is hollowed out. Find the volume of the remaining solid to the nearest cm cube. Find its total surface area.
Last updated date: 19th Mar 2023
•
Total views: 205.2k
•
Views today: 1.84k
Answer
205.2k+ views
Hint: The area can be defined as the space occupied by a flat surface of an object. The area is the number of unit squares closed by figure. Perimeter is the total length of the sides of the two dimensional shape. Perimeter is always less than the area of the given figure. Because the perimeter is outer and the area is inner property. Volume is the capacity which hold by objects
As we know that
$\therefore V = \dfrac{1}{3}\pi {r^2}h$
Here
V=volume
r=radius
h=height
Complete step-by-step solution:
Given,
Diameter of cylinder,$d = 1.4cm$
Radius of cylinder, $r = \dfrac{{1.4}}{2}cm$
Radius of cylinder, $r = 0.7cm$
Height of cylinder, $h = 2.4cm$
Diameter of cone,$d = 1.4cm$
Radius of cone, $r = \dfrac{{1.4}}{2}cm$
Radius of cone, $r = 0.7cm$
Height of cone, $h = 2.4cm$
Volume=?
Now the volume of cylinder
As we know that
$\therefore V = \pi {r^2}h$
Put the value
\[ \Rightarrow V = \dfrac{{22}}{7} \times {(0.7)^2} \times 2.4\]
Simplify
\[ \Rightarrow V = 22 \times 0.7 \times 2.4\]
$ \Rightarrow {V_{cylinder}} = 36.96c{m^3}$
Now volume of cone
$\therefore V = \dfrac{1}{3}\pi {r^2}h$
Put the value
\[ \Rightarrow V = \dfrac{1}{3} \times \dfrac{{22}}{7} \times {(0.7)^2} \times 2.4\]
Simplify
\[ \Rightarrow V = 22 \times 0.7 \times 0.8\]
$ \Rightarrow {V_{cone}} = 12.32c{m^3}$
Volume remaining
$ \Rightarrow {V_{remaining}} = {V_{cylinder}} - {V_{cone}}$
Put the value
$ \Rightarrow {V_{remaining}} = 36.96 - 12.32$
$ \Rightarrow {V_{remaining}} = 24.64c{m^3}$
Now the total curved surface area of cylinder
As we know that
$\therefore 2\pi rh$
Put the value
$ \Rightarrow {A_{cylinder}} = 2 \times \dfrac{{22}}{7} \times 0.7 \times 2.4$
Simplify
$ \Rightarrow {A_{cylinder}} = 44 \times 0.1 \times 2.4$
\[ \Rightarrow {A_{cylinder}} = 10.56c{m^2}\]
Now curved surface area of cone,
$\therefore {A_{cone}} = \pi rl$
Now find the slant height l=
$\therefore l = \sqrt {{r^2} + {h^2}} $
Put the value
$ \Rightarrow l = \sqrt {{{(0.7)}^2} + {{(2.4)}^2}} $
$ \Rightarrow l = \sqrt {0.49 + 5.76} $
$ \Rightarrow l = \sqrt {6.25} $
$ \Rightarrow l = 2.5cm$
As we know that
$\therefore {A_{cone}} = \pi rl$
Put value
$ \Rightarrow {A_{cone}} = \dfrac{{22}}{7} \times 0.7 \times 2.5$
$ \Rightarrow {A_{cone}} = 2.2 \times 2.5$
$ \Rightarrow {A_{cone}} = 5.5c{m^2}$
Area of base is given by
As we know that
$\therefore {A_c} = \pi {r^2}$
Put the value
$ \Rightarrow {A_c} = \dfrac{{22}}{7} \times {(0.7)^2}$
Simplify
$ \Rightarrow {A_c} = \dfrac{{22}}{7} \times 0.7 \times 0.7$
$ \Rightarrow {A_c} = 2.2 \times 0.7$
$ \Rightarrow {A_c} = 1.54c{m^2}$
Total curved surface area
$\therefore {A_T} = {A_{cylinder}} + {A_{cone}} + {A_{circle}}$
Put the value
$ \Rightarrow {A_T} = 10.56 + 5.5 + 1.54$
$ \Rightarrow {A_T} = 17.6c{m^2}$
Hence the total surface area of cylinder is\[17.6c{m^2}\]
Note: Curved surface area means it covered the area or curved surface. It leaves that area of base of top and bottom. When we talk about the total surface area the area of base and curved surface area will include. There is no parameter of 3-D geometry.
As we know that
$\therefore V = \dfrac{1}{3}\pi {r^2}h$
Here
V=volume
r=radius
h=height
Complete step-by-step solution:
Given,

Diameter of cylinder,$d = 1.4cm$
Radius of cylinder, $r = \dfrac{{1.4}}{2}cm$
Radius of cylinder, $r = 0.7cm$
Height of cylinder, $h = 2.4cm$
Diameter of cone,$d = 1.4cm$
Radius of cone, $r = \dfrac{{1.4}}{2}cm$
Radius of cone, $r = 0.7cm$
Height of cone, $h = 2.4cm$
Volume=?
Now the volume of cylinder
As we know that
$\therefore V = \pi {r^2}h$
Put the value
\[ \Rightarrow V = \dfrac{{22}}{7} \times {(0.7)^2} \times 2.4\]
Simplify
\[ \Rightarrow V = 22 \times 0.7 \times 2.4\]
$ \Rightarrow {V_{cylinder}} = 36.96c{m^3}$
Now volume of cone
$\therefore V = \dfrac{1}{3}\pi {r^2}h$
Put the value
\[ \Rightarrow V = \dfrac{1}{3} \times \dfrac{{22}}{7} \times {(0.7)^2} \times 2.4\]
Simplify
\[ \Rightarrow V = 22 \times 0.7 \times 0.8\]
$ \Rightarrow {V_{cone}} = 12.32c{m^3}$
Volume remaining
$ \Rightarrow {V_{remaining}} = {V_{cylinder}} - {V_{cone}}$
Put the value
$ \Rightarrow {V_{remaining}} = 36.96 - 12.32$
$ \Rightarrow {V_{remaining}} = 24.64c{m^3}$
Now the total curved surface area of cylinder
As we know that
$\therefore 2\pi rh$
Put the value
$ \Rightarrow {A_{cylinder}} = 2 \times \dfrac{{22}}{7} \times 0.7 \times 2.4$
Simplify
$ \Rightarrow {A_{cylinder}} = 44 \times 0.1 \times 2.4$
\[ \Rightarrow {A_{cylinder}} = 10.56c{m^2}\]
Now curved surface area of cone,
$\therefore {A_{cone}} = \pi rl$
Now find the slant height l=
$\therefore l = \sqrt {{r^2} + {h^2}} $
Put the value
$ \Rightarrow l = \sqrt {{{(0.7)}^2} + {{(2.4)}^2}} $
$ \Rightarrow l = \sqrt {0.49 + 5.76} $
$ \Rightarrow l = \sqrt {6.25} $
$ \Rightarrow l = 2.5cm$
As we know that
$\therefore {A_{cone}} = \pi rl$
Put value
$ \Rightarrow {A_{cone}} = \dfrac{{22}}{7} \times 0.7 \times 2.5$
$ \Rightarrow {A_{cone}} = 2.2 \times 2.5$
$ \Rightarrow {A_{cone}} = 5.5c{m^2}$
Area of base is given by
As we know that
$\therefore {A_c} = \pi {r^2}$
Put the value
$ \Rightarrow {A_c} = \dfrac{{22}}{7} \times {(0.7)^2}$
Simplify
$ \Rightarrow {A_c} = \dfrac{{22}}{7} \times 0.7 \times 0.7$
$ \Rightarrow {A_c} = 2.2 \times 0.7$
$ \Rightarrow {A_c} = 1.54c{m^2}$
Total curved surface area
$\therefore {A_T} = {A_{cylinder}} + {A_{cone}} + {A_{circle}}$
Put the value
$ \Rightarrow {A_T} = 10.56 + 5.5 + 1.54$
$ \Rightarrow {A_T} = 17.6c{m^2}$
Hence the total surface area of cylinder is\[17.6c{m^2}\]
Note: Curved surface area means it covered the area or curved surface. It leaves that area of base of top and bottom. When we talk about the total surface area the area of base and curved surface area will include. There is no parameter of 3-D geometry.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
