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MATHS [3 marks sum]

Surface Area, Volume and Capacity · ICSE STD 8 (ICSE - English Medium). 15 questions.

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[3 marks sum]

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15 questions · timed · auto-graded

Question 13 Marks
The diameter of a garden roller is $1.4 m$ and it $2 m$ long. Find the maximum area covered by its $50$ revolutions?
Answer
Diameter of the roller $=1.4 m$
Radius $(r)=\frac{1.4}{2}=0.7 m$
and length $(h)=2 m$
Curved surface area $=2 \pi rh =2 \times \frac{22}{7} \times 0.7 \times 2 \ cm ^2=8.8 m ^2$
Area covered in $50$ complete revolutions $=8.8 \times 50 m ^2=440 m ^2$
Area of the playground $=440 m ^2$
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Question 23 Marks
In a building, there are $24$ cylindrical pillars. For each pillar, the radius is $28 m$, and the height is $4 m$. Find the total cost of painting the curved surface area of the pillars at the rate of $₹ 8$ per $m^2.$
Answer
$>$ Radius of cylindrical pillar, $r =28 \ cm =0.28 m$,
height $= h =4 m >>>$
curved surface area of a cylinder $=2 \pi rh >$
curved surface area of a pillar $=2 \times \frac{22}{7} \times 0.28 \times 4=7.04 m ^2>$
curved surface area of $24$ such pillar $=7.04 \times 24=168.96 m ^2>$
cost of painting an area of $1 m ^2=$ Rs. $8>>$
Therefore, cost of painting $1689.6 m ^2=168.96 \times 8= Rs. 1351.68 .$
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Question 33 Marks
Find the capacity of a cylindrical container with an internal diameter of $28\  cm$ and a height of $20\  cm$.
Answer
Diameter $=28 \ cm$
Radius $=\frac{28}{2} \ cm =14 \ cm$
Height $=20 \ cm$
Volume $=\pi r^2 h=\frac{22}{7} \times 14 \times 14 \times 20$
Volume $=12320 \ cm ^3$
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Question 43 Marks
The ratio between the curved surface area and the total surface area of a cylinder is $1: 2$. Find the ratio between the height and the radius of the cylinder.
Answer
Let $r$ be the radius and $h$ be the height of a right circular cylinder, then Curved surface area $=2 \pi r h$ and total surface area $=2 \pi r h \times 2 \pi r^2=2 \pi r(h+r)$
But their ratio is $1: 2$
$\therefore \frac{2 \pi r}{2 \pi r(h+r)}=\frac{1}{2} $
$\Rightarrow \frac{h}{h+r}=\frac{1}{2}$
$ \Rightarrow 2 h=h+r $
$\Rightarrow 2 h-h=r$
$ \Rightarrow h=r=1: 1$
Hence their radius and height are equal
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Question 53 Marks
The curved surface area of a cylinder of height $14\  cm$ is $88\ cm^2.$ Find the diameter of the base of the cylinder.
Answer
Height $(h)=14 \ cm$
Curved surface area $(2 \pi rh )=88 \ cm ^2$
Then, $2 \pi rh =88 \ cm ^2$
$

\Rightarrow 2 \times \frac{22}{7} \times r \times 14=88 \ cm ^2$
$ \Rightarrow 88 r =88

$
$
\Rightarrow r =\frac{88}{88}=1 \ cm
$
Then diameter $=1 \times 2=2 \ cm$
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Question 63 Marks
The height of a circular cylinder is $20\  cm$ and the diameter of its base is $14 \ cm$. Find : $(i)$ the volume $(ii)$ the total surface area.
Answer
Height of cylinder $(h)=20 \ cm$ and diameter of its base $(d) =14 \ cm$ and radius of its base $(r)=\frac{14}{2}=7 \ cm (i)$ Volume $=\pi r^2 h$
$=\frac{22}{7} \times 7 \times 7 \times 20 \ cm ^3=3080 \ cm ^3$

$(ii)$ Total surface area $=2 \pi r(h+r)$
$=2 \times \frac{22}{7} \times 7(20+7) \ cm ^2=44 \times 27=1188 \ cm ^2$
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Question 73 Marks
The capacity of a rectangular tank is $5.2\ m^3$ and the area of its base is $2.6 \times 104\ cm^2;$ find its height $($depth$)$.
Answer
Capacity of a tank $=5.2 m ^3$
and area of its base $=2.6 \times 10^4 \ cm ^2$
$=\frac{2.6 \times 10000}{100 \times 100}=2.6 m ^2$
$ \Rightarrow lb =2.6 m ^2$
and $Ibh =5.2 m ^3$
$\therefore$ Height $(h)=\frac{5.2}{2.6}=2 m$
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Question 83 Marks
The length of the diagonals of a cube is $8\sqrt3 \ cm$.Find its$:(i)$ edge$\ (ii)$ total surface area$\ (iii)$ Volume
Answer
$(i)$ Length of diagonal of a cube $=8 \sqrt{3} \ cm$ Length of edge $=\frac{8 \sqrt{3}}{\sqrt{3}}=8 \ cm$
$(ii)$ Total surface area $=6 a ^2=6 \times 8^2=6 \times 64 \ cm ^2=384 \ cm ^2$
$(iii)$ Volume $=a^3=(8)^3=512 \ cm ^3$
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Question 93 Marks
Find the area of metal-sheet required to make an open tank of length $= 10 m$, breadth $= 7.5 m$ and depth $= 3.8 m$.
Answer
Length of the tank $= 10 m$ Breadth of the tank $= 7.5 m$
Depth of the tank $= 3.8 m$
Area of four walls $= 2[L+B] \times H = 2(10 + 7.5) \times 3.8$
$= 2 \times 17.5 \times 3.8 = 35 \times 3.8 = 133 m^2$
Area of the floor $= L \times B = 10 \times 7.5 = 75 m$
Area of metal sheet required to make the tank $=$Area of four walls $+$ Area of floor $= 133\ m^2 + 75\ m^2 = 208\ m^2$
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Question 103 Marks
The length, breadth, and height of a room are $6\  m, 5.4 \ m$, and $4 \ m$ respectively. Find the area of :$(i)$ its four$-$walls$(ii)$ its roof.
Answer
Length of the room $= 6 \ m$ The breadth of the room $= 5.4\  m$
Height of the room $= 4\  m$
$(i)$ Area of four walls $= 2(L+B) \times H$
$= 2(6 + 5.4) \times 4 = 2 \times 11.4 \times 4 = 91.2\ m^2$
$(ii)$ Area of the roof $= L \times B = 6 \times 5.4 = 32.4\ m^2$
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Question 113 Marks
A tank $30\  m$ long, $24 \ m$ wide, and $4.5\  m$ deep is to be made. It is open from the top. Find the cost of iron$-$sheet required, at the rate of $₹ 65 \ per\  m^2,$ to make the tank.
Answer
Length of the tank $= 30\  m$ Width of the tank $= 24\  m$
Depth of the tank $= 4.5\  m$
Area of four walls of the tank $= 2[L+B] \times H = 2(30 + 24) \times 4.5 = 2 \times 54 \times 4.5 m^2 = 486 \ m^2$
Area of the floor of the tank $= L \times B = 30 \times 24 = 720\ m^2$
Area of Iron sheet required to make the tank $=$ Area of four walls $+$ Area of floor $= 486 + 720 = 1206\ m^2$
Cost of iron sheet required $@\ ₹\ 65$ per $m^2 = 1206 \times 65 = ₹ 78390$
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Question 123 Marks
The total surface area of a cube is $216 \ cm ^2$. Find its volume.
Answer
$6($Edge$)^2=$Total surface area of a cube 
$ 6($Edge$)^2=216 \ cm ^2$
$ =>($Edge$)^2=\frac{216}{6}$
$ =>($Edge$)^2=36$
$ \Rightarrow$ Edge $=\sqrt{36}$
$ \Rightarrow $ Edge $=6 \ cm$
Volume of the given cube $=($Edge$)^3=(6)^3=6 \times 6 \times 6=216 \ cm ^3$
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Question 133 Marks
The volume of a cuboid is $7.68\ m^3$. If its length $= 3.2\ m$ and height $= 1.0\ m$; find its breadth.
Answer
The volume of a cuboid $=7.68 m ^3$
Length of a cuboid $=3.2 m$
Height of a cuboid $=10 m$
We know
Length $x$ Breadth $x$ Height $=$ Volume of a cuboid
$3.2 \times$ Breadth $\times 1.0=7.68$
$\Rightarrow$ Breadth $=\frac{7.68}{3.2 \times 1.0}$
$\Rightarrow$ Breadth $=\frac{7.68}{3.2}$
$\Rightarrow$ Breadth $=2.4 m$
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Question 143 Marks
The volume of a cuboid is $3456 \ cm^3.$ If its length $= 24 \ cm$ and breadth $= 18 \ cm$ ; find its height.
Answer
The volume of the given cuboid $=3456 \ cm ^3$.
Length of the given cuboid $=24 \ cm$.
Breadth of the given cuboid $=18 \ cm$
We know,
Length $\times$ Breadth $\times$ Height $=$ Volume of a cuboid
$\Rightarrow 24 \times 18 \times$ Height$ =3456$
$\Rightarrow$  Height $=\frac{3456}{24 \times 18}$
$ \Rightarrow$ Height $=\frac{3456}{432}$
$\Rightarrow$ Height $=8 \ cm$
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Question 153 Marks
A solid cube of edge $14 \ cm$ is melted down and recast into smaller and equal cubes each of the edge $2\  cm$ ; find the number of smaller cubes obtained.
Answer
Edge of the big solid cube $=14 \ cm$
Volume of the big solid cube $=14 \times 14 \times 14 \ cm ^3=2744 \ cm ^3$
Edge of the small cube $=2 \ cm$
Volume of one small cube $=2 \times 2 \times 2 \ cm ^3=8 \ cm ^3$
Number of smaller cubes obtained $=\frac{\text { Volume of big cube }}{\text { Volume of one small cube }}$
$=\frac{2744}{8}=343$
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[3 marks sum] - MATHS STD 8 Questions - Vidyadip