Question 14 Marks
The area of cross section of a pipe is $5.4$ square / cm and water is pumped out of it at the rate of $27\ km$ per hour. Find, in litres, the volume of water which flows out of the pipe in $2$ minutes.
Answer$ 1 \ km / hr$
$=\frac{1 \ km }{1 hr }$
$=\frac{100000 \ cm }{3600 s }$
$=\frac{250}{9} \ cm / s $
Rate of flow of water
$=$ Area of cross section $x$ rate
$=5.4 \times 27 \ km / hr$
$=5.4 \times 27 \times 250 / 9 \ cm / s$
$=4050 \ cm ^3 / s$
Volume of water flown out in $2$ minutes
$=$ Rate of flow of water $x$ Time
$=4050 \times(2 \times 60$ seconds $)$
$=486000 \ cm ^3$
$=\frac{486000}{1000}$
$=486$ litres $ \ldots\left(\because 1\right.$ litre $\left.=1000 \ cm ^3\right)$
$\therefore$ Volume of water which flows out of the pipe in $2$ minutes is $486$ litres.
View full question & answer→Question 24 Marks
A swimming pool is $50\ m$ long and $15\ m$ wide. Its shallow and deep ends are $1.5\ m$ and $4.5\ m$ respectively. If the bottom of the pool slopes uniformly, find the amount of water in kilolitres required to fill the pool $(1\ m^3 = 1000$ liters$).$
Answer
Area of cross section $=$ Area of trapezium
$=\frac{1}{2} \times($sum of parallel sides $) \times$ height
$=\frac{1}{2} \times(1.5+4.5) \times 50$
$=\frac{1}{2} \times 6 \times 50$
$=150\ m ^2$
$\therefore$ Volume of the pool
$=$area of cross section $\times$ height
$=150 \times 15$
$=2250\ m ^3 \ldots(\because 1\ m ^3=1$ kilolitres$)$
$\therefore$ Volume of the pool is $2250$ kilolitres. View full question & answer→Question 34 Marks
Water flows at the rate of $1.5$ meters per second through a pipe with area of cross section $2.5 \ cm 2$ into a rectangular water tank of length $90 \ cm$ and breadth $50 \ cm$. Find the rise in water level in the tank after $4$ minutes.
AnswerRate of flow of water
$= 15\ m/s$
$= 150\ cm/s$
Rate of volume of water flown
$=$ Rate of flow $\times$ cross section area
$= 150 \times 2.5$
$= 375\ cm^3/s$
Total volume of water flow
$=$ Rate of volume of water flown $\times$ Time
$= 375 \times (4 \times 60 $ seconds$)$
$= 90000\ cm^3$
Volume of water flown
$=$ Volume of water in the tank
$90000 = l \times b \times h$
$90000 = 90 \times 50 \times h$
$h = 20\ cm$
$\therefore $ The rise in the level of water is $20\ cm.$
View full question & answer→Question 44 Marks
The given figure is a cross $-$section of a victory stand used in sports. All measurements are in centimetres. Assume all angles in the figure are right angles. If the width of the stand is $60 \ cm$ , find The total surface area in $m ^2$.
AnswerTotal surface area $=2 \times$ Area of cross section $+$ Area of bottom face $+$ Area of left face $+$ Area of right face $+$ Area of top face
Area of Cross$-$section $=8000 \ cm^2 \dots. . .($ from $a)$
Width of the stand $=60 \ cm \dots. .. ($given$)$
Area of bottom face
$= 130 \times 60$
$= 7800\ cm^2$
Area of the left face
$= 40 \times 60 + 50 \times 60 + 50 \times 60$
$= 8400\ cm^2$
Area of right face
$= 60 \times 60 + 40 \times 60 + 30 \times 60$
$= 7800\ cm^2$
Area of the top face
$= 40 \times 60$
$= 2400\ cm^2$
Total surface area
$= 2 \times 8000 + 7800 + 8400 + 7800 + 2400$
$= 42400\ cm^2$
$= 4.24\ m^2$
$\therefore $ The total surface area is $4.24\ m^2.$
View full question & answer→Question 54 Marks
A rectangular water tank measuring $8\ m \times 6\ m \times 4 \ cm$ is filled from a pipe of cross sectional area $1.5 \ cm^2$, the water emerging at $10\ m/s$. How long does it take to fill the tank?
AnswerTime taken to fill the tank
Volume of the tank
Volume of tank
$=800 \times 600 \times 4 \ cm ^3$
Volume of water flown out of the pipe in $1\ s$
$=$ Area of cross section $\times$ Rate of flow
$=1.5 \times 10( m / s )$
$=1.5 \times 10 \times 100( \ cm / s )$
$ =1500 \ cm$
$\therefore$ The time taken to fill the tank
$=\frac{800 \times 600 \times 4}{1500}$
$=1280\ s$
$\therefore$ The time taken to fill the tank is $1280\ s . $
View full question & answer→Question 64 Marks
The cross section of a piece of metal $2\ m$ in length is shown. Calculate the volume of the piece of metal.
Answer
Length $($height$)$ of the metal
$= 2\ m$
$= 200\ cm$
Volume of the metal
$=$ Area of cross$-$section $\times$ height
$= 57 \times 200$
$= 11400\ cm^3$
$\therefore$ Volume of the metal is $11400\ cm^3$. View full question & answer→Question 74 Marks
$\text{ABCDE}$ is the end view of a factory shed which is $50\ m$ long. The roofing of the shed consists of asbestos sheets as shown in the figure. The two ends of the shed are completely closed by brick walls.
If the cost of asbestos sheet roofing is $Rs. 20 \sim$ per $\sim m^2$, find the cost of roofing. AnswerAsbestos sheets are spread on the area formed by the rectangle with $C D$ and $D E$ as lengths.
$\triangle C D E$, by Pythagoras theorem,
$ D E^2=($perpendicular $)^2+\left(\frac{ Ab }{2}\right)^2$
$D E 2=3^2+\left(\frac{8}{2}\right)^2$
$D E^2=3^2+4^2$
$D E^2=25$
$\therefore DE = CD =5\ m $
Area of asbestos sheets $=D E \times$ length $+D c \times$ length
Area of asbestos sheet
$ =2 \times 5 \times 50$
$=500\ m ^2 $
Cost of roofing
$=$ Area $\times$ rate
$=500 \times 20$
$= Rs. 10,000$
$\therefore$ The cost of roofing is $Rs. 10,000 .$
View full question & answer→Question 84 Marks
The radius of the base of a right circular cylinder is doubled and the height is halved. What is the ratio of volume of the new cylinder to that of the original cylinder?
AnswerLet the radius of base of the original cylinder $=r$
And the height of the cylinder $=h$
Volume of original cylinder $=\pi r^2 h$
Given that, the radius of new cylinder $=2 r$
And height $=\frac{h}{2}$
$\therefore$ Volume of new cylinder
$ =\pi \times(2 r )^2 \times \frac{ h }{2}$
$=2 \times r ^2 h $
Ratio of volume of new cylinder to that of original cylinder
$ =\frac{2 \pi r^2 h}{\pi r^2 h}$
$=2: 1 . $
View full question & answer→Question 94 Marks
The volume of a solid cylinder is $7700\ cm^3$. Find its height and total surface area if the diameter of its base is $35\ cm$.
AnswerVolume of cylinder $=7700 \ cm ^3$
Diameter of base $=35 \ cm$
$\therefore$ Radius $(r)=17.5 \ cm$
Let $h$ be the height of the cylinder
Volume $=7700$
$\pi \times r^2 \times h=7700$
$\frac{22}{7} \times 17.5^2 \times h =7700$
$962.5\ h =7700$
$h =\frac{7700}{9625} \times \frac{10}{10}$
$h =\frac{77000}{9625}$
$h =8 \ cm$
Now,
$\text{T.S.A.}$ of cylinder
$ =(2 \times \pi \times r \times h)+\left(2 \times \pi \times r^2\right)$
$=\left(2 \times \frac{22}{7} \times 17.5 \times 8\right)+\left(2 \times \frac{22}{7} \times 17.5^2\right)$
$=880+1925$
$=2805 \ cm ^2 . $
View full question & answer→Question 104 Marks
The curved surface area of a cylinder is $198\ cm^2$ and its base has diameter $21\ cm$. Find the height and the volume of the cylinder.
AnswerThe curved surface area of a cylinder $=198 \ cm ^2$
Diameter of base $=21 \ cm$
$\therefore$ Radius $(r)=10.5 \ cm$
Let $h$ be the height of the cylinder
$\text{L.S.A. }=198 \ cm ^2$
$2 \times \pi \times r \times h=198 \ldots[\because\text{L.S.A.}$ of cylinder $=2 \times \pi \times r \times h]$
$2 \times \frac{22}{7} \times 10.5 \times h =198$
$\frac{462}{7} h =198$
$h =\frac{198 \times 7}{462}$
$h =\frac{1386}{462}=3 \ cm$
Volume of cylinder
$ =\pi \times r^2 \times h$
$=\frac{22}{7} \times 10.5^2 \times 3$
$=1039.5 \ cm ^3 . $
View full question & answer→Question 114 Marks
A cylindrical tube, open at both ends, is made of metal. The bore $($internal diameter$)$ of the tube is $10.4 \ cm$ and its length is $25 \ cm$. The thickness of the metal is $8 \ mm$ everywhere. Calculate the volume of the metal. Also, find the weight of the tube if $1 \ cm^3$ of the metal weighs $1.42\ g.$
AnswerInternal diameter of tube $= 10.4\ cm$
Internal radius $= 5.2\ cm$
Length of tube $= 25\ cm$
Thickness of metal $= 8\ mm$
Thickness of metal
$= 8\ mm$
$= 0.8\ cm$
Outer radius
$=$ Internal radius $+$ Thickness
$= 5.2 + 0.8$
$= 6\ cm$
Volume of metal
$=$ Volume of material between outer radius and inner radius
$= \pi (R^2 - r^2)h$
$= {\pi (6^2 - (5.2)^2} \times 25$
$= 704\ cm^3$
$\therefore$The volume of the metal used is $704\ cm^3\dots....($Ans $1)$
$1\ cm^3$ of metal$ = 1.42\ g$
$\therefore 704\ cm^3$ of metal
$= 704 \times 1.42$
$= 999.68\ g$
$\therefore $ The weight of the tube is $999.68\ g \dots ...($Ans $2)$
View full question & answer→Question 124 Marks
A well with $6\ m$ diameter is dug. The earth taken out of it is spread uniformly all around it to a width of $2\ m$ to form an embankment of height $2.25\ m.$ Find the depth of the well.
AnswerRadius of the well $( r)$
$ =\frac{6}{2}$
$=3\ m $
Volume of earth dug out of the well
$ =\pi r^2 h \ldots($Volume of cylinder$)$
$=\pi \times(3)^2 \times h$
$=9 \pi h m ^2 $
Area of embankment $=\pi R^2-\pi r^2$
Where $R=r+$ width of the embankment
$ =3+2$
$R=5\ m $
Area of embankment
$ =\pi 5^2-\pi 3^2$
$=16 \pi m ^2 $
Volume of earth dug out for the embankment
$=$area of embankment $\times$ height
$ 9 \pi h=16 \pi \times 2.25$
$h=\frac{16 \pi \times 2.25}{9 \pi}$
$\therefore h=4\ m $
$\therefore$ The depth of the well is $4\ m$.
View full question & answer→Question 134 Marks
The height of a circular cylinder is $4.2\ cm.$ There times the sum of the areas of its two circular faces is twice the area of the curved surface. Find the volume of the cylinder correct to $1$ decimal place.
AnswerHeight of the cylinder $(h)=4.2 \ cm$
Let $r$ be the radius of the cylinder
Sum of the area of $2$ circular faces $=\pi r^2+\pi r^2$
Curved surface area of cylinder $=2 \pi rh$
Given that:
$ 3 \times\left(\pi r^2+\pi r^2\right)=2 \times 2 \pi r h$
$6 \pi r^2=4 \pi r h$
$6 r=4\ h$
$6 r=4 \times 4.2$
$r=\frac{16.8}{6}$
$r=2.8 \ cm $
$\therefore$ Volume of the cylinder
$ =\pi \times r ^2 \times h$
$=\frac{22}{7} \times 2.8^2 \times 4.2$
$=103.5 \ cm ^3 $.
View full question & answer→Question 144 Marks
Find the length of $22\ kg$ copper wire of diameter $0.8\ cm,$ if the weight of $1 \ cm^3$ copper is $4.2\ g.$
AnswerDiameter of the wire $=0.8 \ cm$
Radius of the wire $=0.4 \ cm$
If $4.2\ g$ of copper $=1 \ cm ^3$ of copper
Then $22 \ kg$ copper $=\frac{220000}{4.2} \ cm ^3$
Volume of the copper wire $=$ Area of base $\times$ length of wire
$ \frac{22000}{\frac{420}{2} 000}=\pi r ^2 \times h$
$\frac{22}{4.2} \times 0.4^2 \times h $
$h=\frac{22000 \times 7}{4.2 \times 22 \times 0.4 \times 0.4}$
$ h=10416.7 \ cm$
$\therefore h=104.17\ m $
$\therefore$ The length of the copper wire is $104.17\ m$.
View full question & answer→Question 154 Marks
Find the lateral surface area, total surface area and the volume of the following cylinders: Diameter $= 10\ m,$ High $= 7\ m$
AnswerGiven that:
Diameter $=10\ m$
Radius $(r)=5\ m$
Height $(h)=7\ m$
Now,
$\text{L.S.A}$ of cylinder
$ =2 \times \pi \times r \times h$
$=2 \times \frac{22}{7} \times 5 \times 7$
$=220\ m ^2 $
$\text{T.S.A}$ of cylinder
$ =(2 \times \pi \times r \times h)+\left(2 \times \pi \times r^2\right)$
$=\left(2 \times \frac{22}{7} \times 5 \times 7\right)+\left(2 \times \frac{22}{7} \times 5^2\right)$
$=220+157.14$
$=377.14\ m ^2 $"
Volume of cylinder
$ =\pi \times r ^2 \times h$
$=\frac{22}{7} \times 5^2 \times 7$
$=550\ m ^3 . $
View full question & answer→Question 164 Marks
The radius of the base of a right circular cylinder is tripled and the height is doubled. What is the ratio of volume of the new cylinder to that of the original cylinder?
AnswerLet the radius of base of the original cylinder $=r$
And the height of the cylinder $=h$
Volume of original cylinder $=\pi r^2 h$
Given that, the radius of new cylinder $=3 r$
And, height $=2 r$
$\therefore$ Volume of new cylinder
$ =\pi \times(3 r)^2 \times 2 h$
$=18 \pi r^2 h $
Ratio of volume of new cylinder to that of original cylinder
$ =\frac{18 \pi r^2 h}{\pi r^2 h}$
$=18: 1 $
View full question & answer→Question 174 Marks
The base of a rectangular container is a square of side $12 \ cm$. This container holds water up to $2 \ cm$ from the top. When a cube is placed in the water and is completely submerged, the water rises to the top and $224 \ cm^3$ of water overflows. Find the volume and surface area of the cube.
AnswerWhen the cube is submerged, the level of water is increased and some water flows out of it.
Volume of the cube
$=$Volume of water le
$=12 \times 12 \times 2+224$
$=512 \ cm ^3 $
$=$Volume of water level rise $+$ Volume of water overflown
$\therefore$ The volume of the cube is $512 \ cm ^3\dots...($Ans $1)$
Volume of the cube
$ =s^3$
$512=s^3$
$s=\sqrt[3]{512}$
$s=8 \ cm $
Surface area of cube
$ =6 s ^2$
$=6 \times 8^2$
$=384 \ cm ^2 $
$\therefore$ The surface area of the cube is $384 \ cm ^2\dots....($Ans$2)$
View full question & answer→Question 184 Marks
The length of a cold storage is double its breadth. Its height is $3\ m$. The area of its four walls including doors is $108\ m^2$. Find its volume.
AnswerGiven that:
The length of a cold storage is double its breadth
Height $=3\ m$
Area of its four walls $=108\ m ^2$
Let the Breadth $(b)$ of cold storage $=x m$
Thus, the length$ (l)$ of cold storage $=2 xm$
$\text{L.S.A}$ of cold storage
$ =2 \times h \times(l+b)$
$108=2 \times 3 \times(2 x+x) \ldots[$From$(1)]$
$=6 \times 3 x$
$18 x=108$
$x=\frac{108}{18}$
$x=6 $
$\therefore$ Length $(l)=2 \times 6=12\ m$
Breadth $(b)=1 \times 6=6\ m$
Thus,
Volume of cold storage
$ =1 \times b \times h$
$=12 \times 6 \times 3$
$=216\ m ^3 . $
View full question & answer→Question 194 Marks
The cost of papering the four walls of a room at $Rs. 1$ per $m^2$ is $Rs. 210$. The height of the room is $5\ m$. Find the length and the breadth of the room if they are in the ratio $5:2$.
Answer$\because$ The cost of papering the four walls of the room at $Rs. 1$ per $m ^2$ is $Rs .210$.
$\therefore$ The area of the $4$ walls is $210\ m ^2$,
The length and breadth are in the ratio $5: 2\dots...($given$)$
Let the common multiple be $x$
$ \therefore$ Length$=5 \times m$
breadth$=2 xm$
height $=5\ m \ldots ($given$)$
Surface area of the $4$ walls $=2 hl +2 hb$
$ 210=2 \times 5 \times 5 x+2 \times 5 \times 2 x$
$210=50 x+20 x$
$210=70 x$
$\therefore x=\frac{210}{70}$
$\therefore x=3 $
length$=5 x=5 \times 3=15 \ cm$
beadth$=2 x=2 \times 3=6 \ cm . $
View full question & answer→Question 204 Marks
The length, breadth, and height of a rectangular solid are in the ratio $6 : 4 :3$. If the total surface area is $1728 \ cm^2$. Find its dimensions.
AnswerLet the length, breadth and height of the rectangular solid be $' a\ '$, $' b\ '$ and $' c\ '$ respectively.
$ \therefore a: b: c=6: 4: 3$
Let the common $m$
$a=6 x \ cm$
$b=4 x \ cm$
$c=3 x \ cm $
Let the common multiple be $x$
$ a=6 x \ cm$
$b=4 x \ cm$
$c=3 x \ cm $
The total surface area of the cuboid
$ =1728 \ cm ^2$
$2(l b+b h+h l)=1728$
$2(a b+b c+c a)=1728$
$2[(6 x \cdot 4 x)+(4 x \cdot 3 x)+(3 x \cdot 6 x)]=1728$
$24 x^2+12 x^2+18 x^2=\frac{1728}{2}$
$54 x^2=864$
$x^2=\frac{864}{54}$
$x^2=16$
$x=\sqrt{16}$
$\therefore x=4$
$\therefore a=6 x=6 \times 4=24 \ cm$
$\therefore b=4 x=4 \times 4=16 \ cm$
$\therefore c=3 x=3 \times 4=12 \ cm $
Hence, the dimensions of the rectangular solid are $24 \ cm , 16 \ cm$ and $12 \ cm$.
View full question & answer→Question 214 Marks
A room is $5\ m$ long, $2\ m$ broad and $4\ m$ high. Calculate the number of persons it can accommodate if each person needs $0.16\ m^3$ of air.
AnswerGiven that:
Length $(l)$ of the room $=5\ m$
Breadth $(b)$ of the room $=2\ m$
Height $(h)$ of the room $=4\ m$
$\therefore$ Volume of the air in the room
$=1 \times b \times h$
$=5 \times 2 \times 4$
$=40\ m ^3$
Since, $1$ person needs $=0.16\ m ^3$ of air
i.e., $0.16\ m ^3$ of air $=1$ person
$\therefore 1\ m ^3$ of air $=\frac{1}{0.16}$ person
So, $40\ m ^3$ of air $=\frac{1}{0.16} \times 40$
$=\frac{40}{0,16} \times \frac{100}{100}$
$=\frac{4000}{16}$
$=250$ Persons
Thus, the room can accommodate $250$ persons.
View full question & answer→Question 224 Marks
A closed box is made of wood $5\ mm$ thick. The external length, breadth and height of the box are $21 \ cm$, $13 \ cm$ and $11 \ cm$ respectively. Find the volume of the wood used in making the box.
AnswerThickness of the closed box
$= 5\ mm$
$= 0.5\ cm$
External Dimensions are:
length $= 21\ cm$
breadth $= 13\ cm$
height $= 11\ cm$
Internal dimensions $=$ External dimensions $- 2($thickness$)$
$\therefore $ Internal Dimensions are:
length $= 20\ cm$
breadth $= 12\ cm$
height $= 10\ cm$
Volume ofthe wood used in making the box
$=$ Volume of External cuboid $-$ Volume of internal cuboid
$= (21 \times 13 \times 11) - (20 \times 12 \times 10)$
$= 3003 - 2400$
$= 603\ cm^3$
Hence, the volume of wood used in making the box is $603\ cm^3$.
View full question & answer→Question 234 Marks
Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the resulting cuboid to that of the sum of the total surface areas of the three cubes.
AnswerWe need to find:
Total surface area of cuboid
Sum of total surface areas of $3$ cubes
Cube:
Let the side of the cube be $' a\ '$ units
$\therefore$ Total surface area of $1$ cube
$=6 a ^2 sq.$ units
$\therefore$ Total surface area of $3$ such cubes
$=3 \times 6 a ^2 sq.$ units
$=18 a^2 sq.$ units
The cuboid is formed by joining $3$ cubes:
length $=3 a \ cm$
breadth $= a \ cm$
height $= a m$
$\therefore$ Total surface area of cuboid
$=2( lb + bh + hl )$
$=2(3 a \times a+a \times a+a \times 3 a)$
$=2\left(3 a^2+a^2+3 a^2\right)$
$=2\left(7 a^2\right)$
$=14 a^2 sq.$ units
Total surface area of cuboid
Sum of total surface areas of $3$ cubes
$=\frac{14 a ^2}{18 a ^2}$
$=\frac{7}{9}$
$\therefore$ The ratio of Total surface area of cuboid to the Sum of total surface areas of $3$ cubes is $7: 9$.
View full question & answer→Question 244 Marks
Three cubes of sides $x \ cm, 8\ cm$ and $10\ cm$ respectively are melted and formed into a single cube of edge $12\ cm$, Find $'x\ '.$
AnswerGiven that:
Side $\left( I _1\right)$ of cube $( a )= x \ cm$
Side $\left( l _2\right)$ of cube $( b )=8 \ cm$
Side $\left( l _3\right)$ of cube $( c )=10 \ cm$
Edge length of new formed cube $=12 \ cm$
Volume of cube $(a)=\left(l_1\right)^3=x^ 3 \ cm ^3$
Volume of cube $(b)=\left( I _2\right)^3=8^3=512 \ cm ^3$
Volume of cube $(c)=\left(l_3\right)^3=10^3=1000 \ cm ^3$
Total Volume of all three cubes $-$ Volume of $1$ cube
$ x^3+8^3+10^3=12^3$
$x^3+512+1000=1728$
$x^3=1728-1512$
$x=\sqrt[3]{216}$
$\therefore x=6 \ cm . $
View full question & answer→Question 254 Marks
Three metal cubes with edges $6\ cm, 8\ cm$ and $10\ cm$ respectively are melted together and formed into a single cube. Find the diagonal of this cube.
AnswerGiven that:
Side $\left(l_1\right)$ of metal cube $(a)=6 \ cm$
Side $\left( I _2\right)$ of metal cube $( b )=8 \ cm$
Side $\left( I _3\right)$ of metal cube $( c )=10 \ cm$
Total Volume of all three cubes $=$ Volume of $1$ cube
Volume of cube $(a)=\left( I _1\right)^3=6^3=216 \ cm ^3$
Volume of cube $(b)=\left(I_2\right)^3=8^3=512 \ cm ^3$
Volume of cube $(c)=\left(I_3\right)^3=10^3=1000 \ cm ^3$
Total volume of all three cubes $=1728 \ cm ^3$
$\therefore$ Volume of $1$ cube $=1728 \ cm ^3$
$ \text { i.e., } \beta^3=1728$
$I=\sqrt[3]{1728}$
$\therefore$ Side $(l)=12 \ cm$
Length of diagonal of cube
$=\sqrt{3} \times 1$
$ =\sqrt{3} \times 12$
$=12 \sqrt{3} \ cm .$
View full question & answer→Question 264 Marks
A cuboid is $25\ cm$ long, $15\ cm$ board and $9\ cm$ high. Find the whole surface of a cube having its volume equal to that of the cuboid.
AnswerGiven that:
Length of the cuboid $=25 \ cm$
Breadth of the cuboid $=15 \ cm$
Height of the cuboid $=9 \ cm \ \ \ $
Volume of cube $=$ volume of cuboid
We know that volume of Cuboid
$ =1 \times b \times h$
$=25 \times 15 \times 9$
$=375 \ cm ^3$
$\therefore$ Volume of cube $=3375 \ cm ^3$
But we know that volume of cube $=1^3$
i.e.,$ \left.\right|^3=3375$
$=\sqrt[3]{3375}$
$=15 \ cm $
$\therefore$ Side $( l )=15 \ cm$
Now,
$\text{T.S.A}$ of cube
$=61^2$
$=6 \times 15 \times 15$
$=1350 \ cm ^2$
View full question & answer→Question 274 Marks
A square plate of side $'x\ ' \ cm$ is $4\ mm$ thick. If its volume is $1440 \ cm^3$; find the value of $'x\ '.$
AnswerVolume of the square plate $=$ Volume of a cuboid
$ h =4\ mm$
$=\frac{4}{10} \ cm$
$=0.4 \ cm $
Volume of the square plate
$ =I \times b \times h$
$1440=X \times X \times 0.4$
$1440=x^ 2 \times 0.4$
$x^ 2=\frac{1440}{0.4}=3600$
$x=\sqrt{3600}$
$\therefore x=60 \ cm . $ View full question & answer→