Given below are the surface area and volume class 9 worksheet with solutions
(a) Concepts questions
(b) Calculation problems
(c) Long answer questions
Question 1
A cube and cuboids have the same volume. The dimensions of cuboids are in ratio 1 : 2 : 4. If the difference between the cost of painting the cuboids and cube whole surface area at rate of Rs 5 per m
2 is Rs. 80. Find their volumes.
Solution
Let each side of cube be x m.
Ratio of sides of cuboid is 1:2:4.
Let the sides be y ,2y and 4y m
Now as per question
Volume of cube= volume of cuboid
$x^3=y \times 2y \times 4y=8y^3$
or x=2y
Surface Area of Cube = $6x^2= 24y^2$
Surface Area of Cubiod=$2(lb+lh+bh) =2(2y^2 + 8y^2 + 4y^2)$
As per the question
$[2(2y^2 + 8y^2 + 4y^2) - 24y^2 ] \times 5 =80$
$20y^2 = 80$
y=2
Therefore Side of cube = 4 m
Volume of cube =$(4)^3=64 \ m^3$
Now the sides of cuboid=y,2y,4y=2,4,8m
Therefore Volume of cuboid =$2 \times 4 \times 8=64 \ m^3$
Question 2
A tent is in shape of right circular cylinder up to a height of 3m and a cone above it. The maximum height of tent above ground is 13.5m. Calculate the cost of painting the inner side of tent at rate of Rs. 3/sq m if radius of base is 14m.
Solution
Dimension of Cylinder
$H_1=3 \ m$
r=14 m
Dimension of Cone
$H_2= 13.5 - 3 = 10.5 \ m$
r= 14 cm
Curved Surface Area of Tent = Curved Surface Area of Cylinder + Curved Surface Area of Cone
$=2 \pi rh_1 + \pi r \sqrt {r^2 + H_2^2}$
=1034 sq m
So cost will be = 1034 * 3 =Rs 3102
Question 3
A school provides milk to students daily in cylindrical glasses of diameter 7cm. If glass is filled with milk up to a height of 12cm, find how many liters of milk is needed to serve 1600 students.
Solution
d= 7 cm, r= 3.5 cm
h=12 cm
Volume of Glass = $\pi r^2 h= 462 \ cm^3$
Volume of Glass for 1600 students= 1600 * 462=739200 cm3 = 739.2 litre
Question 4
Eight metallic spheres, each of radius 2 mm, are melted and cast into a single sphere. Calculate the radius of the new sphere
Solution
$\frac {4}{3} \pi R^3= 8 \frac {4}{3} \pi (2)^3$
R=4mm
Question 5
The diameter of a sphere is decreased by 30%. By what percent its surface area decreases?
Solution
Initial radius=r
Original surface area =$4 \pi r^2$
If the radius of the sphere is decreased by 30% then its radius becomes 7r/10
Then the surface area of the new sphere will be
$= 4 \pi (\frac {7r}{10})^2 =\frac {49}{100} \times 4 \pi r^2$
Now the surface area is decreased by
$=\frac {4 \pi r^2 - 4 \pi r^2 \times \frac {49}{100}}{4 \pi r^2}= \frac {51}{100}$
So the percentage of the change in surface area is 51%.
Question 6
In a cylinder radius is doubled and height is halved then find its curved surface area.
Solution
$R= 2r$ and $H=\frac {h}{2}$
$S= 2 \pi RH= 2 \pi (2r) \frac {h}{2}= 2 \pi rh$
So CSA remains same
Question 7
A right- angled ABC with sides 3cm, 4cm & 5cm is revolved about fixed side of 4cm. find the volume of solid generated. Also total surface area of solid.
Solution
Volume of cone =$\frac {1}{3} \pi r^2 h= \frac {1}{3} \times \frac {22}{7} \times 9 \times 4=\frac {264}{7} cm^3 $
total surface area= $\pi rl + \pi r^2=15 \pi + 9 \pi =24 \pi cm^2 $
Question 8
If h,s and v are height, curved surface and volume of cone respectively, prove that
$3v h^3 + 9v^2 -s^2 h^2=0$
Solution
$s= \pi rl$
$v= \frac {1}{3} \pi r^2 h$
Taking LHS
$3v h^3 + 9v^2 -s^2 h^2$
=$3 (\frac {1}{3} \pi r^2 h) h^3 + 9 (\frac {1}{3} \pi r^2 h)^2 - (\pi rl)^2 h^2$
$=\pi ^2 r^2 h^4 + \pi ^2 r^4 h^2 - \pi ^2 r^2 h^2 l^2$
Now $l^2 = h^2 + r^2$
$=0$
Hence Proved
Question 9
An edge of a cube measure r cm. If the largest possible right circular cone is cut out of this cube, then find the volume of cone?
Solution
Diameter of Cone= r, Radius= r/2
Height =r
Volume of cone=$ \frac {1}{3} \pi r^2 h$
$= \frac {1}{3} \pi (r/2)^2 r$
$=\frac {1}{12} \pi r^3$
Question 11
How many planks of dimensions 5m × 25cm × 10cm can be stored in a pit which is 20 m long, 6m wide and 80 cm deep?
Solution
Number of plank that can be stored =$\frac {\text{Volume of pit}}{\text {Volume of each plank}} = \frac {20 \times 6 \times .8}{5 \times .25 \times .1}=768$
Question 12
The surface area of a sphere of radius 5cm is 5 times the area of the curved surface area of cone of radius 4cm. Find the height and volume of cone. ( π= 22/7)
Solution
Surface area of sphere= $4 \pi r^2= 400 \pi \ cm^2$
curved surface area of cone=$ \pi rl \ cm^2 =4 \pi l \ cm^2$
Now As per the question
$400 \pi= 5 \times 4 \pi l$
l=5 cm
Now we know that
$l^2 = h^2 + r^2$
or $h^2 = l^2 - r^2 =5^2 - 4^2 = 3^2$
h= 3 cm
Now Volume of Cone is given by
$V= \frac {1}{3} \pi r^2 h$
Substituting the values
V=50.29 cm3
Question 13
A sweet shop has one spherical ladoo of radius 5cm with the same amount of material, how many ladoos of radius 2.5cm can be made?
Solution
Number of ladoos = $\frac {\text{Volume of Big Ladoo}}{\text{Volume of smal ladoo}}= \frac {\frac {4}{3} \pi (5)^3}{ \frac {4}{3} \pi (2.5)^3}=8$
Question 14
A sphere and right circular cylinder of same radius have equal volumes by what percentage does diameters of cylinder exceeds its height?
Solution
Let the radius of sphere= r= Radius of a right circular cylinder
As per to the question,
Volume of cylinder= volume of a sphere
$\pi r^2 h=\frac {4}{3} \pi r^3$
$h=\frac {4}{3}r$
Now Diameter of the cylinder = 2r
So, Inreased diameter from height of the cylinder =$2r-\frac {4}{3}r=\frac {2}{3} r$
Therefore percentage increase in diameter of the cylinder =$ \frac {(2/3)r \times 100}{(4/3)r=50$ %
Question 15
Curved surface area of an ice cream cone of slate height 12cm is 113.04cm
2. Find the base radius and height of cone ( π= 3.14)
Solution
CSA of cone = π rl
Therefore
$113.04= 3.14 \times r \times 12$
or r=3 cm
Now
$l = \sqrt {r^2 + h^2}$
or
$h = \sqrt {l^2 - r^2} = \sqrt {135} = 3 \sqrt {15}$ cm
Question 16
A semi circular sheet of metal of radius 14cm is to be bent to form an open conical cup. Find the capacity of cup?
Question 17
Two solid spheres made of same metal have weight 5920g and 740g respectively. Determine the radius of larger sphere, if the diameter of smaller one is 5cm.
Question 18
Each edge of cube is increased by 50%. Find the percentage increase in surface area of cube.
Solution
Let a be the edge
Original Surface Area= $6a^2$
If edge of cube is increased by 50%, then edge becomes = 1.5a
New Surface Area = $6 (1.5a)^2= 13.5a^2$
% increase = $\frac { 13.5a^2 - 6a^2}{6a^2} \times 100= 125$%
Question 19
A residential house society is built in 4000 sq m in area. It has an underground thank to collect the rain water the length, breadth and height of which are 50m, 40m and 4m respectively. If it rains at the rate of 2mm per minute for 5 hours, then calculate the depth of water in the tank. What value is depicted in problem?
Solution
Let h be the depth of water in the tank
Height of rain water falling in 1 min = 2mm
Height of rain water falling in 1hr =2*60=120 mm
Height of rain water falling in 5 hr =600 mm=.6m
Therefore Volume of rain water which falls on land =4000 * 0.6=2400 m3
Now Volume of water filled in tank = Volume of rain water which falls on land
So, $50 \times 40 \times \times h=2400$
h = 1.2 m
Question 20
A rectangular tank is 225m × 162m at base with what speed should water flow into it through an aperture 60cm x 45cm so that level of water is raised by 20cm in 2.5 hours.
Solution
Volume of a cuboid tank=LBH
Now after 5 hours, the height of the water is 20cm,
So, Volume of water in the tank in 5 hours =$225 \times 162 \times .20 =7290 m^3$
Let x m/ hour be the speed of the water flowing thought the aperture.
Then, Volume of water flown through the aperture in 5 hours =$.6 \times .45 \times 5x$
Now This should be equal to water volume in tank
$.6 \times .45 \times 5x=7290 $
x=5400 m/hr
Question 21
The pillars of a temple are cylindrical shaped. If each pillar has a circular base of radius 20cm and height 7m, then find the quantity of concrete mixture used to build 20 such pillars. Also find cost of concrete mixture at the rate of Rs. 200 per m
3. ( π=22/7)
Question 22
50 circular plates, each of radius 7cm and thickness 12cm are placed one above the other to form a solid right circular cylinder. Find the total surface area and volume of the cylinder
Question 23
Three solid sphere of iron whose diameters are 2cm, 12cm and 16cm respectively is melted into a single solid sphere. Find the radius of solid sphere.
Solution
Volume of single solid sphere= sum of the volume of small sphere
$\frac {4}{3} \pi R^3=\frac {4}{3} \pi (1)^3 + \frac {4}{3} \pi (6)^3 + \frac {4}{3} \pi (8)^3$
$R^3=1 + 216+512=729$
$R=9$ cm
Question 24
Coins of same size (say 5 rupee coin) are placed one above the other and cylindrical block is obtained. The volume of this block is 67.76cm
3. Find no. of coins arranged in block, if thickness of each coin is 2mm and radius of each coin is 1.4cm.
Question 25
Metal spheres, each radius 2cm, are packed in a rectangular box of internal dimensions 16cm × 8cm × 8cm. When 16 spheres are packed, the box filled with preservative liquid. Find the volume of liquid.
Solution
Volume of Liquid = Volume of Box - 16 × Volume of one sphere
Volume of Box= LBH
Volume of Sphere = $\frac {4}{3} \pi r^3}$
Substituting the values
Volume of Liquid =488 cm3
Question 26
A semi- circular sheet of metal of diameter 14cm is bent to form an open conical container. Find the capacity of container?
Solution
Radius of semi - circular piece= 14 cm
Circumference of Sheet= &pi R=44 cm
Now when semi- circular sheet is bent to form an open conical container.
Circumference of Base of Cone = Circumference of Sheet
$2 \pi r= 44$
r= 7 cm
Now Slant height of cone= Radius of the plate = 14 cm
Now height of cone can be derived from the formula
$l^2 = h^2 + r^2$
$h= 7 \sqrt 3 $ cm
Now Volume of Cone is given by
$V= \frac {1}{3} \pi r^2 h$
Substituting the values
V=622.3 cm3
Question 27
A 4cm cube is cut into 1cm cubes. Calculate the total surface area of the small cubes. What is the ratio of surface area of small cubes to that of large cube?
Solution
Volume of 4 cm cube=64 cm3
Volume of 1 cm cube=1 cm3
Therefore total 64 cube were formed
Surface area of 1 cm cube=6 (1) 2 =6 cm2
Total Surface area of all 64 cube=384 cm2
Surface of Original cube= 6(4) 2 =96 cm2
Ration= 384/96= 4:1
Question 28
A cylindrical pipe opened at both the ends is made of iron sheet which is 2cm thick. If the outer diameter is 12cm and its length is 200cm. Find how many cm
3 of iron has been used in making the pipe.
Question 29
At ramzan Mela, a stall keeper in one of the food stalls has a large cylindrical vessel of base radius 15cm filled up to a height of 32cm with orange juice. The juice is filled in small cylindrical glasses of radius 3cm up to a height of 8cm and sold for Rs 3 each. How much money does the stall keeper receive by selling the juice completely?
Question 30
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7mm and diameter of graphite is 1mm. If the length of the pencil is 14cm, find the volume of wood and that of graphite.
Question 31
A right triangle ABC with sides 5cm, 12cm and 13cm is revolved about the side 12cm. Find the volume of the solid so obtained?
Solution
This is same as question 7
Volume of cone =$\frac {1}{3} \pi r^2 h= \frac {1}{3} \times \frac {22}{7} \times 25 \times 12=\frac {2200}{7} cm^3 $
Question 32
A heap of wheat is in the form of a cone whose diameter is 10.5m and height is 3m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required?
Question 33
A shot- putt is a metallic sphere of radius 4.9cm. If the density of the metal is 7.8g/cm
3, find the mass of shot- putt?
Solution
$V = \frac {4}{3} \pi r^3 = \frac {4}{3} \frac {22}{7} (4.9)^3=493 cm^3 $
Now
Density = mass /volume
or
Mass = Density * volume= 7.8 * 493=3845 g$
Question 34
A hemispherical tank is made up of iron sheet 1cm thick. If the inner radius is 1m, then find the volume of iron used in making the tank.
Question 35
A dome of the building is in form of a hemisphere from inside, it was white – washed at the cost of Rs 498.96. If the cost of white- washing is Rs. 2.00/sq.m find the
(a)inside surface area of dome
(b) volume of air inside dome
Question 36
Twenty seven solid spheres, each of radius r and surface area s are melted to form a sphere with surface area S. Find the
(a)R of new sphere
(b)ratio of s and S
Solution
Volume of 1 Sphere=$\frac {4}{3} \pi r^3$
Volume of 27 Sphere=$27 \times \frac {4}{3} \pi r^3= 36 \pi r^3$
Now Volume of 27 Sphere= Volume of Large Sphere
$36 \pi r^3= \frac {4}{3} \pi R^3$
$R=3r$
Now $s= 4 \pi r^2$
$S= 4 \pi R^2= 36 \pi r^2$
Therefore
$s : S = 1 :9$
Question 37
The sum of length, breadth and height of a cuboid is 21 cm and the length of its diagonal is 12 cm. Find the surface area of the cuboid.
Solution
Here
L+B+H=21
$\sqrt {L^2 + B^2 + H^2} =12$ or $L^2 + B^2 + H^2=144$
Now
$(L+B+H)^2 = L^2 + B^2 + H^2 + 2(LB+BH+ LH)$
or
$2(LB+BH+ LH)= (L+B+H)^2 -(L^2 + B^2 + H^2 )=441- 144=297 cm^2$
Question 38
The dimensions of a cubiod are in the ratio of 1 : 2 : 3 and its total surface area is 88sq m. Find the dimensions.
Solution
Let L,B and H be x,2x,3x
Now Total Surface= 2(LB+BH+ LH)
Therefore
$2(2x^2 + 3x^2 + 6x^2)=88$
$x=2$ m
So dimensions are 2, 4, 6 m
Question 39
Find the volume of the largest right circular cone that can be fitted in a cube whose edge is 14cm.
Summary
This Surface area and volume class 9 worksheet with solutions is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail. You can download this test as pdf also as below
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