Chapter – Surface Area and Volumes
Important questions:
Ques:1) Find the total surface area of a cube of edge 10 cm. (Mark-1)
Solution:
Given,
Edge of the cube = 10 cm
We know that,
Total Surface area of cube = 6a²
Total Surface area of cube = 6×10²
= 600 cm²
Ques:2) Calculate the volume of a cuboid whose dimensions are 3.6 cm, 8.2 cm,11 cm.(Mark-1)
Solution:
Given,
Length of cuboid = 3.6 cm
Breadth of cuboid = 8.2 cm
Height of cuboid = 11 cm
We know that,
Volume of cuboid
= L×B×H
= 3.6 cm × 8.2 cm × 11 cm = 324.72 cm³
Therefore,Volume of cuboid is 324.72 cm³
Ques:3) Find the volume of right circular cylinder whose height is 20 cm and diameter of base is 14 cm.(Mark-1)
Solution:
Given,
Diameter of base of cylinder (d)= 14 cm
Radius of base of cylinder(r)= d/2
= 14/2
r = 7 cm
Height of the cylinder = 20 cm
We know that,
Volume of cylinder = πr²h
= 22/7 × 7² × 20
= 22 × 7 × 20
=22 × 140
= 3080 cm²
Ques:4) The slant height of a conical mountain is 2.5 km and the area of its base is 1.54 km². Find the height of mountain.(Mark-1)
Solution:
Given,
Slant height of a conical mountain (l) = 2.5 km
Area of base of conical mountain (A) = 1.54 km²
Now, We know that,
Area of base of mountain(A)= πr²
1.54 = 22/7 × r²
r² = 1.54 × (22/7)
r = 2.2 km
To find height of the mountain,
Let the height of the mountain be ‘h’.
Now by applying pythagoras theorm,
l² = r² + h²
Or, 2.5² = 2.2² + h²
Or, h² = 2.5² – 2.2²
Or,. h = 1.187 km
Ques:5) The surface area of a sphere is 154 cm². Find its radius.(Marks-2)
Solution:
Given,
Surface area of sphere = 154 cm²
Now we know that,
Surface area of sphere = 4πr²
154 cm² = 4πr²
154 = 4× 22/7 × r²
r² = 154×7/(22×4)
r = √(49/4)
r = 7/2 = 3.5
Ques:6) The height and slant height of a cone are 21 cm and 28 cm respectively.Find the radius of cone and volume of cone.(Marks-2)
Solution:
Given,
Height of the cone (h) = 21 cm
Slant height of the cone (l) = 28 cm
Now,
Let the radius of the come be ‘r’.
Now by applying pythagoras theorm,
l² = r² + h²
Or, 28² = r² + 21²
Or, 28² – 21² = r²
Or, r = √343
Or, r = 18.52
Now, we know that,
Volume of cone = (1/3)πr²h
= 1/3 × 22/7 × 18.52² × 21
= 7545.78 cm³
Ques:7) Calculate the surface area of a sphere whose volume is 4851 cm³ (Marks-2).
Solution:
Given,
Volume of the sphere = 4851 cm³
Now, we know that,
Volume of sphere = (4/3)πr³
4851 cm³ = (4/3)πr³
r³ = 4851×(3/4)×(7/22)
r = 10.5
Now we know that,
Surface area of sphere = 4πr²
= 4× (22/7)× 10.5²
= 1385.44 cm²
Ques:8) The circumference of the base of a cylinder vessel is 132 cm and its height is 25 cm. How many litres of water can it hold?(Marks-3)
Solution:
Given,
Circumerence of base of cylinder =132 cm
Height of the cylinder = 25cm
Now, we know that,
Circumerence of base of cylinder = 2πr
Also,
2πr = 132
r = 132/2π
r = 21 cm
Now, Volume of cylinder = πr²h
= π ×21×21×25
= 34650 cm³
Volume of cylindrical vessel in litres =34650/1000
=34.65 litres
Ques:9) Find the number of spherical lead shots each 2.1 cm in radius that can be obtained from a cuboidal solid lead with dimensions 66cm × 42cm × 21cm.(Marks-3)
Solution:
Given,
Lenth of cuboidal solid lead (l) = 66 cm
Breadth of cuboidal solid lead (b) = 42 cm
Heightof cuboidal solid lead (h) = 21 cm
Radius of spherical lead shots (r)= 2.1 cm
Now,
Let the ‘n’ be the no. of Spherical lead shots formed by cuboidal solid lead.
Therefore,
(Volume of cuboidal solid) =(n×Volume of each lead shots)
l×b×h = n × (4/3)× π × r³
66×42×21 = n × (4/3)× π × 2.1³1
n = 1500
Therefore,No. of Spherical lead shots formed by cuboidal solid lead. are 1500.
Ques:10) A village has a population of 4000, required 150 litre of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will water of this tank last? What value is being showm if all measures are being taken to avoid the wastage of water by overflowing of tank.(Marks-4)
Solution:
Given,
Total population in the village = 4000
Water required per head per day = 150 l
Also, Length of the water tank= 20m
Breadth of the water tank = 15 m
Height of the water tank = 6 m
Now,
Total volume of water required by village
=Total population ×water reqd.(per head)
= 4000×150
= 600000 litre
Also, we know that,
Volume of the tank = L×B×H
= 20×15×6 m³
= 1800 m³
= 1800×1000 litres
= 1800000
Now,
Number of days water will last in the tank=
= Vol. Of tank /reqd. water per day
= 1800000/600000
= 3 days.
Therefore, water will last for 3 days.
Ques:11) The diameter of a roller is 84 cm and it length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of playground in meter square. Find the cost of leveling it at the rate ₹ 7.50 per square meter.(Marks-4)
Solution:
Given,
Diameter Of roller (d) = 84 cm= 0.84 m
Radius of roller(r)= 84/2=42 cm= 0.42m
Length of roller (h) = 120 = 1.20m
No. Of revolutions of roller to level the ground = 500
Now, we know that,
C.S.A of the roller = 2πrh
= 2 × 22/7 × 0.42 × 1.20
= 3.165 m²
Total area of the playground = 3.165×500
= 1582.56 m²
Now, cost of levelling the playground at the rate of ₹7.50/m² = ₹ 7.50 × 1582.56 m²
= ₹ 11869.20
Ques:12) A farmer wants to dig a well either in the form of cuboidal shape of 1.5m × 1.5m × 7m or in the cylindrical shape of radius 75 cm and height 7 m. The rate of digging a well is ₹ 75/m³. The farmer decided to dig cylindrical form of well.(Use π= 22/7)
(a) Calculate the cost of digging in both the cases.
(b) By the decision what value is depicted by the farmer?(Marks-4)
Solution:
Given,
Dimensions of cuboidal well
Lenth of the well (l)= 1.5 m
Breadth of the well(b) = 1.5 m
Height of the well(h)= 7 m
Dimensions of cuboidal well
Radius of cylindrical well (r)= 0.75m
Height of the cylindrical well(H)= 7m
Now,
(i)Volume of the cuboidal well = L×B×H
= 1.5×1.5×7
= 15.75 m³
Now,
Volume of cylindrical well = πr²h
= π×0.75×0.75 × 7
= 12.37 m³
Therefore,
Cost of digging of cuboidal well at the rate of ₹75/m³
= ₹75×15.75 = ₹ 1181.25
Cost of digging of cylindrical well at the rate of ₹75/m³
= ₹75×12.37 = ₹ 927.75
(ii)Choosing to dig a cylindrical well as compared to cuboidal well by farmer, shows that Farmer wants to reduce his labour work and Cost of digging.
Ques:13) A cuboidal box of external dimensions 84cm × 64cm × 17cm (open from the top) is made up of wood of thickness 2 cm. Find the:
(a) Volume of the inner cuboid.
(b) Outer lateral surface area.(Marks-4)
Solution:
Given,
External dimensions of cuboidal box ( Open Top)
Length = 84 cm
Breadth = 64 cm
Height = 17 cm
Thickness of the box = 2 cm
Now,
Internal dimension of cuboidal box ( Open Top)
Length = 84-4 = 80 cm
Breadth =64-4 = 60 cm
Height = 17-2 = 15 cm
Now,
(i)
Inner volume of cuboidal box=L ×B×H
=80×60×15
= 72000 cm³
(i)
Outer lateral surface area =2(L+B)×H
=2(84+64)×17
=2(148)×17
= 5032 cm²
Ques:14) A cubical tank whose side is 2m filled with water. The water from cubical tank is shifted to a cuboidal tank whose length, breadth and height are 250 cm, 200 cm, 2 m respectively. Find the depth of tank which will remain empty.(Marks-4)
Solution:
Ques:15) A residential colony has a population of 5400 and 60 litres of water is required per person per day. For the effective utilization of rain water, they constructed a water reservoir measuring 48m × 27m × 25m to collect the rain water. For how many days, the water of this tank is sufficient, if during rain the height of water level is 5 m?(Marks-4)
Ques:16) A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of pencil is 7mm and the diameter of graphite is 1mm. If the pencil is 14 cm long, find:
(a) volume of graphite.
(b) volume of wood.(Marks-4).
Apex Class 
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