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# Previous Year Questions - Surface Area and Volumes (part-1) Class 10 Notes | EduRev

## Class 10 : Previous Year Questions - Surface Area and Volumes (part-1) Class 10 Notes | EduRev

The document Previous Year Questions - Surface Area and Volumes (part-1) Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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Q.1.  A juice seller was serving his customers using glasses as shown in Fig, The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of the glass was 10 cm, find the apparent capacity of the glass and its actual capacity. (Use π = 3.14).    [CBSE 2019 (30/5/1)]
Ans.
Since, the inner diameter of the glass = 5 cm and height =10 cm, the apparent capacity of the glass = πr2h
= (3.14 x 2.5 x 2.5 x 10) cm3 = 196.25 cm3
But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass.

= 32.71 cm3
So, the actual capacity of the glass
= apparent capacity of glass - volume of the hemisphere
= (196.25 - 32.71) cm3 = 163.54 cm

Q.2. Two cubes have their volumes in the ratio 1: 27. Find the ratio of their surface areas.    [CBSE 2018 (C)]
Ans.
Let r1 and r2 be the edges of the two cubes respectively.
Ratio of their volumes =

Ratio of their surface area

Q.3. A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?    [CBSE 2018]
Ans.
For cone, Height (h) - 3.5 m

Volume of the conical heap of rice

Canvas cloth required to just cover the heap

Q.4. A right circular cylinder and a cone have equal bases and equal heights. If their curved surface areas are in the ratio 8: 5, show that the ratio between the radius of their bases to their height is 3: 4.   [CBSE 2018 (C)]
Ans.
A cylinder and a cone have equal bases and equal heights.
Let r be the radius of both cylinder and cone and h be the height of cylinder and cone
Therefore, slant height of cone
Ratio of their curved surface

Put,

Squaring both sides, we get

Q.5. A circus tent is in the shape of a cylinder surmounted by a conical top of the same diameter. If their common diameter is 56 cm, the height of the cylindrical part is 6 m and the total height of the tent above the ground is 27 m, find the area of canvas used in making the tent.     [Delhi 2017 (C)]
Ans.
Let l be the slant height of the conical part of the tent.

Radius of the conical part (r) = 28 m

Height of conical part (h) = 21 m

Curved surface area of conical part = πrl
= π(28)35 m2 = 980π m2
Radius of cylindrical part = 28 m
Height of cylindrical part = 6 m
Curved surface area of cylindrical part = 2πrh
= 2π(28)6 = 336π m2
Total curved surface area = 980π + 336π

∴ Area of canvas used = 4136 m2

Q.6.The dimensions of a solid iron cuboid are 4.4 m x 2.6 m x 1.0 m. It is melted and recast into a hollow cylindrical pipe of 30 cm inner radius and thickness 5 cm. Find the length of the pipe.    [AI 2017]
Ans.
Let the length of cylindrical pipe = 4 marks
Its inner radius r = 30 cm = 3.0 m
Length of cuboid = 4.4 m
Breadth of cuboid = 2.6 m
and height of cuboid = 1.0 m
Inner radius ‘r’ of pipe = 30 cm = 0.3 m
Width of pipe = 5 cm = 0.05 m
∴ Outer radius ‘R’ of pipe = 0.3 m + 0.05 m = 0.35 m
Let the length of pipe formed = H metres.
A.T.Q.
Solid cuboid is melted to form a hollow cylindrical pipe with thickness.
∴ Volume of cuboid = Volume of outer cylinder - Volume of inner cylinder

H = 112 m, Hence the length of the pipe is 112 m.

Q.7.  The sum of the radius of the base and the height of a solid cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 cm2, find the volume of the cylinder. [π = 22/7]     [Delhi 2016]
Ans.
Let the radius of the base be r
r + h = 37 cm    [Given]
⇒ h = (37 - r) cm
Total surface area = 2πr (r + h)

⇒ h = (37 - 7) cm = 30 cm
Volume o f the cylinder =
= 22 x 7 x 30 cm3
= 4620 cm

Q.8. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.      [NCERT, CBSE (AI) 2017]
Ans.
We have, slant height, l = 4 cm
Let R and r be the radii of two circular ends respectively. Therefore, we have

∴ Curved surface area of the frustum = (πR + πr)l
= (9 + 3) x 4 = 12 x 4 = 48 cm

Q.9. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius on its circular face. The total height of the toy is 15.5 cm. Find the total surface area of the toy.    [CBSE(AI) 2017]
Ans.
We have, CD = 15.5 cm and OB = OD = 3.5 cm
Let r be the radius of the base of cone and h be the height of conical part of the toy.
Then, r = OB = 3.5 cm
h = OC = CD - OD = (15.5 - 3.5) cm = 12 cm

Also, radius of the hemisphere, r = 3.5 cm
∴ Total surface area of the toy
= surface area of cone + surface area of hemisphere

Q.10. Water is flowing at the rate of 5 km/hour through a pipe of diameter 14 cm into a rectangular tank of dimensions 50 m X 44 m. Find the time in which the level of water in the tank will rise by 7 cm.    [CBSE Delhi 2017 (C)]
Ans. Let the time taken by pipe be t hours.
∴ Speed = 5 km/h
∴ Length in t hours = 5000 t m.
According to the question,
Volume of water flown through pipe = Volume of water in tank
πr2k = l x b x h

Hence required time is 2 hours.

Q.11.  Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere?     [Delhi 2017]
Ans.
Let, radius of hemisphere = r
A.T.Q.
Volume of solid hemisphere = Surface area of a solid hemisphere

∴ Diameter of hemisphere = 2 x r = 2 x 9/2 = 9 unit

Q.12.  The radius and height of a solid right circular cone are in the ratio of 5: 12. If its volume is 314 cm3, find its total surface area. [Take π = 3.14]     [CBSE (F) 2017]
Ans.
Given r : h = 5 : 12
Let r = 5x and h = 12x
Volume of cone = 1/3 πr2h

⇒ x3 = 1
⇒ x = 1
So, the value of r = 5 cm and h = 12 cm

TSA of cone = πr(l + r) = 3.14 x 5(13 + 5)
= 3.14 x 90 = 282.6 cm

Q.13. A wire of diameter 3 mm is wound about a cylinder whose height is 12 cm and radius 5 cm so as to cover the curved surface of the cylinder completely. Find the length of the wire.     [CBSE (F) 2017]
Ans. CSA of cylinder = 2π(5) x 12 = 120 πcm
Let length of wire = h cm and radius of wire = 3/20cm
According to question
CSA of wire = CSA of cylinder

⇒ h = 400 cm

Q.14. A hemispherical tank, of diameter 3 m, is full of water. It is being emptied by a pipe at the rate of   litre per second. How much time will it take to make the tank half empty?     [CBSE (F) 2016]
Ans.
Radius of hemispherical tank = 3/2m = 150 cm
Volume of water in the hemispherical tank = 2/3πr3

Q.15. A cylindrical tub, whose diameter is 12 cm and height 15 cm is full of ice cream. The whole ice cream is to be divided into 10 children in equal ice cream cones, with a conical base surmounted by a hemispherical top. If the height of the conical portion is twice the diameter of the base, find the diameter of the conical part of an ice cream cone.    [CBSE (F) 2016]
Ans.
Volume of ice-cream in the cylinder = πr2h = π(6)2 x 15 cm3
Volume of ice-cream in one ice-cream cone =

∴ Volume of ice-cream in 10 such cones = 10 x 2πr3 = 20 πr
20πr3 = π x 36 x 15

Diameter of conical ice-cream cup = 6 cm

Q.16. A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by  Find the diameter of the cylindrical vessel.     [CBSE (AI) 2016]Ans. Volume of sphere = 4/3 π(6)3 cm
Volume of water rise in cylinder = ∴ ⇒ r = 9 cm
Q.17. Two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the new sphere.    [CBSE (F) 2015]
Ans.
Volume of the smaller sphere Volume of smaller sphere x density = mass∴ 36π (density of metal) = 1
Density of metal = 1/36π
∴ Volume of bigger sphere x density = mass
Volume of new sphere = volume of smaller sphere + volume of bigger sphere(R')3 = [33 + 7 x 33]
(R')3 = 33(l + 7)
(R')3 = 33 x 8
(R')3 =  33 x 23
R' =3 x 2
R' = 6 cm
∴ Diameter of new sphere =12 cm.

Q.18. A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of each bottle, if 10% liquid is wasted in this transfer.   [CBSE (AI) 2015]
Ans.
Radius of hemispherical bowl, R = 36/2 = 18 cm
Radius of cylindrical bottle, r = 6/2 = 3 cm
Let height of cylindrical bottle = h
Since 10% liquid is wasted, therefore only 90% liquid is filled into 72 cylindrical bottles.
∴ The volume of 72 cylindrical bottles = 90% of the volume in the bowl
⇒

Q.1.  A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the solid, (use π = 22/7)     [Allahabad 2019]
Ans.
Diameter of cylinder
= diameter of the hemisphere = 7 cm
∴ Radius of cylinder = 7/2 cm
Total height of the solid = 20 cm
Height of the cylinder
Volume of the solid= volume of the cylinder + 2 x volume of one hemisphere
Q.2. A bucket open at the top is in the form of a frustum of a cone with a capacity of 12308.8 cm3. The radii of the top and bottom of circular ends of the bucket are 20 cm and 12 cm respectively. Find the height of the bucket and also the area of the metal sheet used in making it. (Use π = 3.14)    [CBSE 2019 (30/1/2)]
Ans.
r = 12 cm, R = 20 cm, V = 12308.8 cm
Area of metal sheet used = π(R + r)l + πr
= π(20 + 12) x 17 + π x 144
Hence, the height of the bucket is 15 cm and the area of the metal sheet used is 2160.32 cm2.
Q.3.  An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet (Fig. 13.37). The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also, find the volume of water the bucket can hold.       [CBSE 2019(30/5/1)]
Ans.
The total height of the bucket = 40 cm, which includes the height of the base. So, the height of the frustum of the cone = (40 - 6) cm = 34 cm.
Therefore, the slant height of the frustum,
So, Area of the metallic sheet usedCurved surface area of frustum of cone + Area of circular base + Curved surface area of cylinder
= [π x 35.44 (22.5 + 12.5) + π x (12.5)2 + 2π x 12.5 x 6] cm
= 4860.9 cm2
Now, the volume of water that the bucket can hold (also, known as the capacity of the bucket)
Q.4. A cylindrical bucket, 32 cm high and with a radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
OR
A girl empties a cylindrical bucket, full of sand, of base radius 18 cm and height 32 cm, on the floor to form a conical heap of sand. If the height of this conical heap is 24 cm, then find its slant height correct up to one place of decimal.

[CBSE (E) 2014, 2019 (30/5/1)]
Ans. We have,
Radius of cylindrical bucket =18 cm
Height of cylindrical bucket = 32 cm
And the height of conical heap = 24 cm
Let the radius of the conical heap be r cm
Volume of the sand = volume of the cylindrical bucket
= πr2h = π x (18)2 x 32
Now, volume of conical heap Here, volume of the conical heap will be equal to the volume of sand∴ 8πr2 = π x (18)2 x 32
⇒ r2 = 18 x 18 x 4 = (18)2 x (2)2
⇒ r2 = (36)2 or r = 36 cm
Q.5. The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm, find the area of the metal sheet used to make the bucket. [Use π = 3.14]     [CBSE 2018]
Ans.
For bucket,
Upper diameter (D) = 30 cm
∴ Lower diameter (d) = 10 cmHeight of bucket (h) = 24 cm∴ The area of the metal sheet used = CSA of the frustum (bucket) + Area of bottom part (base)
Q.6. A metallic right circular cone 20 cm high whose vertical angle is 60° which is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.    [NCERT, CBSE (F) 2017]
Ans.
Let VAB be the metallic right circular cone of height 20 cm. Suppose this cone is cut by a plane parallel to its base at a point O' such that VO' = O' O i.e., O' is the midpoint of VO.
Let r1 and r2 be the radii of circular ends of the frustum ABB' A'.
Now, in ΔVOA and VO' A', we have
⇒ ⇒ ∴ Volume of the frustum = Let the length of wire of diameter 1/16  cm be l cm. Then,Volume of metal used in wire Since the frustum is recast into a wire of length l cm and diameter 1/16 cm.∴ Volume of the metal used in wire = volume of the frustum
⇒ ⇒ Q.7. A solid metallic cylinder of diameter 12 cm and height 15 cm is melted and recast into toys each in the shape of a cone of radius 3 cm and height 9 cm. Find the number of toys so formed.     [AI 2017 (C)]
Ans.  Diameter of metallic cylinder = 12 cm
∴ Radius of metallic cylinder (r) = 6 cm
Height (h) = 15 cm
Volume of cylinder = πr2h = π(6)2 x 15 cm3
Radius of cone = 3 cm
Height of cone = 9 cm
Volume of cone = 1/3 π(3)2 x 9 = 3 x 9π cm3
Number of toys so formed = If the question is “A tent is in the form of a cylinder surmounted by a cone. Find the capacity of the tent and the cost of canvas for making the tent at Rs 100 per sq.m.”, then the solution is given as “Tent is a combination of a cylinder and a cone. For capacity∴ Volume (capacity) of the tent = Volume of the cylindrical part + Volume of the conical part
For the cost of the canvas, we find the total surface area.
Total surface area = Curved area of the cylindrical part + Curved surface area of the conical part
Now, proceed to find its cost.”
Note: Don’t solve as “Total surface area of canvas = Total surface area of cylinder + Total surface area of a cone and proceed further"
This is the wrong solution.

Q.8. The 3/4th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with an internal radius of 10 cm. Find the height of water in a cylindrical vessel.    [Delhi 2017]
Ans.
Radius of conical vessel (r) = 5 cm
Height of conical vessel (h) = 24 cm
Radius of cylindrical vessel (R) = 10 cm
Let H be the height of water in the cylindrical vessel Now, the total volume of the conical vesselAccording to the question,3/4  of the volume of water from the conical vessel is emptied into the cylindrical vessel.
⇒ 3/4  x Volume of conical vessel =Volume of water in the cylindrical vessel.
⇒ 25 x 6 = 10 x 10 x H⇒ ⇒ H = 1.5 cm∴ Height of water in the cylindrical vessel = 1.5 cm

Q.9. In a rain-water harvesting system, the rainwater from a roof of 22 m x 20 m drains into a cylindrical tank having a diameter of base 2 m and height of 3.5 m.
If the tank is full, find the rainfall in cm.    [AI 2017]
Ans.
Length of roof (l) = 22 m
Breadth of roof (b) = 20 m
Let the height of water column collected on roof= hm
∴ Volume of standing water on rooftop = lbh
= (22 x 20 x h) m3
This water is taken into a cylindrical tank of diameter of base 2 m and height 3.5 m.
Tank gets completely filled with this amount of water.
⇒ Volume of water from roof-top = Volume of cylinder
Diameter of tank = 2 m
∴ Radius of tank = 2/2 = 1 m
Hence rainfall is 2.5 cm.
Q.10. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.    [CBSE (AI) 2017]
Ans.
We have,
Radius of the cylinder = Height of the cylinder = 2.4 cmAlso, radius of the cone = 0.7 cm and height of the cone = 2.4 cm
Now, slant height of the cone =∴ Total surface area of the remaining solid= curved surface area of cylinder + curved surface area of the cone + area of upper circular base of the cylinder.
Q.11. In Fig. a cone of radius 10 cm is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base. Compare the volumes of the two parts.
Ans.
Let.BC = r cm, DE = 10 cm
Since B is the mid-point of AD and BC is parallel to DE, therefore C is the mid-point of AE.
∴ AC = CE
∴ The required ratio = 1 : 7.

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## Mathematics (Maths) Class 10

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