Class 10 Maths Chapter 12 Previous Year Questions - Surface Area and Volumes

Ans:
Here, the height of small cylinder (h₁) = 50 cm
Radius of small cylinder r =  7 cm
Height of longer cylinder (h₂) = 200 cm
Radius of longer cylinder (R) = 14 cm
Volume of figure = volume of small cylinder + volume of big cylinder

Mass = volume × Density
= 1,30,900
= 10,47,200 g
≈ 1047.2 kg 

Ans:
Here, two figures are combined, 2 hemisphere and cylinder.
Radius of cylinder = radius of hemisphere
= 4/2 = 2 mm
Height of cylinder (h) = 14 – (2 + 2)
= 14 – 4 = 10 mm.
Surface area of capsule = curved surface area of two hemispheres + curved surface area of cylinder

Volume of capsule = volume of 2 hemisphere + volume of cylinder

[∵ Volume of 2 hemisphere = volume of a sphere]

Ans: (c)
We have, the height of cone. h = 24 cm and radius, r = 7 cm.

We know that,
=
= √625

= 25
Now. curved surface area = πrl
= 22/7 x 7 x 25
= 550 cm²

Ans: (b)
Curved surface area of cylinder = 2πrh
⇒ 94.2 = 2 x 3 .14 x r x 5

⇒ r = 3 cm

Ans: 

Radius of the cylinder (r) = 3.5 cm
Height of the cylinder (h) = 10 cm
Curved surface area = 2πrh
=
= 220 cm²
Curved surface area of a hemisphere = 2πr²
∴ Curved surface area of both hemispheres
=
= 154 cm²
Total surface area of the Article
= (220 + 154) cm²
= 374 cm².

Ans: Let r be the radius and h be the height of the cylindrical part and R be the radius of hemispherical part

volume of air = volume of cylinder = 1408/21 
∴
Now, volume of air = 



⇒ h = 4
Now, radius of hemispherical part R = 1/2h = 2m
∴ Total height of the room = R + h = 2 + 4 = 6m

Ans: Radius of cone, r = 3 cm 
Height of cone, h = 12 cm 
Let x be the volume of unfilled part of cone.
Now, volume of cone, =
Volume of filled part of cone = Volume of cone - Volume of unfilled part of cone

Now, volume of ice-cream = volume of filled part of cone + volume of hemisphere
=
= 150.72 cm³

Ans: Given ratio of radium and height of the right circular cylinder = 2:3
Let radius (r) of the base be 2x and height(h) be 3x.
Volume of cylinder, V = πr²h

Total surface area of cylinder = 2πr (h + r)

= 770 cm²

Ans: Paper required to make four caps is 2,200 sq.cm.
Weight of the cake for required dimensions is 1kg.
Step-by-step explanation:
(a) Given the base circumference of the cone, c = 44 cm
Height of a cone, h = 24 cm.
Base circumference of the cone, c = 2πr = 44 cm
Thus, the radius of the cone is

The curved surface area of the cone is given by

Substituting the values of h and r,

Thus, to make one cap, 550 sq.cm of paper is required.
Then to make four caps, the required paper is
500 x 4 = 2000 sq. cm
Therefore, 2,200 sq.cm of paper is required to make four caps.
(b) Given the diameter of cylindrical shape cake, d = 24 cm
Height of cylindrical shape cake, h = 14 cm.
Radius of the cylindrical shape cake,

Volume of the cylinder is given byV = πr2h
Substituting the values of h and r,


The required volume of the cylindrical shape cake is 6,336 cu.cm.
Given 650 cu.cm equals 100 g of cake.
Then the required weight of the cake is

Given the bakery shop sells cakes by weight of 0.5 kg, 1 kg, 1.5 kg, etc.
Since, ,therefore, the cake of 1kg should be ordered for required dimensions.

Ans: The dimension of the cuboids so formed are
length = 18 cm
breath = 6 cm and height = 6 cm.
Surface area of cuboids = 2 (l× b + b × h + l × h)
= 2 × (18 × 6 + 6 × 6+ 18 × 6)
= 504 cm²

Ans: According to given information, we have the following figure.

Clearly, the radius of conical part = radius of cylindrical part = 30/2 = 15 m = r  ...(say)
Let h and H be the height of conical and cylindrical part respectively.
Then h = 8 m and H = 9 m

= 17 m

(i) The area of the canvas used in making the tent
= Curved surface area of cone + Curved surface area of cylinder
= πrl + 2πrH
= πr(l + 2H)

= 1650 m²
(ii) Area of canvas bought for the tent
= (1650 + 30) m²
= 1680 m²
Now, this cost of the canvas height for the tent
= ₹ (1680 × 200)
= ₹ 3,36,000

Ans: Given diameter of circular pipe = 8 cm
So, radius of circular pipe = 4cm
Length of flow of water in one sec = 80 cm
length of flow of water in one hour =  80 x 60 x 60 cm=288000 cm=h
Volume of cylinderical pipe in one hour = πr²h

= 14482.28 litre [approx.]
14482.28 litres of water being pumped out through this pipe in 1 hr.

Ans:  (d)
Diameter of sphere = Distance between opposite faces of cube = 2a
radius of sphere = a
So, volume of spherical ball =
=

Ans: (c)
Let radius of the sphere be r.
According to the question,
⇒

Ans: Let height of one cone be h and height of another cone be 3h. Radius, of one cone is 3r and radius of another cone is r.
∴ Ratio o f their volumes =
= 3 : 1

Ans: Let n be the number of solid cubes of  2cm made from a solid cube of side 10 cm.

∴ n x Volume of one small cube = Volume of big cube
⇒ n x (2)³ = (10)³
⇒ 8n = 1000
⇒ n = 1000/8
= 125
Thus, the number of solid cubes formed of side 2 cm each is 125.

Ans: Let the radius and the height of the cylinder are r and h respectively.
So, radios of t he cone is r and height of the cone is 3h.
∴ Volume of the cylinder = πr2h
So Volume of cone =
So, require ratio =
= 1 : 1

Ans: Given that, the depth of the well is 14 m and the diameter is 3 m.
The width of the circular ring of the embankment is 4 m.
A figure is drawn below to visualize the shapes.

From the above figure, we can observe that the shape of the well will be cylindrical, and earth evenly spread out to form an embankment around the well in a circular ring will be cylindrical in shape (Hollow cylinder) having outer and inner radius.

Volume of the earth taken out from well = Volume of the earth used to form the embankment

Hence, Volume of the cylindrical well = Volume of the hollow cylindrical embankment

Let us find the volume of the hollow cylindrical embankment by subtracting volume of inner cylinder from volume of the outer cylinder.

Volume of the cylinder = πr²h where r and h are the radius and height of the cylinder respectively.

Depth of the cylindrical well, = h₁ = 14 m

Radius of the cylindrical well, = r = 3/2 m = 1.5 m

Width of embankment = 4 m

Inner radius of the embankment, r = 3/2 m = 1.5 m

Outer radius of the embankment, R = Inner radius + Width

R = 1.5 m + 4 m

= 5.5 m

Let the height of embankment be h

Volume of the cylindrical well = Volume of the hollow cylindrical embankment

πr²h₁ = πR²h - πr²h

πr²h₁ = πh (R² - r²)

r²h₁ = h (R - r )(R + r)

h = [(r²h₁)/(R - r)(R + r)]

h = [((1.5 m)² × 14 m)/(5.5 m - 1.5 m)(5.5 m + 1.5 m)]

= (2.25 m² × 14 m)/(4m × 7 m)

= 1.125 m

Therefore, the height of the embankment will be 1.125 m.

Ans: Given diameter of conical part = Diameter of hemispherical part = 4cm
∴ Radios of conical part (r) = Radius of hemispherical part (r) = 4/2 = 2 cm
Height of conical part (h) = 2 cm
∴ Volume of toy = Volume of hemisphere + volume of cone

= 3 .14 x 4(1.33 + 0.66)= 3.14 x 4 x 1.99 cm³
Volume of the toy = 24.99 cm³

Ans: Radius of the cone = Radius of the hemisphere = r = 7cm 
Height of the cone, h = 10 cm 
Now, volume of the toy = volume of hemisphere + volume of cone

=
=  1232 cm³
Curved surface area of the toy = Curved surface area of cone + Curved surface area of hemisphere
= πrl + 2πr²

= 22(12.2 + 14)
= 22 x 26.2
= 576.4 cm³

Ans: Radius of cylinderical part (r) = Radius of each spherical part(r) = 7/2 cm
Height of cylinderical part (h) = 20 - 7/2 - 7/2 = 13 cm

Now. Volume of the solid = Volume of cylinderical part + Volume of two hemispherical endsVolume of the solid =

Volume of the solid = 680.17 cm³.

Ans: Base radius = 5/2 = 2.5 cm
Apparent capacity of glass = Volume of cylindrical portion
= πr²h
= 3.14 x (2.5)² x 10
= 196.25 cm³
Actual capacity of the glass = Volume of cylinder - Volume of hemisphere

= 196.25 - 32.71
= 163.54 cm³   

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

Ans: r = 12 cm, R = 20 cm, V = 12308.8 cm³ 


Area of metal sheet used = π(R + r)l + πr² 
= 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.

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 used
Curved 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] cm² 

= 4860.9 cm²
Now, the volume of water that the bucket can hold (also, known as the capacity of the bucket)

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

Ans: For bucket,
Upper diameter (D) = 30 cm
∴
Lower diameter (d) = 10 cm

Height of bucket (h) = 24 cm
∴ The area of the metal sheet used = CSA of the frustum (bucket) + Area of bottom part (base)

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
⇒ 
⇒ 

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.

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 vessel

According 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

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.

Ans: We have,
Radius of the cylinder = Height of the cylinder = 2.4 cm
Also, 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.

Ans: Volume of sphere = 4/3 π(6)³ cm³ 
Volume of water rise in cylinder =
∴
⇒ r = 9 cm

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.

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
⇒

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
Also, ΔABC ~ ΔADE




∴ The required ratio = 1 : 7.