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

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.