Surface Areas & Volumes (Exercise 12.3) NCERT Solutions - Mathematics (Maths) Class 10

[Take π = 22/7, (Unless stated otherwise)]
Q 1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Sol. Radius of the sphere (r₁) = 4.2 cm
∴ Volume of the sphere
Radius of the cylinder (r₂) = 6 cm
Let ‘h’ be the height of the cylinder
∴ Volume of the cylinder =
Since, Volume of the metallic sphere = Volume of the cylinder.


Q 2. Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Sol. Radii of the given spheres are:
r₁ = 6 cm
r₂ = 8 cm
r₃ = 10 cm
⇒ Volume of the given spheres are:

∴ Total volume of the given spheres
= V₁ + V₂ + V₃


Let the radius of the new big sphere be R.
∴ Volume of the new sphere

Since the two volumes must be equal.


⇒ R³ = 1728
⇒ R³ =2³ × 2³ × 3³
⇒ R³ = (2 × 2 × 3)³
⇒ R = 2 × 2 × 3
⇒ R = 12 cm
Thus, the required radius of the resulting sphere = 12 cm.

Q 3. A 20 m deep well with a diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Sol. Given, diameter of the well = 7m
⇒ Radius = 7/2m
and depth of the well = 20 m
Volume of the earth taken out from the well = πr²h

Earth is evenly spread on a platform 22 m by 14 m
Let the height of platform be h
∴ Volume of platform = 22 x 14 x h
According to the question,
Volume of the platform = Volume of the earth taken out from the well
⇒ 22 x 14 x h = π x 49 x 5
⇒ 22 x 14 x h = 22/7 x 49 x 5


Q 4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Sol. Diameter of cylindrical well (d) = 3 m
⇒ Radius of the cylindrical well (r) = 3/2 m = 1.5m
Depth of the well (h) = 14 m

Let the height of the embankment = ‘H’ meter
Internal radius of the embankment (r) = 1.5 m.
External radius of the embankment R = (4 + 1.5) m = 5.5 m.
∴ Volume of the embankment
= πR²H − πr²H = πH [R² − r²] = πH (R + r) (R − r)
= 22/7 × H (5.5 + 1.5) (5.5 − 1.5) = 22/7 × H × 7 × 4 m³
Since, Volume of the embankment = Volume of the cylindrical well

Thus, the required height of the embankment = 1.125 m.

Q 5. A container shaped like a right circular cylinder having a diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Sol. For the circular cylinder:
Diameter = 12 cm ⇒ Radius = 12/2 =  6 cm
Height (h) = 15 cm
∴ Volume = πr²h
⇒ Volume of total ice cream = 22/7 × 6 × 6 × 15 cm³
For conical + hemispherical ice-cream cone:
Diameter = 6 cm ⇒ radius (R) = 3 cm
Height of ice cream cone (H) = 12 cm
Volume = (Volume of the conical part) + (Volume of the hemispherical part)

Let number of ice-cream cone required to fill the total ice cream = n.

Thus, the required number of cones is 10.

Q 6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Sol. For a circular coin:
Diameter = 1.75 cm

   |∵ A coin is like a cylinder
For a cuboid:
Length (l) = 10 cm, Breadth (b) = 5.5 cm
and Height (h) = 3.5 cm

Number of coins
Let the number of coins need to melt be ‘n’

Thus, the required number of coins = 400.

Q 7. A cylindrical bucket, 32 cm high and with 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.
Sol. For the cylindrical bucket:
Radius (r) = 18 cm
Height (h) = 32 cm
Volume = πr²h

⇒ Volume of the sand =

For the conical heap:
Height (H) = 24 cm
Let radius of the base be (R).
∴ Volume of conical heap =  

Radius of the conical heap of sand:
∵ Volume of the conical heap of sand = Volume of the sand

Slant Height
Let ‘l’ be the slant height of the conical heap of the sand.

∴ l² = R² + H²
⇒ l² =24² + 36²
⇒ l² = (12 × 2)² + (12 × 3)²
⇒ l² =12² [2² + 3²]
⇒ l² =12² × 13

Thus, the required height = 36 cm and slant height  

Q 8. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Sol. Width of the canal = 6 m
Depth of the canal = 1.5 m
Length of the water column in 1 hr = 10 km
∴ Length of the water column in 30 minutes  

Since the above amount (volume) of water is spread in the form of a cuboid of height as

Let the area of the cuboid = a
∴ Volume of the cuboid=Area × Height
Thus,

Thus, the required area = 56.25 hectares.

Q 9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Sol.  Given, diameter of pipe = 20 cm
⇒ radius = 10 cm
Water flowing from the pipe at rate = 3 km/h
Let it filled the tank in t hours.
Volume of water flowing in t hours

Volume of the water in cylindrical tank = πr²h = π(5)² x 2 m³
  
According to the question,