These NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Exercise 13.3
Unless stated otherwise, take π = \(\frac { 22 }{ 7 }\)
Question 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.
Solution:
Given: radius of metallic sphere = 4.2 cm
∴ Volume = \(\frac { 4 }{ 3 }\)π(4.2)³ …. (i)
∵ Sphere is melted and recast into a cylinder of radius 6 cm and height h.
∴ Volume of the cylinder =πr²h = π(6)² x h … (ii)
According to question,
Volume of the cylinder = Volume of the sphere
310.464 cm³ = \(\frac{22 \times 36}{7}\) h
310.464 cm³ = 113.142 cm³h
h = \(\frac{310.464}{113.142}\) cm³
h = 2.74 cm.
Height of cylinder = 2.74 cm.
Question 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.
Solution:
Radius of sphere (r1) = 6 cm.
So the volume = \(\frac { 4 }{ 3 }\)π x (r1)³
= \(\frac { 4 }{ 3 }\)π (6)³ cm.
Radius of sphere (r2) = 8 cm.
So the volume = \(\frac { 4 }{ 3 }\)π x (r2)²
= \(\frac { 4 }{ 3 }\)π (6)² cm.
Radius of sphere (r3 ) = 10 cm.
So the volume = \(\frac { 4 }{ 3 }\)π x (r3)²
= \(\frac { 4 }{ 3 }\)π (10)² cm.
Now, spheres are melted and form a single
Sphere of radii R. Volume of single sphere = \(\frac { 4 }{ 3 }\)πR³
Radius of single sphere = 12 cm.
Question 3.
A 20 m deep well with 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.
Solution:
Given: diameter of the well = 7 m Radius = \(\frac { 7 }{ 2 }\)m
and depth of the well = 20 m
Volume of the earth taken out from the well = πr²
= \(\frac { 22 }{ 7 }\) x \(\frac { 7 }{ 2 }\) x \(\frac { 7 }{ 2 }\) x 20
= 770 cm³.
To form a platform of 22m , 14 m ,and h height.
∴ Volume = l x b x h
Volume = 22 x 14 x h
Volume of earth = Volume of platform
770m³ = 22 x 14 x h
So, h = \(\frac{770 m^{3}}{308 m^{3}}\)
h= 2.5 m
Height of the platform = 2.5 m
Question 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.
Solution:
Diameter of the well = 3 m
Inner radius of well = \(\frac { 3 }{ 2 }\)m
Volume of the earth dug out = (πr²h) m³
Width of circular ring = 4 m
Outer radius of well = \(\frac { 3 }{ 2 }\) + 4 = \(\frac { 11 }{ 2 }\) m
Volume of the earth = \(\frac { 22 }{ 1 }\) x \(\frac { 3 }{ 2 }\) x \(\frac { 3 }{ 2 }\) x 14 m
= 99m³
Area of shaded region = π(R² – r²)
Question 5.
A container shaped like a right circular cylinder having 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.
Solution:
We have
Radius of the cylinder = 6 cm
Height of the cylinder = 15 cm
∴ Volume of the cylinder = πr²h
= π x 6² x 15 cm.
= 540π cm³
Radius of the ice cream cone = 3 cm
Height of the ice cream cone = 12 cm
Question 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 x 10 cm x 3.5 cm?
Solution:
Given: diameter of each coin = 1.75 cm ⇒ radius = \(\frac { 1.75 }{ 2 }\)cm
and thickness of each coin = 2 mm
Let the number of coins be n. So volume of coins of
= \(\frac { 22 }{ 7 }\) x (0.875)2 x (0.2) cm x n
= \(\left(\frac{4.4}{7} \times 0.765\right) n\)
= \(\frac{(3.3687) n}{7}\)
= (0.48125) n
Coins melted to form a cuboid of dimensions 5.5 cm, 10 cm, 3.5 cm,
= 5.5 x 10 cm x 3.5 cm
= 192.5 cm³.
∴ 0.48125 n = 192.5
∴ n = \(\frac{192.5}{0.48125}\) = 400
No. of coins = 400
Question 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.
Solution:
Given: radius of the cylindrical bucket = 18 cm
and height = 32 cm
Height of conical heap = 24 cm. Let the radius of conical heap be r1 and slant height, l then the volume of conical heap
Volume of cylinder = Volume of conical heap
3258.14 cm³ = 25.1428 r² cm².
Slant height of conical heap = 12\(\sqrt{13}\) cm.
Question 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?
Solution:
Given: width of canal = 6m, depth = 1.5 m
Rate of flowing water – 10 km/h
Volume of the water flowing in 30 minutes = \(\frac { 6×1.5×30×10 }{ 60 }\)km³
= \(\frac { 6×1.5×10×1000×30 }{ 10×60 }\)km³ = 45000 m³
We require water for standing up to height = 8 cm = \(\frac { 8 }{ 100 }\) m
Let the required area he A
∴ Volume of water required = A(\(\frac { 8 }{ 100 }\))m³
According to question. 45000 = \(\frac { A×8 }{ 100 }\)
⇒ \(\frac { 45000×100 }{ 8 }\) = A ⇒ A = 562500 m²
Area will it irrigate in 30 minutes = 562500 m² or 56.25 hectares.
Question 9.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his Held, 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?
Solution:
Suppose the tank is filled in x hours. Since water is flowing at the rate of 3 km/hr. Therefore, length of the water of the water column in x hours – 3x km = 3000x meters. Clearly, the water column forms a cylinder of radius r = \(\frac { 20 }{ 2 }\) cm = 10 cm = \(\frac { 1 }{ 10 }\) m and h = height (length) = 300x meters.
∴ Volume of the water that flows in the tank in x hours
Since the tank is filled in x hours
∴ Volume of the water that flows in the tank in x hours = Volume of the tank
