Three Dimensional Solids Class 10 OP Malhotra Exe-15D ICSE Maths Solutions

Three Dimensional Solids Class 10 OP Malhotra Exe-15D ICSE Maths Solutions  Ch-15 questions as latest prescribe guideline for upcoming exam. In this article you would learn how to solve problems on Surface Area and Volume of Sphere. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

Three Dimensional Solids Class 10 OP Malhotra Exe-15D ICSE Maths Solutions

Three Dimensional Solids Class 10 OP Malhotra Exe-15D ICSE Maths Solutions  Ch-15

Board ICSE
Publications  S Chand
Subject Maths
Class 10th
Chapter-15 Three Dimensional Solids
Writer OP Malhotra
Exe-15D Surface Area and Volume of Sphere
Edition 2024-2025

Surface Area and Volume of Sphere

  • Total surface area of a sphere is 4 π r
  •  Volume of a sphere is  = (4 ⁄ 3) π r3  
  • Total Surface Area of Hemisphere = 3 π r2
  • Curve Surface Area of Hemisphere = 2 π r2
  • Total Surface Area of Hollow Hemisphere: = 2π (r2+ r12) + π(r22 – r12)  If r1 and  and r2 is the radius of internal and external hemisphere.

Exercise- 15D (Surface Area and Volume of Sphere)

Three Dimensional Solids Class 10 OP Malhotra Exe-15D ICSE Maths Solutions  Ch-15

Que-1: Find the volume and curved surface area of the sphere if (i) radius = 1cm  (ii) radius = 14cm  (iii) radius = 28cm (iv) radius = {5*(1/4)}cm  (v) diameter = 14cm   (vi) diameter = 35cm

Sol:  (i) Volume of sphere = (4/3)πr³
= (4/3)×(22/7)×(1)³
= (4/3)×(22/7)×1
= 88/21 cm³
Curved surface area of sphere = 4πr2
= 4 × (22/7) × 1² (∵r = 1 cm)
= 4 × (22/7) × 1
= 4 × (22/7)
= 88/7 cm²

(ii) Volume of sphere = (4/3)πr³
= (4/3)×(22/7)×(14)³
= (4/3)×22×392
= 34496/3 = 11498*(2/3) cm³
Curved surface area of sphere = 4πr2
= 4 × (22/7) × 14² (∵r = 14 cm)
= 4 × 22 × 28
= 2464 cm²

(iii) Volume of sphere = (4/3)πr³
= (4/3)×(22/7)×(28)³
= (4/3)×22×3136
= 275968/3 = 91989*(1/3) cm³
Curved surface area of sphere = 4πr2
= 4 × (22/7) × 28² (∵r = 28 cm)
= 4 × 22 × 112
= 9856 cm²

(iv) Volume of sphere = (4/3)πr³
= (4/3)×(22/7)×(21/4)³
= (4/3)×(22/7)×(9261/64)
= 11×(441/8)
= 4851/8 = 606*(3/8) cm³
Curved surface area of sphere = 4πr2
= 4 × (22/7) × (21/4)² (∵r = 21/4 cm)
= 4 × (22/7) × (441/16)
= 11 × (63/2)
= 693/2 = 346*(1/2) cm²

(v) diameter = 14cm
r = 14/2 = 7cm
Volume of sphere = (4/3)πr³
= (4/3)×(22/7)×(7)³
= (4/3)×22×49
= 4312/3 = 1437.3 cm³
Curved surface area of sphere = 4πr2
= 4 × (22/7) × 7² (∵r = 7 cm)
= 4 × (22/7) × 49
= 4 × 22 × 7
= 616 cm²

(vi) diameter = 35 cm
r = 35/2 cm
Volume of sphere = (4/3)πr³
= (4/3)×(22/7)×(35/2)³
= (4/3)×(22/7)×(42875/8)
= (1/3)×11×6125
= 67375/3 = 22458*(1/3) cm³
Curved surface area of sphere = 4πr2
= 4 × (22/7) × (35/2)² (∵r = 35/2 cm)
= 4 × (22/7) × (1225/4)
= 175 × 22
= 3850 cm²

Que-2: Find the radius of the sphere whose surface area is equal to volume.

Sol:  Surface area = volume
⇒ 4πr² = (4/3) πr³
⇒ 3r² = r³
⇒ r = 3
Radius of the sphere = 3 units

Que-3: Find the diameter of the sphere whose surface area is equal to the area of circle of diameter 2.8 cm.

Sol:  Diameter of circle = 2.8 cm ⇒ radius = r = 1.4 cm
Area of a circle = πr²
= π(1.4)²
= 1.96π
Surface area of sphere = 4πr²
Given ,
Surface area of sphere = Area of the circle
⇒ 4πr² = 1.96π
⇒ r² = 1.96/4
⇒ r² = 0.49
⇒ r = 0.7cm
d = 2r = 2×0.7 = 1.4 cm
Diameter of the sphere = 1.4 cm.

Que-4: Find the radius of the sphere whose volume is {32π/3}cm³.

Sol:  Let r be the radius of the spherical tank.
Given volume of the spherical tank = 32π/3
(4/3)πr³ = 32π/3
4r3 = 32
r3 = 324 = 8
r3 = 23
r = 2
∴ Radius of the spherical tank r = 2 cm.

Que-5: The volume of the sphere is 4.851 cm³. Find its surface area.

Sol:  Let the radius of the sphere be r.
As,
Volume of the sphere = 4851 cm3
⇒ (4/3) πr³ = 4851
⇒ (4/3)×(22/7)×r³ = 4851
⇒ r³ = (4851×3×7)/(4×22)
⇒ r³ = 9261/8
= r = 3√9261/8
⇒ r = 21/2 cm
Now,
Curved surface area of the sphere = 4πr2
= 4×(22/7)×(21/2)×(21/2)
= 1386 cm2
So, the curved surface area of the sphere is 1386 cm2.

Que-6: Find the volume of a hollow sphere whose outer diameter is 10cm and the thickness of the material of which it is made is 1 cm.

Sol:  Inner diameter = 8 cm
Inner radius = r = 4 cm
Outer radius = R = 4cm + 1cm thick material = 5 cm
Volume of hemisphere = (2/3)πr³
Required Volume = (4/3) π(R³-r³)
= (4/3)×(22/7)×(5³-4³)
= (4/3)×(22/7)×61
= 255.6 cm3
Required volume= 255.6 cm3

Que-7: The surface areas of two spheres are in the ratio 4:9. Find the ratio of their volumes.

Sol:  The ratio of curved surface area of two spheres is 4 : 9.
π = 22/7
According to the question,
The ratio of curved surface area of two spheres is 4 : 9.
⇒ 4π R2/4π r2 = 4/9
⇒ R2/r2 = 4/9
⇒ R/r = 2/3
Ratio of Volume of sphere
⇒ (4/3)π R3/(4/3)π r3
⇒ 23/33
⇒ 8/27
∴ The ratio of their volume 8 : 27.

Que-8: The volumes of two spheres are in the ratio 64 : 27. Find their radii if the sum of their radii is 21 cm.

Sol:  ratio of volumes = V/v = 64/27
(4/3πR³)/(4/3πr³) = 4³/3³
R³/r³ = 4³/3³
R/r = 4/3
R = 4r/3—–(1)
sum of the radii = 21
R+r = 21
4r/3 +r = 21 [ from (1)]
(4r+3r)/3 = 21
7r/3 = 21
r = 21*3/7
r = 3*3
r = 9
substitute r = 9 in (2)
R+r = 21
R+9 = 21
R = 21-9
R = 12
therefore
R = 12, r = 9

Que-9: Find the cost of painting a hemispherical dome of diameter 10m at the rate of Rs1.40 per m².

Sol:  Diameter of the hemispherical dome = 10 m
Therefore, radius of dome = 5 m
Curved surface area = 2πr²
= 2×(22/7)×5×5
= 157.14 m2
Cost of painting one sq. metre = Rs. 1.40
Cost of painting 157.14 m2 = Rs.(1.40 x 157.14)
= Rs. 219.99 = Rs 220
Therefore, cost of painting the dome = Rs 220

Que-10: A certain spherical ball of diameter 4cm weights 8kg. Find the weight of a spherical ball of the same material whose inner and outer diameter are 8 cm and 10 cm respectively.

Sol:  A certain spherical ball of diameter 4 cm has a mass of 8 kg
Volume of Spherical Ball = (4/3) π R³
R = Radius = Diameter/2 = 4/2  = 2 cm
Volume of Spherical Ball = (4/3) π 2³   = 32π/3  cm³
Density = Mass/ Volume   =  8 /(32π/3)  kg / cm³
= 3/4π kg / cm³
Volume of spherical shell whose outer and inner diameters are 10 cm and 8cm  => radius
are 5 cm & 4cm
= (4/3)π5³  – (4/3)π4³
=  (4/3)π ( 125 – 64)
= (4/3)π * 61 cm³
Mass = density * volume
= (3/4π)  * (4/3)π  * 61
= 61 kg

Que-11: (i) Prove that the surface area of a sphere of diameter d is πd² and the volume is (1/6)πd³.  (ii) The volumes and diameters of a cone and sphere are equal. Prove that the height of the cone is twice the diameter of the sphere.

Sol: Surface area of sphere = 4πr²
r = d/2
A = 4π(d/2)²
A = 4π(d²/4)
A = πd².
Hence Proved.
Volume of sphere = (4/3)πr³
V = (4/3)π(d/2)³
V = (4/3)π(d³/8)
V = πd³/6.
Hence Proved.

(ii) r = d/2
Volume of sphere = (4/3)πr³
V = (4/3)π(d/2)³
V = (4/3)π(d³/8)
V = πd³/6
Volume of cone = (1/3)πr²h
V = (1/3)π(d/2)²h
V = (1/3)π(d²/4)h
V = πd²h/12
Since the volume of the cones and sphere is equal :
πd³/6 = πd²h/12
2d³ = d²h
2d = h.
Hence Proved.

Que-12: The external diameter of a hollow metal sphere is 2cm, and its thickness is 2cm. Find the radius of a solid sphere made of the same material and having the same weight as the hollow sphere.

Sol:  External diameter of hollow sphere = 12 cm
External radius = R = 6 cm
Internal diameter of hollow sphere =  ( 12 – 4) cm= 8 cm
Internal radius = r = 4 cm
Volume of metal used = (4/3) π(R³-r³)
= (4/3)×(22/7)×(6³-4³)
= (4/3)×(22/7)×152
= 636.95 cmcm3
Volume of metal used = 636.95 cm3 = volume of sdid sphere
⇒ (4/3)πr³ = 636.95
⇒ (4/3)×(22/7)×r³ = 636.95
⇒ r³ = (636.95×3×7)/(4×22)
⇒ r³ = 151.99 = 152
⇒ r = 5.34 cm
Radius of the solid sphere = 5.34 cm

Que-13: A hollow copper ball has an external diameter of 12cm, and a thickness of 0.1cm. Find : (i) the outer surface area of the ball (ii) the weight of the ball if 1 cm³ of copper weighs 8.88 g.

Sol:  (i) Outer radius of the sphere, R = 12/2 = 6 cm
Thickness of the sphere = 0.01 m = 0.01 × 100 cm = 1 cm           (1 m = 100 cm)
∴ Inner radius of the sphere, r = R − 1 = 6 − 1 = 5 cm
Outer surface area of the sphere = 4πR²
= 4×3.14×(6)²
= 4×3.14×36
= 452.16 cm2

(ii) Volume of metal in the sphere = (4/3)π(R³−r³)
= (4/3)×3.14×(6³−5³)
= (4/3)×3.14×(216−125)
= (4/3)×3.14×91
= 380.97 cm3
Density of the metal = 8.88 g/cm3
∴ Mass of the sphere = Volume of metal in the sphere × Density of the metal = 380.97 × 8.88 = 3383.01 g.

Que-14: Marbles of diameter 1.4cm, are dropped into a beaker containing some water and are fully submerged. The diameter of the beaker is 7cm. Find how many marbles have been dropped in it if the water rises by 5.6 cm.

Sol:  Diameter of each marble = 1.4cm
r = d/2 = 1.4/2 = 0.7 cm
Diameter of beaker = 7cm
r = d/2 = 7/2 = 3.5 cm
Height of water rise = 5.6 cm
Volume of one marble = (4/3)πr³
V = (4/3)π(0.7)³
V = (4/3)π(0.343)
V = (4/3)(3.14)(0.343)
V = 1.439 cm³
Volume displaced = πr²h
V = (3.14)(3.5)²(5.6)
V = (3.14)(12.25)(5.6)
V = 215.22 cm³
Number of marble = Volume displaced/Volume Marble
n = 215.22/1.439
n = 149.5 = 150.

Que-15: A cylindrical tub of radius 5cm and length 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed into the tub. If the radius of the hemisphere is 3.5 cm and height of the cone outside the hemisphere is 5cm, find the volume of water left in the tub.

Sol:  For right circular cylinder, we have
r = 5cm
h = 9.8 cm
The volume of the cylinder is
V1 = πr2h
= (22/7)×5²×9.8
= 770cm
For hemisphere and cone, we have
r = 3.5 cm
h = 5 cm
Therefore the total volume of the cone and hemisphere is
V2 = {(1/3)πr²h} + (2/3)πr³
= {(1/3)×(22/7)×3.5²×5} + {(2/3)×(22/7)×3.5³}
= 154 cm3
The volume of the water left in the tube is
V = V1 – V2
V = 770 – 154
Hence, the volume of the water left in the tube is V = 616cm3

Que-16: A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm. The total height of the toy is 15.5 cm. Find the total surface area.

Sol:  Radius of hemisphere = 3.5 cm
total height of the toy = 15.5 cm.
Surface area of cone = πrl
l = √{(12)²+(3.5)²} = √156.25
= 12.5cm
Therefore,
Surface area of cone
= (22/7)×3.5×12.5
= 137.5 cm²
Surface area of hemisphere = 2πr²
= 2×(22/7)×3.5×3.5
= 77 cm²
Therefore,
Total surface area of the toy
= 137.5+77
= 214.5 cm²

Que-17: The largest sphere is carved out of a cube of side 7cm. Find the volume of the sphere.

Sol:  The largest sphere is carved out of a cube of side 7 cm.
⇒ Diameter of largest sphere = side of cube = 7 cm
= (4/3)×(22/7)×(7/2)³
= (11/3)×49
= 539/3 = 179.7 cm³.

Que-18: A cylindrical bucket, whose base has a radius of 15cm, is filled with water up to a height of 20cm. A heavy iron spherical ball of a radius 10cm is dropped to submerge completely in water in the bucket. Find the increase in the level of water.

Sol: Volume of sphere = (4/3)πr³
= (4/3)πr³ = (4/3)π(1000) = 4000π/3 cm³
Volume of displaced water = volume of sphere = 4000π/3 cm³
Volume of displaced water = πr²h
4000π/3 = π(15)²h
4000/3 = 225h
h = 4000/(3×225)
h = 5.93 cm.

Que-19: The radius of the base of a cone and the radius of a sphere are the same, each being 8cm. Given that the volumes of these two solids are also the same, calculate the slant height of the cone, correct to one place of decimal.

Sol:  Volume of sphere = Volume of cone
(4/3)πr³ = (1/3)πr²h
4r = h
4(8) = h
h = 32 cm
l = √(r²+h²)
l = √(8²+32²)
l = √(64+1024)
l = √1088 = 32.96 = 33 cm.

Que-20: A vessel is in the form of an inverted cone. Its height is 8cm and the radius of its top which is open is 5cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.5cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Sol:  Volume of cone = (1/3)πr²h
= (1/3)π(5)²(8) = 200π/3 cm³
Volume displaced = (1/4) Volume of cone
= (1/4) (200π/3) = 50π/3 cm³
Volume of sphere = (4/3)πr³
= (4/3)π(0.5)³= = (4/3)π(0.125)
= 4π/24 = π/6 cm³
n = Volume displaced/Volume sphere
n = {50π/3}/{π/6}
n = 100.

Que-21: A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 metres and its volume is (2/3) of the hemisphere. Calculate the height of the cone and the surface area of the buoy correct to 2 places of decimals.

Sol:  Radius of hemisphere and cone = 3.5 m
Volume of cone = (2/3) of the volume of hemisphere.
Volume of hemisphere = (2/3)πr³
= (2/3)π(3.5)³ = 85.75π/3 m³
Volume of cone = (2/3) Volume of sphere
Volume of cone = (2/3) × (85.75π/3) = 171.5π/5 m³
Volume of cone = (1/3)πr²h
171.5π/5 = (1/3)π(3.5)²h
171.5π/5 = (1/3)π(12.25)h
12.25h = (171.5×)/9
h = 57.1667/12.25
h = 4.67 m
CSA of cone = πrl
l = √(r²+h²) = √(3.5²+4.67²) = √(12.25+21.81) = √34.06 = 5.84 m.
CSA of cone = π(3.5)(5.84)
= 64.2 m²
CSA hemisphere = 2πr²
= 2π(3.5)² = 2π(12.25) = 76.96 m²
Total SA = CSA of cone + CSA of hemisphere
= 64.2 + 76.96
= 141.16 m²

Que-22: A cylinder, whose height is equal to its diameter, has the same volume as a sphere of radius 4cm. Calculate the radius of the base of cylinder, correct to 1 decimal places.

Sol:  Given that,
Height of cylinder = diameter = h
We know that,
Diameter = 2(radius)
h = 2r
Volume of the cylinder = volume of the sphere
⇒ hr²×2r = (4/3)h(4)³
⇒ r³ = 128/3
⇒ r = 3.5 cm

–: End of Three Dimensional Solids Class 10 OP Malhotra Exe-15D ICSE Maths Solutions :–

Return to :  OP Malhotra S Chand Solutions for ICSE Class-10 Maths

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