Three Dimensional Solids Class 10 OP Malhotra Exe-15E ICSE Maths Solutions Ch-15 questions as latest prescribe guideline for upcoming exam. In this article you would learn how to Solve problems on Conversion of Solid by Melting and Recasting. easily. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Three Dimensional Solids Class 10 OP Malhotra Exe-15E ICSE Maths Solutions Ch-15
Board | ICSE |
Publications | S Chand |
Subject | Maths |
Class | 10th |
Chapter-15 | Three Dimensional Solids |
Writer | OP Malhotra |
Exe-15E | Conversion of Solids by Melting and Recasting |
Edition | 2024-2025 |
Conversion of Solids by Melting and Recasting
- According to law of conservation of mass the volume of combining solids must be same volume of product solids.
- Each and every solid that exists occupies some volume. When you convert one solid shape to another, its volume remains the same, no matter how different the new shape is
- Hence, when you convert one solid shape to another, all you need to remember is that the volume of the original, as well as the new solid, remains the same
Exe-15E (Conversion of Solids by Melting and Recasting)
Que-1: The diameter of a metallic sphere is 6cm. The sphere is melted and drawn into a wire of uniform circular cross section. If the length of the wire is 36 cm, find its radius.
Sol: Let the wire’s radius be a.
Given, sphere is melted into the wire.
The wire formed is a cylinder, hence the volume of wire will be equal to the volume of sphere.
Radius of sphere (r) = Diameter/2
= 6/2 = 3 cm.
Volume of sphere (V) = (4/3)πr³
Putting values we get,
V = (4/3)π×(3)³
= 4π×3²
= 36π cm³.
Given, length of wire = 36 cm.
So, height of cylinder = 36 cm.
Volume of cylinder = V = 36π cm3.
∴ πr²h = 36π
⇒ r² = 36π/πh
⇒ r² = 36/36
⇒ r² = 1
⇒ r = 1.
Que-2: How many lead balls, each of radius 1cm, can be made from a sphere whose radius is 8 cm?
Sol: Radius of the sphere = R = 8 cm
Volume of the sphere = (4/3)πR³ = (4/3)π×8×8×8
= (4/3)π×512 cm³
Radius of each new ball = r = 1 cm
Volume of each ball = (4/3)πr³ = (4/3)π×1×1×1
= (4/3)π×1 cm³
Total number of new balls that can be made = Volume of sphere/Volume of each ball
= {(4/3)π×512}/{(4/3)π×1}
= 512.
Que-3: A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones, each of diameter 7cm and height 3cm. Find the number of cones so formed.
Sol: Diameter of sphere = 21cm
radius = 21/2 cm
Height of the cone = 3 cm
Radius of the cone = 7/2 cm
Volume of the sphere = (4/3)πr³
= (4/3)π(21/2)³
= 37044π/24 cm³
Volume of cone = (1/3)πr²h
= (1/3)π(7/2)²(3)
= 49π/4 cm³
Let n be the number of cone formed
n = Volume of sphere/Volume of cone
n = {37044π/24}/{49π/4}
n = 126.
Que-4: A metallic disc, in the shape of a right circular cylinder, is of height 2.5 mm, and base radius 12 cm. Metallic disc is melted and made into a sphere. Calculate the radius of the sphere.
Sol: Height = 2.5 mm = 25/(10×10) cm = 1/4 cm Base radius = 12 cm
We know that the volume of right circular cylinder = πr²h
Put the value in this concept
= (22/7)×12×12×(1/4)
= 792/7 cm³ …….… (1) Suppose that
The sphere’s radius represented by = (r)
Volume of sphere = (4/3)πr³
Put the value in this formula we obtain
= (4/3)×(22/7)×r³
= (88/21) r³ ……(2)
Now we have calculated the equation (1) and (2)
(88/21) r³ = 792/7
r³ = (792/7)×(21/88)
r³ = 27
r = 3 cm.
Que-5: A hollow sphere of internal and external diameters is 6cm and 10cm respectively is melted and recast into a cone of base diameter 14 cm. Find the height of the cone.
Sol: Internal diameter of the hemispherical shell = 6 cm
Therefore, internal radius of the hemispherical shell = 3 cm
External diameter of the hemispherical shell = 10 cm
External radius of the hemispherical shell = 5 cm
Volume of hemispherical shell = (4/3)π(5³-3³) = (4/3)π(98) = 392π/3 cm³
Radius of cone = 7 cm
Let the height of the cone be h cm.
Volume of cone = (1/3)πr²h = (1/3)π×7²h = 49πh/3 cm³
The volume of the hemispherical shell must be equal to the volume of the cone therefore
392π/3 = 49πh/3
49h = 392
h = 8.
Que-6: A hollow metallic cylindrical tube has an internal radius of 3cm and height 21 cm. The thickness of the metal tube is (1/2)cm. The tube is melted and cast into a right circular cone of height 7cm. Find the radius of the cone to 1 decimal place.
Sol: lnternal radius of the hollow cylinder = r = 3 cm
Height = h = 21 cm
Thickness of the metal = 0.5 cm
Therefore, Outer radius= R = (3+0.5) cm = 3.5 cm
Now, Volume of metal used = πh(R²-r²)
= (22/7)×21×(3.5²-3²)
= (22/7)×21×(12.25-9)
= (22/7)×21×3.25
= 214.5 cm3
Volume of metal used = 214.5 cm3
Therefore, Volume of cone= 214.5 cm3 and height = 7 cm
Let r1 be the radius of cone.
∴ Volume = (1/3)πr²1h
⇒ (1/3)πr²1h = 214.5
⇒ (1/3)×(22/7)×r²1×7 = 214.5
⇒ r²1 = (214.5×3×7)/(22×7)
⇒ r²1 = 29.25
⇒ r1 = 5.4 cm
Radius of the cone = 5.4 cm.
Que-7: Three solid glass balls of radius 1,6 and 8 cm respectively are melted into a solid sphere. Find its radius.
Sol: Three solid metallic spheres of radii 1 cm, 6 cm, and 8 cm, respectively, are melted and recast into a single solid sphere.
Let the radius be R
As per the question,
4/3 π (13 + 63 + 83 ) = 4/3πR3
R3 = 729
R3 = 93
R = 9 cm
Que-8: A hemispherical bowl of internal radius 9cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3cm and height 4cm. How many bottles are necessary to empty the bowl?
Sol: Internal radius of the hemispherical bowl R = 9 cm
Radius of the cylindrical bottles r = 3/2 cm
Height of the cylindrical bottles h = 4 cm
Let number of cylindercal bottes be n, then
According to the question
π r2 h × n = (2/3) × πR3
⇒ 3/2 × 3/2 × 4 × n = 2/3 × 9 × 9 × 9
⇒ n = (2 × 3 × 9 × 9)/(3 × 3)
⇒ n = 54
So, the number of bottles is 54.
Que-9: How many spherical lead shots, each 4.2 cm in diameter, can be obtained from a rectangular solid of lead of dimensions 66cm × 42cm × 21cm.
Sol: The dimensions of the solid rectangular lead piece is
66cm × 42cm × 21cm.
Diameter of the spherical lead shots = 4.2 cm Let n spherical lead shots be obtained from the rectangular piece.
n × volume of spherical lead shot = Volume of the rectangular lead piece
⇒ Volume of the rectangular lead piece/ volume of spherical lead shot = n
⇒ (66×42×21)/{(4/3)πr³} = n
⇒ (66×42×21)/{(4/3)π(4.2/2)³} = n
⇒ 58212/38.808 = n
⇒ n = 1500
Hence, 1500 lead shots can be formed.
Que-10: The radius of a solid iron sphere is 8cm. Eight rings of iron plates of external radius {6*(2/3)}cm, thickness 3cm are made by melting this sphere. Find the internal diameter of each rings.
Sol: Volume of solid iron sphere (4/3)π×8³ cm³ = (2048/3) π cm³
External radius of each iron ring = 6*(2/3) cm = 20/3 cm
Let the internal radius of each ring be r cm
Since each ring forms a hollow cylindrical shell of external and internal radii 20/3 cm and r cm respectively and height 3 cm
Volume of each ring = π{(20/3)²−r²} × 3 cm³
volume of 8 such rings
= 8π(400/9−r²) × 3 cm³ = 24π(400/9−r²) cm³
Clearly volume of 8 rings= volume of the sphere
⇒ 24π(400/9−r²) = (2048/2) π
⇒ (400/9−r²) = (2048/2)π × 1/24π
⇒ r² = 16
⇒ r = 4cm.
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