West Bengal Board Class 10 Math Book Solution In English
Chapter 19 Real-Life Problems Related To Different Solid Objects Exercise 19.1
Application 1. But if the length of the outer diameter of a hollow iron cylinder with a height of 25 cm is 14 cm and the length of its inner diameter of it is 10 cm and if by melting the cylinder, a solid right circular cone having half of its height is made, then let us determine the length of the base diameter of the cone.
Solution:
Given
The length of the outer diameter of a hollow iron cylinder with a height of 25 cm is 14 cm and the length of its inner diameter of it is 10 cm and if by melting the cylinder, a solid right circular cone having half of its height is made
The length of the outer radius of the hollows cylinder (R) = 14/2 cm = 7 cm
the length of the inner radius (r) = 10/2 cm = 5cm,
and the height of the cylinder = 25cm.
∴ The quantity of iron in the cylinder = π(R² – r²).
height = (7² – 5²) x 25 cm3
=π x 24 x 25 cm3
The height of the solid cone = 25/2 cm.
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Let the length of the base radius of the cone = r cm.
According, to the condition, 1/3 π x r² x 25/2 = π x 24 25
r² = 24 x 2 x 3 = 144
∴ r = ±12
But the length of the radius can not be negative.
∴ r-12, so, r = 12
∴ The base diameter of the cone = 2 x r cm. = 24 cm.
Application 2. The length of the external diameter of the face of a hemispherical bowl made of the silver sheet is 8 cm and the length of the internal diameter of it is 4 cm. By melting the bowl a solid cone is made whose diameter is 8 cm in length. Let us O calculate the height of the cone.
Solution:
Given
The length of the external diameter of the face of a hemispherical bowl made of the silver sheet is 8 cm and the length of the internal diameter of it is 4 cm. By melting the bowl a solid cone is made whose diameter is 8 cm in length.
External radius of the hemispherical bowl = 8/2 cm = 4 cm.
& Internal radius of the hemispherical bowl = 4/2 cm = 2 cm.
The radius of the base of the solid cone = 8/2 cm = 4cm
Let the height of the cone = h cm.
According to the problem, the volume of the cone = Volume of the hemispherical bowl.
or, 1/3 π x (4)² x h = 2/3 π x (4³-2³)
= 1/3 π x 16 x h = 2/3 x π x (34-8)
16h = 2 x 56
∴ h = 2×56 / 16
= 7
∴ height of the cone = 7 cm.
Application 3. There is some water in the right circular drum. A conical iron piece having a diameter of 2.8 dcm and a height of 3 dcm is immersed completely in this water. For this, the water level of the drum is raised by 0.64 dcm. Let us determine the base diameter of the drum.
Solution:
Given
There is some water in the right circular drum. A conical iron piece having a diameter of 2.8 dcm and a height of 3 dcm is immersed completely in this water. For this, the water level of the drum is raised by 0.64 dcm.
The length of the base radius of the cone = 28/2 dcm = 1.4 dcm
and height= 3 dcm.
∴ Volume of the cone = 1/3 π x (1.4)² x 3 dcm³
Let the length of the base radius of the drum be r dcm.
For immersing the conical iron piece into the water of the drum, the water level is raised by 0.64 dcm.
∴ The volume of increased water in the drum = Volume of the conical iron piece.
∴ πr² x 0.64 = (1.4)² x 3 or, r² = ± 1.4×1.4/0.64
or, r± 3.0625
But the length of the radius can not be negative. ∴ r= -3.0625
∴ r = 3.0625
∴ The length of the base diameter of the drum = 2 x 3.0625
= 6.125 dcm.
The length of the base diameter of the drum = 6.125 dcm.
Application 4. There is some water in the right circular cylinder with a diameter of 28 cm. Three solid iron spheres can be immersed completely into it. After immersing the spheres, the height of the water level is increased by 7 cm than it was before the immersion of the spheres into water. Let us determine the length of the diameter of the spheres.
Solution:
Given
There is some water in the right circular cylinder with a diameter of 28 cm. Three solid iron spheres can be immersed completely into it. After immersing the spheres, the height of the water level is increased by 7 cm than it was before the immersion of the spheres into water.
Let the radius of the sphere = r cm.
∴ The volume of the 3 spheres = 3 x 4/3 πr³ cu cm.
The radius of the base of the cylinder = 28/2 cm = 14 cm.
After immersing the spheres, the height of the water level is increased by 7 cm.
∴ The volume of the 3 spheres volume of increased water in the cylinder
3 x 4/3 πr³ = π x 14 x14x7
∴ r³=7x7x7
∴ The diameter of each sphere = 2 x 7 cm = 14 cm.
∴ r=7
The length of the diameter of the spheres. = 14 cm.
Application 5. If recasting a solid metallic sphere of radius 2.1 cm into a right circular rod, then let us write by calculating, the length of diameter of the rod.
Solution:
Given
If recasting a solid metallic sphere of radius 2.1 cm into a right circular rod,
Let us determine the ratio of total surface areas of the rod and sphere.
Let the length of the right circular rod be r cm.
According to the condition, r² x7 = 4/3π x (21)³ [ 2.1 dcm = 21 cm]
r² = 4/3 x 21x21x21/7 or, r = ±42
so, r ≠ -42
∴ r = +42
∴ The length of the radius of the rod is 42 cm.
∴ The length of diameter of the rod is 84cm.
∴ The ratio of total surface areas of the rod and the sphere is = 1: 3696.
Question 1. There is a solid iron pillar in front of Anowar’s house whose lower portion is right circular cylinder shaped and upper portion is cone shaped. The length of their base diameter is 20 cm, the height of the cylindrical portion is 2.8 m and the height of the conical portion is 42 cm. If the weight of 1 cm3 iron is 7.2 gm, then let us calculate the weight of the iron pillar.
Solution:
Given
There is a solid iron pillar in front of Anowar’s house whose lower portion is right circular cylinder shaped and upper portion is cone shaped. The length of their base diameter is 20 cm, the height of the cylindrical portion is 2.8 m and the height of the conical portion is 42 cm. If the weight of 1 cm3 iron is 7.2 gm,
Radius of the base of solid iron sphere = 20/2 cm = 10 cm
Height of the cylindrical portion = 2.8m = 280 cm.
∴ The volume of the cylindrical portion = лr²h cu cm.
= 22/7 x 10 x 10 x 280 cu cm = 88000 cu cm.
& Volume of the conical portion = 1/3 лr2h cu cm.
= 1/3 x 22/7 x 10 x 10 x 42 cu cm = 4400 cu cm.
∴ The total volume of the iron pillar = (88000+ 4400) cu cm.
= 92400 cu cm.
The total volume of the iron pillar = 92400 cu cm.
Question 2. The height of a solid right circular cone is 20 cm. and its slant height is 25 cm. If the height of a solid right circular cylinder, having as much volume as that of the cone, is 15 cm., then let us calculate the base diameter of the cylinder.
Solution:
Given
The height of a solid right circular cone is 20 cm. and its slant height is 25 cm. If the height of a solid right circular cylinder, having as much volume as that of the cone, is 15 cm
Let the radius of the base of the cone = r1 cm
∴ \((\text { Slant height })^2=(\text { radius })^2+(\text { height })^2\)
or, \((25)^2=r_1^2+(20)^2\)
or, \(r_1{ }^2=625-400=225\)
∴ \(r_1=\sqrt{225}=15\)
∴ Volume of the cone = \(\frac{1}{3} \pi r_2^2 \times \text { height }\)
= \(\frac{1}{3} \times \pi \times 15 \times 15 \times 20 \mathrm{cucm}\)
Again let the radius of the base of the cylinder = r2 cm
∴ Volume of the cone = \(\pi r_2^2 \times \text { height }\)
∴ According to the problem,
\(\frac{1}{3} \times \pi \times 15 \times 15 \times 20=\pi \times r^2 \times 15\) [As height of the cylinder = 15 cm].
\(r_2{ }^2=100\)∴ \(r_2=\sqrt{100}=10\)
∴ Diameter of the base of the cylinder = 2r2 = 2 x 10 cm = 20 cm.
Question 3. There is some water in the right circular cylindrical can with a diameter of 24 cm. If 60 solid conical iron pieces with a base diameter of 6 cm and height of 4 cm are immersed completely in the water, then let us write by calculating, the increased height of the water level.
Solution:
Given
There is some water in the right circular cylindrical can with a diameter of 24 cm. If 60 solid conical iron pieces with a base diameter of 6 cm and height of 4 cm are immersed completely in the water,
Radius of the base of the cone = 6/2 cm = 3cm.
Volume of 60 cones = 60 xπ (3)² x 4 cu cm.
Let after immersing the cones, the height of water in the cylinder increased by h cm.
∴ The volume of increased water in the cylinder = (Tr²h)
∴ According to the problem,
The volume of increased water = Volume of 60 cones
π x 12 x 12 x h = 60×1/3π×3×3×4
∴ h= 60×3×4 / 12×12
= 5
∴ Increased height of water = 5 cm.
Question 4. If the ratio of the curved surface areas of a solid cone and a solid right circular cylinder having the same base radii and same height is 5: 8, then let us determine the ratio of their base radii and heights.
Solution:
Given
If the ratio of the curved surface areas of a solid cone and a solid right circular cylinder having the same base radii and same height is 5: 8
Let the height & radius of the base of the cone be h units & r units respectively.
∴ Slant height (l) = √h² +2 units.
∴ Area of the curved surface of cone = πrl = πr(h² + r²) sq units.
Area of the curved surface of cylinder = 2лrh where, height
& radius of the base of the cylinder = h & r respectively
πrx √h²+r²/ 2πrh = 5/8
∴ √h² +r² = 5/8
or,h² +r² /2h = 25/16
Or,1 + r²/h² = 25/16
r²/h² = 25/16 – 1
= 25-16 / 16
= 9/16
∴ r/h = √9/16
= 3/4
∴The ratio of the radius of base: height = 3: 4.
Question 5. By melting a solid iron sphere with 8 cm of radius, how many marble balls of 1 cm diameter can be obtained let us write by calculating it.
Solution: Volume of the sphere of radius 8 cm = 4/3 π (8)³
= 4/3 π x 8 x 8 x 8 cu cm.
Now the radius of a small (marble) sphere = 1/2 cm.
The volume of marble balls = 4/3 π(1/2)³
= 4/3π x 1/2 x 1/2 x 1/2 cu cm.
∴ Number of marble balls = 4/3π x 8 x 8 x 8 / 4/3π x 1/2 x 1/2 x 1/2
= 8 x 2 x 8 x 2 x 8 x 2
= 4096.
Number of marble balls = 4096.
Question 6. The base radius of a solid right circular iron rod is 32 cm and its length is 35 cm. Let us calculate the number of solid cones of 8 cm radius and 28 cm height that can be made by melting this rod.
Solution:
Given
The base radius of a solid right circular iron rod is 32 cm and its length is 35 cm.
Let the number of cones to be made = n
∴ Volume of n cones = n x 1/3 π²h
=n x 1/3 π (8)² x 28 cu cm. [Since radius = 8 cm & height = 28 cm.]
Again volume of a big cylinder with a radius of 32 cm & height = is 35 cm.
= π (32)² × 35
According to the problem,
n x 1/3π x 8 x 8 x 28
= 1/3 π x 32 x 32 x35
∴ n = 20 x 3 = 60
∴ The number of cones = 60.
Question 7. Let us determine the volume of a solid right circular cone that can be made from a solid wooden cube of 4.2 dcm edge length by wasting a minimum quantity of wood.
Solution: To make a cone from a wooden cube of 4.2 dcm edge length by wasting a minimum quantity of wood.
∴ Diameter of base of cone = One edge of cube 4.2 dcm.
∴ The radius of the base of cone R = 4.2/2
= 2.1 dcm.
∴ The volume of the cone = 1/3 πr²h
= 1/3 x 22/7 2.1 x 2.1 x 4.2 cu dcm
= 1/3 x 22/7 x 21/10 x 21/ x 42/10 cu dcm.
= 22x21x42 / 10×10×10
= 19.4 cu dcm.
The volume of the cone = 19.4 cu dcm.
Question 8. If the radii and the volumes of a solid sphere and a solid right circular cylinder are equal then let us calculate the ratio of the radius and height of the cylinder.
Solution: Let the radius of the sphere & cylinder = r units
& height cylinder = h units.
∴ The volume of the sphere = Volume of the cylinder
or, 4/3 π = r²h
∴ 4/3r = h
∴ r/h = 3/4
∴ Ratio of radius & height = 3: 4
Question 9. Let us determine the number of solid spheres with a diameter of 2.1 dcm. can be made by melting a solid copper rectangular parallelopiped piece with a length of 6.6 dcm, and a breadth of 4.2 dcm. and thickness of 1.4 dcm and also calculate the quantity of metal in dcm3 in each sphere.
Solution: Volume of solid rectangular parallelopiped = Area of base x height = 6.6 x 4.2 x 1.4 cu dcm.
The volume of n sphere = nx4/3 πx (2.1 / 2)³ cu dcm,
where radius = 2.1/2 dcm.
According to the problem,
= n x 4/3 x 22/7 x 21/20 x 21/20 x 21/20
or, n x 21/2
= 3 x 2 x 14
N = 3 x 2 x 14 x 2 / 21
= 8.
∴ 8 solid spheres can be made & a volume of metal in each sphere
4/3 x 22/7 x 21/20 x 21/20 cu dcm
= 4.851 cu dcm.
4.851 cu dcm quantity of metal in dcm³ in each sphere.
Question 10. Let us determine the length of a right circular rod of 2.8 cm diameter made by recasting a solid gold sphere of 4.2 cm radius.
Solution: Let the height of the cylinder rod = h = cm
∴ The volume of the cylinder rod = π (2.8/2)² x h cu cm. [as radius = 1.4 cm.]
The volume of the sphere = (4.2)³ cu cm [as the radius of the sphere = 4.2 cm.]
According to the problem,
π(1.4)²x h = 4/3 π(4.2)³
or, 14/10 x 14/10 h = 4/3 x 42/10 x 42/10 x42/10
∴ Height of cylinder rod = 50.4 cm.
Question 11. If a solid silver sphere of a diameter of 6 dom le melted and recast Into a solid right circular rod, then let us determine the length of the diameter of the rod. 6
Solution:
Given
A solid silver sphere of a diameter of 6 dom le melted and recast Into a solid right circular rod
Radius of sphere = 6/2 dcm = 3 dcm.
∴ The volume of the sphere = 4/3 π(3)³ cu dom.
If the radius of the cylinder rod = r dcm.
∴ Its volume = πr²h = 22/7 x r2 x 1 cu dcm.
According to the condition,
πr²x1 = 4/3 π x 3 x 3 x 3
r² = 36 ∴ r = √36 = 6 dcm.
Diameter of the rod = 2r
= 2 x 6 dcm
= 12 dcm.
The length of the diameter of the rod = 12 dcm.
Question 12. The length of the radius of the cross-section of a solid right circular rod is 3.2 dcm. By melting the rod 21 solid spheres are made. If the radius of the sphere is 8cm., then let us determine the length of the rod.
Solution:
Given
The length of the radius of the cross-section of a solid right circular rod is 3.2 dcm. By melting the rod 21 solid spheres are made. If the radius of the sphere is 8cm.,
The radius of each sphere = 8 cm.
∴ Volume of 21 spheres = 21 x 4/3π(8)³ cu cm.
Let the height of the cylindrical rod = h cm.
& Radius of base = 3.2 dcm = 32 cm.
∴ The volume of the cylindrical rod = πr²h
= π(32)²h
According to the problem,
π(32 x 32) x h = 21 × (8×8×8)
h = 4x8x8x8x7/32 x 32
= 14 cm.
∴ Height of the rod = 14 cm.
Question 13. Half of a tank of 21 dcm in length, 11 dcm. the breadth and 6 dcm depth is full of water. Now if 100 iron spheres of 21 cm diameter is immersed completely into the water of this tank, then let us calculate the rise of water level in dcm.
Solution:
Given
Half of a tank of 21 dcm in length, 11 dcm. the breadth and 6 dcm depth is full of water. Now if 100 iron spheres of 21 cm diameter is immersed completely into the water of this tank
Diameter of the sphere = 2.1 dcm
∴ Radius (r) = 2.1/2 dcm.
∴ The volume of each sphere = 4/3 πx (2.1 / 2) cu dcm.
∴The volume of 100 spheres = 100 x 4/3 x 22/7 x 21/20 x 21/20 x 21/20 cu dcm.
Let after immersing 100 spheres, the height of the water level rises by h dcm.
∴ The volume of increased water = 21 x 11 x h cu dcm.
According to the problem,
21 x 11 x h = 100 x 4/3 x 22/7 x 21/20 x 21/20 x 21/10
h = 21/10 = 2.1 dcm.
∴ The water level will rise by 2.1 dcm.
Question 14. Let us determine the ratio of volumes of a solid cone, a solid hemisphere, and a solid cylinder of the same base diameter and same height.
Solution: Diameter & height of cone, hemisphere & cylinder are equal
∴ The length of their radius is also equal
Let the radius of cone = radius of hemisphere = radius of cylinder = r unit
Now the height of hemisphere = radius of hemisphere = r unit.
.. Volume of cone: Volume of the hemisphere: Volume of cylinder
= 1/3πr² r:πr³ : πr² .
= 1/3 : 2/3 : 1
= 1:2:3
The ratio of volumes = 1:2:3
Question 15. The external radius of a hollow sphere made of a lead sheet of 1 cm thickness is 6 cm If melting the sphere, a solid right circular rod of 2 cm radius is made, then let us calculate the length of the rod.
Solution:
Given
The external radius of a hollow sphere made of a lead sheet of 1 cm thickness is 6 cm If melting the sphere, a solid right circular rod of 2 cm radius is made,
External radius of the hollow sphere is 6 cm.
& Internal radius of the hollow sphere = (6-1) = 5 cm.
∴ Volume of the hollow sphere = 4/3π {6³ – 5³}
= 4/3π (216 – 125) cu cm.
= 4/3 π x 91 cu cm.
Let the height of the rod = h cm.
The volume of a solid cylindrical rod = π(2)2 x h [as the radius of the rod = 2 cm.]
∴ According to the problem,
π(2) h = 4/3 π x 91
4h = 4 x 31 / 3
∴ h = 91/3
= 30.33 cm.
The length of the rod = 30.33 cm.
Question 16. The cross-section of a rectangular parallelopiped wooden log of 2m in length is a square and each of its sides is 14 dcm in length. If this log can be converted into a right circular log by wasting a minimum amount of wood, then let us calculate what amount of wood (in m3) will be wasted.
[Hints: The circumcircle is inscribed in a rectangular figure, the length of the diameter of the circle is equal to the length of the side of the square.]
Solution:
Given
The cross-section of a rectangular parallelopiped wooden log of 2m in length is a square and each of its sides is 14 dcm in length. If this log can be converted into a right circular log by wasting a minimum amount of wood,
Length of each side of the square base of the rectangular parallelopiped of wooden log 14 dcm = 1.4 m.
& the height of wooden log = 2 m.
∴ The volume of the wooden log = Area of base x height
= 1.4 x 1.4 x 2 cu m = 3.92 cu m.
As the diameter of the wooden log = 1.4 m.
∴ radius of the wooden log = 1.4/2 = 0.7 m.
∴ The volume of wood in the remaining portion of the log
= Area of circular base x height
= π(-7)² x 2 cu m. = 3.08 cu.m.
∴ The volume of wood that remained in it = 3.08 cu m.
The amount of wood (in m³) will be wasted = (3.92 -3.08) cu m
= 0.84 cu m.
0.84 cu m amount of wood (in m³) will be wasted.
Class 10 WBBSE Math Solution In English Chapter 19 Real-Life Problems Related To Different Solid Objects Exercise 19.1 Multiple Choice Question
Question 1. A solid sphere of r unit radius is melted and from it, a solid right circular cone is made. The base radius of the cone is
1. 2r unit
2. 3r unit
3. r unit
4. 4r unit
Solution:
Given
A solid sphere of r unit radius is melted and from it, a solid right circular cone is made.
Let the radius of the cone (R)
∴ According to the problem,
1/3 πR² x r = 4/3πr³r²
∴ R² = 4r²
∴ R = 2r.
The base radius of the cone is 2r.
Answer. 1. 2r unit
Question 2. By melting a solid right circular cone, a solid right circular cylinder of same radius is made whose height is 5 cm. The height of the cone is
1. 10 cm
2. 15 cm
3. 18 cm
4. 24 cm
Solution:
By melting a solid right circular cone, a solid right circular cylinder of same radius is made whose height is 5 cm.
Given
Let the height of the cone = h
∴ 1/3 πr²h = πr² x 5
∴ h=5 x 3 = 15 cm.
The height of the cone is 15 cm.
Ans. 2. 15 cm
Question 3. The length of the radius of a right circular cylinder is r unit and the height is 2r unit. The length of diameter of the largest sphere that can be kept in the cylinder is
1. r unit
2. 2r unit
3. r/2 unit
4. 4r unit
Solution: Length of the diameter of the largest sphere of height 2r unit = 2r unitr.
Ans. 2. 2r unit
Question 4. The volume of the largest solid that can be cut out from a solid hemisphere of r unit radius is
1. 4πr³ unit³
2. 3πr³ unit³
3. πr³/4 unit³
4. πr³ unit³
Solution: Volume of the solid cone that can be cut out from the solid hemisphere = πr³/3
Answer. 4.πr³ unit³
Question 5. If from a solid cube with the edge of x unit length, the largest solid sphere is cut out, then the length of the diameter of the sphere is
1. x unit
2. 2x unit
3. x/2 unit
4. 4x unit
Solution: The diameter of the sphere is x unit
Answer. 1. x unit
WBBSE Class 10 Maths Solutions Chapter 19 Real-Life Problems Related To Different Solid Objects Exercise 19.1 Multiple Choice Question True Or False
1. If two solid hemispheres of the same type whose base radii are r units each and are connected along the base, then the total surface area of the connected solid is 67r² sq. unit.
Answer: False
2. The base radius of a solid circular cone is r unit, height is h unit and slant height is unit. The base of the cone is joined along a base of a right circular cylinder. If the base radii and heights of the cylinder and cone are the same, then the total surface area of the connected solid is (πrl+2πrh + 2πr²) sq. unit.
Answer: True
Maths WBBSE Class 10 Solutions Chapter 19 Real-Life Problems Related To Different Solid Objects Exercise 19.1 Fill In The Blanks
1. The base radii of a solid right circular cylinder and two hemispheres are the same. If two hemispheres are fixed with two surfaces of the cylinder, then the total surface area of the new solid = curved surface area of one hemisphere + curved surface area of Cylinder + curved surface area of another hemisphere.
2. The shape of a pencil cutting one face is a combination of cone and Cylinder
3. By melting a solid sphere, a solid right circular cylinder is made. The volume of the sphere and cylinder Equal
Chapter 19 Real-Life Problems Related To Different Solid Objects Exercise 19.1 Short Answers
Question 1. A solid right circular cylinder is made by melting a solid right circular cone. The radii of both are equal. If the height of the cone is 15 cm, then let us determine the height of the solid cylinder.
Solution.
Given
A solid right circular cylinder is made by melting a solid right circular cone. The radii of both are equal. If the height of the cone is 15 cm
Let the radius of cone & cylinder = r cm
& height of cylinder = h cm
& height of cone = 15 cm.
According to the problem,
= πr²h = 1/3 πr² x 15
∴ h = 15/3
= 5 cm.
The height of the solid cylinder = 5 cm.
Question 2. The radii and volumes of a solid circular cone and a solid sphere are equal. Let us determine the ratio of the sphere and the height of the cone.
Solution:
Given
The radii and volumes of a solid circular cone and a solid sphere are equal.
Let the radius of the sphere = r unit & height of the cone = h unit
∴ 4/3πr³ = 1/3πr²h
∴ 4r = 4
or, 2r/h = 1/2
Ratio of the diameter of the sphere: height of cone = 1:2.
Question 3. Let us determine the ratio of the volumes of a solid right circular cone and a solid sphere of equal diameter and equal height.
Solution. Let the diameters of the cylinder, cone & sphere be equal.
∴ Length of their radius is also equal.
Let the radius of cylinder, cone & sphere: = r unit.
Height of sphere =’Diameter of sphere
& height of cylinder & cone = 2r unit.
∴ The ratio of the volume of a cylinder, cone & sphere
=πr² x 2 r:1/3πr². 2r:4/3πr³
= 2:2/3:4/3
= 3:1:2
The ratio of the volumes = 3:1:2
Question 4. The shape of the lower portion of a solid is the hemisphere and the shape of the upper portion of it is a right circular cone. If the surface areas of the two parts are equal, then let us write by calculating, the ratio of the radius and height of the cone.
Solution.
Given
The shape of the lower portion of a solid is the hemisphere and the shape of the upper portion of it is a right circular cone. If the surface areas of the two parts are equal
Let the radius of hemisphere & cone = r unit
& height of the cone = h unit.
The surface area of the solid hemisphere = 3πr² sq unit.
The curved surface area of the cone = πr(r + l)sq unit.
= πr (r+ √h² + r² ) sq unit.
.. πr² + πr √h² + r²
= 3πг²
or, πr √h²+r² = 2πr²
or, √h²+r² = 2r
or, h² + r² = 4r²
or, h² = 3r²
or, h = √3r
∴ h:r = 1: √3
The ratio of the radius and height of the cone are √3 : 1
Question 5. The base radius of a solid right circular cone is equal to the length of the radius of a solid sphere. If the volume of the sphere is twice of that of the cone, then let us write by calculating the ratio of the height and base radius of the cone.
Solution.
Given
The base radius of a solid right circular cone is equal to the length of the radius of a solid sphere. If the volume of the sphere is twice of that of the cone
Let the radius of cone & sphere = r unit
& height of the cone = h unit
∴ 4/3 πr³ 2 x 1/3 πr2h
or, 2r= h
∴ h/r = 2/1
∴ The ratio of height & radius of the cone = 2:1