WBBSE Solutions Guide Class 10 Chapter 12 Sphere Exercise 12.2
Question 1. If the length of the radius of a sphere is 10.5 cm, let us write by calculating the whole surface area of the sphere.
Solution: Radius = 10.5 cm
Total surface area = 4πr²= 4 x 22/7 X (10.5)² sq cm.
= 4 x 22/7 x 10.5 10.5 sq cm
= 1386 sq cm.
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Question 2. If the cost of making a leather ball is Rs. 431.20 at Rs. 17.50 per square cm, let us write by calculating the length of the diameter of the ball.
Solution: Surface area of the leather ball = 4πr² (r= radius)
According to the problem,
4πr² x 17.50 = 431.20
.. 4πг² = 431.20 /17.50
= 24.64
or, 4 x 22/7 x r² = 24.64
r² = 24.64×7 / 2×22
r² = 1.96
∴ r=√1.96
r = 1.4
∴ Diameter = 2 x 1.4 cm = 2.8 cm.
WBBSE Solutions Guide Class 10
Question 3. If the length of diameter of the ball used for playing shotput in our school is 7 cm, let us write by calculating how many cubic cm of iron is there in the ball.
Solution: Radius of the ball /r=7/2 cm (as. Diameter = 7 cm)
Volume of the ball = 4/3 πг²
= 4/3 x 22/7 x 7/2 x 7/2 x 7/2
=539/2 cu cm.
=179 2/3 cu cm.
∴ 179 2/3 cu cm iron is in the ball.
Question 4. If the length the diameter of a solid sphere is 28 cm and it is completely immersed in the water, let us calculate the volume of water displaced by the sphere.
Solution: Diameter = 28 cm.
Radius (r) = 28/2
= 14 cm.
Volume of the sphere = 4/3 πr³
= 4/3 x 22/7 x 14 x 14 x 14 cu cm.
= 11498.67 cu cm.
.. Volume of displaced water = 11498.67 cu cm.
WBBSE Solutions Guide Class 10
Question 5. The length of the radius of a spherical gas balloon increases from 7 cm to 21 cm as air is being pumped into it; let us find the ratio of surface areas of the balloon in two cases.
Maths WBBSE Class 10 Solutions
Question 6. 127 sq cm of the sheet is required to make a hemispherical bowl. Let us write by calculating the length of the diameter of the forepart of the bowl.
Solution: Surface area of the hemispherical bowl = 127 2/7 sq
cm = 891/7 sq cm.
If the radius = r cm.
∴2πr² = 891/7
2 x 22/7 r²= 291/7
∴ r² = 891/7 x 7/22×2
=81/4
∴ r = 9/2 cm.
∴ Diameter = 2 x²
= 9 cm.
Question 7. The length of the radius of a solid spherical ball is 2.1 cm; let us write by calculating how much cubic cm iron is there and let us find the curved surface area of the iron ball.
Solution: The radius of the spherical ball = 2.1 cm.
The volume of the sphere = 4/3 π x (2.1)³ cu cm.
= 4/3 x 22/7 x 2.1 x 2.1 cu cm.
= 38.81 cu cm.
Total surface area = 4πr² = 4 x 22/7 x (2.1)2 sq cm.
= 4 x 22/7 x 2.1 x 2.1 sq cm.
= 55.44 sq cm.
Maths WBBSE Class 10 Solutions
Question 8. The length of the diameter of a solid sphere of lead is 14 cm. If the sphere is melted, let us write by calculating how many spheres with a length of 3.5 cm radius can be made.
Solution: Diameter of the solid sphere = 14 cm.
∴ Radius (r) of the solid sphere = 14/2 π = 7 cm.
The volume of lead in the solid sphere = 4/3 π (7)³ cu cm.
The radius of the small sphere = 3.5 cm.
∴ Volume of small sphere =4/3 π (3.5)³ cu cm.
∴ No. of small sphere = 4/3π x7x7x7 /4/3π x3.5×3.5×3.5
=2x2x2
= 8.
Question 9. Three spheres made of copper having lengths of 3 cm, 4 cm and 5 cm radii are melted and a large sphere is made. Let us write by calculating the length of the radius of the large sphere.
Solution: Volume of 1st sphere with radius 3 cm = 4/3 π (3)³ cu cm.
The volume of the 2nd sphere with a radius of 4 cm = 4/3π (4)³ cu cm.
& Volume of 3rd sphere with radius 5 cm = 4/3π (5)³ cu cm.
Total volume of 3 small spheres= 4/3π(3)³ + 4/3 π(4)³ + 4/3π (5)³
= 4/3π(33 +43 +53) cu cm.
= 4/3π(27 +64 +125) cu cm.
= 4/3π x 216 cu cm.
Now, let the radius of the new big sphere = R cm.
.. Its volume = 4/3πR³
According to the problem,
4/3πR³= 4/3π x 216
.. R³ = 216
R=3√216
=6 cm.
Maths WBBSE Class 10 Solutions
Question 10. The length of diameter of the base of a hemispherical tomb is 42 dm. Let us write by calculating the cost of colouring the upper surface of the tomb at the rate of Rs. 35 per square metre.
Solution: Diameter of base of a hemispherical tomb = 42 dm.
.. Radius of base r = 42/2
= 21 dm.
Surface area = 2πr² = 2 x 22/7 x (21)² sq dm.
= 2 x 22/7 x 21 x 21 sq dm.
= 2772 sq dm.
= 27.72 sqm.
The total cost for colouring the upper surface of the tomb is Rs. 35 x 27,72= Rs. 970.20.
Question 11. Two hollow spheres with lengths of the diameter of 21 cm and 17.5 cm respec- tively are made from sheets of the same metal. Let us calculate the volumes of sheets of metal required to make the two spheres.
Solution: The radius of the 1st sphere = 21/2 cm and the radius of the 2nd sphere = 17.5/2 cm.
The total surface area of the 2nd sphere = 4π (21/2)2 sq cm.
The total surface area of the 2nd sphere = 4π (17.5/2)2 sq cm.
.. Ratio of metal sheet to make the two spheres
= 4π (21/2)2 : 4π (17.5/2)2
= 21×21 / 2×2 : 17.5×17.5 / 2xx
= 441/4: 5359.375 /4
= 441 / 4
= 4/306.25
= 36: 25.
Question 12. The curved surface of a solid metallic sphere is cut in such a way that the curved surface area of the new sphere is half of the previous one. Let us calculate the ratio of the volumes of the portion cut off and the remaining portion of the sphere.
Solution: Let the radius of the old sphere = R cm and the radius of the new sphere = r cm.
Surface area of the old sphere = 4πR² sq cm.
The surface area of the new sphere = 4πr²
According to the question, 1/2 4πR² = 4лг²
R=2r²
∴ R= √2r.
The volume of the old sphere =4/3 лг³
=4л(√2r)³ cu cm.
The volume of the new sphere = 4/3 лг³
Volume of the remaining sphere = 4л(√2r)³ – 4лr³
=4лr³ (2√2-1)
.. Ratio of the volumes of the portion cut off and the remaining portions of sphere
=4лr³ (2√2-1): 4лr³
= (√2-1):1
West Bengal Board Class 10 Math Book Solution In English
Question 13. On the curved surface of the axis of a globe with a length of 14 cm radius, two circular holes are made, each of which has a length of radius 0.7 cm. Let us calcu- late the area of the metal sheet surrounding its curved surface.
Solution: Radius of the globe = 14 cm
Surface area of the globe = 4π x 14 x 14 = 4 x 22/7 x 14 x 14
= 2464 sq cm.
Radius of each hole = 0.7 cm.
∴ Area of two holes = 2 × π (0.7)³ = 2 × 22/7 x 7/10 x 7/10
= 308/100
= 3.08 sq cm.
∴ Area of the metal sheet surrounding its curved surface = (2464 – 3.08) sq cm
= 2460.92 sq cm.
Question 14. Let us write by calculating how many marbles with lengths of 1 cm radius may be formed by melting solid sphere of iron having an 8 cm length of radius.
Solution: Radius of the solid sphere = 8 cm.
The volume of the solid sphere = 4/3 (8)³ cu cm.
Radius of the small sphere = 1 cm.
Volume of each small sphere = 4/3 (1)³ cu cm.
.. No. of small solid spheres may be formed from the big solid sphere
= 4/3π (8)³ / 4/3π (1)³
= 8x8x8/1x1x1.
= 512.
West Bengal Board Class 10 Math Book Solution In English Chapter 12 Sphere Exercise 12.2 Multiple Choice Question
Question 1. The volume of a solid sphere having a radius of 2r units length is
1. 32nr³/3 cubic unit
2. 16nr³/3 cubic unit
3. 8nr³/3 cubic unit
4. 64nr³/3 cubic unit
Solution: Volume of a solid sphere having the radius of 2r unit
= 4/3 π(2r)³ = 4π x 2rx 2rx2r/3
= 32nr³/3 cu units
Answer: 1. 32πr³/3 cubic unit
Question 2. If the ratio of the volumes of two solid spheres is 1: 8, the ratio of their curved surface areas is
1. 1:2
2. 1:4
3. 1:8
4. 1: 16
Solution: If the ratio of the volumes of two solid spheres = 1: 8
The ratio of their curved surface area = 1:4
Ans. 2. 1:4
Question 3. The whole surface area of a solid hemisphere with a length of 7 cm radius is
1. 588π sq cm.
2. 392π sq.cm.
3. 147π sq cm.
4. 98π sq cm.
Solution: Whole surface area of a solid hemisphere of radius 7cm
= 3π(7)² sq cm
= 147π sq cm.
Answer. 3. 147π sq cm.
Question 4. If the ratio of curved surface areas of two solid spheres is 16: 9, the ratio of their volumes is
(a) 64:27
(b) 4:3
(c) 27: 64
(d) 3:4
Solution: If the ratio of curved surface areas of two solid sphere is 16: 9.
∴ Ratio of their volumes = 64: 27
Answer. 1. 64: 27
Question 5. If the numerical value of curved surface area of a solid sphere is three times of its volume, the length of radius is
1. 1 unit
2. 2 unit
3. 3 unit
4. 4 unit
Answer. 1. 1 unit
Class 10 WBBSE Math Solution In English Chapter 12 Sphere Exercise 12.2 True Or False
1. If we double the length of radius of a solid sphere, the volume of sphere will be doubled.
False
2. If the ratio of curved surface areas of two hemispheres is 4 9, the ratio of their lengths of radii is 2: 3.
True
Fill In The Blanks
1. The name of a solid which is composed of only one surface is Sphere.
2. The number of curved surfaces of a solid hemisphere is One.
3. Solid hemisphere is 2r units, its whole surface area is 2r units, and its whole sur- face area is 12πr²r² sq units.
Chapter 12 Sphere Exercise 12.2 Short Answers.
Question 1. The numerical values of volume and whole surface area of a solid hemisphere are equal; let us write the length of radius of the hemisphere.
Solution: Let the radius of the hemisphere = r unit.
Volume of the hemisphere=2/3 π r3 cu unit
and total surface area = 3πr2 sq unit.
According to the problem,
=2/3 π r3 = 3πr2
or, 2r=9
..r=9/2
= 4.5 unit.
Radius of the hemisphere is 4.5 units.
Question 2. The curved surface area of a solid sphere is equal to the surface area of a solid right circular cylinder. The lengths of both height and diameter of the cylinder are 12 cm. Let us write the length of radius of the sphere.
Solution. Let the radius of solid sphere = r cm.
Surface area of solid sphere = 47r² sq cm²
Again, height of the cylinder is 12 cm.
Radius of the cylinder = 6 cm.
Curved surface area of the cylinder = 2лrh = 2π x 6 x 12 sq cm.
= 144π sq cm.
According to the problem,
4πr² = 144π
r² = 36,
or, r= 6
∴The radius of the sphere = 6 cm.
Class 10 WBBSE Math Solution In English
Question 3. The whole surface area of a solid hemisphere is equal to the curved surface area of the solid sphere. Let us write the ratio of lengths of radii of the hemisphere and the sphere.
Question 4. If the curved surface area of a solid sphere is S and the volume is V, let us write the value of S3/V2 [not putting the value of π]
Maths WBBSE Class 10 Solutions
Question 5. If the length of the radius of a sphere is increased by 50%, let us write how much per cent will be increased of its curved surface area.
