Maths WBBSE Class 10 Solutions Chapter 12 Sphere Exercise 12.1
Application 1. If the diameter of a ball is 42 cm, let us calculate how much leather is required for making the ball.
Solution: Diameter of the ball = 42 cm.
∴ Radius of the ball = 42/2 cm.
= 21. cm.
∴The whole surface area of the ball = 4 x (21)² sq cm.
= 4π x 22/7 x 21 21 sq cm.
= 5544 sq cm.
∴The ball contains 5544 sq cm of leather.
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Application 2. The diameter of a sphere made of the iron sheet is 14 cm, let us calculate what is the cost of coloring the sphere at Rs. 2.50 per square cm.
Solution Radius of the sphere (r) = 14/2
=7 cm.
Area of the curved surface = 4πr²
= 4 x 22/7 x 7 x 7 Sq cm. = 616 sq cm.
Cost of coloring the sphere at Rs. 2.50 per sq cm.
= Rs. 2.50 616
= Rs. 1540.
Maths WBBSE Class 10 Solutions
Application 3. The length of the diameter of a solid spherical ball is 14 cm. Let us write by calculating how much stone is there in the spherical solid ball.
Solution: The length of the radius of the solid spherical ball of stone = 14/2 cm = 7 cm.
∴ A solid spherical ball of stone contains a stone of volume = 4/3 x 22/7 x 73 cubic cm.
= 1437 1/3 cu cm.
Application 4. My spherical stone ball having a radius of 0.7 dcm in length is completely immersed in the water of the reservoir. Let us write by calculating how much water will be overflowed from the reservoir.
Solution: Radius of the spherical stone ball (r) = 0.7 dcm = 7/10 dcm.
The volume of the stone ball = 4/3 r²
= 4/3 x 22/7 (7/10)³ cu dcm.
= 4/3 x 22/7 x 7x7x7/10x10x10 cu dcm.
= 1.44 cu dcm.
WBBSE Solutions Guide Class 10
Application 6. If the length of the diameter of a hemispherical solid object is 14 cm, let us write by calculating its whole surface area.
Solution: Whole surface area = 3 x 22/7 x 14 x 14 sq units
= 1848 sq cm.
Application 7. If it requires 173.25 sq cm of the sheet to make a hemispherical bowl they let us write by calculating the length of the diameter of the forepart of the bowl.
Answer hints: As the bowl is not a solid object, a curved surface area of the sheet will only be required.
Solution: Let the diameter of the hemispherical bowl = zπr cm.
.. Radius = r cm. Area of on road surface = 2πr²
.. 2πr² = 173.25
or, 2 x 22/7
r² = 17325/100
r² = 17325 / 100 x 7/22×2
=1575×7/400
r° = ^ ^ 1575×7 / 400
= 105/20
= 5.25
∴Radius 5.25 cm & Diameter = 10.5 cm.
WBBSE Solutions Guide Class 10
Application 8. If the length of the radius of a solid hemispherical object is 14 cm, let us calculate the ratio of their volumes.
Solution: Stone contained in hemispherical paper weight is = 2/3 x 22/7 x 14 x 14 x 14 cubic units.
=17248 cu cm.
Application 10. If the ratio of lengths of radii of two spheres is 1:2, let us write by calculating the ratio of their whole surface areas.
WBBSE Solutions Guide Class 10
Application 12. If two spheres with radii of 1 cm and 6 cm lengths are melted and a hollow sphere with a thickness of 1 cm is made, let us write by calculating the outer curved surface area of the hollow sphere.
Solution: Let the length of the outer radius of the sphere is r cm.
∴ The length of the inner radius of that sphere is = (r-1) cm.
By condition, 4/3 πr³ – 4/3 π(r-1)³
= 4/3π(1)³+4/3π(6)³
or, 4/3 π {r³ (r− 1)³} = 3π (1+216)
or, 3+3r²-3r+ 1 = 217
or, 32-3r-216=0
or, r²-r-72 = 0.
or, r²-9r+8r-72=0
or, r(r-9)+8(r-9)=0
or, (r-9) (r+8)= 0
Either, r-9=0 ..r=9
or, r+8=0 r=-8
∴Since r = not -8, as r is the length of radius, i.e., it can not be negative.
∴r=9
∴The length of the outer radius of the new hollow sphere is 9 cm.
∴ Outer surface area = 4 x 22/7 x 9 x9 sq cm = 1018 2/7 sq cm.
