Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions -Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q1. Find the volume of a sphere whose radius is : (i) 2 cm (ii) 3.5 cm (iii) 10.5 cm.

Sol.

(i)  Radius(r) = 2cm

Therefore volume = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsπr3

= Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×(2)3

= 33.52cm3

(ii) Radius(r) = 3.5cm

Therefore volume = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsπr3

= Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×(3.5)3  = 179.666cm3

(iii) Radius(r) = 10.5cm

Therefore volume = 43πr3

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 4851cm3

 

Q2. Find the volume of a sphere whose diameter is : (i) 14 cm (ii) 3.5 dm (iii) 2.1 m

Sol.

(i)  Diameter = 14cm, Radius(r) =  Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 7cm

Therefore volume = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsπr3

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics  = 1437.33cm3

(ii) Diameter = 3.5dm, Radius(r) =  Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 1.75dm

Therefore volume = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsπr3

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= 22.46dm3

(iii) Diameter = 2.1m, Radius(r) =  Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 1.05m

Therefore volume = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsπr3

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 4.851m3

 

Q3. A hemispherical tank has the inner radius of 2.8 m.  Find its capacity in liters.

Sol.

Radius of the tank = 2.8m

Therefore Capacity = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsπr3

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 45.994m3

1m3 = 1000l

Therefore capacity in litres = 45994 litres

 

Q4. A hemispherical bowl is made of steel 0.25 cm thick. The inside radius of the bowl is 5 cm. Find the volume of steel used in making the bowl.

Sol.

Inner radius = 5cm

Outer radius = 5 + 0.25 = 5.25

Volume of steel used = Outer volume - Inner volume

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

= 41.282cm3

 

Q5. How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?

Sol.

Cube edge = 22cm

Therefore volume of the cube =  (22)3  = 10648cm3

And,

Volume of each bullet = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsπr=   Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 MathematicsEx-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 MathematicsEx-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics 

Number of bullets = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics  Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 2541

 

Q6. A shopkeeper has one laddoo of radius 5 cm. With the same material, how many laddoos of radius 2.5 cm can be made?

Sol.

Volume of laddoo having radius = 5cm

i.e Volume(V1)  = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsπr3

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Also Volume of laddoo having radius 2.5cm

i.e Volume(V2)  = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsπr3

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Therefore number of laddoos = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 8

 

Q7. A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be 32cm and 2 cm, find the diameter of the third ball.

Sol.

Volume of lead ball =  Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsπr3

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Diameter of first balld =  Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicscm

Radius of first ball r1   Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicscm

Diameter of second ball d2 = 2cm

Radius of second ball r2  =  Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicscm = 1cm

Diameter of third balld3  = d

Radius of third ball r3  =  Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicscm

Volume of lead ball = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Volume of lead ball =  Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 MathematicsEx-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 MathematicsEx-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 MathematicsEx-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

d = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

d = 2.5cm

 

Q8. A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicscm. Find the radius of the cylinder.

Sol.

Radius of cylinder = r

Radius of sphere = 5cm

Volume of sphere = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsπr3

= Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics×π×(5)3

Height of water rised = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicscm

Volume of water rised in cylinder = πr2h

Therefore, Volume of water rises in cylinder = Volume of sphere

Let r be the radius of the cylinder

πr2h = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsπr3

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

r2 = 20×5

r = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

r = 10cm

 

Q9. If the radius of a sphere is doubled, what is the ratio of the volume of the first sphere to that of the second sphere?

Sol.

Let v1 and v2 be the volumes of the first and second sphere respectively

Radius of the first sphere = r

Radius of the second sphere = 2r

Therefore, Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

 

Q10. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

Sol.

Given that

Volume of the cone = Volume of the hemisphere

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

r2h = 2r3

h = 2r

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 2

Therefore

Ratio of their heights = 2 : 1

 

Q11. A vessel in the form of a hemispherical bowl is full of water. Its contents are emptied in a right circular cylinder. The internal radii of the bowl and the cylinder are 3.5 cm and 7 cm respectively. Find the height to which the water will rise in the cylinder.

Sol.

Given that

Volume of water in the hemispherical bowl = Volume of water in the cylinder

Let h be the height to which water rises in the cylinder

Inner radii of the bowl =  r = 3.5cm

Inner radii of the bowl =  r2  = 7cm

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

 

Q12. A cylinder whose height is two thirds of its diameter has the same volume as a sphere of radius 4 cm. Calculate the radius of the base of the cylinder.

Sol.

Given that

Height of the cylinder =  Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsdiameter

We know that

Diameter = 2(radius)

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Volume of the cylinder = Volume of the sphere

πr2h = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicsπr3

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

(r)3 = (4)3

r = 4cm

 

Q13. A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of water in the cylinder.

Sol.

It is given that

Volume of water in hemispherical bowl = Volume of cylinder

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

h = 9cm

 

Q14. A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball?

Sol.

Let r be the radius of the iron ball

Radius of the cylinder = 16cm

Then,

Volume of iron ball = Volume of water raised in the hub

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = πr2h

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = (16)2×9

r=Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

r3 = 1728

r = 12cm

Therefore radius of the ball = 12cm.

 

Q15. A cylinder of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. (Use = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics).

Sol.

Given that :

Radius of the cylinder = 12cm =  r1

Raised in  raised = 6.75 cm =  r2

Volume of water raised = Volume of the sphere

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

=  12×12×6.75 = Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 729

=  r2  = 9cm

Radius of the sphere is 9cm

 

Q16. The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross - section. If the length of the wire is 108 m, find its diameter.

Sol.

Given that diameter of a copper sphere = 18cm

Radius of the sphere = 9cm

Length of the wire = 108m = 10800cm

Volume of cylinder = Volume of sphere

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics ×9×9×9

Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 0.009

r1 = 0.3cm

Therefore Diameter = 2×0.3 = 0.6cm

The document Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on Ex-21.2 (Part - 1), Surface Area And Volume Of Sphere, Class 9, Maths RD Sharma Solutions - RD Sharma Solutions for Class 9 Mathematics

1. What is the formula for calculating the surface area of a sphere?
Ans. The formula for calculating the surface area of a sphere is 4πr², where "r" represents the radius of the sphere.
2. How do you find the volume of a sphere?
Ans. The volume of a sphere can be calculated using the formula (4/3)πr³, where "r" represents the radius of the sphere.
3. What is the difference between the surface area and volume of a sphere?
Ans. The surface area of a sphere represents the total area covered by its outer surface, while the volume of a sphere represents the amount of space enclosed within its boundaries.
4. Can the surface area of a sphere be greater than its volume?
Ans. No, the surface area of a sphere cannot be greater than its volume. In fact, for any given sphere, the volume will always be greater than the surface area.
5. How can the surface area and volume of a sphere be useful in real-life applications?
Ans. The surface area and volume of a sphere have numerous real-life applications. For example, in architecture, the surface area of a dome can help determine the amount of materials needed for its construction. The volume of a sphere is useful in calculating the capacity of containers, such as water tanks or storage bins.
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