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 = πr³

= ××(2)³

= 33.52cm³

(ii) Radius(r) = 3.5cm

Therefore volume = πr³

= ××(3.5)³  = 179.666cm³

(iii) Radius(r) = 10.5cm

Therefore volume = 4³πr³

 = 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) =   = 7cm

Therefore volume = πr³

  = 1437.33cm³

(ii) Diameter = 3.5dm, Radius(r) =   = 1.75dm

Therefore volume = πr³

= 22.46dm³

(iii) Diameter = 2.1m, Radius(r) =   = 1.05m

Therefore volume = πr³

 = 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 = πr³

 = 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

= 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 = πr³ =    

Number of bullets =    = 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)  = πr³

Also Volume of laddoo having radius 2.5cm

i.e Volume(V₂)  = πr³

Therefore number of laddoos =  = 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 =  πr³

Diameter of first balld₁  =  cm

Radius of first ball r₁   cm

Diameter of second ball d₂ = 2cm

Radius of second ball r₂  =  cm = 1cm

Diameter of third balld₃  = d

Radius of third ball r₃  =  cm

Volume of lead ball =

Volume of lead ball =  

d =

d = 2.5cm

Q8. A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises cm. Find the radius of the cylinder.

Sol.

Radius of cylinder = r

Radius of sphere = 5cm

Volume of sphere = πr³

= ×π×(5)³

Height of water rised = cm

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 = πr³

r² = 20×5

r =

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, 

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

r²h = 2r³

h = 2r

 = 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 =  r₁  = 3.5cm

Inner radii of the bowl =  r₂  = 7cm

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 =  diameter

We know that

Diameter = 2(radius)

Volume of the cylinder = Volume of the sphere

πr²h = πr³

(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

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

 = πr²h

 = (16)²×9

r³ =

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 = ).

Sol.

Given that :

Radius of the cylinder = 12cm =  r₁

Raised in  raised = 6.75 cm =  r₂

Volume of water raised = Volume of the sphere

=  12×12×6.75 =

 = 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

 ×9×9×9

 = 0.009

r₁ = 0.3cm

Therefore Diameter = 2×0.3 = 0.6cm