RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions for Class 9 Mathematics

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Class 9 : RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

The document RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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Q17. A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimeters?

Sol.

Given that,

Radius of the cylinder jar = 6cm = r1

Level to be rised = 2cm

Radius of each iron sphere = 1.5cm = r2

Numberofsphere = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

Numberofsphere = 16

 

Q18. A measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm each are dropped in it and they sink down in water completely. What will be the change in the level of water in the jar?

Sol.

Given that,

Diameter of jar = 10cm

Radius of jar = 5cm

Let the level of water be raised by h

Diameter of the spherical bowl = 2cm

Radius of the ball = 1cm

Volume of jar = 4(Volume of spherical ball)

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

5×5×h = 4×RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

5×5×h = 4×RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev×1×1×1

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

Heightofwaterinjar = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRevcm

 

Q19. The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of the wire.

Sol.

Given that

Diameter of sphere = 6cm

Radiusofsphere = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev= 3cm = r1

Diameterofthewire = 0.2cm

Radiusofthewire = 0.1cm = r2

Volume of sphere = Volume of wire

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

 = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev×3×3×3 = 0.1×0.1×h

h = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

h = 3600cmh = 36m

Therefore length of wire = 36m

 

Q20. The radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively. If it is melted and recast into a solid cylinder of height 22/3cm. Find the diameter of the cylinder.

Sol.

Given that,

Internal radius of the sphere = 3cm = r1

External radius of the sphere = 5cm = r2

Height of the cylinder = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRevcm = h

Volume of the spherical shell = Volume of cylinder

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

r = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

r= 7cm

Therefore diameter of the cylinder = 2(radius) = 14cm

 

Q21. A hemisphere of the lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.

Sol. Given

Radius of the hemisphere = Volume of cone

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

r2 = 3.47cm

Therefore radius of the base = 3.74cm

 

Q22. A hollow sphere of internal and external radii 2cm and 4 cm respectively is melted into a cone of base radius 4cm. Find the height and slant height of the cone.

Sol.

Given that

Hollow sphere external radii = r= 4cm

Internal radii = r1 = 2cm

Cone base radius(R) = 4cm

Height = h

Volume of cone = Volume of sphere

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

h = 14cm

Slantheight(l) = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

Slantheight(l) = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

l =RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

l= RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

l =RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

l = 14.56cm

 

Q23. A metallic sphere of radius 10.5 cm is melted and thus recast into small cones, each of radius 3.5 cm and height 3 cm. Find how many cones are obtained.

Sol.

Given that

Metallic sphere of radius = 10.5cm

Cone radius = 3.5cm

Height of radius = 3cm

Let the number of cones obtained be x

vs = x×vcone

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

x=  RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

x = 126

Therefore number of cones = 126

 

Q24. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

Sol.

Given that

A cone and a hemisphere have equal bases and volumes

vcone = vhemisphere

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

r2h = 2r3

h = 2r

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

h : r = 2 : 1

Therefore the ratio is 2 : 1

 

Q25. A cone, a hemisphere, and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1 : 2 : 3.

Sol.

Given that

A cone, a hemisphere and a cylinder stand on one equal bases and have the same weight

We know that

vcone : vhemisphere : vcylinder

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

multiplying by 3

πr2h : 2πr3 : 3πr2hπr3 : 2πr3 : 3πr(∵ r = h and r2h = r3)

1 : 2 : 3

Thereforetheratiois1 : 2 : 3.

 

Q26. A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?

Sol.

Radius of cylindrical tub = 12cm

Depth = 20cm

Let r be the radius of the ball

Then

Volume of the ball = Volume of water raised

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRevπr3 = πr2h

r3 =RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

r3 = 729

r= RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

r = 9cm

Therefore radius of the ball = 9cm

 

Q27. The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of the sphere.

Sol.

Side of cube = 10.5cm

Volume of sphere = v

Diameter of the largest sphere = 10.5cm

2r = 10.5

r = 5.25cm

Volumeofsphere =RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev×5.25×5.25×5.25

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

v = 606.375cm3

 

Q28. A sphere, a cylinder, and a cone have the same diameter. The height of the cylinder and also the cone are equal to the diameter of the sphere. Find the ratio of their volumes.

Sol.

Let r be the common radius

Height of the cone = height of the cylinder = 2r

Let

v1 = Volume of sphere = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRevπr3

v= Volume of cylinder = πr2h =  πr2×2r

v= Volume of cone = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

Now

v1 : v2 : v3 = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

= 4 : 6 : 2 = 2 : 3 : 1

 

Q29. A cube of side 4 cm contains a sphere touching its side. Find the volume of the gap in between.

Sol.

It is given that

Cube side = 4cm

Volume of cube = (4cm)3 = 64cm3

Diameter of the sphere = Length of the side of the cube = 4cm

Therefore radius of the sphere = 2cm

Volume of the sphere = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev×(2)= 33.52cm3

Volume of gap = Volume of cube - Volume of sphere

= 64cm3 - 33.52cm3 = 30.48cm3

 

Q30. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Sol.

Given that,

Inner radius of the hemispherical tank = 1m = r1

Thickness of the hemispherical tank = 1cm = 0.01m

Outer radius of hemispherical tank = (1+0.01) = 1.01m = = r2

Volume of iron used to make the tank = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

= 0.06348m3.

 

Q31. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (mm3) is needed to fill this capsule?

Sol.

Given that

Diameter of capsule = 3.5mm

Radius =RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev= 1.75mm

Volume of spherical sphere = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRevπr3  

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev×(1.75)3

= 22.458mm3

Therefore 22.46mm3 of medicine is required

 

Q32. The diameter of the moon is approximately one - fourth of the diameter of the earth. What is the earth the volume of the moon?

Sol.

Diameter of moon =  RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRevthdiameter of earth

Let the diameter of earth be d, so radius = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

Then diameter of moon = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

Radius = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

Volume of moon =RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

Volume of earth = RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev

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

Thus the volume of the moon is RD Sharma Solutions -Ex-21.2 (Part - 2), Surface Area And Volume Of Sphere, Class 9, Maths Class 9 Notes | EduRev of volume of earth

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