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 = r_{1}

Level to be rised = 2cm

Radius of each iron sphere = 1.5cm = r_{2}

Numberofsphere =

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)

5Ã—5Ã—h = 4Ã—

5Ã—5Ã—h = 4Ã—Ã—1Ã—1Ã—1

Heightofwaterinjar = cm

**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 = = 3cm = r_{1}

Diameterofthewire = 0.2cm

Radiusofthewire = 0.1cm = r_{2}

Volume of sphere = Volume of wire

= Ã—3Ã—3Ã—3 = 0.1Ã—0.1Ã—h

h =

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 = r_{1}

External radius of the sphere = 5cm = r_{2}

Height of the cylinder = cm = h

Volume of the spherical shell = Volume of cylinder

r_{3 } =

r_{3 }= 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

r_{2} = 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_{2 }= 4cm

Internal radii = r_{1} = 2cm

Cone base radius(R) = 4cm

Height = h

Volume of cone = Volume of sphere

h = 14cm

Slantheight(l) =

Slantheight(l) =

l =

l=

l =

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

v_{s} = xÃ—v_{cone}

x=

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

v_{cone }= v_{hemisphere}

r^{2}h = 2r^{3}

h = 2r

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

multiplying by 3

Ï€r^{2}h : 2Ï€r^{3} : 3Ï€r^{2}hÏ€r^{3} : 2Ï€r^{3} : 3Ï€r^{3 }(âˆµ r = h and r^{2}h = r^{3})

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

Ï€r^{3} = Ï€r^{2}h

r^{3} =

r^{3} = 729

r=

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 =Ã—5.25Ã—5.25Ã—5.25

v = 606.375cm^{3}

**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

v_{1} = Volume of sphere = Ï€r^{3}

v_{1 }= Volume of cylinder = Ï€r^{2}h = Ï€r^{2}Ã—2r

v_{1 }= Volume of cone =

Now

v_{1} : v_{2} : v_{3} =

= 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} = 64cm^{3}

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

Therefore radius of the sphere = 2cm

Volume of the sphere = Ã—(2)^{3 }= 33.52cm^{3}

Volume of gap = Volume of cube - Volume of sphere

= 64cm^{3} - 33.52cm^{3} = 30.48cm^{3}

**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 = r_{1}

Thickness of the hemispherical tank = 1cm = 0.01m

Outer radius of hemispherical tank = (1+0.01) = 1.01m = = r_{2}

Volume of iron used to make the tank =

= 0.06348m^{3}.

** **

**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 == 1.75mm

Volume of spherical sphere = Ï€r^{3}

Ã—(1.75)^{3}

= 22.458mm^{3}

Therefore 22.46mm^{3} 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 = ^{th}diameter of earth

Let the diameter of earth be d, so radius =

Then diameter of moon =

Radius =

Volume of moon =

Volume of earth =

=

Thus the volume of the moon is of volume of earth