Class 9 RD Sharma Solutions – Chapter 21 Surface Area and Volume of a Sphere – Exercise 21.2 | Set 2
Question 17. 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?
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Radius of the cylinder jar = r1 = 6 cm,
Level to be rised = 2 cm,
Radius of each iron sphere = r2 = 1.5 cm,
Number of sphere = (Volume of Cylinder) / (Volume of Sphere)
= (π x (r1)^2 x h) / (4/3 x π x (r2)^3)
= ((r1)^2 x h) / (4/3 x (r2)^3) = (6 x 6×2) / (43 x 1.5 x 1.5 x 1.5)
Hence Number of Sphere is 16.
Question 18. 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?
Diameter of jar = 10 cm,
Radius of jar = 5 cm,
Diameter of the spherical bowl = 2 cm,
Radius of the ball = 1 cm.
Let the level of water be raised by h
Volume of jar = 4 x (Volume of spherical ball)
π x (r1)^2 x h = 4 x (4/3 x π x (r2)^3)
(r1)^2 x h = 4 x (4/3 x (r2)^3)
5 x 5 x h = 4 x 4/3 x 1 x 1 x 1
h = 16/75 cm
Hence Height of water in jar = 16/75 cm.
Question 19. 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.
Diameter of sphere = 6 cm,
Radius of sphere = d/2 = r1 = 6/2 = 3 cm,
Diameter of the wire = 0.2 cm,
Radius of the wire = r2 = 0.1 cm,
Volume of sphere = Volume of wire
4/3 x π x (r1)^3 = π(r2)^2 x h
4/3 x 3 x 3 x 3 = 0.1 x 0.1 x h
h = 3600 cm = 36m
Hence length of wire = 36 m.
Question 20. 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 22/3 cm. Find the diameter of the cylinder.
Internal radius of the sphere = r1 = 3 cm,
External radius of the sphere = r2 = 5 cm,
Height of the cylinder = h = 8/3cm,
Volume of the spherical shell = Volume of cylinder
4/3π(r23 – r13) = π x (r3)^2 x h
4/3(53 – 33) = 8/3 x (r3)^2
(r3)^2 = √49
r3 = 7 cm
Hence Diameter of the cylinder is 2 x radius = 14 cm.
Question 21. 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.
Radius of the hemisphere = Volume of cone
2/3 π(r1)^3 = 1/3 π(r2)^2 x h
2/3 x 7^3 = 1/3 x (r2)^2 x 49
(r2)^2 = (2x7x7x7x3) / (3×49)
(r2)^2 = 2058147
r2 = 3.47 cm
Hence radius of the base is 3.74 cm.
Question 22. A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone.
Hollow sphere external radius = r2 = 4 cm,
Internal radii = r1 = 2 cm,
Cone base radius (R) = 4 cm.
Height = h
Volume of cone = Volume of sphere (Given)
1/3 π(r)^2 x h = 4/3 x π x ((r2)^2 – (r1)^2)
4^2 x h = 4(43 – 23)
h = 14 cm
As we know that Slant Height = √(r)2 + (h)2
l = √(4)2 + (14)2 = √16 + 196 = √212 = 14.56 cm
Hence Height of cone is 14 cm and Slant Height is 14.56 cm.
Question 23. 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.
Metallic sphere of radius = 10.5 cm,
Cone radius = 3.5 cm,
Height of radius = 3 cm,
Let the number of cones obtained be x
Vs = X x Volume of cone
4/3 x π x (r)^3 = X x 1/3 π (r)^2 h
X = (4 x 10.5 x 10.5 x 10.5) / (3.5 x 3.5 x 3)
X = 126
Hence number of cones is 126.
Question 24. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.
A cone and a hemisphere have equal bases and volumes.
Volume of cone = Volume of hemisphere
1/3πr2h = 2/3πr3
r2h = 2r3
h = 2r
h x r = 2/1
h : r = 2:1
Therefore the ratio is 2 : 1
Question 25. 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.
A cone, a hemisphere and a cylinder stand on one equal bases and have the same weight
Volume of cone : Volume of hemisphere : Volume of cylinder
1/3 πr2h : 2/3 πr3 : πr2h
Multiplying by 3
πr2h : 2πr3 : 3πr2h
πr3 : 2πr3 : 3πr3 (Given that r = h and r2h = r3)
Hence, the ratio is 1:2:3.
Question 26. 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?
Radius of cylindrical tub = 12 cm,
Depth = 20 cm,
Let r be the radius of the ball then,
Volume of the ball = Volume of water raised
4/3 x π x (r)^3 = π x (r)^2 x h
r^3 = (3.14x(12)2 x 6.75 x 3) / 4
r^3 = 729
r = 9 cm
Hence radius of the ball is 9 cm.
Question 27. The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of the sphere.
Side of cube = 10.5 cm,
Diameter of the largest sphere = 10.5 cm so r = 5.25 cm
Volume of sphere = 4/3πr3 = 4/3 × 22/7 × 5.25 × 5.25 × 5.25 = 606.375 cm3
Hence Volume of Sphere is 606.375 cm3.
Question 28. 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.
Let us assume that r be the common radius and,
Height of the cone = Height of the cylinder = 2r (Given)
Let’s assume that,
v1 = Volume of sphere = 4/3 πr3
v2 = Volume of cylinder = πr2h = πr2 × 2r
v3 = Volume of cone = 1/3πr2h = 1/3πr3
v1 : v2 : v3 = 4/3πr3 : 2πr3 : 2/3πr3
= 4 : 6 : 2 = 2 : 3 : 1
Hence the ratio of their volumes is 2 : 3 : 1.
Question 29. A cube of side 4 cm contains a sphere touching its side. Find the volume of the gap in between.
Cube side = 4cm
Volume of cube = (4 cm)3 = 64 cm3
Diameter of the sphere = Length of the side of the cube = 4 cm
Therefore radius of the sphere = 2cm
Volume of the sphere = 4/3πr3 = 4/3 × 22/7 × (2)3 = 33.52 cm3
Volume of gap = Volume of cube – Volume of sphere
= 64 – 33.52 = 30.48 cm3
Hence the volume of the gap in between is 30.48 cm3.
Question 30. 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.
Inner radius of the hemispherical tank = 1 m = r1,
Thickness of the hemispherical tank = 1 cm = 0.01 m,
Outer radius of hemispherical tank = r2 = (1 + 0.01) = 1.01 m,
Volume of iron used to make the tank = 2/3 π x ((r2)3 – (r1)3)
= 2/3 x 22/7 x [(1.01)3 – (1)3]
= 0.06348 m^3
Hence volume of the iron used to make the tank is 0.06348 m3.
Question 31. 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?
Diameter of capsule = 3.5 mm
Radius = 3.5/2 = 1.75 mm
Volume of spherical sphere = 4/3πr3
= 4/3 × 22/7 × (1.75)3
= 22.458 mm3
Therefore 22.46 mm3 of medicine is required.
Question 32. The diameter of the moon is approximately one-fourth of the diameter of the earth. What is the earth the volume of the moon?
Diameter of moon = 1/4th Diameter of earth.
Let us assume, diameter of earth is d, so radius = d/2
Then diameter of moon = d/4
Radius = d/8
Volume of moon = 4/3πr3 = 4/3π(d/8)3 = 4/3 x (1/512) x πd3
Volume of earth = 4/3πr3 = 4/3π(d/2)3 = 4/3 x (1/8) x πd3
(Volume of moon) / (Volume of earth) = 1/64
Hence the volume of the moon is 1/64 of volume of earth.