Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
Find the volume of a sphere whose radius is:
(i) 2 cm
(ii) 3.5 cm
(iii) 10.5 cm.
(i) Radius (r) = 2 cm
Therefore volume = 4/3πr^{3}
= 4/3 × 22/7 × (2)^{3}
= 33.52 cm^{3}
(ii) Radius (r) = 3.5 cm
= 4/3 × 22/7 × (3.5)^{3} = 179.666 cm^{3}
(iii) Radius (r) = 10.5 cm
= 4/3 × 22/7 × (10.5)^{3} = 4851 cm^{3}
Find the volume of a sphere whose diameter is:
(i) 14 cm
(ii) 3.5 dm
(iii) 2.1 m
(i) Diameter = 14 cm, Radius(r) = 14/2 = 7 cm
= 4/3 × 22/7 × (7)^{3} = 1437.33 cm^{3}
(ii) Diameter = 3.5 dm, Radius (r) = 3.52 = 1.75 dm
= 4/3 × 22/7 × (1.75)^{3}
= 22.46 dm^{3}
(iii) Diameter = 2.1m, Radius(r) = 2.1/2 = 1.05 m
= 4/3 × 22/7 × (1.05)^{3} = 4.851m^{3}
A hemispherical tank has the inner radius of 2.8 m. Find its capacity in liters.
Radius of the tank = 2.8 m
Therefore Capacity = 2/3πr^{3}
= 2/3 × 22/7 × (2.8)^{3} = 45.994 m^{3}
1m^{3} = 1000l
Therefore capacity in litres = 45994 litres
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.
Inner radius = 5 cm
Outer radius = 5 + 0.25 = 5.25
Volume of steel used = Outer volume-Inner volume
= 2/3 × π × ((5.25)^{3} − (5)^{3})
= 2/3 × 22/7 × ((5.25)^{3} − (5)^{3})
= 41.282 cm^{3}
How many bullets can be made out of a cube of lead, whose edge measures 22 cm, each bullet being 2 cm in diameter?
Cube edge = 22 cm
Therefore volume of the cube = (22)^{3} = 10648 cm^{3}
And,
Volume of each bullet = 4/3πr^{3}
= 4/3 × 22/7 × (1)^{3}
= 4/3 × 22/7
= 88/21cm^{3}
A shopkeeper has one laddoo of radius 5 cm. With the same material, how many laddoos of radius 2.5 cm can be made?
Volume of laddoo having radius = 5 cm
i.e Volume (V_{1}) = 4/3πr^{3}
(V_{1}) = 4/3 × 22/7 × (5)^{3}
(V_{1}) = 11000/21 cm^{3}
Also Volume of laddoo having radius 2.5 cm
i.e Volume (V_{2}) = 4/3πr^{3}
(V_{2}) = 4/3 × 22/7 × (2.5)^{3}
(V_{2}) = 1375/21 cm^{3}
Therefore number of laddoos = V_{1}/V_{2 }= 11000/1375 = 8
A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. If the diameters of two balls be 3/2cm and 2 cm, find the diameter of the third ball.
Volume of lead ball = 4/3πr^{3}
= 4/3 × 22/7 × (3/2)^{3}
Diameter of first ball d_{1} = 3/2 cm
Radius of first ball r_{1}
Diameter of second ball d_{2} = 2 cm
Radius of second ball r_{2} = 2/2cm = 1 cm
Diameter of third ball d_{3} = d
Radius of third ball r_{3} = d/2 cm
d = 2.5 cm
A sphere of radius 5 cm is immersed in water filled in a cylinder, the level of water rises 5/3 cm. Find the radius of the cylinder.
Radius of cylinder = r
Radius of sphere = 5cm
Volume of sphere = 4/3πr^{3}
= 4/3 × π × (5)^{3}
Height of water rised = 5/3cm
Volume of water rised in cylinder = πr^{2}h
Therefore, Volume of water rises in cylinder = Volume of sphere
Let r be the radius of the cylinder
πr^{2}h = 4/3πr^{3}
r^{2} × 5/3 = 4/3 × π × (5)^{3}
r^{2} × 5/3 = 4/3 × 22/7 × 125
r^{2} = 20 × 5
r = √100
r = 10 cm
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?
Let v_{1} and v_{2} be the volumes of the first and second sphere respectively
Radius of the first sphere = r
Radius of the second sphere = 2r
= 1/8
A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.
Given that
Volume of the cone = Volume of the hemisphere
1/3πr^{2}h = 2/3πr^{3}
r^{2}h = 2r^{3}
h = 2r
h/r = 1/1 × 2 = 2
Therefore
Ratio of their heights = 2:1
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.
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_{1} = 3.5 cm
Inner radii of the bowl = r_{2} = 7 cm
h = 7/12 cm
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.
Height of the cylinder = 2/3 diameter
We know that
Diameter = 2(radius)
h = 2/3 × 2r = 4/3r
Volume of the cylinder=Volume of the sphere
π × r^{2} × (4/3r) = 4/3π(4)^{3}
(r)^{3} = (4)^{3}
r = 4 cm
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.
It is given that
Volume of water in hemispherical bowl = Volume of cylinder
h = 9 cm
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?
Let r be the radius of the iron ball
Radius of the cylinder = 16 cm
Then,
Volume of iron ball = Volume of water raised in the hub
r^{3} = 1728
r = 12 cm
Therefore radius of the ball = 12 cm.
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 = 227).
Given that:
Radius of the cylinder = 12cm = r_{1}
Raised in raised = 6.75 cm = r_{2}
Volume of water raised = Volume of the sphere
= r_{2} = 9 cm
Radius of the sphere is 9 cm
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.
Given that diameter of a copper sphere = 18 cm
Radius of the sphere = 9 cm
Length of the wire = 108 m = 10800 cm
Volume of cylinder = Volume of sphere
r_{1} = 0.3 cm
Therefore Diameter = 2 × 0.3 = 0.6 cm
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?
Given that,
Radius of the cylinder jar = 6 cm = r_{1}
Level to be rised = 2 cm
Radius of each iron sphere = 1.5 cm = r_{2}
Number of sphere = 16
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
Let the level of water be raised by h
Diameter of the spherical bowl = 2 cm
Radius of the ball = 1 cm
Volume of jar = 4 (Volume of spherical ball)
Height of water in jar = 16/75 cm
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 = 6/2 = 3 cm = r_{1}
Diameter of the wire = 0.2 cm
Radius of the wire = 0.1 cm = r_{2}
Volume of sphere = Volume of wire
h = 3600 cm
h = 36 m
Therefore length of wire = 36 m
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 = 3 cm = r_{1}
External radius of the sphere = 5 cm = r_{2}
Height of the cylinder = 8/3cm = h
Volume of the spherical shell = Volume of cylinder
r_{3} = 7cm
Therefore diameter of the cylinder = 2(radius) = 14 cm
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.
Given
Radius of the hemisphere = Volume of cone
r_{2} = 3.47 cm
Therefore radius of the base = 3.74 cm
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 radii = r_{2} = 4 cm
Internal radii = r_{1} = 2 cm
Cone base radius (R) = 4 cm
Height = h
Volume of cone = Volume of sphere
h = 14 cm
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
x = 126
Therefore number of cones = 126
A cone and a hemisphere have equal bases and volumes
v_{cone} = v_{hemisphere}
hr = 2/1
h:r = 2:1
Therefore the ratio is 2:1
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
v_{cone}:v_{hemisphere}:v_{cylinder}
1/3 πr^{2}h: 2/3 πr^{3}: πr^{2}h
Multiplying by 3
πr^{2}h: 2πr^{3}: 3πr^{2}h
πr^{3}: 2πr^{3}: 3πr^{3 }(∴ r = h and r^{2}h = r^{3})
1:2:3 Therefore the ratio is 1: 2: 3.
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
r = 9 cm
Therefore radius of the ball = 9 cm
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
Volume of sphere = v
Diameter of the largest sphere = 10.5 cm
2r = 10.5
r = 5.25 cm
Volume of sphere = 4/3πr^{3} = 4/3 × 22/7 × 5.25 × 5.25 × 5.25
v = 606.375 cm^{3}
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 r be the common radius
Height of the cone = height of the cylinder = 2r
Let
v_{1} = Volume of sphere = 4/3 πr^{3}
v_{1} = Volume of cylinder = πr^{2}h = πr^{2} × 2r
v_{1} = Volume of cone = 1/3πr^{2}h = 1/3πr^{3}
Now
v_{1}:v_{2}:v_{3} = 4/3πr^{3}:2πr^{3}: 2/3πr^{3}
= 4:6:2 = 2:3:1
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 cm^{3}
Diameter of the sphere = Length of the side of the cube = 4cm
Therefore radius of the sphere = 2cm
Volume of the sphere = 4/3πr^{3 }= 4/3 × 22/7 × (2)^{3} = 33.52 cm^{3}
Volume of gap = Volume of cube - Volume of sphere
= 64 cm^{3} - 33.52 cm^{3} = 30.48 cm^{3}
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 = r_{1}
Thickness of the hemispherical tank = 1 cm = 0.01 m
Outer radius of hemispherical tank = (1 + 0.01) = 1.01 m = r_{2}
Volume of iron used to make the tank
= 0.06348 m^{3}
A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (mm^{3}) is needed to fill this capsule?
Diameter of capsule = 3.5 mm
Radius = 3.5/2 = 1.75 mm
Volume of spherical sphere = 4/3πr^{3}
= 22.458 mm^{3}
Therefore 22.46 mm^{3} of medicine is required
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 the diameter of earth be d, so radius = d/2
Then diameter of moon = d/4
= 1/64
Thus the volume of the moon is 1/64 of volume of earth
Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
Chapter 21: Surface Area and Volume of a Sphere...