Problem 1 :
A hollow cylindrical pipe is of length 40 cm. Its internal and external radii are 4 cm and 12 cm respectively. It is melted and cast into a solid cylinder of length 20 cm. Find the radius of the new solid.
length of cylindrical pipe (h) = 40 cm
internal radius of the pipe (r) = 4 cm
external radius of the pipe (R) = 12 cm
Height of cylinder = 20 cm
Volume of hollow cylindrical pipe = Volume of cylinder
Π h (R2 - r2) = Π r2 h
h (R2 - r2) = r2 h
(40) (122 - 42) = r2 (20)
(40) (144 - 16) = r2 (20)
(40) (128)/20 = r2
r2 = (40) (128)/20
r2 = 2 (128)
r = √256
r = 16 cm
Therefore radius of cylinder = 16 cm
Problem 2 :
An iron right circular cone of diameter 8 cm and height 12 cm is melted and recast into spherical lead shots each of radius 4 mm. How many lead shots can be made?
diameter of right circular cone = 8 cm
radius of right circular cone (r) = 4 cm
Height of right circular cone (h) = 12 cm
Radius of spherical lead shot (r) = 4 mm
10 mm = 1 cm
4 mm = (4/10) cm
Volume of right circular cone
= n x (Number of spherical lead shots)
n = Volume of cone/Number of spherical lead shots
n = (1/3)Π r2 h/(4/3)Π r3
n = (1/3) (4)2 (12) x (3/4) (10/4)3
n = 3 x 5 x 5 x 10
n = 750 lead shots
Therefore 750 lead shots can be made.
Problem 3 :
A container with a rectangular base of length 4.4 m and breadth 2 m is used to collect rain water. The height of the water level in the container is 4 cm and water is transferred into a cylindrical vessel with radius 40 cm. What will be the height of the water level in the cylinder?
Volume of water in cuboidal container = Volume of water in cylindrical container
lwh = Π r2h
l = 4.4 m, w = 2 m, h = 4 cm ==> 4/100 ==> 1/25 m
r = 40 cm ==> 40/100
4.4(2) (1/25) = Π (2/5)2h
h = 4.4 (2) (1/25) (1/Π) (25/4)
h = 2.2/Π
h = 0.7 m
Problem 4 :
A cylindrical bucket of height 32 cm and radius 18 cm is filled with sand. The bucket is emptied on the ground and a conical heap and sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Volume of sand in the cylindrical bucket = Volume of sand of conical heap
Let r1, h1 be the radius and height of cylindrical tank
Let r2, h2 be the radius and height of conical tank
Π r12h1 = (1/3)Π r22h2
r1 = 18 cm, h1 = 32 cm and h2 = 24 cm
182(32) = (1/3) r22(24)
182(32) = 8 r22
r22 = (18⋅18⋅32)/8
r2 = 18(2)
r2 = 36 cm
So, the radius of the conical heap is 36 cm.
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