Problem 1 :
An oil funnel of tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion be 8 cm and the diameter of the top of the funnel be 18 cm, then find the area of the tin sheet required to make the funnel.
Solution :
Area of sheet required to make the funnel
= Curved surface area of frustum cone + Curved surface area of cylinder
= π(R + r)l + 2πrh
= π [(R + r)l + 2rh] ----(1)
Total height of funnel = 22
Height of cylinder + Height of frustum cone = 22
10 + h = 22
h = 22 - 10
h = 12
Height of frustum cone = 12 cm
Radius of cylindrical portion (r) = 8/2 = 4 cm
Radius of the top of the frustum cone (R) = 18/2 = 9 cm
l = √[h2 + (R −r)2]
l = √[122 + (9 − 4)2]
l = √[122 + 52]
l = √(144 + 25) = √169
l = 13 cm
By applying these values in (1), we get
= π [(R + r)l + 2rh]
= (22/7) [(9 + 4) (13) + 2 (4) (10)]
= (22/7) [169 + 80]
= 782.57 cm2
Problem 2 :
Find the number of coins, 1.5 cm in diameter and 2 mm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Solution :
Radius of coin (r1) = (1.5/2) cm
height of coin (h1) = thickness of coin
= 2 mm = (2/10) cm
Radius of cylinder (r2) = (4.5/2) cm
height of cylinder (h2) = 10 cm
n x Volume of 1 coin = Volume of cylinder
n x πr12h1 = πr22h2
n = (4.5/2)2 ⋅ 10 / (1.5/2)2 ⋅ (2/10)
= 450 coins
Problem 3 :
A hollow metallic cylinder whose external radius is 4.3 cm and internal radius is 1.1 cm and whole length is 4 cm is melted and recast into a solid cylinder of 12 cm long. Find the diameter of solid cylinder.
Solution :
Volume of hollow cylinder = Volume of right circular cylinder
R = 4.3 cm, r = 1.1 cm, height = 4 cm
Height of right circular cylinder = 12 cm
π h(R2 - r2) = π r2 h
4(4.32 - 1.12) = r2 (12)
r2 = (4/12)(4.3 + 1.1) (4.3 - 1.1)
r2 = (1/3) (5.4) (3.2)
r2 = 5.76
r = 2.4
Diameter of cylinder = 2(2.4) = 4.8 cm
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