basic_engineering_mathematics0

184 Basic Engineering Mathematics
Radius of cylindrical head = 1 cm = 0.5 cm and
2
height of cylindrical head = 2mm= 0.2cm
Hence, volume of cylindrical head
= πr 2 h = π(0.5) 2 (0.2) = 0.1571 cm 3
Volume of cylindrical shaft
( ) 0.2 2
= πr 2 h = π (1.5) = 0.0471 cm 3
2
Total volume of 1 rivet = 0.1571 + 0.0471 = 0.2042 cm 3
Volume of metal in 2000 such rivets = 2000 × 0.2042
= 408.4 cm 3
Problem 13. A solid metal cylinder of radius 6 cm and
height 15 cm is melted down and recast into a shape comprising
a hemisphere surmounted by a cone. Assuming that
8% of the metal is wasted in the process, determine the
height of the conical portion, if its diameter is to be 12 cm.
Volume of cylinder = πr 2 h = π × 6 2 × 15 = 540π cm 3
If 8% of metal is lost then 92% of 540π gives the volume of the
new shape (shown in Fig. 24.7).
h
i.e. height of conical portion,
h =
496.8 − 144
12
= 29.4cm
Problem 14. A block of copper having a mass of 50 kg is
drawn out to make 500 m of wire of uniform cross-section.
Given that the density of copper is 8.91 g/cm 3 , calculate
(a) the volume of copper, (b) the cross-sectional area of the
wire, and (c) the diameter of the cross-section of the wire.
(a) A density of 8.91 g/cm 3 means that 8.91 g of copper has a
volume of 1 cm 3 , or 1 g of copper has a volume of (1/8.91) cm 3
Hence 50 kg, i.e. 50 000 g, has a volume
(b) Volume of wire
50 000
8.91 cm3 = 5612 cm 3
= area of circular cross-section × length of wire.
Hence 5612 cm 3 = area × (500 × 100 cm),
from which, area = 5612
500 × 100 cm2 = 0.1122 cm 2
(c) Area of circle = πr 2 or πd2
4
from which
√ ( 4 × 0.1122
d =
π
, hence 0.1122 =
πd2
4
)
= 0.3780 cm
i.e. diameter of cross-section is 3.780 mm.
Fig. 24.7
r
12 cm
Hence the volume of (hemisphere + cone) = 0.92 × 540π cm 3 ,
i.e. 1 ( ) 4
2 3 πr3 + 1 3 πr2 h = 0.92 × 540π
Dividing throughout by π gives:
2
3 r3 + 1 3 r2 h = 0.92 × 540
Since the diameter of the new shape is to be 12 cm, then radius
r = 6 cm,
hence 2 3 (6)3 + 1 3 (6)2 h = 0.92 × 540
144 + 12h = 496.8
Problem 15. A boiler consists of a cylindrical section of
length 8 m and diameter 6 m, on one end of which is surmounted
a hemispherical section of diameter 6 m, and on
the other end a conical section of height 4 m and base diameter
6 m. Calculate the volume of the boiler and the total
surface area.
The boiler is shown in Fig. 24.8.
Volume of hemisphere,
P = 2 3 πr3 = 2 3 × π × 33 = 18π m 3
Volume of cylinder,
Q = πr 2 h = π × 3 2 × 8 = 72π m 3
Volume of cone,
R = 1 3 πr2 h = 1 3 × π × 32 × 4 = 12π m 3