Basic Engineering Mathematics, Fifth Edition

250 Basic Engineering Mathematics

Hence, the volume of (hemisphere+cone)

= 0. 92 × 540 πcm^3

i.e.^1

2

(

4

3

πr^3

)

+

1

3

πr^2 h= 0. 92 × 540 π

Dividing throughout byπgives

2

3

r^3 +

1

3

r^2 h= 0. 92 × 540

Sincethediameterofthenewshapeistobe12cm,radius

r=6cm,

hence^2

3

( 6 )^3 +

1

3

( 6 )^2 h= 0. 92 × 540

144 + 12 h= 496. 8

i.e.height of conical portion,

h=

496. 8 − 144

12

=29.4cm

Problem 22. A block of copper having a mass of

50kg is drawn out to make 500m of wire of

uniform cross-section. Given that the density of

copper is 8.91g/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 8.91g/cm^3 means that 8.91g of cop-

per has a volume of 1cm^3 , or 1g of copper has a

volume of( 1 ÷ 8. 91 )cm^3.

Density=

mass

volume

from which volume=

mass

density

Hence, 50kg, i.e. 50000g, has a

volume=

mass

density

=

50000

8. 91

cm^3 =5612cm^3

(b) Volume of wire=area of circular cross-section

×length of wire.

Hence,5612cm^3 =area×( 500 ×100cm)

from which,area=

5612

500 × 100

cm^2

=0.1122cm^2

(c) Area of circle=πr^2 or

πd^2

4

hence, 0. 1122 =

πd^2

4

from which,d=

√(

4 × 0. 1122

π

)

= 0 .3780cm

i.e.diameter of cross-section is 3.780 mm.

Problem 23. A boiler consists of a cylindrical

section of length 8m and diameter 6m, on one end

of which is surmounted a hemispherical section of

diameter 6m and on the other end a conical section

of height 4m and base diameter 6m. Calculate the

volume of the boiler and the total surface area

The boiler is shown in Figure 27.17.

8m

4m

C

I

B

R

A

Q

P

3m

6m

Figure 27.17

Volume of hemisphere, P=

2

3

πr^3

=

2

3

×π× 33 = 18 πm^3

Volume of cylinder, Q=πr^2 h=π× 32 × 8

= 72 πm^3

Volume of cone, R=

1

3

πr^2 h=

1

3

×π× 32 × 4

= 12 πm^3

Total volume of boiler= 18 π+ 72 π+ 12 π

= 102 π=320.4m^3

Surface area of hemisphere,P=

1

2

( 4 πr^2 )

= 2 ×π× 32 = 18 πm^2