Basic Engineering Mathematics, Fifth Edition

78 Basic Engineering Mathematics

Using a calculator,gas constant,R=4160J/(kg K),

correct to 4 significant figures.

Problem 20. A rectangular box with square ends

has its length 15cm greater than its breadth and the

total length of its edges is 2.04m. Find the width of

the box and its volume

Letxcm=width=height of box. Then the length of

the box is(x+ 15 )cm, as shown in Figure 11.1.

(x^1

15)

x

x

Figure 11.1

The length of the edges of the box is 2( 4 x)+ 4 (x+ 15 )

cm, which equals 2.04m or 204cm.

Hence, 204 = 2 ( 4 x)+ 4 (x+ 15 )

204 = 8 x+ 4 x+ 60

204 − 60 = 12 x

i.e. 144 = 12 x

and x=12cm

Hence,the width of the box is 12cm.

Volume of box=length×width×height

=(x+ 15 )(x)(x)=( 12 + 15 )( 12 )( 12 )

=( 27 )( 12 )( 12 )

=3888cm^3

Problem 21. The temperature coefficient of

resistanceαmay be calculated from the formula

Rt=R 0 ( 1 +αt).Findα,givenRt= 0 .928,

R 0 = 0 .80 andt= 40

SinceRt=R 0 ( 1 +αt),then

0. 928 = 0 .80[1+α( 40 )]

0. 928 = 0. 80 +( 0. 8 )(α)( 40 )

0. 928 − 0. 80 = 32 α

0. 128 = 32 α

Hence, α=

0. 128

32

= 0. 004

Problem 22. The distancesmetres travelled

in timetseconds is given by the formula

s=ut+

1

2

at^2 ,whereuis the initial velocity in m/s

andais the acceleration in m/s^2 .Findthe

acceleration of thebody if it travels 168m in 6s,

with an initial velocity of 10m/s

s=ut+

1

2

at^2 ,ands= 168 ,u=10 andt= 6

Hence, 168 =( 10 )( 6 )+

1

2

a( 6 )^2

168 = 60 + 18 a

168 − 60 = 18 a

108 = 18 a

a=

108

18

= 6

Hence,the acceleration of the body is 6 m/s^2.

Problem 23. When three resistors in an electrical

circuit are connected in parallel the total resistance

RTis given by

1

RT

=

1

R 1

+

1

R 2

+

1

R 3

.Findthe

total resistance whenR 1 = 5 ,R 2 = 10 and

R 3 = 30 

1

RT

=

1

5

+

1

10

+

1

30

=

6 + 3 + 1

30

=

10

30

=

1

3

Taking the reciprocal of both sides givesRT= 3 

Alternatively, if

1

RT

=

1

5

+

1

10

+

1

30

, the LCM of the

denominators is 30RT.

Hence, 30 RT

(

1

RT

)

= 30 RT

(

1

5

)

+ 30 RT

(

1

10

)

+ 30 RT

(

1

30

)

.

Cancelling gives 30= 6 RT+ 3 RT+RT

i.e. 30 = 10 RT

and RT=

30

10

= 3 ,as above.