Question

Iron has a density of 7.87 g/cm3. What mass of iron would be required to cover a football playing surface of 120 yds × 60 yds to a depth of 1.0 mm? (1 inch = 2.54cm)

Answer 1

The answer is option (A) 1.0x10⁵ lb

Given,

Density of Iron = 7.87 g/cm³

The required conversions are:

1 inch =2.54cm

1lb = 453.6g

1 yard = 0.9144m = 91.44cm

Convert the given yard and depth measurement into cm, then you get

120 yards = 120*91.44 = 10972.8 cm

60 yards = 60*91.44 = 5486.4 cm

1mm = 0.1cm

Volume of the football = (10972.8)*(5486.4)*(0.1) cm = 6,020,116.992 cm³ (because the formula for voulume=length*breadth*depth)

We need to find mass of Iron.

We know that Density = Mass / Volume

So formula for Mass = Density * Volume

So now substitue the values of volume and density in the above formula. Then we get

Mass = (7.87 g/cm³ ) * (6,020,116.992 cm³)

Mass = 47,378,320.73g = 47 Mg

Now convert grams into lb, 1 lb = 453.6g

So Mass = 47,378,320.73g = 47 Mg = 1.0x 10⁵ lb

So the answer is option (A) 1.0x10⁵ lb