Question

An iron casting containing a number of cavities weight 6000N in air and 4000N in water. What is the volume of the cavities in the casting? Density of iron is 7.87g//cm^(3). Take g = 9.8 m//s^(2) and density of water = 10^(3) kg//m^(3).

Answer 1

The volume of the cavities in the casting = 0.127 m³

Given:

An iron casting containing a number of cavities weight 6000N in air and 4000N in water.

Density of iron is 7.87 g/cm³

g = 9.8 m/s²

Density of water = 10³ kg/m³ =1 gm/cm³

To find:

The volume of the cavities in the casting.

Explanation:

$V_{casting}$ = $V_{cav.} +V_{iron}$

Weight in air = $V_{iron}$ $\rho_{iron}$ g

$V_{iron}$ = $\frac{6000}{7.87\times 10^3\times 10}$ = 0.07625 m³

When we put it in water there is buoyant force is acting

So Weight in water = Weight in air - buoyant force

      Weight in water =  $V_{iron}$ $\rho_{iron}$ g - $V_{casting}$ $\rho_{water}$ g

   $V_{casting}$ =  (Weight in air -  Weight in water)  ÷ $\rho_{water}$ g

   $V_{casting}$  = (6000-4000) ÷ $10^{3}$ × 9.81

   $V_{casting}$ = 0.2038

$V_{cav.}$ = $V_{casting}$ - $V_{iron}$

$V_{cav.}$ = 0.1275 m³

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