Question

A cube 10.0 cm on each side has a density of 2.053 x 10^4 kg/m^3. It’s apparent weight in fresh water is 192 N. Find the buoyant force.

Answer 1

Answer:

9 N

Explanation: Given that L = side of the cube = 10cm = 0.1m

Mass = Density * Volume

Let’s determine the volume of the cube in m^3.

V = L^3

V = 0.1^3 = 0.001m^3

Mass = 0.001 * 2.053 * 10^4 = 20.53kg

Weight = 20.53 * 9.8 = 201.194

Buoyant force = 201.194 – 192 = 9.194 N

This is approximately 9 N.

Answer 2

The buoyant force will be "9 N".

Given:

Length,

• L = 10 cm

          = 0.1 m

Density,

• d = $2.053\times 10^4 \ kg/m^3$

Let,

The volume be in "$m^3$"

then,

→ $V = L^3$

       $= 0.1^3$

       $= 0.001 \ m^3$

Now,

→ $Mass = Density\times Volume$

By putting the values,

             $= 0.001\times 2.053\times 10^4$

             $= 20.53 \ kg$

then,

Weight = $20.53\times 9.8$

           = $201.194$

hence,

The Buoyant force is:

= $201.194-192$

= $9.194 \ N$

Thus the above answer is correct.

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