Question

A piece of metal weights 45 N in air and 28.3 N when fully submerged in water. Find the specific gravity of the metal

Answer 1

Answer:

A metal piece weighs in air = 45 N

When submerged in water it weighs 28.3 N

So, specific gravity of the metal = 45/45-28.3 = 45/16.7 = 2.694

Explanation:

Answer 2

Answer:

The metal's specific gravity was determined to be is 2.694.

Explanation:

Data given,

Weight of the metal within the air = 45 N

Weight of the metal when completely submerged within the water = 28.3 N

The specific gravity of metal, S =?

• Specific gravity is the ratio of the density of an object and a reference substance, i.e., water.

As we all know,

According to Archimedes's Principle, the buoyant force of an object is equal to the weight of the liquid which is displaced or to the apparent loss of weight in water.

Now,

• The difference in weight = Buoyant force of water = 45 - 28.3 = 16.7 N

Now,

• The buoyant force of water = ρ ( water) × g × V (object)

Here,

• ρ ( water) = density of water = 1000Kg/m³
• g = acceleration due to gravity = 9.81 N/Kg
• V (object) = volume of the object

Therefore,

• 16.7 = 1000 × 9.81 × V ( object )
• Volume of the object = 0.0017 m³

Now, in air

• Weight of the metal within the air = mass of metal (m) × g
• 45 = m × 9.81
• m = 4.58 Kg

Therefore, the density of the metal, d,

• $d = \frac{m}{v}$
• $d = \frac{4.58}{0.0017}$  = 2694.1 Kg/m³

Now, as we all know,

• $S = \frac{\rho_{w} }{\rho_{m} }$

Here,

• S = specific gravity of the metal
• ρ(w) = density of water
• ρ(m) = density of metal

Therefore,

• $S = \frac{2694.1}{1000}$ = 2.694

Hence, the specific gravity of the metal = 2.694.