APSC AE WRD 2019 PYQ: Specific Gravity

Specific Gravity Multiple Choice Question

Question: A piece weighing 3 kg in air was found to weigh 2.5 kg when submerged in water. Its specific gravity is:
(A) 1
(B) 5
(C) 7
(D) 6

Answer: (D) 6

Explanation

The specific gravity of an object is the ratio of its density to the density of water, determined using its weight in air and the loss of weight when submerged in water, based on Archimedes' principle.

Calculation Steps

Given:

• Weight in air = 3 kg
• Weight in water = 2.5 kg

1. Calculate the loss of weight in water (equal to the weight of water displaced):

Loss of weight = 3 kg - 2.5 kg = 0.5 kg

2. Use the formula for specific gravity:

Specific Gravity = Weight in air / Loss of weight in water = 3 / 0.5 = 6

Key Notes

• Specific gravity is a dimensionless quantity, as it is the ratio of the object's weight to the weight of an equal volume of water.
• The loss of weight in water represents the buoyant force, which equals the weight of the displaced water.
• This calculation is fundamental in civil engineering for determining material properties in applications like concrete mix design and buoyancy analysis.
• The specific gravity of 6 indicates the object is six times denser than water.

Note: The specific gravity of 6, calculated using Archimedes' principle, is critical for material characterization in civil engineering projects.

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