JAMB 2018 Physics · Question 31

A solid weighs 45N and 15N respectively in air and water. Determine the relative density of the solid

Options

A.0.33
B.0.50
C.1.50
D.3.00

Answer: C. 1.50

Explanation written with AI and checked against the answer key

1. Analyze the question

We are given the weight of a solid in two different media:

Weight in air ( $W_a$ ) = 45 N
Weight in water ( $W_w$ ) = 15 N We need to find the Relative Density (RD) of the solid.

2. Explain the concept

Relative density is defined as the ratio of the density of a substance to the density of water. When dealing with weights, the formula is:

\text{Relative Density} = \frac{\text{Weight of object in air}}{\text{Loss in weight in water}}

The "Loss in weight in water" is the difference between the weight in air and the weight in water (this is also equal to the upthrust experienced by the object).

\text{Loss in weight} = W_a - W_w

3. Show the solution

Step 1: Calculate the loss in weight.

\text{Loss in weight} = 45\text{ N} - 15\text{ N} = 30\text{ N}

Step 2: Apply the Relative Density formula.

\text{RD} = \frac{\text{Weight in air}}{\text{Loss in weight in water}}

\text{RD} = \frac{45}{30}

Step 3: Simplify the fraction.

\text{RD} = \frac{45}{30} = \frac{3}{2} = 1.5

4. Identify the answer

Comparing our result to the given options: A. 0.33 B. 0.50 C. 1.50 D. 3.00

The correct option is C.

5. Tip

When solving Archimedes' Principle problems, always remember that the "Loss in weight" is the weight of the fluid displaced. If the question had mentioned a liquid other than water, you would multiply the result by the relative density of that liquid. Since we are using water (which has a relative density of 1), the calculation remains straightforward!

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