A solid weighs 10.00N in air, 6N when forcefully immersed in water, and 7.0N when fully immersed in a liquid, X. Calculate the relative density of the liquid X.

Analysis

To solve this problem, we need to apply Archimedes' Principle, which states that the upthrust (loss in weight) experienced by an object immersed in a fluid is equal to the weight of the fluid displaced.

Concept

The relative density (RD) of a liquid is defined as:

Since the object is fully immersed in both cases, the volume of liquid displaced is equal to the volume of the object. Therefore:

1. Upthrust in water = Weight in air - Weight in water
2. Upthrust in liquid X = Weight in air - Weight in liquid X

The ratio of these upthrusts gives the relative density of the liquid.

Step-by-Step Solution

Step 1: Calculate the upthrust in water.

• Weight in air = $10.0\text{ N}$
• Weight in water = $6.0\text{ N}$
• Upthrust in water = $10.0\text{ N} - 6.0\text{ N} = 4.0\text{ N}$

Step 2: Calculate the upthrust in liquid X.

• Weight in air = $10.0\text{ N}$
• Weight in liquid X = $7.0\text{ N}$
• Upthrust in liquid X = $10.0\text{ N} - 7.0\text{ N} = 3.0\text{ N}$

Step 3: Calculate the relative density of liquid X.

Conclusion

Comparing our result to the given options, we find that the calculated value matches option C.

Exam Tip

Always remember that "Relative Density" is a ratio of weights (or upthrusts). If you ever get a value greater than 1 for a liquid that is less dense than water (like oil), double-check your subtraction! In this case, since the object weighed more in liquid X ($7\text{N}$) than in water ($6\text{N}$), liquid X must be less dense than water, so your answer must be less than 1.