An object weighs 30N in air and 21N in water. The weight of the object when completely immersed in a liquid of relative density 1.4 is

1. Analyze the question

We are given the weight of an object in air and its weight when immersed in water. We need to find its weight when immersed in a different liquid with a known relative density.

2. Explain the concept

This problem is based on Archimedes' Principle, which states that the upthrust on an object immersed in a fluid is equal to the weight of the fluid displaced.

• Upthrust in water = Weight in air - Weight in water
• Upthrust in any liquid = Weight in air - Weight in liquid
• Relative Density (RD) = $\frac{\text{Upthrust in liquid}}{\text{Upthrust in water}}$

3. Show the solution

Step 1: Calculate the upthrust in water.

• Weight in air ($W_a$) = 30 N
• Weight in water ($W_w$) = 21 N
• Upthrust in water ($U_w$) = $W_a - W_w = 30 - 21 = 9\text{ N}$

Step 2: Calculate the upthrust in the new liquid. We know the Relative Density (RD) of the liquid is 1.4.

• $RD = \frac{\text{Upthrust in liquid } (U_l)}{\text{Upthrust in water } (U_w)}$
• $1.4 = \frac{U_l}{9}$
• $U_l = 1.4 \times 9 = 12.6\text{ N}$

Step 3: Calculate the weight of the object in the liquid.

• Weight in liquid ($W_l$) = Weight in air - Upthrust in liquid
• $W_l = 30 - 12.6$
• $W_l = 17.4\text{ N}$

4. Identify the answer

Comparing our result to the given options, the correct value is 17.4N.

5. Tip

When dealing with buoyancy problems, always remember that Upthrust = Weight in air - Weight in liquid. If you are given the Relative Density of a liquid, it is simply the ratio of the upthrust in that liquid to the upthrust in water. Don't let the "Relative Density" term confuse you—it is just a multiplier for the upthrust!