A cube having a side of 10 cm with unknown mass and 200 gm mass were hung at two ends of an uniform rigid rod of 27 cm long. The rod along with masses...| Doubtify JEE

 

🧠 Physics in Real Life: How to Find the Unknown Mass of a Cube Using Archimedes’ Principle

Imagine this:

You have a cube of side 10 cm, and it’s hanging on one side of a rod, with a 200-gram mass hanging on the other side. The rod is 27 cm long and uniform in nature. It is placed on a wedge at a point 25 cm away from the 200 g mass. The setup is not in balance initially.

Now, something interesting happens. You place a beaker under the cube and slowly add water. As the cube starts to dip, there comes a point where exact balance is achieved, and half the cube is submerged in water. The question is:

What is the mass of the cube in kilograms?
(Given: Density of water = 1 g/cm³, and the cube's material does not absorb water)

 

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🧪 Step-by-Step Concept Breakdown

Let’s break it down using Archimedes’ Principle and Torque Balance.

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1️⃣ Volume of Cube:

Since each side = 10 cm
So, Volume = $10 \times 10 \times 10 = 1000 \text{ cm}^3$

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2️⃣ Buoyant Force Acting (Half Submerged):

If half the cube is inside the water, the displaced volume = $\frac{1000}{2} = 500 \text{ cm}^3$

Buoyant force = Weight of displaced water
= $500 \times 1 = 500 \text{ grams} = 0.5 \text{ kgf}$

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3️⃣ Torque Balance Around Wedge (Fulcrum):

Let the unknown mass be M (in kg)

Distance from fulcrum:

• Distance between 200g mass and wedge: 25 cm

• Rod length = 27 cm ⇒ Distance from cube side to wedge = 2 cm

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Now apply Torque balance:

Clockwise Torque = Anticlockwise Torque
$(0.2 \, \text{kg}) \times 25 = (M - 0.5) \times 2$

(Remember: Buoyant force reduces effective weight of the cube)

⇒
$5 = (M - 0.5) \times 2$
⇒
$M - 0.5 = 2.5$
⇒
M = 3.0 kg

✅ Final Answer: 3.0 kg

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🔍 Key Physics Concepts Used:

• Archimedes’ Principle: Buoyant force = Weight of displaced fluid

• Torque Equilibrium: For balance, net torque = 0

• Effective Weight: Actual weight – Buoyant force

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🧠 Why This Problem Is Important for JEE:

This question beautifully combines fluids with rotational mechanics, two frequently asked topics in JEE Physics. It tests your conceptual clarity and calculation skills.

📝 Topics Covered:

• Laws of Equilibrium

• Center of Mass

• Buoyancy

• Linear Mechanics

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📢 Conclusion:

This is a classic conceptual JEE problem where mechanics and fluid dynamics intersect. It shows how real-life physics can help us solve tricky balance problems using simple logic and laws.

Keep practicing such problems to master JEE Physics!