Balance of Rod with Floating Cube Buoyancy Concept

❓ Question

A cube having a side of:

$$10\text{ cm}$$

with unknown mass and:

$$200\text{ gm}$$

mass were hung at two ends of a uniform rigid rod of:

$$27\text{ cm}$$

long.

The rod was placed on a wedge keeping the balance between wedge point and:

$$200\text{ gm}$$

weight as:

$$25\text{ cm}$$

Initially the masses were not at balance.

A beaker is placed beneath the unknown mass and water is added slowly to it.

At given point the masses were in balance and half volume of unknown mass was inside water.

(Take density of unknown mass equal to density of water, the rod mass negligible and water density: 1 g/cm³)

The unknown mass is ______ kg.

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🖼 Question Image

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✍️ Short Explanation

This problem is based on: 

 👉 Torque equilibrium 

 👉 Buoyant force 

 👉 Archimedes principle. 

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🔷 Step 1 — Understand Geometry 💯

Rod length:

$$27\text{ cm}$$

Distance between wedge and:

$$200\text{ gm}$$

mass:

$$25\text{ cm}$$

Hence distance of unknown cube from wedge:

$$27-25=2\text{ cm}$$

So lever arms are:

$$200\text{ gm}\rightarrow25\text{ cm}$$
$$\text{Unknown mass}\rightarrow2\text{ cm}$$

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🔷 Step 2 — Initial Balance Condition

Initially system is not balanced.

After adding water:

✔ Half cube submerged

✔ Buoyant force acts upward

✔ Effective weight decreases

Then equilibrium occurs.

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🔷 Step 3 — Volume of Cube

Side of cube:

$$10\text{ cm}$$

Volume:

$$V=10^3$$
$$=1000\text{ cm}^3$$

Half submerged volume:

$$V_s=500\text{ cm}^3$$

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🔷 Step 4 — Buoyant Force

Density of water:

$$1\text{ gm/cm}^3$$

Buoyant force equals displaced water weight.

So buoyant force corresponds to:

$$500\text{ gm}$$

Thus effective mass of cube becomes:

$$(M-500)\text{ gm}$$

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🔷 Step 5 — Apply Torque Equilibrium

At balance:

$$(M-500)\times2 = 200\times25$$
$$2M-1000=5000$$
$$2M=6000$$
$$M=3000\text{ gm}$$

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🔷 Step 6 — Convert into kg

$$3000\text{ gm}=3\text{ kg}$$

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🔷 Step 7 — JEE Trap Alert 🚨

❌ Full cube volume use kar lena instead of half submerged volume

❌ Lever arm distances galat lena

❌ Buoyant force ko directly Newton me unnecessarily convert karna

Remember:

$$\boxed{ F_b=\rho Vg }$$

and in balance:

$$\boxed{ \tau_{clockwise}=\tau_{anticlockwise} }$$

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✅ Final Answer

$$\boxed{3\text{ kg}}$$

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