What is the key principle for balancing the rod?

The key principle is the Law of Moments (Torque Balance), where Clockwise Torque = Anticlockwise Torque about the pivot.

How is the distance from the wedge calculated?

Total rod length is 27 cm. One mass is at 25 cm from the wedge, so the other mass at the opposite end is 2 cm from the wedge.

What is the volume of the cube?

The volume $V$ of a cube with side 10 cm is $10^3 = 1000$ cm³.

How much buoyant force acts on the cube?

Since half volume is submerged, $V_{sub} = 500$ cm³. Buoyant force = $500 imes 1 imes g$, equivalent to 500 g mass.

Why is the rod mass neglected?

In this problem, the 'uniform rigid rod' setup focuses on the suspended masses; typically, if rod mass isn't given, it's considered negligible.

What is the unit of the final answer?

The final answer must be converted from grams to kilograms as per the question requirement.

What does density > water imply?

It ensures the mass would naturally sink without support, allowing it to displace water effectively.

What is the effective mass calculation?

Effective mass $M_{eff} = ext{Actual Mass} - ext{Mass of displaced water}$.

Can we use meters instead of cm?

Yes, but using CGS (cm, gm) is simpler here as all values are provided in those units.

Why did the system balance only after adding water?

Initially, the torque from the unknown mass was too high; water provided an upward force to reduce that torque to a balanced state.

A cube having a side of 10 cm with unknown mass and 200 gm mass were hung at two ends of a uniform rigid rod of 27 cm long. The rod along with masses was placed on a wedge keeping the distance between wedge point and 200 gm weight as 25 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 the unknown mass was inside the water.

A cube having a side of 10 cm with unknown mass and 200 gm mass were hung at two ends of a uniform rigid rod of 27 cm long. The rod along with masses was placed on a wedge keeping the distance between wedge point and 200 gm weight as 25 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 the unknown mass was inside the water. (Take the density of unknown mass is more than that of the water, the mass did not absorb water and water density is 1 gm/cm³.) The unknown mass is ____ kg. | JEE Main Physics

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Solution Steps

Determine Lever Arms

Total rod length = 27 cm. The 200 gm mass is at 25 cm from the wedge. Since the unknown mass $M$ is at the other end, its distance from the wedge is:

$$d_M = 27 – 25 = 2 \text{ cm}$$

Distance for 200 gm mass, $d_{200} = 25 \text{ cm}$.

Calculate Cube Volume

Side of the cube $a = 10 \text{ cm}$.

Total Volume $V = a^3 = 10 \times 10 \times 10 = 1000 \text{ cm}^3$.

Calculate Buoyant Force

The problem states half the volume is submerged in water:

$$V_{sub} = \frac{1000}{2} = 500 \text{ cm}^3$$

Buoyant force $F_B = V_{sub} \times \rho_{water} \times g = 500 \times 1 \times g = 500g \text{ dynes}$. This is equivalent to an upward mass of 500 gm.

Setup Torque Equilibrium

At equilibrium, Torque about wedge = 0. Let the unknown mass be $M$ grams.

$$\text{Torque}_{left} = \text{Torque}_{right}$$

$$(M – 500) \times g \times 2 = 200 \times g \times 25$$

Solve for Mass M

Cancel $g$ and solve the equation:

$$2(M – 500) = 5000$$

$$M – 500 = 2500$$

$$M = 3000 \text{ gm}$$

Convert to Final Units

The question asks for the mass in kilograms:

$$M = \frac{3000}{1000} = 3 \text{ kg}$$

Final Answer: 3

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Theory

1. Principle of Moments

The principle of moments states that when an object is in rotational equilibrium, the sum of the clockwise moments about any pivot point is equal to the sum of the anticlockwise moments about that same point. In this problem, the rigid rod acts as a lever with the wedge serving as the fulcrum. The unknown mass creates a counter-clockwise torque, while the 200 gm mass creates a clockwise torque. By measuring the distances (lever arms) from the fulcrum, we can establish an equation to find unknown forces or masses. This concept is fundamental to statics and mechanical advantage in physics.

2. Archimedes’ Principle

Archimedes’ principle explains the upward buoyant force exerted on a body immersed in a fluid. It states that the buoyant force is equal to the weight of the fluid displaced by the body. For the cube in this problem, the upward force $F_B$ reduces its effective weight acting on the rod. Since the density of water is 1 gm/cm³, the weight of the displaced water in grams is numerically equal to the volume submerged in cm³. This reduction in effective weight is what allows the system to reach a state of balance that was not possible in air.

3. Apparent Weight in Fluids

The apparent weight of an object submerged in a fluid is the difference between its actual weight (in vacuum or air) and the buoyant force acting upon it. Mathematically, $W_{app} = W_{real} – F_{buoyant}$. In torque problems involving immersion, we use the apparent weight at the point of suspension. If an object is denser than the fluid, it will have a positive apparent weight and exert a downward force. This principle is crucial in determining the density of solids using hydrostatic balances and understanding how objects behave when partially or fully submerged.

4. Rotational Equilibrium Conditions

A body is said to be in mechanical equilibrium if it satisfies two conditions: translational equilibrium (net force is zero) and rotational equilibrium (net torque is zero). In this problem, the rod is supported by a wedge, which provides a normal reaction force to satisfy translational equilibrium. However, the positioning of weights determines rotational stability. For the rod to remain horizontal without rotating, the line of action of all forces must result in zero net moment about the fulcrum. This balance is sensitive to both the magnitude of the masses and their respective distances from the pivot point.

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FAQs

Why did we use 2 cm as the distance for M?

The rod is 27 cm long. One end is 25 cm from the wedge, so the other end must be $27 – 25 = 2$ cm from the wedge.

What is the significance of the cube side?

The side of 10 cm allows us to calculate the volume ($10^3 = 1000$ cm³), which is necessary to find the buoyant force.

Why did we subtract 500 from M?

M is the downward mass, and the water provides an upward buoyant force equal to 500g (half the volume). The net downward mass is $M – 500$.

Does the density of the unknown mass matter?

Yes, the problem states it is more than water, ensuring the cube sinks and displaces water rather than floating on the surface.

What if the rod was not uniform?

If non-uniform, the center of gravity would not be at the midpoint, and we would need additional information about its mass distribution.

Is the answer exactly 3?

Yes, based on the calculation $2(M-500) = 5000$, $M = 3000$ gm, which is exactly 3 kg.

Can this be solved using SI units?

Yes, but CGS is easier here. In SI, $1000$ cm³ becomes $10^{-3}$ m³ and $1$ gm/cm³ becomes $1000$ kg/m³.

What does ‘rigid rod’ mean?

It means the rod does not bend under the weights, keeping the distances from the wedge constant and the torque calculations accurate.

How do we know which mass is on which side?

Usually, the larger distance (25 cm) corresponds to the lighter weight to balance a heavier weight at a shorter distance (2 cm).

Does the shape of the beaker matter?

No, only the depth of the water and the submerged volume of the cube determine the buoyant force.

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