Physics Q50 – JEE-MAIN 2026

Home Physics Mechanical Properties of FluidsQuestion #255

A tub is filled with water and a wooden cube

10 cm \times 10 cm \times 10 cm

is placed in the water. The wooden cube is found to float on the water with a part of it submerged in water. When a metal coin is placed on the wooden cube, the submerged part is increased by

3.87 cm

. The mass of the metal coin is _______ gram. (Take water density as

1 {g/cm}^{3}

and density of wood as

0.4 {g/cm}^{3}

)

Enter Calculated Numerical Value:

Clue 1 of 3: Archimedes' Principle

When an object floats, the buoyant force acting on it is equal to its total weight.

🧠 Full Solution Path

Step 1: Understand the principle of flotation✦ Active

When an object floats, the buoyant force equals the total weight of the object. When a coin is placed on the wooden cube, the total weight increases. To maintain flotation, the buoyant force must also increase. This increase in buoyant force is achieved by an increase in the submerged volume of the wooden cube.

💡 Teacher's Secret Hint

Remember that the buoyant force is directly proportional to the volume of the fluid displaced.

Step 2: Relate additional mass to additional buoyant force○ Expand

The additional weight of the metal coin ( $M_{c o i n} \times g$ ) is balanced by the additional buoyant force. This additional buoyant force is generated by the increase in the submerged volume of the wooden cube. The increase in submerged volume is the cross-sectional area of the cube ( $A$ ) multiplied by the increase in submerged depth ( $Δ h$ ). Therefore, we can write:

M_{c o i n} \times g = ρ_{w} \times (A \times Δ h) \times g

The acceleration due to gravity ( $g$ ) cancels out, simplifying the equation to:

M_{c o i n} = ρ_{w} \times A \times Δ h

💡 Teacher's Secret Hint

The acceleration due to gravity $g$ cancels out, simplifying the calculation.

Step 3: Calculate the mass of the metal coin○ Expand

Given the side of the cube $L = 10 cm$ , the cross-sectional area is $A = L^{2} = (10 cm)^{2} = 100 {cm}^{2}$ . The increase in submerged depth is $Δ h = 3.87 cm$ . The density of water is $ρ_{w} = 1 {g/cm}^{3}$ . Substitute these values into the simplified equation:

M_{c o i n} = 1 {g/cm}^{3} \times 100 {cm}^{2} \times 3.87 cm

Calculating the mass:

M_{c o i n} = 387 g

💡 Teacher's Secret Hint

Ensure all units are consistent (grams and centimeters in this case) to get the correct result.

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Physics Question 50 – JEE-MAIN 2026

Clue 1 of 3: Archimedes' Principle

🧠 Full Solution Path