A 400g solid cube having an edge of length 10cm floats in water. How much volume of the cube is outside the water?…

Year	2026	2025	2024	Total
Questions	9	15	8	32

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

By the law of flotation, the weight of a floating body must exactly balance the buoyant force exerted by the displaced fluid volume:

M \cdot g = ρfluid \cdot Vsubmerged \cdot g

Core Logic

Given parameters:

Mass of the cube, $M = 400 g = 0.4 kg$ $M = 400 \text{ g} = 0.4 \text{ kg}$
Total volume of the cube, $Vtotal = (10 cm)³ = 1000 cm³ = 10⁻³ m³$
Density of water, $ρwater = 1000 kg/m³$

Equating weight to buoyant force to find the submerged volume $Vd$ $V_d$ :

0.4 = 1000 \times Vsubmerged

Vsubmerged = (0.4)/(1000) = 4 \times 10⁻⁴ m³ = 400 cm³

Calculate the volume remaining outside the water surface :

Voutside = Vtotal - Vsubmerged

Voutside = 1000 cm³ - 400 cm³ = 600 cm³

Pattern Recognition

The fraction of a floating body's volume that is submerged equals the ratio of the body's density to the fluid's density: $VsubmergedVtotal = ρbodyρfluid$ . Here, the cube's effective density is $0.4 g/cm³$ $0.4 \text{ g/cm}^3$, meaning $40%$ $40\%$ is submerged and $60%$ $60\%$ stays outside.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

More Mechanical Properties of Fluids Previous-Year Questions — Page 5

Q jee_main_2025_24_jan_morning Surface Energy

The amount of work done to break a big water drop of radius 'R' into 27 small drops of equal radius is 10 J. The work done required to break the same big drop into 64 small drops of equal radius will be :-

A. 15 J
B. 10 J
C. 20 J
D. 5 J

Solution

Related Formula

The work done $W$ $W$ in expanding surface area with surface tension $S$ $S$ is given by:

W = S \cdot Δ A

Core Logic

By volume conservation during splitting :

(4)/(3)π R³ = n ((4)/(3)π r³) r = Rn1/3

Total change in area gives work expression:

W = S (n \cdot 4π r² - 4π R²) = 4π R²S (n1/3 - 1)

Step 1: Set up Proportions

For $n = 27$ $n = 27$ drops:

W = 4π R²S (271/3 - 1) = 4π R²S (3 - 1) = 2(4π R²S) = 10 J

4π R²S = 5 J

For $n = 64$ $n = 64$ drops:

W' = 4π R²S (641/3 - 1) = 4π R²S (4 - 1) = 3(4π R²S)

W' = 3 \times 5 = 15 J

Pattern Recognition

Work scales scaling-wise linearly with the key multiplier index $(n1/3 - 1)$ $(n^{1/3} - 1)$. Taking ratios between targets directly avoids calculations of constants.

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q3 jee_main_2025_24_jan_morning Excess Pressure and Surface Tension

An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density

1000 kg/m³

If the pressure inside the bubble is 2100 N/m² greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use

g=10 m/s²)

A. 0.02
B. 0.1
C. 0.25
D. 0.04

Solution

Related Formula

The absolute pressure inside an air bubble submerged in a liquid is given by :

Pᵢₙ = P₀ + ρ gh + (2T)/(R)

where $P₀$ $P_0$ is atmospheric pressure, $ρ$ $\rho$ is the liquid density, $g$ $g$ is gravity, $h$ $h$ is depth, $T$ $T$ is surface tension, and $R$ $R$ is the radius.

Core Logic

We are given that the difference between the inside pressure and atmospheric pressure is $2100 N/m²$ $2100\text{ N/m}^2$:

Pᵢₙ - P₀ = ρ gh + (2T)/(R) = 2100

Step 1: Numerical Evaluation

Substitute the given values into the relation :

ρ gh = 1000 \times 10 \times 0.20 = 2000 N/m²

Now find the excess pressure from surface tension:

(2T)/(R) = 2100 - 2000 = 100 N/m²

T = (100 \times R)/(2) = 50 \times (0.1 \times 10⁻²) = 0.05 N/m

Pattern Recognition

Total inside pressure accounts for both the hydrostatic pressure of the fluid column ( $ρ gh$ $\rho gh$) and the spherical geometry boundary constraint pressure ( $(2T)/(R)$ $\frac{2T}{R}$).

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q jee_main_2025_29_jan_morning Pascal\'s Law

In a hydraulic lift, the surface area of the input piston is

6cm²

and that of the output piston is

1500cm²

. If

100N

force is applied to the input piston to raise the output piston by

20cm

, then the work done is ________ kJ.

Numerical Answer. Answer: 5 to 5

Solution

Related Formula

W = F₁ \cdot s₁ = F₂ \cdot s₂

Core Logic

By conservation of liquid volume displacement during output piston elevation :

A₁ \cdot s₁ = A₂ \cdot s₂ 6 \cdot s₁ = 1500 \cdot 20 s₁ = 5000 cm = 50 m

Work performed on input boundary matches :

W = F₁ \cdot s₁ = 100 N \cdot 50 m = 5000 J = 5 kJ

Alternatively via output force profile calculation:

F₂ = F₁ (A₂)/(A₁) = 100 (1500)/(6) = 25000 N

W = F₂ \cdot s₂ = 25000 \cdot 0.2 = 5000 J = 5 kJ

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

Q58 jee_main_2024_01_february_morning Bernoulli's Principle

A plane is in level flight at constant speed and each of its two wings has an area of

40~m²

. If the speed of the air is

180~km/h

over the lower wing surface and

252~km/h

over the upper wing surface, the mass of the plane is _______

kg

. (Take air density to be

1~kg m⁻³

and

g = 10~ms⁻²

)

Numerical Answer. Answer: 9600 to 9600

Solution

Related Formula

Bernoulli's pressure balance equation for aerofoils:

Δ P = P₁ - P₂ = (1)/(2)ρ(v₂² - v₁²)

Dynamic Lift force balancing plane weight:

Flift = Δ P \cdot Atotal = mg

Core Logic

Convert velocity limits to SI units:

v₁ = 180~km/h = 180 \times (5)/(18) = 50~ms⁻¹

v₂ = 252~km/h = 252 \times (5)/(18) = 70~ms⁻¹

Total effective wing area layout (2 wings):

Atotal = 2 \times 40 = 80~m²

Step 1: Calculate Mass Balance

Substitute these values into the dynamic lift equation:

mg = (1)/(2) ρ (v₂² - v₁²) Atotal

m(10) = (1)/(2) \times 1 \times (70² - 50²) \times 80

10m = 40 \times (4900 - 2500) = 40 \times 2400 = 96000

m = 9600~kg

Pattern Recognition

Remember to multiply individual wing areas by 2 for standard multi-wing lift structures ( $Atotal = 2A$ $A_{\text{total}} = 2A$).

Chapter Mix

Class 11 Physics: Mechanical Properties of Fluids

A 400g$400\mathrm{g}$ solid cube having an edge of length 10cm$10\mathrm{cm}$ floats in water. How much volume of the cube is outside the water? (Given: density of water = 1000kgm⁻³$= 1000\mathrm{kg}\mathrm{m}^{-3}$ )

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Mechanical Properties of Fluids Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Set up Proportions

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Numerical Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Mass Balance

Pattern Recognition

Chapter Mix

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A $400g$ $400\mathrm{g}$ solid cube having an edge of length $10cm$ $10\mathrm{cm}$ floats in water. How much volume of the cube is outside the water? (Given: density of water $= 1000kgm⁻³$ $= 1000\mathrm{kg}\mathrm{m}^{-3}$ )