Given below are two statements: Statement I: If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in hot water. Statement II: If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in cold water. In the light of the above statements, choose the most appropriate from the options given below:

Q: Given below are two statements: Statement I: If a capillary tube is immersed first in cold water and then in hot water, the height of capillary rise will be smaller in hot water. Statement II: If a ca

The height of capillary rise (h) is given by: where, T = surface tension of the liquid theta = angle of contact = density of the liquid r = radius of the capillary tube As the temperature of water increases, its intermolecular cohesive forces decrease. This leads to a decrease in surface tension (T). Since h propto T (assuming and theta remain relatively constant), height of capillary rise decreases with increase in temperature. Because Thot Tcold, it follows that hhot hcold. * **Statement I** s

A. Both Statement I and Statement II are true
B. Both Statement I and Statement II are false
C. Statement I is true but Statement II is false
D. Statement I is false but Statement II is true

Solution & Explanation

### Related Formula The height of capillary rise (

h

) is given by:

h = frac2T cos thetarho g r

where,

T

= surface tension of the liquid

theta

= angle of contact

rho

= density of the liquid

r

= radius of the capillary tube ### Core Logic As the temperature of water increases, its intermolecular cohesive forces decrease. This leads to a decrease in surface tension (

T

). Since

h propto T

(assuming

rho

and

theta

remain relatively constant), height of capillary rise decreases with increase in temperature. ### Step 1: Evaluate Statement I and II Because

T_texthot lt T_textcold

, it follows that

h_texthot lt h_textcold

. * **Statement I** states: the height of capillary rise will be smaller in hot water. This is **True**. * **Statement II** states: the height of capillary rise will be smaller in cold water. This is **False**. ### Pattern Recognition Remember the key physical dependence: **Temperature Up

implies

Surface Tension Down

implies

Capillary Rise Down**. This basic trend of cohesive/adhesive properties versus thermal energy frequently appears in competitive conceptual physical chemistry/fluid physics questions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

Reference Study Guides

Mechanical Properties Of Fluids

More Mechanical Properties of Fluids Previous-Year Questions

Q4 jee_main_2025_02_april_evening Surface Tension and Surface Energy

Two water drops each of radius

r

coalesce to form a bigger drop. If

T

is the surface tension, the surface energy released in this process is:

A. $4pi r^2 T left[2 - 2^frac23right]$
B. $4pi r^2 T left[2 - 2^frac13right]$
C. $4pi r^2 T left[1 + sqrt2right]$ $4\pi r^2 T \left[1 + \sqrt{2}\right]$
D. $4pi r^2 T left[sqrt2 - 1right]$ $4\pi r^2 T \left[\sqrt{2} - 1\right]$

Solution

### Related Formula 1. Surface Energy:

U = T cdot A = T cdot (4pi R^2)

2. Conservation of Volume during coalescence of drops:

2 times left(frac43pi r^3right) = frac43pi R^3

### Core Logic When two drops of radius

r

coalesce into a single larger drop of radius

R

, volume is conserved:

R^3 = 2r^3 implies R = 2^1/3 r

- Initial surface area of the two separate drops:

A_i = 2 times 4pi r^2 = 8pi r^2

- Final surface area of the combined single drop:

A_f = 4pi R^2 = 4pi (2^1/3 r)^2 = 4pi r^2 2^2/3

- Surface energy released:

Delta E = U_i - U_f = T(A_i - A_f)

Delta E = T left( 8pi r^2 - 4pi r^2 2^2/3 right) = 4pi r^2 T left[ 2 - 2^2/3 right]

### Pattern Recognition Sees: Coalescence of

N

identical drops. Trap: Forgetting to conserve volume first, or confusing initial and final surface areas. Shortcut: Energy released when

N

drops coalesce into one big drop is:

Delta E = 4pi r^2 T left[ N - N^2/3 right]

Here, substituting

N = 2

directly gives

4pi r^2 T left[ 2 - 2^2/3 right]

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

Q21 jee_main_2025_02_april_morning Hydrostatic Pressure

A vessel with square cross-section and height of

6mathrm~m

is vertically partitioned. A small window of

100mathrm~cm^2

with hinged door is fitted at a depth of

3mathrm~m

in the partition wall. One part of the vessel is filled completely with water and the other side is filled with the liquid having density

1.5times 10^3mathrm~kg/m^3

. What force one needs to apply on the hinged door so that it does not get opened? (Acceleration due to gravity

= 10mathrm~m/s^2

)

Numerical Answer. Answer: 150 to 150

Solution

### Related Formula

P = P_0 + rho g h

F = Delta P cdot A

### Core Logic Let's analyze the pressure acting on the window at depth

h = 3mathrm~m

from both sides: - Water side (density

rho_w = 1.0 times 10^3mathrm~kg/m^3

P_w = P_0 + rho_w g h

- Denser liquid side (density

rho_l = 1.5 times 10^3mathrm~kg/m^3

P_l = P_0 + rho_l g h

The net pressure difference pushing on the window is:

Delta P = P_l - P_w = (rho_l - rho_w) g h

Delta P = (1.5 times 10^3 - 1.0 times 10^3) times 10 times 3 = 500 times 30 = 15000mathrm~N/m^2

The window has area

A = 100mathrm~cm^2 = 100 times 10^-4mathrm~m^2 = 10^-2mathrm~m^2

. To keep the door from opening, we must apply a force balancing the pressure difference:

F = Delta P cdot A = 15000 times 10^-2 = 150mathrm~N

### Step 1: Final Conclusion The required force is

150\mathrm{~N}

. ### Pattern Recognition For a vertical interface between two fluids, the net pressure difference is simply given by

\Delta P = \Delta \rho \cdot g h

. The ambient atmospheric pressure

P_0$ cancels out since it acts on both sides of the window. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

Q jee_main_2025_03_april_evening Work Done by Gravity in Connecting Vessels

Two cylindrical vessels of equal cross sectional area of

2mathrm~m^2

contain water upto height

10mathrm~m

and

6mathrm~m

, respectively. If the vessels are connected at their bottom then the work done by the force of gravity is : (Density of water is

10^3mathrm~kg/m^3

and

g=10mathrm~m/s^2

)

A. $1 times 10^5mathrm~J$ $1 \times 10^{5}\mathrm{~J}$
B. $4 times 10^4mathrm~J$ $4 \times 10^{4}\mathrm{~J}$
C. $6 times 10^4mathrm~J$ $6 \times 10^{4}\mathrm{~J}$
D. $8 times 10^4mathrm~J$ $8 \times 10^{4}\mathrm{~J}$

Solution

### Related Formula The gravitational potential energy

U

of a liquid column of mass

m

and height

h

is evaluated relative to its bottom by placing its total mass at its center of mass (

h/2

U = m g left(frach2right) = (rho A h) g left(frach2right) = frac12 rho A g h^2

Work done by the force of gravity (

W

) equals the negative change in potential energy:

W = -Delta U = U_i - U_f

### Core Logic Since the vessels are identical and connected at the bottom, water flows from the higher column to the lower one until their final heights equalize at:

h_f = frac10 + 62 = 8mathrm~m

### Step 1: Calculate Initial Potential Energy (

U_i

) Let the reference level

U=0

be at the bottom:

U_i = U_1 + U_2 = frac12 rho A g h_1^2 + frac12 rho A g h_2^2

U_i = frac12 rho A g left(10^2 + 6^2right) = frac12 rho A g (100 + 36) = 68 rho A g

Work Done by Gravity in Connecting Vessels

### Step 2: Calculate Final Potential Energy (

U_f

) Both vessels equalize to

h_f = 8mathrm~m

U_f = 2 times left[ frac12 rho A g h_f^2 right] = rho A g (8^2) = 64 rho A g

### Step 3: Work Done by Gravity (

W

)

W = U_i - U_f = 68 rho A g - 64 rho A g = 4 rho A g

Substitute the given values (

ho = 10^3mathrm~kg/m^3

A = 2mathrm~m^2

g = 10mathrm~m/s^2

W = 4 times 10^3 times 2 times 10 = 8 times 10^4mathrm~J

### Pattern Recognition For leveling liquids between two identical connected columns, the shift in center of mass always simplifies. The loss in potential energy is given by

\Delta U = \frac{1}{4} \rho A g (h_1 - h_2)^2

. Applying this directly:

Delta U = frac14 times 10^3 times 2 times 10 times (10 - 6)^2 = 5000 times 16 = 8 times 10^4mathrm~J

$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

Q12 jee_main_2025_03_april_evening Viscosity and Terminal Velocity

A solid steel ball of diameter

3.6mathrm~mm

acquired terminal velocity

2.45times10^-2mathrm~m/s

while falling under gravity through an oil of density

925mathrm~kg~m^-3

. Take density of steel as

7825mathrm~kg~m^-3

and g as

9.8mathrm~m/s^2

. The viscosity of the oil in SI unit is :

A. 2.18
B. 2.38
C. 1.68
D. 1.99

Solution

### Related Formula Terminal velocity

v_t

of a spherical body falling through a viscous fluid is given by Stokes' Law:

v_t = frac29 fracr^2 g (rho_textbody - rho_textfluid)eta

Hence, the viscosity coefficient

eta

is:

eta = frac29 fracr^2 g (rho_textbody - rho_textfluid)v_t

### Core Logic Given parameters: - Diameter

d = 3.6mathrm~mm Rightarrow

radius

r = 1.8mathrm~mm = 1.8 times 10^-3mathrm~m

- Terminal velocity

v_t = 2.45 times 10^-2mathrm~m/s

- Liquid density

rho_textfluid = 925mathrm~kg/m^3

- Steel density

rho_textbody = 7825mathrm~kg/m^3

- Acceleration due to gravity

g = 9.8mathrm~m/s^2

### Step 1: Substitute parameters into formula

eta = frac29 times frac(1.8 times 10^-3)^2 times 9.8 times (7825 - 925)2.45 times 10^-2

eta = frac29 times frac3.24 times 10^-6 times 9.8 times 69002.45 times 10^-2

### Step 2: Solve the numeric calculation

eta = frac2 times 0.36 times 10^-6 times 9.8 times 69002.45 times 10^-2

eta = frac0.72 times 9.8 times 6900 times 10^-62.45 times 10^-2

eta = frac48700.8 times 10^-62.45 times 10^-2 = frac0.04870080.0245 approx 1.99mathrm~Pacdot s

### Pattern Recognition Be careful when converting diameter to radius (

r = d/2$). Always ensure all numerical parameters are converted cleanly to SI base units (meters, kilograms, seconds) before applying Stokes' formula. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

Q22 jee_main_2025_03_april_evening Excess Pressure in Soap Bubbles

The excess pressure inside a soap bubble A in air is half the excess pressure inside another soap bubble B in air. If the volume of the bubble A is

n

times the volume of the bubble B, then the value of

n

is ________.

Numerical Answer. Answer: 8 to 8

Solution

### Related Formula Excess pressure inside a soap bubble in air (which has two liquid-gas interfaces) is given by:

Delta P = frac4TR Rightarrow R propto frac1Delta P

The volume

V

of a spherical bubble of radius

R

is:

V = frac43pi R^3 Rightarrow V propto R^3 propto left(frac1Delta Pright)^3

### Core Logic Given state:

Delta P_A = frac12 Delta P_B Rightarrow fracDelta P_BDelta P_A = 2

### Step 1: Calculate the ratio of radii

fracR_AR_B = fracDelta P_BDelta P_A = 2

### Step 2: Calculate volume scaling parameter (

n

)

V_A = n V_B Rightarrow n = fracV_AV_B = left(fracR_AR_Bright)^3 = (2)^3 = 8

### Pattern Recognition Volume scale factors depend on the cube of linear scale factors (

V propto R^3

). Since radius is inversely proportional to excess pressure, a halving of excess pressure leads to doubling of radius, scaling volume by

2^3 = 8

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

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