A capillary tube of radius 0.1 mm is partly dipped in water (surface tension 70~dyn/cm$70~dyn/cm$ and glass water contact angle simeq0^circ$simeq0^{circ}$) with 30^circ$30^{circ}$ inclined with vertical.
The length of water risen in the capillary is cm.
(Take g=9.8~m/s^2$g=9.8~m/s^{2}$) [cite: 83, 84]
A.frac825$\frac{82}{5}$ [cite: 85]
B.frac572$\frac{57}{2}$ [cite: 88]
C.frac715$\frac{71}{5}$ [cite: 86]
D.frac685$\frac{68}{5}$ [cite: 88]
Solution & Explanation
### Related Formula
h = frac2Tcosthetarho g r$h = \frac{2T\cos\theta}{\rho g r}$ [cite: 694]
### Core Logic
First, calculate the vertical height h$h$ using CGS units: [cite: 694]
* T = 70\ textdyn/cm$T = 70\ \text{dyn/cm}$ [cite: 83]
* costheta = cos 0^circ = 1$\cos\theta = \cos 0^{\circ} = 1$ [cite: 83]
* rho = 1\ textg/cm^3$\rho = 1\ \text{g/cm}^3$ [cite: 694]
* g = 980\ textcm/s^2$g = 980\ \text{cm/s}^2$ [cite: 694]
* r = 0.1\ textmm = 10^-2\ textcm$r = 0.1\ \text{mm} = 10^{-2}\ \text{cm}$ [cite: 83, 694]
h = frac2 times 70 times 11 times 980 times 10^-2 = frac1409.8 = frac1007\ textcm$h = \frac{2 \times 70 \times 1}{1 \times 980 \times 10^{-2}} = \frac{140}{9.8} = \frac{100}{7}\ \text{cm}$ [cite: 694]
Since the tube is inclined at 30^circ$30^{\circ}$ to the vertical, the angle it makes with the horizontal is 60^circ$60^{\circ}$[cite: 83, 694]. The relationship between vertical height h$h$ and slant length l$l$ is: [cite: 694]
sin 60^circ = frachl implies l = frachsin 60^circ = frac2hsqrt3$\sin 60^{\circ} = \frac{h}{l} \implies l = \frac{h}{\sin 60^{\circ}} = \frac{2h}{\sqrt{3}}$ [cite: 694]
l = frac1007 times frac2sqrt3 = frac2007sqrt3 approx 16.49\ textcm$l = \frac{100}{7} \times \frac{2}{\sqrt{3}} = \frac{200}{7\sqrt{3}} \approx 16.49\ \text{cm}$ [cite: 696, 697, 698]
Checking option values, frac825 = 16.4\ textcm$\frac{82}{5} = 16.4\ \text{cm}$, which is closest[cite: 85, 698].
### Pattern Recognition
When a capillary tube is tilted, the vertical height of the fluid column remains constant to maintain hydrostatic pressure balance[cite: 694]. Thus, the length of the liquid along the slant always scales as l = frachcosalpha$l = \frac{h}{\cos\alpha}$ where alpha$\alpha$ is the tilt angle relative to the vertical line[cite: 83, 694].
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Capillarity diagram for Q8 - JEE Main 2025 Evening
More Mechanical Properties of Fluids Previous-Year Questions
Q42025Surface Tension and Surface Energy
Two water drops each of radius r$r$ coalesce to form a bigger drop. If T$T$ is the surface tension, the surface energy released in this process is:
A.4pi r^2 T left[2 - 2^frac23right]$4\pi r^2 T \left[2 - 2^{\frac{2}{3}}\right]$
B.4pi r^2 T left[2 - 2^frac13right]$4\pi r^2 T \left[2 - 2^{\frac{1}{3}}\right]$
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)$U = T \cdot A = T \cdot (4\pi R^2)$
2. Conservation of Volume during coalescence of drops:
2 times left(frac43pi r^3right) = frac43pi R^3$2 \times \left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3$
### Core Logic
When two drops of radius r$r$ coalesce into a single larger drop of radius R$R$, volume is conserved:
R^3 = 2r^3 implies R = 2^1/3 r$R^3 = 2r^3 \implies R = 2^{1/3} r$
- Initial surface area of the two separate drops:
A_i = 2 times 4pi r^2 = 8pi r^2$A_i = 2 \times 4\pi r^2 = 8\pi r^2$
- 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$A_f = 4\pi R^2 = 4\pi (2^{1/3} r)^2 = 4\pi r^2 2^{2/3}$
- Surface energy released:
Delta E = U_i - U_f = T(A_i - A_f)$\Delta E = U_i - U_f = T(A_i - A_f)$Delta E = T left( 8pi r^2 - 4pi r^2 2^2/3 right) = 4pi r^2 T left[ 2 - 2^2/3 right]$\Delta E = T \left( 8\pi r^2 - 4\pi r^2 2^{2/3} \right) = 4\pi r^2 T \left[ 2 - 2^{2/3} \right]$
### Pattern Recognition
Sees: Coalescence of N$N$ identical drops.
Trap: Forgetting to conserve volume first, or confusing initial and final surface areas.
Shortcut: Energy released when N$N$ drops coalesce into one big drop is:
Delta E = 4pi r^2 T left[ N - N^2/3 right]$\Delta E = 4\pi r^2 T \left[ N - N^{2/3} \right]$
Here, substituting N = 2$N = 2$ directly gives 4pi r^2 T left[ 2 - 2^2/3 right]$4\pi r^2 T \left[ 2 - 2^{2/3} \right]$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q212025Hydrostatic Pressure
A vessel with square cross-section and height of 6mathrm~m$6\mathrm{~m}$ is vertically partitioned. A small window of 100mathrm~cm^2$100\mathrm{~cm}^2$ with hinged door is fitted at a depth of 3mathrm~m$3\mathrm{~m}$ 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$1.5\times 10^{3}\mathrm{~kg/m}^{3}$. 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$= 10\mathrm{~m/s}^2$)
Numerical Answer.Answer: 150 to 150
Solution
### Related Formula
P = P_0 + rho g h$P = P_0 + \rho g h$F = Delta P cdot A$F = \Delta P \cdot A$
### Core Logic
Let's analyze the pressure acting on the window at depth h = 3mathrm~m$h = 3\mathrm{~m}$ from both sides:
- Water side (density rho_w = 1.0 times 10^3mathrm~kg/m^3$\rho_w = 1.0 \times 10^3\mathrm{~kg/m}^3$):
P_w = P_0 + rho_w g h$P_w = P_0 + \rho_w g h$
- Denser liquid side (density rho_l = 1.5 times 10^3mathrm~kg/m^3$\rho_l = 1.5 \times 10^3\mathrm{~kg/m}^3$):
P_l = P_0 + rho_l g h$P_l = P_0 + \rho_l g h$
The net pressure difference pushing on the window is:
Delta P = P_l - P_w = (rho_l - rho_w) g h$\Delta P = P_l - P_w = (\rho_l - \rho_w) g h$Delta P = (1.5 times 10^3 - 1.0 times 10^3) times 10 times 3 = 500 times 30 = 15000mathrm~N/m^2$\Delta P = (1.5 \times 10^3 - 1.0 \times 10^3) \times 10 \times 3 = 500 \times 30 = 15000\mathrm{~N/m}^2$
The window has area A = 100mathrm~cm^2 = 100 times 10^-4mathrm~m^2 = 10^-2mathrm~m^2$A = 100\mathrm{~cm}^2 = 100 \times 10^{-4}\mathrm{~m}^2 = 10^{-2}\mathrm{~m}^2$.
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$F = \Delta P \cdot A = 15000 \times 10^{-2} = 150\mathrm{~N}$
### Step 1: Final Conclusion
The required force is $
### 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 $.
### 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 $. 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
Q2025Work Done by Gravity in Connecting Vessels
Two cylindrical vessels of equal cross sectional area of 2mathrm~m^2$2\mathrm{~m}^{2}$ contain water upto height 10mathrm~m$10\mathrm{~m}$ and 6mathrm~m$6\mathrm{~m}$, 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$10^{3}\mathrm{~kg/m}^{3}$ and g=10mathrm~m/s^2$g=10\mathrm{~m/s}^{2}$)
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$U$ of a liquid column of mass m$m$ and height h$h$ is evaluated relative to its bottom by placing its total mass at its center of mass (h/2$h/2$):
U = m g left(frach2right) = (rho A h) g left(frach2right) = frac12 rho A g h^2$U = m g \left(\frac{h}{2}\right) = (\rho A h) g \left(\frac{h}{2}\right) = \frac{1}{2} \rho A g h^2$
Work done by the force of gravity (W$W$) equals the negative change in potential energy:
W = -Delta U = U_i - U_f$W = -\Delta U = U_i - U_f$
### 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$h_f = \frac{10 + 6}{2} = 8\mathrm{~m}$
### Step 1: Calculate Initial Potential Energy (U_i$U_i$)
Let the reference level U=0$U=0$ be at the bottom:
U_i = U_1 + U_2 = frac12 rho A g h_1^2 + frac12 rho A g h_2^2$U_i = U_1 + U_2 = \frac{1}{2} \rho A g h_1^2 + \frac{1}{2} \rho A g h_2^2$U_i = frac12 rho A g left(10^2 + 6^2right) = frac12 rho A g (100 + 36) = 68 rho A g$U_i = \frac{1}{2} \rho A g \left(10^2 + 6^2\right) = \frac{1}{2} \rho A g (100 + 36) = 68 \rho A g$Work Done by Gravity in Connecting Vessels
### Step 2: Calculate Final Potential Energy (U_f$U_f$)
Both vessels equalize to h_f = 8mathrm~m$h_f = 8\mathrm{~m}$:
U_f = 2 times left[ frac12 rho A g h_f^2 right] = rho A g (8^2) = 64 rho A g$U_f = 2 \times \left[ \frac{1}{2} \rho A g h_f^2 \right] = \rho A g (8^2) = 64 \rho A g$
### Step 3: Work Done by Gravity (W$W$)
W = U_i - U_f = 68 rho A g - 64 rho A g = 4 rho A g$W = U_i - U_f = 68 \rho A g - 64 \rho A g = 4 \rho A g$
Substitute the given values (
ho = 10^3mathrm~kg/m^3$
ho = 10^3\mathrm{~kg/m}^3$, A = 2mathrm~m^2$A = 2\mathrm{~m}^2$, g = 10mathrm~m/s^2$g = 10\mathrm{~m/s}^2$):
W = 4 times 10^3 times 2 times 10 = 8 times 10^4mathrm~J$W = 4 \times 10^3 \times 2 \times 10 = 8 \times 10^4\mathrm{~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 $
### 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:
$. Applying this directly:
$Delta U = frac14 times 10^3 times 2 times 10 times (10 - 6)^2 = 5000 times 16 = 8 times 10^4mathrm~J$\Delta U = \frac{1}{4} \times 10^3 \times 2 \times 10 \times (10 - 6)^2 = 5000 \times 16 = 8 \times 10^4\mathrm{~J}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
Q122025Viscosity and Terminal Velocity
A solid steel ball of diameter 3.6mathrm~mm$3.6\mathrm{~mm}$ acquired terminal velocity 2.45times10^-2mathrm~m/s$2.45\times10^{-2}\mathrm{~m/s}$ while falling under gravity through an oil of density 925mathrm~kg~m^-3$925\mathrm{~kg~m}^{-3}$. Take density of steel as 7825mathrm~kg~m^-3$7825\mathrm{~kg~m}^{-3}$ and g as 9.8mathrm~m/s^2$9.8\mathrm{~m/s}^{2}$. 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$v_t$ 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$v_t = \frac{2}{9} \frac{r^2 g (\rho_{\text{body}} - \rho_{\text{fluid}})}{\eta}$
Hence, the viscosity coefficient eta$\eta$ is:
eta = frac29 fracr^2 g (rho_textbody - rho_textfluid)v_t$\eta = \frac{2}{9} \frac{r^2 g (\rho_{\text{body}} - \rho_{\text{fluid}})}{v_t}$
### Core Logic
Given parameters:
- Diameter d = 3.6mathrm~mm Rightarrow$d = 3.6\mathrm{~mm} \Rightarrow$ radius r = 1.8mathrm~mm = 1.8 times 10^-3mathrm~m$r = 1.8\mathrm{~mm} = 1.8 \times 10^{-3}\mathrm{~m}$
- Terminal velocity v_t = 2.45 times 10^-2mathrm~m/s$v_t = 2.45 \times 10^{-2}\mathrm{~m/s}$
- Liquid density rho_textfluid = 925mathrm~kg/m^3$\rho_{\text{fluid}} = 925\mathrm{~kg/m}^3$
- Steel density rho_textbody = 7825mathrm~kg/m^3$\rho_{\text{body}} = 7825\mathrm{~kg/m}^3$
- Acceleration due to gravity g = 9.8mathrm~m/s^2$g = 9.8\mathrm{~m/s}^2$
### 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$\eta = \frac{2}{9} \times \frac{(1.8 \times 10^{-3})^2 \times 9.8 \times (7825 - 925)}{2.45 \times 10^{-2}}$eta = frac29 times frac3.24 times 10^-6 times 9.8 times 69002.45 times 10^-2$\eta = \frac{2}{9} \times \frac{3.24 \times 10^{-6} \times 9.8 \times 6900}{2.45 \times 10^{-2}}$
### Step 2: Solve the numeric calculation
eta = frac2 times 0.36 times 10^-6 times 9.8 times 69002.45 times 10^-2$\eta = \frac{2 \times 0.36 \times 10^{-6} \times 9.8 \times 6900}{2.45 \times 10^{-2}}$eta = frac0.72 times 9.8 times 6900 times 10^-62.45 times 10^-2$\eta = \frac{0.72 \times 9.8 \times 6900 \times 10^{-6}}{2.45 \times 10^{-2}}$eta = frac48700.8 times 10^-62.45 times 10^-2 = frac0.04870080.0245 approx 1.99mathrm~Pacdot s$\eta = \frac{48700.8 \times 10^{-6}}{2.45 \times 10^{-2}} = \frac{0.0487008}{0.0245} \approx 1.99\mathrm{~Pa\cdot s}$
### Pattern Recognition
Be careful when converting diameter to radius ($
### 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
Q222025Excess 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$n$ times the volume of the bubble B, then the value of n$n$ 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$\Delta P = \frac{4T}{R} \Rightarrow R \propto \frac{1}{\Delta P}$
The volume V$V$ of a spherical bubble of radius R$R$ is:
V = frac43pi R^3 Rightarrow V propto R^3 propto left(frac1Delta Pright)^3$V = \frac{4}{3}\pi R^3 \Rightarrow V \propto R^3 \propto \left(\frac{1}{\Delta P}\right)^3$
### Core Logic
Given state:
Delta P_A = frac12 Delta P_B Rightarrow fracDelta P_BDelta P_A = 2$\Delta P_A = \frac{1}{2} \Delta P_B \Rightarrow \frac{\Delta P_B}{\Delta P_A} = 2$
### Step 1: Calculate the ratio of radii
fracR_AR_B = fracDelta P_BDelta P_A = 2$\frac{R_A}{R_B} = \frac{\Delta P_B}{\Delta P_A} = 2$
### Step 2: Calculate volume scaling parameter (n$n$)
V_A = n V_B Rightarrow n = fracV_AV_B = left(fracR_AR_Bright)^3 = (2)^3 = 8$V_A = n V_B \Rightarrow n = \frac{V_A}{V_B} = \left(\frac{R_A}{R_B}\right)^3 = (2)^3 = 8$
### Pattern Recognition
Volume scale factors depend on the cube of linear scale factors (V propto R^3$V \propto R^3$). Since radius is inversely proportional to excess pressure, a halving of excess pressure leads to doubling of radius, scaling volume by 2^3 = 8$2^3 = 8$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Mechanical Properties of Fluids
More Mechanical Properties of Fluids Questions — jee_main_2025_07_april_evening
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