Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between $P$ and $Q$ is $15\ \text{N m}^{-2}$. The area of cross-section at $P$ and $Q$ are $40\ \text{cm}^2$ and $20\ \text{cm}^2$, respectively. The rate of flow of water through the pipe, in $\text{cm}^3\text{s}^{-1}$, is :
[Take density of water $= 1000\ \text{kg m}^{-3}$]
In the measurement of viscosity of liquids using terminal velocity experiment, spherical balls of same radius but having different densities are used. The variation of the terminal velocity ($v$) with the ratio of density of spherical ball ($\sigma$) to density of the liquid ($\rho$), is best represented by :
Step 1. In terms of $\dfrac{\sigma}{\rho}$. $$v = \frac{2r^2\rho g}{9\eta}\left(\frac{\sigma}{\rho} - 1\right)$$
Step 2. Shape. A straight line, zero at $\dfrac{\sigma}{\rho} = 1$ and rising beyond it (lighter balls do not sink): graph (1).
Answer. Graph (1) (Option A).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-003NEETNEET 2025Top questionHard
Consider a water tank shown in the figure. It has one wall at $x = L$ and can be taken to be very wide in the $z$ direction. When filled with a liquid of surface tension $S$ and density $\rho$, the liquid surface makes angle $\theta_0$ ($\theta_0 << 1$) with the $x$-axis at $x = L$. If $y(x)$ is the height of the surface then the equation for $y(x)$ is :
(take $\theta(x) = \sin\theta(x) = \tan\theta(x) = \dfrac{dy}{dx}$, $g$ is the acceleration due to gravity)
Concept. Balance the vertical forces on a thin strip of the surface: the net surface-tension pull against the pressure deficit $\rho gy$ below a surface at height $y$.
Step 1. Surface tension. Net vertical component on a strip $dx$: $$S\sin\theta(x + dx) - S\sin\theta(x) = S\frac{d^2y}{dx^2}dx$$
Step 2. Pressure. Just below the surface at height $y$ the liquid is $\rho gy$ below atmospheric, so the net downward push is $\rho gy\,dx$.
A balloon is made of a material of surface tension $S$ and its inflation outlet (from where gas is filled in it) has small area $A$. It is filled with a gas of density $\rho$ and takes a spherical shape of radius $R$. When the gas is allowed to flow freely out of it, its radius $r$ changes from $R$ to $0$ (zero) in time $T$. If the speed $v(r)$ of gas coming out of the balloon depends on $r$ as $r^a$ and $T \propto S^\alpha A^\beta \rho^\gamma R^\delta$ then
A thin flat circular disc of radius $4.5$ cm is placed gently over the surface of water. If surface tension of water is $0.07\ \text{N m}^{-1}$, then the excess force required to take it away from the surface is :
The amount of energy required to form a soap bubble of radius $2$ cm from a soap solution is nearly : (surface tension of soap solution $= 0.03\ \text{N m}^{-1}$)
Concept. A venturimeter finds the flow speed from the pressure difference between the wide and narrow parts.
Step 1. Principle. That pressure difference follows from Bernoulli's principle (with continuity).
Answer. Bernoulli's principle (Option B).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-008NEETNEET 2021Top questionMedium
The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is $\dfrac{d}{2}$, then the viscous force acting on the ball will be :
A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary is $5$ g. Another capillary tube of radius $2$r is immersed in water. The mass of water that will rise in this tube is :
Concept. $h = \dfrac{2S\cos\theta}{r\rho g} \propto \dfrac1r$, so the raised mass $\pi r^2h\rho \propto r$.
Step 1. Double $r$. The mass doubles: $10$ g.
Answer. $10.0$ g (Option C).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-010NEETJEE MainEasy
The absolute pressure at a depth of $10$ m in a freshwater lake is about (atmospheric pressure $= 10^5$ Pa, density of water $= 1000$ kg m$^{-3}$, $g = 10$ m s$^{-2}$)
In a hydraulic lift the small and large pistons have areas $10\ \text{cm}^2$ and $500\ \text{cm}^2$. A force of $200$ N on the small piston can support a mass of ($g = 10$ m s$^{-2}$)
Concept. Pascal's law: equal pressure on both pistons, so $F_2 = F_1\dfrac{A_2}{A_1}$.
Step 1. Force. $200 \times 50 = 10^4$ N.
Step 2. Mass. $\dfrac{10^4}{10} = 1000$ kg.
Answer. $1000$ kg (Option B).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-012NEETJEE MainMedium
A U-tube contains water. Oil is poured into one arm until the oil column is $12$ cm long, while the water level in the other arm stands $10$ cm above the oil–water interface. The density of the oil is about
A wooden block floats in water with $\dfrac{2}{3}$ of its volume submerged. In a liquid X it floats with $\dfrac{5}{6}$ of its volume submerged. The density of liquid X is
Concept. Loss of weight $=$ weight of an equal volume of water; relative density $= \dfrac{\text{weight in air}}{\text{loss of weight}}$.
Step 1. Loss. $60 - 40 = 20$ N.
Step 2. Relative density. $\dfrac{60}{20} = 3$.
Answer. $3$ (Option B).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-016NEETJEE MainEasy
Water flows through a horizontal pipe whose diameter narrows from $4$ cm to $2$ cm. If the speed in the wide part is $1$ m s$^{-1}$, the speed in the narrow part is
Water flows through a horizontal pipe. At point P its speed is $2$ m s$^{-1}$ and at point Q it is $4$ m s$^{-1}$. The pressure at P exceeds that at Q by
A tank standing on the ground is filled with water to a height $H$. A small hole is to be made in its side so that the water jet lands as far from the tank as possible. The hole should be at a height above the ground of
Water (density $1000$ kg m$^{-3}$, viscosity $10^{-3}$ Pa s) flows at $0.1$ m s$^{-1}$ through a pipe of diameter $2$ cm. Find the Reynolds number of the flow.
Concept. Reynolds number $Re = \dfrac{\rho vD}{\eta}$.
Step 1. Substitute. $$Re = \frac{1000 \times 0.1 \times 0.02}{10^{-3}} = 2000$$ (Close to the onset of turbulence.)
Answer. $2000$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-022NEETJEE MainMedium
A soap bubble in air and an air bubble inside the same soap solution have the same radius. The ratio of the excess pressure inside the soap bubble to that inside the air bubble is
Step 2. Solve. $R^2 = 9 + 16 = 25 \Rightarrow R = 5$ cm.
Answer. $5$ cm (Option A).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-024NEETJEE MainMedium
Water (surface tension $0.07$ N m$^{-1}$, contact angle $0^\circ$) rises in a glass capillary of radius $0.5$ mm. The height of rise is ($g = 10$ m s$^{-2}$)
Answer: (C) water rises to the top and stops, with a meniscus of larger radius
Concept. $hR = \dfrac{2T}{\rho g}$ is fixed, where $R$ is the meniscus radius of curvature.
Step 1. Short tube. With $h$ limited to $4$ cm, the meniscus flattens (its radius grows to $\dfrac64$ of normal). Nothing supplies energy for a continuous flow, so there is no overflow.
Answer. Water rises to the top and stops, with a meniscus of larger radius (Option C).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-026NEETJEE MainHard
A liquid drop of radius $R$ and surface tension $T$ is split into $1000$ identical small drops. The work done is
Answer: (A) falls below the outside level, because the contact angle is obtuse
Concept. $h = \dfrac{2T\cos\theta}{\rho gr}$; an obtuse contact angle makes $h$ negative.
Step 1. Mercury on glass. $\theta \approx 140°$, $\cos\theta < 0$: the level is depressed and the meniscus is convex.
Answer. Falls below the outside level, because the contact angle is obtuse (Option A).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-028JEE MainJEE Main 2026 (8 Apr, Shift 2)Medium
A liquid of density $600\ \text{kg/m}^3$ flowing steadily in a tube of varying cross-section. The cross-section at a point $A$ is $1.0\ \text{cm}^2$ and that at $B$ is $20\ \text{mm}^2$. Both the points $A$ and $B$ are in same horizontal plane, the speed of the liquid at $A$ is $10\ \text{cm/s}$. The difference in pressures at $A$ and $B$ points is ______ Pa.
Q 11-09-029JEE MainJEE Main 2026 (8 Apr, Shift 2)Medium
A spherical liquid drop of radius $R$ acquires the terminal velocity $v_1$ when falls through a gas of viscosity $\eta$. Now the drop is broken into $64$ identical droplets and each droplet acquires terminal velocity $v_2$ falling through the same gas. The ratio of terminal velocities $v_1/v_2$ is ______.
Q 11-09-030JEE MainJEE Main 2026 (6 Apr, Shift 2)Hard
A cylindrical vessel of $40$ cm radius is completely filled with water and its capacity is $528\ \text{dm}^3$ (dm : decimeter). The vessel is placed on a solid block of exactly same height as vessel. If a small hole is made at $70$ cm below the top of water level, then horizontal range of water falling on the ground in the beginning is ______ cm.
Concept. Efflux speed $v=\sqrt{2gh}$ at depth $h$ below the surface; then the water is a horizontal projectile. Range $=2\sqrt{h\,H}$, with $H$ the hole's height above the ground.
Step 1. Vessel height. $V=\pi r^2H_v$: $0.528=\dfrac{22}{7}(0.16)H_v\Rightarrow H_v=1.05$ m; the block is also $1.05$ m.
Step 2. Hole. $h=0.70$ m below the surface; height above the ground $=2.10-0.70=1.40$ m.
Step 3. Range. $2\sqrt{0.70\times1.40}=2\sqrt{0.98}=1.4\sqrt2$ m $=140\sqrt2$ cm.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-031JEE MainJEE Main 2026 (5 Apr, Shift 2)Easy
Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is ______ ($S$ is the surface tension of mercury).
Step 3. Energy released. $S(32-16)\pi r^2=16\pi r^2S$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-032JEE MainJEE Main 2026 (4 Apr, Shift 2)Easy
A water spray gun is attached to a hose of cross sectional area $30\ \text{cm}^2$. The gun comprises of $10$ perforations each of cross sectional area of $15\ \text{mm}^2$. If the water flows in the hose with the speed of $50\ \text{cm/s}$, calculate the speed at which the water flows out from each perforation. (Neglect any edge effects)
Q 11-09-033JEE MainJEE Main 2026 (4 Apr, Shift 1)Easy
The surface tension of a soap solution is $3.5 \times 10^{-2}$ N/m. The work required to increase the radius of a soap bubble from $1$ cm to $2$ cm is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is ______. ($\pi = 22/7$)
Concept. A soap bubble has two surfaces: $W=2S\times4\pi(r_2^2-r_1^2)$.
Step 1. Substitute. $W=8\times\dfrac{22}{7}\times3.5\times10^{-2}\times(4-1)\times10^{-4}=264\times10^{-6}$ J.
Answer. $264$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-034JEE MainJEE Main 2026 (2 Apr, Shift 1)Medium
A liquid drop of diameter $2$ mm breaks into $512$ droplets. The change in surface energy is $\alpha\times10^{-6}$ J. The value of $\alpha$ is ______. (Take surface tension of liquid $=0.08$ N/m)
Concept. Volume is conserved: $n$ drops of radius $r=R/n^{1/3}$. The surface energy change is $S\times\Delta A$.
Step 1. Radii. $R=1$ mm, $n=512$: $r=R/8$.
Step 2. Area increase. $4\pi(nr^2-R^2)=4\pi R^2(8-1)=28\pi\times10^{-6}$ m$^2$.
Step 3. Energy. $0.08\times28\pi\times10^{-6}\approx7.0\times10^{-6}$ J.
Answer. $7$ (Option B).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-035JEE MainJEE Main 2026 (2 Apr, Shift 1)Easy
A tub is filled with water and a wooden cube $10\ \text{cm}\times10\ \text{cm}\times10\ \text{cm}$ 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 $1\ \text{g/cm}^3$ and density of wood as $0.4\ \text{g/cm}^3$)
Q 11-09-036JEE MainJEE Main 2026 (2 Apr, Shift 2)Easy
The surface tension of a soap bubble is $0.03$ N/m. The work done in increasing the diameter of bubble from $2$ cm to $6$ cm is $\alpha\pi\times10^{-4}$ J. The value of $\alpha$ is ______. (Take $\pi=3.14$)
Concept. Two surfaces: $W=8\pi S(r_2^2-r_1^2)$, with the radii half the diameters.
Step 1. Substitute. $W=8\pi(0.03)(9-1)\times10^{-4}=1.92\pi\times10^{-4}$ J.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-037JEE MainJEE Main 2026 (2 Apr, Shift 2)Medium
If an air bubble of diameter $2$ mm rises steadily through a liquid of density $2000\ \text{kg/m}^3$ at a rate of $0.5$ cm/s, then the coefficient of viscosity of liquid is ______ Poise. (Take $g=10\ \text{m/s}^2$)
Concept. A rising air bubble (negligible mass) at terminal speed: buoyancy $=$ viscous drag, $\eta=\dfrac{2r^2\rho g}{9v}$ with $\rho$ the liquid density.
Q 11-09-038JEE MainJEE Main 2026 (24 Jan, Shift 2)Easy
A cubical block of density $\rho_b = 600\ \text{kg/m}^3$ floats in a liquid of density $\rho_e = 900\text{kg/m}^3$. If the height of block is $H = 8.0$ cm then height of the submerged part is
Concept. A floating body: weight $=$ buoyancy, so the submerged fraction is $\dfrac{\rho_b}{\rho_l}$.
Step 1. Submerged height. $8.0\times\dfrac{600}{900}=5.3$ cm.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-039JEE MainJEE Main 2026 (24 Jan, Shift 2)Medium
A soap bubble of surface tension $0.04\ \text{N/m}$ is blown to a diameter of 7 cm . If $(15000 - x)\mu\text{J}$ of work is done in blowing it further to make its diameter 14 cm , then the value of $x$ is
Step 2. Solve. $15000 - x = 3696 \Rightarrow x = 11304$.
Answer. $11304$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-040JEE MainJEE Main 2026 (24 Jan, Shift 1)Easy
Sixty four rain drops of radius 1 mm each falling down with a terminal velocity of $10\ \text{cm/s}$ coalesce to form a bigger drop. The terminal velocity of bigger drop is
Q 11-09-041JEE MainJEE Main 2025 (22 Jan, Shift 1)Easy
Two soap bubbles of radius $2\ \text{cm}$ and $4\ \text{cm}$, respectively, are in contact with each other. The radius of curvature of the common surface, in cm, is ______.
Concept. The smaller bubble has the higher pressure, so the common film bulges into the larger one. The film's excess pressure equals the difference of the two bubbles' excess pressures: $\dfrac{4T}{r}=\dfrac{4T}{r_1}-\dfrac{4T}{r_2}$.
Step 1. Radius of curvature. $$\frac1r=\frac1{r_1}-\frac1{r_2}=\frac12-\frac14=\frac14$$
Answer. $r=4$ cm.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-042JEE MainJEE Main 2025 (22 Jan, Shift 2)Easy
A small rigid spherical ball of mass $M$ is dropped in a long vertical tube containing glycerine. The velocity of the ball becomes constant after some time. If the density of glycerine is half of the density of the ball, then the viscous force acting on the ball will be (consider $g$ as acceleration due to gravity)
Q 11-09-043JEE MainJEE Main 2025 (22 Jan, Shift 2)Easy
A tube of length $L$ is shown in the figure. The radius of cross section at the point (1) is $2\ \text{cm}$ and at the point (2) is $1\ \text{cm}$, respectively. If the velocity of water entering at point (1) is $2\ \text{m/s}$, then velocity of water leaving the point (2) will be
Q 11-09-044JEE MainJEE Main 2025 (22 Jan, Shift 2)Medium
A tube of length $1\ \text{m}$ is filled completely with an ideal liquid of mass $2M$, and closed at both ends. The tube is rotated uniformly in a horizontal plane about one of its ends. If the force exerted by the liquid at the other end is $F$, then angular velocity of the tube is $\sqrt{\dfrac{F}{\alpha M}}$ in SI unit. The value of $\alpha$ is ______.
Concept. The liquid at distance $r$ needs a centripetal force; the end wall must supply the total, $\displaystyle\int\omega^2r\,dm$ over the whole liquid.
Step 1. Integrate ($\lambda=2M/1$ per metre). $F=\displaystyle\int_0^1\omega^2r\,(2M)\,dr=M\omega^2$.
Step 2. Solve. $\omega=\sqrt{\dfrac FM}$, so $\alpha=1$.
Answer. $1$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-045JEE MainJEE Main 2025 (23 Jan, Shift 1)Easy
Given below are two statements:
Statement I: Hot water flows faster than cold water.
Statement II: Soap water has higher surface tension as compared to fresh water.
In the light of the above statements, choose the correct answer from the options given below.
Answer: (A) Statement I is true but Statement II is false
Concept. Liquid viscosity falls with temperature; soap lowers surface tension.
Step 1. Statement I. Hot water is less viscous, so it flows faster → true.
Step 2. Statement II. Soap lowers surface tension → false.
Answer. Statement I is true but Statement II is false (Option A).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-046JEE MainJEE Main 2025 (23 Jan, Shift 2)Easy
Water flows in a horizontal pipe whose one end is closed with a valve. The reading of the pressure gauge attached to the pipe is $P_1$. The reading of the pressure gauge falls to $P_2$ when the valve is opened. The speed of water flowing in the pipe is proportional to
Q 11-09-047JEE MainJEE Main 2025 (23 Jan, Shift 2)Medium
An air bubble of radius $1.0\ \text{mm}$ is observed at a depth of $20\ \text{cm}$ below the free surface of a liquid having surface tension $0.095\ \text{J/m}^2$ and density $10^3\ \text{kg/m}^3$. The difference between the pressure inside the bubble and atmospheric pressure is ______ $\text{N/m}^2$. (Take $g = 10\ \text{m/s}^2$)
Q 11-09-048JEE MainJEE Main 2025 (24 Jan, Shift 1)Medium
The amount of work done to break a big water drop of radius $R$ into 27 small drops of equal radius is $10\ \text{J}$. The work done required to break the same big drop into 64 small drops of equal radius will be
Concept. Breaking a drop of radius $R$ into $n$ equal drops: each has $r=\dfrac{R}{n^{1/3}}$, and the area increases by $4\pi R^2(n^{1/3}-1)$. Work $=T\times$ increase in area.
Step 1. Into $27$ drops. $W_{27}=4\pi R^2T(3-1)=2\times4\pi R^2T=10$ J.
Step 2. Into $64$ drops. $W_{64}=4\pi R^2T(4-1)=3\times4\pi R^2T$.
Step 3. Ratio. $W_{64}=\dfrac32\times10=15$ J.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-049JEE MainJEE Main 2025 (24 Jan, Shift 1)Easy
An air bubble of radius $0.1\ \text{cm}$ lies at a depth of $20\ \text{cm}$ below the free surface of a liquid of density $1000\ \text{kg/m}^3$. If the pressure inside the bubble is $2100\ \text{N/m}^2$ greater than the atmospheric pressure, then the surface tension of the liquid in SI unit is (use $g = 10\ \text{m/s}^2$)
Concept. $v_t=\dfrac{2r^2(\rho-\sigma)g}{9\eta}$: quadratic in $r$, depends on $\eta$ (which depends on temperature) and on the liquid's density.
Step 1. Check. A true ($v\propto R^2$, a parabola); B false; C true; D true (the formula can give $\sigma$); E false ($\eta$ is a property of the liquid).
Answer. A, C and D only (Option C).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-051JEE MainJEE Main 2025 (28 Jan, Shift 1)Medium
Consider the following statements:
A. Surface tension arises due to extra energy of the molecules at the interior as compared to the molecules at the surface of a liquid.
B. As the temperature of a liquid rises, the coefficient of viscosity increases.
C. As the temperature of a gas increases, the coefficient of viscosity increases.
D. The onset of turbulence is determined by Reynolds number.
E. In a steady flow two streamlines never intersect.
Choose the correct answer from the options given below:
Concept. Check each statement against the basic facts of surface tension, viscosity and flow.
Step 1. Check each.
A: surface molecules have *extra* energy compared with interior ones (they have fewer neighbours). The statement has it backwards. False.
B: for liquids, viscosity *falls* as temperature rises. False.
C: for gases, viscosity rises with temperature. True.
D: turbulence sets in when the Reynolds number is large enough. True.
E: in steady flow the velocity at a point is unique, so streamlines cannot cross. True.
Answer. C, D, E only (Option A).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-052JEE MainJEE Main 2025 (28 Jan, Shift 2)Easy
A $400\ \text{g}$ solid cube having an edge of length $10\ \text{cm}$ floats in water. How much volume of the cube is outside the water? (Given: density of water $= 1000\ \text{kg m}^{-3}$)
Concept. A floating body displaces its own weight of liquid: $\rho_{\text{body}}V=\rho_wV_{\text{sub}}$.
Step 1. Submerged volume. The cube's mass is $400$ g, so it displaces $400$ g of water, i.e. $400$ cm$^3$.
Step 2. Volume outside. Total $1000$ cm$^3$; outside $=1000-400=600$ cm$^3$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-053JEE MainJEE Main 2025 (29 Jan, Shift 1)Easy
In a hydraulic lift, the surface area of the input piston is $6\ \text{cm}^2$ and that of the output piston is $1500\ \text{cm}^2$. If a $100\ \text{N}$ force is applied to the input piston to raise the output piston by $20\ \text{cm}$, then the work done is ______ kJ.
Concept. A hydraulic lift multiplies force but not work. The output force is $F_2=F_1\dfrac{A_2}{A_1}$, and the work done is $F_2\times$ the rise of the output piston (equal to $F_1\times$ the input piston's longer stroke).
Step 1. Output force. $F_2=100\times\dfrac{1500}{6}=25\,000$ N.
Q 11-09-054JEE MainJEE Main 2024 (1 Feb, Shift 1)Medium
A plane is in level flight at constant speed and each of its two wings has an area of $40\ \text{m}^2$. If the speed of the air is $180\ \text{km h}^{-1}$ over the lower wing surface and $252\ \text{km h}^{-1}$ over the upper wing surface, the mass of the plane is ______ kg. (Take air density to be $1\ \text{kg m}^{-3}$ and $g = 10\ \text{m s}^{-2}$)
Concept. Lift on the wings $=$ pressure difference $\times$ area, with $\Delta P=\tfrac12\rho(v_{\text{up}}^2-v_{\text{low}}^2)$ from Bernoulli. In level flight lift $=$ weight.
Step 1. Speeds. $180$ km/h $=50$ m/s, $252$ km/h $=70$ m/s.
Concept. $n$ equal drops merging give radius $R=n^{1/3}r$. Total area becomes $\dfrac{4\pi R^2}{n\cdot4\pi r^2}=\dfrac{n^{2/3}}{n}=n^{-1/3}$ of the original.
Step 1. Apply with $n=1000$: $1000^{-1/3}=\dfrac1{10}$.
Answer. The surface energy becomes $\dfrac1{10}$th (Option D).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-056JEE MainJEE Main 2024 (4 Apr, Shift 1)Medium
Given below are two statements:
Statement I: When the speed of a liquid is zero everywhere, the pressure difference between any two points depends on the equation $P_1 - P_2 = \rho g(h_2 - h_1)$.
Statement II: In the venturi tube shown, $2gh = v_1^2 - v_2^2$.
In the light of the above statements, choose the most appropriate answer from the options given below.
Answer: (B) Statement I is correct but Statement II is incorrect.
Concept. At rest, $P_1-P_2=\rho g(h_2-h_1)$ from hydrostatics. In a venturi tube, Bernoulli gives $P_1-P_2=\tfrac12\rho(v_2^2-v_1^2)$, and the manometer gives $P_1-P_2=\rho gh$.
Step 1. Statement I. The hydrostatic relation. Correct.
Step 2. Statement II. $\rho gh=\tfrac12\rho(v_2^2-v_1^2)\Rightarrow2gh=v_2^2-v_1^2$, not $v_1^2-v_2^2$ (the throat is faster). Incorrect.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-057JEE MainJEE Main 2024 (4 Apr, Shift 1)Medium
A soap bubble is blown to a diameter of $7\ \text{cm}$. $36960\ \text{erg}$ of work is done in blowing it further. If the surface tension of the soap solution is $40\ \text{dyne/cm}$, then the new radius is ______ cm. (Take $\pi = \dfrac{22}{7}$)
Answer: (A) Statement I is true but Statement II is false.
Concept. The contact angle depends on both the solid and the liquid. Capillary rise $h\propto\dfrac1r$.
Step 1. Statement I. True.
Step 2. Statement II. False: a narrower tube gives a higher rise.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-059JEE MainJEE Main 2024 (4 Apr, Shift 2)Medium
Mercury is filled in a tube of radius $2\ \text{cm}$ up to a height of $30\ \text{cm}$. The force exerted by the mercury on the bottom of the tube is ______ N. (Given: atmospheric pressure $= 10^5\ \text{N m}^{-2}$, density of mercury $= 1.36\times10^4\ \text{kg m}^{-3}$, $g = 10\ \text{m s}^{-2}$, $\pi = \dfrac{22}{7}$)
Answer: (C) Statement I is false but Statement II is true
Concept. $h\propto\cos\theta$: no rise or fall means $\cos\theta=0$, i.e. $\theta=90^\circ$, not $0^\circ$.
Step 1. Statement I. False.
Step 2. Statement II. True.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-061JEE MainJEE Main 2024 (5 Apr, Shift 2)Easy
A hydraulic press containing water has two arms with diameters $14\ \text{cm}$ and $1.4\ \text{cm}$. A force of $10\ \text{N}$ is applied on the surface of water in the thinner arm. The force required to be applied on the surface of water in the thicker arm to maintain equilibrium of the water is ______ N.
Q 11-09-062JEE MainJEE Main 2024 (6 Apr, Shift 1)Easy
A small ball of mass $m$ and density $\rho$ is dropped in a viscous liquid of density $\rho_0$. After some time, the ball falls with a constant velocity. The viscous force on the ball is
Concept. As above, $F_v=mg-\rho_0Vg=mg\left(1-\dfrac{\rho_0}{\rho}\right)$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-063JEE MainJEE Main 2024 (6 Apr, Shift 1)Easy
A big drop is formed by coalescing 1000 small droplets of water. The ratio of the surface energy of the 1000 droplets to that of the big drop is $\dfrac{10}{x}$. The value of $x$ is ______.
Q 11-09-064JEE MainJEE Main 2024 (6 Apr, Shift 2)Easy
The pressure inside a soap bubble is greater than the pressure outside by an amount (given: $R$ = radius of bubble, $S$ = surface tension of the soap solution)
Concept. Excess pressure across a curved surface is $\dfrac{2S}{R}$ per surface. A soap bubble has two surfaces (inside and outside of the film); a drop or an air bubble in liquid has one.
Answer. $\dfrac{4S}{R}$ (Option D).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-065JEE MainJEE Main 2024 (27 Jan, Shift 1)Easy
Given below are two statements:
Statement (I): Viscosity of gases is greater than that of liquids.
Statement (II): Surface tension of a liquid decreases due to the presence of insoluble impurities.
In the light of the above statements, choose the most appropriate answer from the options given below:
Answer. Statement I is incorrect but Statement II is correct (Option B).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-066JEE MainJEE Main 2024 (27 Jan, Shift 2)Easy
The reading of a pressure metre attached with a closed pipe is $4.5\times10^4\ \text{N m}^{-2}$. On opening the valve, water starts flowing and the reading of pressure metre falls to $2.0\times10^4\ \text{N m}^{-2}$. The velocity of water is found to be $\sqrt V\ \text{m s}^{-1}$. The value of $V$ is ______.
Concept. With the valve closed, the water is at rest and the gauge reads the full pressure. When it flows, Bernoulli says part of that pressure turns into kinetic energy per volume: $P_1-P_2=\tfrac12\rho v^2$.
Answer: (C) Statement I is true but Statement II is false
Concept. Surface tension falls as temperature rises, so $h\propto S$ is smaller in hot water.
Step 1. Check. Statement I is true; Statement II says the opposite, false.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-068JEE MainJEE Main 2024 (29 Jan, Shift 1)Easy
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wings are $70\ \text{m s}^{-1}$ and $65\ \text{m s}^{-1}$ respectively. If the wing area is $2\ \text{m}^2$, the lift of the wing is ______ N. (Given density of air $= 1.2\ \text{kg m}^{-3}$)
Concept. Air moves faster over the upper surface, so its pressure is lower there (Bernoulli). The pressure difference times the wing area is the lift: $F=\tfrac12\rho(v_u^2-v_l^2)A$.
Q 11-09-071JEE MainJEE Main 2024 (8 Apr, Shift 1)Easy
A liquid column of height $0.04\ \text{cm}$ balances the excess pressure of a soap bubble of certain radius. If the density of the liquid is $8\times10^3\ \text{kg m}^{-3}$ and the surface tension of the soap solution is $0.28\ \text{N m}^{-1}$, then the diameter of the soap bubble is ______ cm. (if $g = 10\ \text{m s}^{-2}$)
Q 11-09-072JEE MainJEE Main 2024 (8 Apr, Shift 2)Easy
A cube of ice floats partly in water and partly in kerosene oil (kerosene forms a layer above the water). The ratio of volume of ice immersed in water to that in kerosene oil is (specific gravity of kerosene oil $= 0.8$, specific gravity of ice $= 0.9$):
Q 11-09-073JEE MainJEE Main 2024 (8 Apr, Shift 2)Easy
Small water droplets of radius $0.01\ \text{mm}$ are formed in the upper atmosphere and are falling with a terminal velocity of $10\ \text{cm/s}$. Due to condensation, if 8 such droplets coalesce and form a larger drop, the new terminal velocity will be ______ cm/s.
Concept. Terminal velocity $\propto r^2$; volume is conserved.
Step 1. New radius. $R = 2r$.
Step 2. New speed. $4 \times 10 = 40$ cm/s.
Answer. $40$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-074JEE MainJEE Main 2024 (9 Apr, Shift 1)Medium
A sphere of relative density $\sigma$ and diameter $D$ has a concentric cavity of diameter $d$. The ratio of $\dfrac Dd$, if it just floats on water in a tank, is:
Concept. "Just floats" means fully submerged with weight equal to buoyancy. The material fills only the shell between $d$ and $D$; the buoyancy acts on the whole outer volume.
Q 11-09-075JEE MainJEE Main 2024 (9 Apr, Shift 2)Easy
The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. The ratio between the volume of the first and the second bubble is:
Concept. Excess pressure in a soap bubble $\dfrac{4S}{r}$, so $r\propto\dfrac1{\Delta P}$ and $V\propto r^3$.
Step 1. Radii. $r_1=\dfrac{r_2}{3}$.
Step 2. Volumes. $1:27$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-076JEE MainJEE Main 2024 (9 Apr, Shift 2)Medium
A spherical ball of radius $1\times10^{-4}\ \text{m}$ and density $10^5\ \text{kg/m}^3$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering the water the velocity of the ball does not change, then the value of $h$ is approximately: (The coefficient of viscosity of water is $9.8\times10^{-6}\ \text{N s/m}^2$)
Concept. The ball must enter the water at its terminal velocity; free fall gives $v^2 = 2gh$. (The ball is $100$ times denser than water, so buoyancy is neglected as the options intend.)
Q 11-09-077JEE MainJEE Main 2024 (30 Jan, Shift 2)Medium
A big drop is formed by coalescing 1000 small identical drops of water. If $E_1$ is the total surface energy of the 1000 small drops and $E_2$ is the surface energy of the single big drop, then $E_1 : E_2$ is $x : 1$, where $x = $ ______.
Concept. Surface energy $\propto$ area; volume is conserved.
Step 1. Big radius. $R = 10r$.
Step 2. Ratio. $\dfrac{1000r^2}{100r^2} = 10$.
Answer. $10$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-078JEE MainJEE Main 2024 (31 Jan, Shift 1)Easy
A small steel ball is dropped into a long cylinder containing glycerine. Which one of the following is the correct representation of the velocity–time graph for the transit of the ball?
Concept. At first only gravity (less buoyancy) acts, so the ball speeds up; drag grows with speed until it balances, and the speed levels off at $v_t$.
Step 1. Choose. A curve rising from zero and flattening to a constant value.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-079JEE MainJEE Main 2024 (31 Jan, Shift 2)Medium
A small spherical ball of radius $r$, falling through a viscous medium of negligible density, has terminal velocity $v$. Another ball of the same mass but of radius $2r$, falling through the same viscous medium, will have terminal velocity:
Concept. In a medium of negligible density there is no buoyancy, so at terminal velocity $mg=6\pi\eta rv$. Here the mass is fixed, so $v\propto\dfrac1r$ (the usual $r^2$ law assumed fixed density).
Step 1. Apply. Radius doubles at the same mass, so $v$ halves.
Answer. $\dfrac v2$ (Option A).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-080JEE MainJEE Main 2023 (29 Jan, Shift 1)Medium
Surface tension of a soap bubble is $2.0\times10^{-2}\ \text{N m}^{-1}$. Work done to increase the radius of soap bubble from $3.5\ \text{cm}$ to $7\ \text{cm}$ will be [Take $\pi=\frac{22}{7}$]
Q 11-09-081JEE MainJEE Main 2023 (29 Jan, Shift 2)Medium
A fully loaded boeing aircraft has a mass of $5.4\times10^5\ \text{kg}$. Its total wing area is $500\ \text{m}^2$. It is in level flight with a speed of $1080\ \text{km h}^{-1}$. If the density of air $\rho$ is $1.2\ \text{kg m}^{-3}$, the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in percentage will be ($g=10\ \text{m s}^{-2}$)
Q 11-09-082JEE MainJEE Main 2023 (29 Jan, Shift 2)Easy
A metal block of base area $0.20\ \text{m}^2$ is placed on a table, as shown in figure. A liquid film of thickness $0.25\ \text{mm}$ is inserted between the block and the table. The block is pushed by a horizontal force of $0.1\ \text{N}$ and moves with a constant speed. If the viscosity of the liquid is $5.0\times10^{-3}\ \text{Pl}$, the speed of block is ______ $\times10^{-3}\ \text{m s}^{-1}$.
Concept. Newton's law of viscosity for a thin film: the drag is $F=\eta A\dfrac{v}{d}$, where $\dfrac vd$ is the velocity gradient across the film. At constant speed the push equals the drag.
Step 1. Solve for $v$. $$v=\frac{Fd}{\eta A}=\frac{0.1\times0.25\times10^{-3}}{5\times10^{-3}\times0.2}=\frac{2.5\times10^{-5}}{10^{-3}}=0.025\ \text{m s}^{-1}$$
Answer. $25\times10^{-3}$ m s$^{-1}$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-083JEE MainJEE Main 2023 (30 Jan, Shift 1)Easy
The height of liquid column raised in a capillary tube of certain radius when dipped in liquid $A$ vertically is $5\ \text{cm}$. If the tube is dipped in a similar manner in another liquid $B$ of surface tension and density double the values of liquid $A$, the height of liquid column raised in liquid $B$ would be ______ m.
Concept. Capillary rise $h=\dfrac{2S\cos\theta}{r\rho g}\propto\dfrac{S}{\rho}$ for the same tube.
Step 1. Scale. $S$ and $\rho$ both doubled: $h$ unchanged, $5$ cm $=0.05$ m.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-085JEE MainJEE Main 2023 (24 Jan, Shift 2)Medium
A spherical ball of radius $1\ \text{mm}$ and density $10.5\ \text{g cc}^{-1}$ is dropped in glycerine of coefficient of viscosity $9.8$ poise and density $1.5\ \text{g cc}^{-1}$. Viscous force on the ball when it attains constant velocity is $3696\times10^{-x}$ N. The value of $x$ is ______. (Given, $g=9.8\ \text{m s}^{-2}$ and $\pi=\frac{22}{7}$)
Step 2. Compare. $3696 \times 10^{-7}$ N, so $x = 7$.
Answer. $7$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-086JEE MainJEE Main 2023 (25 Jan, Shift 2)Easy
A spherical drop of liquid splits into $1000$ identical spherical drops. If $u_i$ is the surface energy of the original drop and $u_f$ is the total surface energy of the resulting drops (ignoring evaporation), $\dfrac{u_f}{u_i}=\left(\dfrac{10}{x}\right)$. Then value of $x$ is ______.
Q 11-09-087JEE MainJEE Main 2023 (13 Apr, Shift 2)Medium
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: A spherical body of radius $(5\pm0.1)$ mm having a particular density is falling through a liquid of constant density. The percentage error in the calculation of its terminal velocity is $4\%$.
Reason R: The terminal velocity of the spherical body falling through the liquid is inversely proportional to its radius.
In the light of the above statements, choose the correct answer from the options given below.
Step 2. Reason. $v_t\propto r^2$, not $1/r$. False.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-088JEE MainJEE Main 2023 (15 Apr, Shift 1)Easy
There is an air bubble of radius $1.0\ \text{mm}$ in a liquid of surface tension $0.075\ \text{N m}^{-1}$ and density $1000\ \text{kg m}^{-3}$ at a depth of $10\ \text{cm}$ below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is ______ Pa ($g=10\ \text{m s}^{-2}$).
Concept. An air bubble inside a liquid has one surface. Pressure inside $=$ pressure of the liquid at that depth $+\dfrac{2T}{r}$, and the liquid's pressure is $P_0+\rho gh$.
Step 3. Total excess over atmosphere. $1000+150=1150$ Pa.
Answer. $1150$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-089JEE MainJEE Main 2023 (10 Apr, Shift 2)Easy
Figure below shows a liquid being pushed out of the tube by a piston having area of cross section $2.0\ \text{cm}^2$. The area of cross section at the outlet is $10\ \text{mm}^2$. If the piston is pushed at a speed of $4\ \text{cm s}^{-1}$, the speed of outgoing fluid is ______ $\text{cm s}^{-1}$.
Concept. Volume is conserved when drops merge: $8$ drops make one drop of radius $R=8^{1/3}r=2r$. Terminal speed $\propto r^2$.
Step 1. New radius. $R=2r$.
Step 2. New speed. $10\times2^2=40$ cm s$^{-1}$.
Answer. Option B. In general, $n$ drops merging: $v=n^{2/3}v_0$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-091JEE MainJEE Main 2023 (11 Apr, Shift 2)Medium
The surface tension of soap solution is $3.5\times10^{-2}\ \text{N m}^{-1}$. The amount of work done required to increase the radius of soap bubble from $10\ \text{cm}$ to $20\ \text{cm}$ is ______ $\times10^{-4}$ J. (take $\pi=\frac{22}{7}$)
Q 11-09-092JEE MainJEE Main 2023 (12 Apr, Shift 1)Medium
Glycerin of density $1.25\times10^3\ \text{kg m}^{-3}$ is flowing through the conical section of pipe. The area of cross-section of the pipe at its ends are $10\ \text{cm}^2$ and $5\ \text{cm}^2$ and pressure drop across its length is $3\ \text{N m}^{-2}$. The rate of flow of glycerin through the pipe is $x\times10^{-5}\ \text{m}^3\ \text{s}^{-1}$. The value of $x$ is ______.
Q 11-09-093JEE MainJEE Main 2023 (13 Apr, Shift 1)Medium
The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross-section. Cross-sectional areas at $A$ is $1.5\ \text{cm}^2$ and $B$ is $25\ \text{mm}^2$. If the speed of liquid at $B$ is $60\ \text{cm s}^{-1}$ then $(P_A-P_B)$ is (Given $P_A$ and $P_B$ are liquid pressures at $A$ and $B$ points. Density $\rho=1000\ \text{kg m}^{-3}$. $A$ and $B$ are on the axis of tube)
Concept. Continuity gives the speed at $A$; Bernoulli for a horizontal tube gives $P_A-P_B=\tfrac12\rho(v_B^2-v_A^2)$. Higher speed means lower pressure.
Q 11-09-094JEE MainJEE Main 2023 (6 Apr, Shift 1)Easy
A small ball of mass $M$ and density $\rho$ is dropped in a viscous liquid of density $\rho_0$. After some time, the ball falls with a constant velocity. What is the viscous force on the ball?
Q 11-09-095JEE MainJEE Main 2023 (6 Apr, Shift 2)Easy
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: When you squeeze one end of a tube to get toothpaste out from the other end, Pascal's principle is observed.
Reason R: A change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of its container.
In the light of the above statements, choose the most appropriate answer from the options given below.
Answer: (D) Both A and R are correct and R is the correct explanation of A
Concept. Squeezing a closed tube raises the pressure, and Pascal's law passes that rise to every part of the paste, pushing it out at the open end.
Answer. Both correct and R is the correct explanation of A (Option D).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-096JEE MainJEE Main 2023 (8 Apr, Shift 1)Easy
An air bubble of volume $1\ \text{cm}^3$ rises from the bottom of a lake $40$ m deep to the surface at a temperature of $12^\circ\text{C}$. The atmospheric pressure is $1\times10^5$ Pa, the density of water is $1000\ \text{kg m}^{-3}$ and $g=10\ \text{m s}^{-2}$. There is no difference of the temperature of water at the depth of $40$ m and on the surface. The volume of air bubble when it reaches the surface will be
Concept. At constant temperature (Boyle's law) $P_1V_1=P_2V_2$, with $P_1=P_0+\rho gh$ at the bottom.
Step 1. Pressure at 40 m. $10^5+1000\times10\times40=5\times10^5$ Pa.
Step 2. Volume at the top. $V_2=\dfrac{5\times10^5\times1}{10^5}=5$ cm$^3$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-097JEE MainJEE Main 2023 (8 Apr, Shift 1)Medium
An air bubble of diameter $6$ mm rises steadily through a solution of density $1750\ \text{kg m}^{-3}$ at the rate of $0.35\ \text{cm s}^{-1}$. The coefficient of viscosity of the solution (neglect density of air) is ______ Pa s (given $g=10\ \text{m s}^{-2}$).
Concept. A light air bubble rises at a steady speed when buoyancy equals viscous drag (its weight is negligible): $\tfrac43\pi r^3\rho g=6\pi\eta rv$. So $\eta=\dfrac{2r^2\rho g}{9v}$.
Step 1. Values. $r=3\times10^{-3}$ m, $v=3.5\times10^{-3}$ m s$^{-1}$.
Q 11-09-098JEE MainJEE Main 2023 (8 Apr, Shift 2)Easy
A hydraulic automobile lift is designed to lift vehicles of mass $5000$ kg. The area of cross section of the cylinder carrying load is $250\ \text{cm}^2$. The maximum pressure the smaller piston would have to bear is [Assume $g=10\ \text{m s}^{-2}$]
Answer: (D) Both Statement I and Statement II are true
Concept. In a liquid at rest, pressure depends only on depth (so it is the same at one level), and Pascal's law says applied pressure is passed on equally in all directions.
Answer. Both true (Option D).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-100JEE MainJEE Main 2023 (1 Feb, Shift 1)Medium
A mercury drop of radius $10^{-3}$ m is broken into $125$ equal size droplets. Surface tension of mercury is $0.45\ \text{N m}^{-1}$. The gain in surface energy is
Q 11-09-101JEE MainJEE Main 2023 (1 Feb, Shift 2)Easy
The surface of water in a water tank of cross section area $750\ \text{cm}^2$ on the top of a house is $h$ m above the tap level. The speed of water coming out through the tap of cross section area $500\ \text{mm}^2$ is $30\ \text{cm s}^{-1}$. At that instant, $\dfrac{dh}{dt}$ is $x\times10^{-3}\ \text{m s}^{-1}$. The value of $x$ will be ______.
Concept. At terminal velocity, viscous drag $6\pi\eta rv$ balances weight minus buoyancy $\tfrac43\pi r^3(\rho-\sigma)g$. So $v_t=\dfrac{2r^2(\rho-\sigma)g}{9\eta}\propto r^2$.
Answer. $v_t\propto r^2$ (Option B).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-103JEE MainJEE Main 2022 (25 Jun, Shift 1)Easy
The velocity of the upper layer of water in a river is $36\ \text{km h}^{-1}$. Shearing stress between horizontal layers of water is $10^{-3}\ \text{N m}^{-2}$. Depth of the river is ______ m. (Coefficient of viscosity of water is $10^{-2}\ \text{Pa s}$)
Concept. With a linear velocity profile from the bed (zero) to the surface, shear stress $=\eta\dfrac{v}{d}$.
Step 1. Speed. $36$ km/h $=10$ m/s.
Step 2. Depth. $d=\dfrac{\eta v}{\tau}=\dfrac{10^{-2}\times10}{10^{-3}}=100$ m.
Answer. $100$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-104JEE MainJEE Main 2022 (26 Jun, Shift 1)Medium
An ideal fluid of density $800\ \text{kg m}^{-3}$ flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from $a$ to $\dfrac a2$. The pressure difference between the wide and narrow sections of the pipe is $4100\ \text{Pa}$. At the wider section, the velocity of the fluid is $\dfrac{\sqrt x}{6}\ \text{m s}^{-1}$ for $x$ = ______. (Given $g = 10\ \text{m s}^{-2}$)
Q 11-09-105JEE MainJEE Main 2022 (26 Jun, Shift 2)Easy
If $\rho$ is the density and $\eta$ is the coefficient of viscosity of a fluid which flows with a speed $v$ in a pipe of diameter $d$, the correct formula for Reynolds number $R_e$ is
Concept. Reynolds number compares inertial and viscous effects: $R_e=\dfrac{\rho vd}{\eta}$ (dimensionless).
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-106JEE MainJEE Main 2022 (25 Jul, Shift 2)Medium
A drop of liquid of density $\rho$ is floating half immersed in a liquid of density $\sigma$ and surface tension $7.5\times10^{-4}\ \text{N cm}^{-1}$. The radius of the drop in cm will be: (Take $g = 10\ \text{m s}^{-2}$)
Step 2. Substitute. $S=7.5\times10^{-2}$ N/m: $r^2=\dfrac{0.0225}{2\rho-\sigma}$ m$^2$, so $r=\dfrac{0.15}{\sqrt{2\rho-\sigma}}$ m $=\dfrac{15}{\sqrt{2\rho-\sigma}}$ cm.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-107JEE MainJEE Main 2022 (26 Jul, Shift 1)Medium
A water drop of radius $1\ \text{cm}$ is broken into $729$ equal droplets. If the surface tension of water is $75\ \text{dyne cm}^{-1}$, then the gain in surface energy up to first decimal place will be [Given $\pi = 3.14$]
Concept. Same method: $729=9^3$, so each droplet has radius $\dfrac R9$, and the area grows by $4\pi R^2(9-1)$.
Step 1. Units. $T=75$ dyne cm$^{-1}=0.075$ N m$^{-1}$; $R=10^{-2}$ m.
Step 2. Energy gained. $$\Delta E=0.075\times4\times3.14\times10^{-4}\times8=7.5\times10^{-4}\ \text{J}$$
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-108JEE MainJEE Main 2022 (27 Jun, Shift 1)Medium
The velocity of a small ball of mass $m$ and density $d_1$, when dropped in a container filled with glycerine, becomes constant after some time. If the density of glycerine is $d_2$, then the viscous force acting on the ball, will be
Q 11-09-109JEE MainJEE Main 2022 (27 Jun, Shift 1)Medium
The area of cross-section of a large tank is $0.5\ \text{m}^2$. It has a narrow opening near the bottom having area of cross-section $1\ \text{cm}^2$. A load of $25$ kg is applied on the water at the top in the tank. Neglecting the speed of water in the tank, the velocity of the water, coming out of the opening at the time when the height of water level in the tank is $40$ cm above the bottom, will be ______ $\text{cm s}^{-1}$.
Concept. A load on the surface adds pressure $\dfrac{mg}{A}$; Bernoulli from the top to the hole: $v^2=2\left(gh+\dfrac{P_{\text{load}}}{\rho}\right)$.
Step 1. Extra pressure. $\dfrac{250}{0.5}=500$ Pa.
Step 2. Speed. $v^2=2(10\times0.4+0.5)=9\Rightarrow v=3$ m/s $=300$ cm/s.
Answer. $300$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-110JEE MainJEE Main 2022 (27 Jul, Shift 2)Easy
A spherical soap bubble of radius $3$ cm is formed inside another spherical soap bubble of radius $6$ cm. If the internal pressure of the smaller bubble of radius $3$ cm in the above system is equal to the internal pressure of the another single soap bubble of radius $r$ cm. The value of $r$ is ______
Concept. Each soap film adds $\dfrac{4T}{R}$ of excess pressure.
Step 1. Inside the small bubble. $$P = P_0 + \frac{4T}{6} + \frac{4T}{3} = P_0 + \frac{4T}{2}$$
Step 2. Equivalent bubble. $r = 2$ cm.
Answer. $2$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-111JEE MainJEE Main 2022 (28 Jul, Shift 1)Medium
The diameter of an air bubble which was initially $2$ mm, rises steadily through a solution of density $1750\ \text{kg m}^{-3}$ at the rate of $0.35\ \text{cm s}^{-1}$. Coefficient of viscosity of the solution is ______ (Assume mass of the bubble to be negligible) (Answer in Poise to the nearest integer)
Step 1. Substitute. $\eta=\dfrac{2\times10^{-6}\times1750\times10}{9\times3.5\times10^{-3}}=1.11$ Pa s $=11.1$ Poise.
Answer. $11$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-112JEE MainJEE Main 2021 (27 Jul, Shift 1)Medium
A light cylindrical vessel is kept on a horizontal surface. Area of the base is $A$. A hole of cross-sectional area $a$ is made just at its bottom side. The minimum coefficient of friction necessary to prevent sliding the vessel due to the impact force of the emerging liquid is
Concept. The emerging jet carries momentum away at rate $\rho av^2$, pushing the vessel back with that force. With $v^2=2gh$ the thrust is $2\rho agh$. Friction (at most $\mu Mg=\mu\rho Ahg$ for a light vessel) must hold it.
Q 11-09-113JEE MainJEE Main 2021 (27 Jul, Shift 2)Easy
A raindrop with radius $R = 0.2$ mm falls from a cloud at a height $h = 2000$ m above the ground. Assume that the drop is spherical throughout its fall and the force of buoyancy may be neglected, then the terminal speed attained by the raindrop is: [Density of water $f_w = 1000\ \text{kg m}^{-3}$ and density of air $f_a = 1.2\ \text{kg m}^{-3}$, $g = 10\ \text{m/s}^2$, coefficient of viscosity of air $= 1.8\times10^{-5}\ \text{N s m}^{-2}$]
Q 11-09-114JEE MainJEE Main 2021 (27 Jul, Shift 2)Easy
The water is filled up to a height of $12$ m in a tank having vertical sidewalls. A hole is made in one of the walls at a depth $h$ below the water level. The value of $h$ for which the emerging stream of water strikes the ground at the maximum range is ______ m.
Concept. Torricelli: water leaves a hole at depth $h$ with $v=\sqrt{2gh}$. It then falls the remaining height $H-h$ as a horizontal projectile. Range $R=v\sqrt{\dfrac{2(H-h)}{g}}=2\sqrt{h(H-h)}$.
Step 1. Maximise $h(H-h)$. This product is largest when $h=\dfrac H2$.
Step 2. Value. $h=\dfrac{12}{2}=6$ m, giving $R_{\max}=H=12$ m.
Answer. $6$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-115JEE MainJEE Main 2021 (24 Feb, Shift 1)Medium
A hydraulic press can lift 100 kg when a mass $m$ is placed on the smaller piston. It can lift ______ kg when the diameter of the larger piston is increased by 4 times and that of the smaller piston is decreased by 4 times keeping the same mass $m$ on the smaller piston.
Concept. Pascal's law: the pressure added at the small piston reaches the large piston unchanged. So $\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}$, i.e. the load lifted is $F_2=F_1\dfrac{A_2}{A_1}=F_1\left(\dfrac{D}{d}\right)^2$.
Step 1. The new area ratio. Large diameter $\times4$ (area $\times16$); small diameter $\div4$ (area $\div16$). The ratio $\dfrac{A_2}{A_1}$ grows by $16\times16=256$.
Step 2. New load (same $m$ on the small piston): $100\times256=25\,600$ kg.
Answer. $25\,600$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-116JEE MainJEE Main 2021 (18 Mar, Shift 2)Medium
Consider a water tank as shown in the figure. It's cross-sectional area is $0.4\ \text{m}^2$. The tank has an opening $B$ near the bottom whose cross-section area is $1\ \text{cm}^2$. A load of 24 kg is applied on the water at the top when the height of the water level is 40 cm above the bottom, the velocity of water coming out the opening $B$ is $v\ \text{m s}^{-1}$. The value of $v$, to the nearest integer, is ______. [Take the value of $g$ to be $10\ \text{m s}^{-2}$]
Q 11-09-118JEE MainJEE Main 2021 (26 Feb, Shift 1)Medium
A large number of water drops, each of radius $r$, combine to have a drop of radius $R$. If the surface tension is $T$ and mechanical equivalent of heat is $J$, the rise in heat energy per unit volume will be:
Concept. Energy released $= T \times$ decrease in area; express areas through the total volume $V$: $n4\pi r^2 = \dfrac{3V}{r}$, $4\pi R^2 = \dfrac{3V}{R}$.
Step 1. Energy released. $3TV\left(\dfrac1r - \dfrac1R\right)$.
Step 2. Heat per unit volume. $$\frac{Q}{V} = \frac{3T}{J}\left(\frac1r - \frac1R\right)$$
Concept. In vacuum the gas pressure inside a bubble is all excess pressure: $P=\dfrac{4T}{r}$. Isothermal combination conserves $PV$ (the amount of gas): $P_1V_1+P_2V_2=PV$.
Step 1. $PV$ for one bubble. $\dfrac{4T}{r}\times\tfrac43\pi r^3\propto r^2$.
Step 2. Conserve. $r_1^2+r_2^2=r^2$.
Answer. $r=\sqrt{r_1^2+r_2^2}$ (Option C). Equivalently, the total surface area is conserved.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-120JEE MainJEE Main 2021 (26 Aug, Shift 1)Medium
Two narrow bores of diameter $5.0$ mm and $8.0$ mm are joined together to form a $U$-shaped tube open at both ends. If this $U$-tube contains water, what is the difference in the level of two limbs of the tube. [Take surface tension of water $T = 7.3\times10^{-2}$ N m$^{-1}$, angle of contact $= 0$, $g = 10$ m s$^{-2}$ and density of water $= 1.0\times10^3$ kg m$^{-3}$]
Concept. Each limb is a capillary: $h=\dfrac{2T}{r\rho g}$ (contact angle $0$). The narrower limb rises higher; the difference in levels is the difference of the two rises.
Q 11-09-121JEE MainJEE Main 2021 (26 Aug, Shift 1)Easy
A soap bubble of radius $3$ cm is formed inside another soap bubble of radius $6$ cm. The radius of an equivalent soap bubble that has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is ______ cm.
Concept. Excess pressures add across nested films. Inside the small bubble the pressure exceeds the space between the bubbles by $\dfrac{4T}{r_1}$, which in turn exceeds the outside by $\dfrac{4T}{r_2}$.
Step 1. Total excess. $\dfrac{4T}{3}+\dfrac{4T}{6}=4T\left(\dfrac13+\dfrac16\right)=\dfrac{4T}{2}$.
Step 2. Equivalent radius. $\dfrac{4T}{r}=\dfrac{4T}{2}\Rightarrow r=2$ cm.
Answer. $2$. In general $\dfrac1r=\dfrac1{r_1}+\dfrac1{r_2}$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-122JEE MainJEE Main 2021 (27 Aug, Shift 1)Easy
In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius $2.0\times10^{-5}$ m and density $1.2\times10^3$ kg m$^{-3}$? Take viscosity of liquid $= 1.8\times10^{-5}$ N s m$^{-2}$. (Neglect buoyancy due to air).
Q 11-09-123JEE MainJEE Main 2020 (8 Jan, Shift 2)Medium
Two liquids of densities $\rho_1$ and $\rho_2$ $(\rho_2 = 2\rho_1)$ are filled up behind a square wall of side $10$ m as shown in figure. Each liquid has a height of $5$ m. The ratio of the forces due to these liquids exerted on upper part MN to that at the lower part NO is (Assume that the liquids are not mixing):
Concept. Force on a vertical strip of wall $=$ average pressure $\times$ area. For each part the pressure varies linearly with depth, so use the pressure at the middle of that part. The lower part also feels the weight of the upper liquid.
Step 1. Upper part $MN$ (liquid $\rho_1$, $5$ m deep, width $10$ m): average pressure $\rho_1g\times2.5$: $$F_1=\rho_1g(2.5)\times(5\times10)=125\rho_1g$$
Step 2. Lower part $NO$. At $N$ the pressure is already $\rho_1g(5)$; through $NO$ it rises by $\rho_2g$ per metre. Average pressure $=5\rho_1g+2.5\rho_2g=5\rho_1g+5\rho_1g=10\rho_1g$ (using $\rho_2=2\rho_1$): $$F_2=10\rho_1g\times50=500\rho_1g$$
Q 11-09-124JEE MainJEE Main 2020 (9 Jan, Shift 1)Medium
Water flows in a horizontal tube. The pressure of water changes by $700$ N m$^{-2}$ between $A$ and $B$ where the areas of cross section are $40$ cm$^2$ and $20$ cm$^2$, respectively. Find the rate of flow of water through the tube. (density of water $= 1000$ kg m$^{-3}$)
Q 11-09-126JEE MainJEE Main 2020 (5 Sep, Shift 2)Easy
In an experiment to verify Stokes' law, a small spherical ball of radius $r$ and density $\rho$ falls under gravity through a distance $h$ in air before entering a tank of water. If the terminal velocity of the ball inside water is the same as its velocity just before entering the water surface, then the value of $h$ is proportional to: (ignore viscosity of air)
Concept. Falling freely through $h$ gives $v=\sqrt{2gh}$. This must equal the terminal speed in water, which is $\propto r^2$.
Step 1. Equate. $\sqrt{2gh}\propto r^2$.
Step 2. Square. $h\propto r^4$.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-127JEE MainJEE Main 2020 (6 Sep, Shift 2)Easy
A fluid is flowing through a horizontal pipe of varying cross-section, with speed $v\ \text{m s}^{-1}$ at a point where the pressure is $P$ pascal. At another point where the pressure is $\dfrac P2$ pascal, its speed is $V\ \text{m s}^{-1}$. If the density of the fluid is $\rho\ \text{kg m}^{-3}$ and the flow is streamline, then $V$ is equal to:
Q 11-09-128JEE MainJEE Main 2020 (7 Jan, Shift 2)Easy
An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipe are $6.4$ cm and $4.8$ cm, respectively. The ratio of the minimum and the maximum velocities of fluid in this pipe is:
Concept. Continuity: an incompressible fluid has the same volume flow everywhere, $A_1v_1=A_2v_2$. So $v\propto\dfrac1A\propto\dfrac1{d^2}$: slowest where the pipe is widest.
Q 11-09-129JEE MainJEE Main 2020 (8 Jan, Shift 1)Medium
A leak proof cylinder of length $1$ m, made of a metal which has a very low coefficient of expansion, is floating vertically in water at $0\ ^\circ\text{C}$ such that its height above the water surface is $20$ cm. When the temperature of water is increased to $4\ ^\circ\text{C}$, the height of the cylinder above the water surface becomes $21$ cm. The density of water at $T = 4\ ^\circ\text{C}$, relative to the density at $T = 0\ ^\circ\text{C}$, is close to:
Q 11-09-130JEE MainJEE Main 2020 (8 Jan, Shift 1)Medium
Consider a solid sphere of radius $R$ and mass density $\rho(r) = \rho_0\left(1 - \dfrac{r^2}{R^2}\right)$, $0 < r \le R$. The minimum density of a liquid in which it will float is:
Step 2. Mean density. $\dfrac{8\pi\rho_0R^3/15}{4\pi R^3/3}=\dfrac{2\rho_0}{5}$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-131JEE MainJEE Main 2020 (2 Sep, Shift 1)Medium
A cylindrical vessel containing a liquid is rotated about its axis so that the liquid rises at its sides as shown in the figure. The radius of vessel is $5\ \text{cm}$ and the angular speed of rotation is $\omega\ \text{rad s}^{-1}$. The difference in the height, $h$ (in cm) of liquid at the centre of vessel and at the sides of the vessel will be:
Concept. A liquid spinning with its container has a paraboloidal surface; the rim stands $h=\dfrac{\omega^2R^2}{2g}$ above the centre (from energy along the surface, or $\tan\theta=\omega^2r/g$).
Step 1. Substitute $R=5$ cm. $h=\dfrac{25\omega^2}{2g}$ (in cm with $g$ in matching units).
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-132JEE MainJEE Main 2020 (2 Sep, Shift 2)Medium
A capillary tube made of glass of radius $0.15\ \text{mm}$ is dipped vertically in a beaker filled with methylene iodide (surface tension $= 0.05\ \text{N m}^{-1}$, density $= 667\ \text{kg m}^{-3}$) which rises to height $h$ in the tube. It is observed that the two tangents drawn from liquid-glass interfaces (from opposite sides of the capillary) make an angle of $60^\circ$ with one another. Then $h$ is close to $(g = 10\ \text{m s}^{-2})$
Concept. Capillary rise $h=\dfrac{2T\cos\theta}{r\rho g}$. The tangent to the meniscus at each wall makes the contact angle $\theta$ with that wall. The two walls are parallel, so the two tangents meet at $2\theta=60^\circ$.
Step 3. Volumes. $\left(\dfrac{r_1}{r_2}\right)^3=8$, so $8:1$.
Answer. Option C. Using the total pressures $1.01:1.02$ is the trap.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-134JEE MainJEE Main 2020 (3 Sep, Shift 1)Medium
When a long glass capillary tube of radius $0.015\ \text{cm}$ is dipped in a liquid, the liquid rises to a height of $15\ \text{cm}$ within it. If the contact angle between the liquid and glass is close to $0^\circ$, the surface tension of the liquid, in milliNewton $\text{m}^{-1}$, is ______. $[\rho_{\text{liquid}} = 900\ \text{kg m}^{-3},\ g = 10\ \text{m s}^{-2}]$ (Give answer in closest integer)
Q 11-09-135JEE MainJEE Main 2020 (4 Sep, Shift 1)Medium
An air bubble of radius $1\ \text{cm}$ in water has an upward acceleration of $9.8\ \text{cm s}^{-2}$. The density of water is $1\ \text{g cm}^{-3}$ and water offers negligible drag force on the bubble. The mass of the bubble is $(g = 980\ \text{cm s}^{-2})$
Q 11-09-136JEE MainJEE Main 2020 (4 Sep, Shift 2)Medium
Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density $d$. The area of the base of both vessels is $S$ but the height of liquid in one vessel is $x_1$ and in the other $x_2$. When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is:
Concept. A column of height $x$ and base $S$ has PE $\tfrac12dgSx^2$ (centre of mass at $x/2$). When joined, both end at the mean level $\dfrac{x_1+x_2}{2}$.
Step 1. Initial PE. $\tfrac12dgS(x_1^2+x_2^2)$.
Step 2. Final PE. $2\times\tfrac12dgS\left(\dfrac{x_1+x_2}{2}\right)^2=\tfrac14dgS(x_1+x_2)^2$.
Step 3. Change. $\tfrac14dgS\left[2x_1^2+2x_2^2-(x_1+x_2)^2\right]=\tfrac14dgS(x_2-x_1)^2$ (lost as heat).
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-137JEE MainJEE Main 2020 (5 Sep, Shift 1)Medium
A hollow spherical shell of outer radius $R$ floats just submerged under the water surface. The inner radius of the shell is $r$. If the specific gravity of the shell material is $\dfrac{27}{8}$ with respect to water, the value of $r$ is:
Q 11-09-138JEE MainJEE Main 2019 (9 Jan, Shift 2)Medium
The top of a water tank is open to air and its water level is maintained. It is giving out $0.74\ \text{m}^3$ water per minute through a circular opening of $2\ \text{cm}$ radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to:
Q 11-09-139JEE MainJEE Main 2019 (10 Jan, Shift 1)Easy
Water flows into a large tank with flat bottom at the rate of $10^{-4}\ \text{m}^3\text{s}^{-1}$. Water is also leaking out of a hole of area $1\ \text{cm}^2$ at its bottom. If the height of the water in the tank remains steady then this height is:
Concept. The level stays steady when outflow equals inflow. The outflow through the hole is $a\sqrt{2gh}$ (Torricelli).
Step 1. Outflow speed. $\sqrt{2gh}=\dfrac{Q}{a}=\dfrac{10^{-4}}{10^{-4}}=1$ m s$^{-1}$.
Step 2. Height. $h=\dfrac{1}{2g}=\dfrac{1}{19.6}=0.051$ m $=5.1$ cm.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-140JEE MainJEE Main 2019 (10 Apr, Shift 1)Medium
The ratio of surface tensions of mercury and water is given to be $7.5$, while the ratio of their densities is $13.6$. Their contact angles, with glass, are close to $135^\circ$ and $0^\circ$, respectively. If it is observed that mercury gets depressed by an amount $h$ in a capillary tube of radius $r_1$, while water rises by the same amount $h$ in a capillary tube of radius $r_2$, then the ratio $\dfrac{r_1}{r_2}$ is close to
Q 11-09-141JEE MainJEE Main 2019 (10 Apr, Shift 2)Easy
A cubical block of side $0.5\ \text{m}$ floats on water with $30\%$ of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water? [Take, density of water $= 10^3\ \text{kg/m}^3$]
Concept. The extra load equals the weight of the extra water displaced when the rest of the block (70%) goes under.
Step 1. Extra volume. $0.7\times(0.5)^3=0.0875$ m$^3$.
Step 2. Load. $1000\times0.0875=87.5$ kg.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-142JEE MainJEE Main 2019 (10 Apr, Shift 2)Easy
Water from a tap emerges vertically downwards with an initial speed of $1.0\ \text{m s}^{-1}$. The cross-sectional area of the tap is $10^{-4}\ \text{m}^2$. Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The cross-sectional area of the stream, $0.15\ \text{m}$ below the tap would be: (Take $g = 10\ \text{m s}^{-2}$)
Q 11-09-143JEE MainJEE Main 2019 (10 Apr, Shift 2)Easy
A submarine experiences a pressure of $5.05\times10^6\ \text{Pa}$ at a depth of $d_1$ in a sea. When it goes further to a depth of $d_2$, it experiences a pressure of $8.08\times10^6\ \text{Pa}$. Then $d_2 - d_1$ is approximately (density of water $= 10^3\ \text{kg/m}^3$ and acceleration due to gravity $= 10\ \text{m s}^{-2}$):
Concept. In a liquid at rest, pressure grows linearly with depth: $P=P_0+\rho gd$. The difference between two depths depends only on the depth difference.
Step 2. Depth difference. $\Delta P=\rho g\,\Delta d\Rightarrow\Delta d=\dfrac{3.03\times10^6}{10^3\times10}=303$ m.
Answer. About $300$ m (Option C).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-144JEE MainJEE Main 2019 (8 Apr, Shift 1)Medium
Water from a pipe is coming at a rate of $100$ litres per minute. If the radius of the pipe is $5\ \text{cm}$, the Reynolds number for the flow is of the order of (density of water $= 1000\ \text{kg/m}^3$, coefficient of viscosity of water $= 1\ \text{mPa s}$)
Concept. The column is held up by the surface tension pull around the rim: weight $=2\pi rT\cos\theta$. So the mass raised is proportional to $r$ (height falls as $\dfrac1r$, but area grows as $r^2$).
Q 11-09-146JEE MainJEE Main 2019 (9 Apr, Shift 2)Easy
A wooden block floating in a bucket of water has $\frac45$ of its volume submerged. When certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is
Concept. First find the block's density from the water-only case. Then balance its weight against the buoyancy from both liquids.
Step 1. Block density. $\dfrac45$ submerged in water: $\rho_b=0.8\rho_w$.
Step 2. Two liquids. Half in water and half in oil: $$0.8=0.5(1)+0.5\rho_{\text{oil}}\ \Rightarrow\ \rho_{\text{oil}}=0.6$$
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-147JEE MainJEE Main 2019 (12 Jan, Shift 2)Medium
A long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of vessel is $5$ cm and its rotational speed is $2$ rotations per second, then the difference in the heights between the centre and the sides, in cm, will be:
Concept. In a rotating liquid the surface becomes a paraboloid. In the rotating frame, the surface is perpendicular to the effective gravity, which gives $y=\dfrac{\omega^2x^2}{2g}$. The rise at the wall is $h=\dfrac{\omega^2R^2}{2g}$.
Step 1. Angular speed. $2$ rps: $\omega=2\pi\times2=4\pi$ rad s$^{-1}$.
Q 11-09-148JEE MainJEE Main 2019 (12 Jan, Shift 2)Medium
A soap bubble, blown by a mechanical pump at the mouth of a tube, increases in volume with time at a constant rate. The graph that correctly depicts the time dependence of pressure inside the bubble is given by:
Concept. The volume grows at a constant rate, $V\propto t$, so $r\propto t^{1/3}$. The pressure inside is $P=P_0+\dfrac{4S}{r}=P_0+\dfrac{k}{t^{1/3}}$.
Step 1. Straight-line plot. $P$ against $\dfrac{1}{t^{1/3}}$ is a straight line with intercept $P_0$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-149JEE MainJEE Main 2019 (12 Apr, Shift 2)Easy
A solid sphere, of radius $R$ acquires a terminal velocity $v_1$ when falling (due to gravity) through a viscous fluid having a coefficient of viscosity $\eta$. The sphere is broken into $27$ identical solid spheres. If each of these spheres acquires a terminal velocity, $v_2$, when falling through the same fluid, the ratio $\left(\dfrac{v_1}{v_2}\right)$ equals:
Q 11-09-150JEE MainJEE Main 2018 (15 Apr, Shift 1)Hard
A thin uniform tube is bent into a circle of radius $r$ in the vertical plane. Equal volumes of two immiscible liquids, whose densities are $\rho_1$ and $\rho_2$ $(\rho_1 > \rho_2)$, fill half the circle. The angle $\theta$ between the radius vector passing through the common interface and the vertical is:
Concept. The liquids are at rest. Each fills a quarter circle, and the interface sits at angle $\theta$ from the vertical. The pressure forces of the tube wall act along radii, so they have no moment about the centre: the two weights must balance in moment.
Step 1. Method. Treat the liquid-filled half as a rigid arc in equilibrium: the weights of the two quarter-arcs must have equal moments about the centre of the circle.
Step 2. Torque balance. The centre of mass of each quarter-arc lies on its bisector. The heavier arc spans from the interface ($\theta$ on one side of the vertical) through $90^\circ$; its bisector is at $45^\circ-\theta$ on the other side. The lighter arc's bisector is at $45^\circ+\theta$ on the first side. Equal arc lengths, so masses $\propto\rho$: $$\rho_1\sin(45^\circ-\theta)=\rho_2\sin(45^\circ+\theta)$$
Q 11-09-151JEE MainJEE Main 2018 (15 Apr, Shift 2)Medium
When an air bubble of radius $r$ rises from the bottom to the surface of a lake, its radius becomes $\dfrac{5r}{4}$. Taking the atmospheric pressure to be equal to $10\ \text{m}$ height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature):
Q 11-09-152JEE MainJEE Main 2018 (16 Apr, Shift 1)Medium
A small soap bubble of radius $4\ \text{cm}$ is trapped inside another bubble of radius $6\ \text{cm}$ without any contact. Let $P_2$ be the pressure inside the inner bubble and $P_0$, the pressure outside the outer bubble. Radius of another bubble with pressure difference $P_2 - P_0$ between its inside and outside would be:
Concept. Each soap film adds $\dfrac{4T}{r}$ of excess pressure.
Step 1. Across both bubbles. $$P_2 - P_0 = \frac{4T}{4} + \frac{4T}{6} = 4T\left(\frac14 + \frac16\right) = \frac{4T}{12/5}$$
Step 2. Equivalent radius. $r = \dfrac{12}{5} = 2.4$ cm.
Answer. $2.4$ cm (Option A).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-153JEE MainJEE Main 2017 (9 Apr)Medium
Two tubes of radii $r_1$ and $r_2$ and lengths $l_1$ and $l_2$, respectively, are connected in series and a liquid flows through each of them in stream line conditions. $P_1$ and $P_2$ are pressure differences across the two tubes. If $P_2$ is $4P_1$ and $l_2$ is $\dfrac{l_1}{4}$, then the radius $r_2$ will be equal to:
Concept. Inside a soap bubble $P_{\text{in}} = P_0 + \dfrac{4T}{r}$.
Step 1. Expanding. As $r$ grows, $\dfrac{4T}{r}$ falls, so the inside pressure decreases (staying above atmospheric).
Answer. Decreases (Option B).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-155NEETNEET 2022Top questionEasy
A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball ($v$) as a function of time ($t$) is
Concept. Viscous drag $6\pi\eta rv$ grows with speed, so the acceleration falls as the ball speeds up.
Step 1. Shape. The speed rises from zero, quickly at first and then more slowly, levelling off at the terminal velocity: curve B.
Answer. Curve B (Option C).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-156JEE MainJEE Main 2025 (2 Apr, Shift 1)Medium
A vessel with square cross-section and height of $6\ \text{m}$ is vertically partitioned. A small window of $100\ \text{cm}^2$ with a hinged door is fitted at a depth of $3\ \text{m}$ in the partition wall. One part of the vessel is filled completely with water and the other side is filled with a liquid of density $1.5\times10^3\ \text{kg/m}^3$. What force (in N) needs to be applied on the hinged door so that it does not get opened? (Acceleration due to gravity $= 10\ \text{m/s}^2$)
Concept. Both sides are filled to the top, so at the door (3 m deep) each liquid presses with $\rho gh$ (the atmosphere acts on both sides and cancels). The door needs the difference times its area.
Q 11-09-158JEE MainJEE Main 2025 (3 Apr, Shift 1)Medium
Consider a completely full cylindrical water tank of height $1.6\ \text{m}$ and cross-sectional area $0.5\ \text{m}^2$. It has a small hole in its side at a height $90\ \text{cm}$ from the bottom. Assume the cross-sectional area of the hole to be negligibly small as compared to that of the water tank. If a load of $50\ \text{kg}$ is applied on the top surface of the water in the tank, then the velocity of the water coming out at the instant when the hole is opened is: ($g = 10\ \text{m/s}^2$)
Concept. A load on the water surface adds a pressure $\dfrac{mg}{A}$ everywhere (Pascal). Bernoulli between the top surface and the hole then gives $\tfrac12\rho v^2=\rho gh+\dfrac{mg}{A}$.
Step 1. Depth of the hole. $1.6-0.9=0.7$ m.
Step 2. Extra pressure. $\dfrac{50\times10}{0.5}=1000$ Pa.
Q 11-09-159JEE MainJEE Main 2025 (3 Apr, Shift 2)Medium
Two cylindrical vessels of equal cross-sectional area $2\ \text{m}^2$ contain water up to heights $10\ \text{m}$ and $6\ \text{m}$, respectively. If the vessels are connected at their bottom, then the work done by the force of gravity is: (Density of water is $10^3\ \text{kg/m}^3$ and $g = 10\ \text{m/s}^2$)
Concept. When the vessels are joined, water flows until the levels are equal. Gravity's work equals the drop in potential energy: the water that moved went down by the fall of its centre of mass.
Step 1. Final level. Same area, so the level is the average: $\dfrac{10+6}{2}=8$ m.
Step 2. Water moved. The top $2$ m of the first vessel (from $8$ to $10$ m, centre at $9$ m) ends up as the layer from $6$ to $8$ m in the second (centre at $7$ m). Mass $=1000\times2\times2=4000$ kg; it falls $2$ m.
Step 3. Work by gravity. $W=4000\times10\times2=8\times10^4$ J.
Answer. Option D. In general the loss is $\tfrac14\rho gS(x_1-x_2)^2$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-160JEE MainJEE Main 2025 (3 Apr, Shift 2)Medium
A solid steel ball of diameter $3.6\ \text{mm}$ acquired a terminal velocity of $2.45\times10^{-2}\ \text{m/s}$ while falling under gravity through an oil of density $925\ \text{kg m}^{-3}$. Take the density of steel as $7825\ \text{kg m}^{-3}$ and $g$ as $9.8\ \text{m/s}^2$. The viscosity of the oil in SI units is:
Q 11-09-161JEE MainJEE Main 2025 (3 Apr, Shift 2)Easy
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 ______.
Concept. $\Delta P=\dfrac{4S}{r}$: half the excess pressure means twice the radius.
Step 1. Volume ratio. $2^3=8$.
Answer. $8$.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-162JEE MainJEE Main 2025 (4 Apr, Shift 1)Easy
Two liquids A and B have $\theta_A$ and $\theta_B$ as contact angles in a capillary tube. If $K = \cos\theta_A/\cos\theta_B$, then identify the correct statement:
Answer: (C) K is negative, then liquid A has concave meniscus and liquid B has convex meniscus.
Concept. $\cos\theta>0$ (acute angle) gives a concave meniscus; $\cos\theta\lt0$ (obtuse) gives a convex meniscus. $K$ negative means the two cosines have opposite signs.
Step 1. Check the options. Opposite signs: one meniscus concave, the other convex. Option C (A concave, B convex) fits; A and B (both the same shape) need $K>0$; D ($K=0$) means $\theta_A=90^\circ$, a flat meniscus.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-09-163JEE MainJEE Main 2025 (7 Apr, Shift 2)Medium
A capillary tube of radius $0.1\ \text{mm}$ is partly dipped in water (surface tension $70\ \text{dyn/cm}$ and glass–water contact angle $\approx 0^\circ$), inclined at $30^\circ$ with the vertical. The length of water risen in the capillary is ______ cm. (Take $g = 9.8\ \text{m/s}^2$)
Step 2. Length. $$l = \frac{100}{7} \times \frac{2}{\sqrt3} \approx 16.5\ \text{cm}$$ The closest option is $\dfrac{82}{5} = 16.4$ cm.
Answer. $\dfrac{82}{5}$ (Option A).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-164JEE MainMedium
Given below are two statements:
Statement I: A small steel ball falling through glycerine at its terminal velocity has zero net force acting on it, yet the viscous force on it continues to do work.
Statement II: The terminal velocity of a ball falling through a viscous liquid is directly proportional to its radius.
In the light of the above statements, choose the correct answer from the options given below:
Answer: (B) Statement I is true but Statement II is false
Concept. At terminal velocity the forces balance, but the ball still moves, so each force does work. Stokes' law gives $v_t = \dfrac{2r^2(\rho - \sigma)g}{9\eta}$.
Step 1. Statement I. Net force is zero (weight $=$ buoyancy $+$ viscous force), so the speed is constant. The viscous force acts opposite to the motion while the ball moves, so it does negative work, turning gravitational energy into heat. True.
Step 2. Statement II. $v_t \propto r^2$, not $r$. False.
Answer. Statement I is true but Statement II is false (Option B).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-165JEE MainHard
An open tank contains two immiscible liquids, one above the other. The graph shows the gauge pressure in the tank as a function of depth below the free surface. A solid block of density $1000$ kg/m³ floats in equilibrium at the interface of the two liquids, completely submerged. The fraction of its volume that lies in the lower liquid is (Take $g = 10\ \text{m/s}^2$)
Concept. The slope of the pressure–depth graph in each liquid is $\rho g$. For a block floating at the interface, its weight equals the total buoyant force from both liquids.
Step 1. Upper liquid. From $0$ to $0.4$ m the pressure rises by $3.2$ kPa: $$\rho_1 = \frac{3200}{0.4 \times 10} = 800\ \text{kg/m}^3$$
Step 2. Lower liquid. From $0.4$ m to $1.0$ m it rises by $10.4 - 3.2 = 7.2$ kPa: $$\rho_2 = \frac{7200}{0.6 \times 10} = 1200\ \text{kg/m}^3$$
Step 3. Floating condition. Let a fraction $f$ be in the lower liquid: $$1000V = 800(1 - f)V + 1200fV \;\Rightarrow\; 200 = 400f \;\Rightarrow\; f = \frac12$$
Answer. $\dfrac12$ (Option C).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-166JEE MainMedium
A soap bubble of radius $r$ is formed inside another soap bubble of radius $6$ cm, the two being concentric. The pressure inside the inner bubble exceeds the atmospheric pressure by the same amount as the pressure inside an isolated soap bubble of radius $2$ cm. The value of $r$ is
Concept. A soap bubble has two surfaces, so the excess pressure across it is $\dfrac{4T}{R}$. For nested bubbles the excess pressures add.
Step 1. Pressure in the inner bubble. $$P_{in} - P_0 = \frac{4T}{6} + \frac{4T}{r}$$
Step 2. Equate to a single bubble of radius 2 cm. $$\frac{4T}{6} + \frac{4T}{r} = \frac{4T}{2} \;\Rightarrow\; \frac1r = \frac12 - \frac16 = \frac13$$
Answer. $3$ cm (Option A).
Solution by Sreeraj P, M.Sc Physics
Q 11-09-167JEE MainHard
An open rectangular tank $3$ m long and $1$ m deep contains water to a depth of $0.7$ m. It is accelerated horizontally along its length, as shown. The maximum acceleration for which no water spills is (Take $g = 10\ \text{m/s}^2$)
Concept. In the frame of the tank each fluid element feels $g$ downward and a pseudo acceleration $a$ backward. The free surface becomes a plane perpendicular to the effective gravity, so its slope is $\tan\theta = \dfrac ag$. The volume of water is unchanged, so the surface pivots about the middle of the tank.
Step 1. Allowed rise. The water can rise at the rear wall by $1 - 0.7 = 0.3$ m.
Step 2. Slope. The rise at the wall is half the length times the slope: $$\frac32\tan\theta = 0.3 \;\Rightarrow\; \tan\theta = 0.2$$
Step 3. Acceleration. $$a = g\tan\theta = 10 \times 0.2 = 2\ \text{m/s}^2$$ At the front wall the depth falls to $0.7 - 0.3 = 0.4$ m, so the bottom stays covered and the tilted-plane picture holds.
Solution by Sreeraj P, M.Sc Physics
