Q.An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters 2.5 cm and 3.75 cm. The ratio of the velocities in the two pipes is

Showing the 12 most recent of 18 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Water is flowing through a horizontal pipe AB of non-uniform cross section. Water is entering the pipe at the end A having cross-sectional area of $4\ \mathrm{cm}^2$ at a pressure of $10^5\ \mathrm{Nm}^{-2}$ with a velocity of $20\ \mathrm{ms}^{-1}$ and water is leaving the pipe at the end B having cross-sectional area $8\ \mathrm{cm}^2$. The pressure at the end B of the pipe is (A) $0.5 \times 10^5\ \mathrm{Nm}^{-2}$ (B) $2.5 \times 10^5\ \mathrm{Nm}^{-2}$ (C) $3.5 \times 10^5\ \mathrm{Nm}^{-2}$ (D) $4.5 \times 10^5\ \mathrm{Nm}^{-2}$

   ›Reveal solutionSolution

   Using the equation of continuity to find the velocity at B, then applying Bernoulli's equation for horizontal flow, gives the pressure at B as $2.5 \times 10^5\ \mathrm{Nm}^{-2}$, so the correct option is (B).

   The problem is a direct application of Bernoulli's principle combined with the equation of continuity. Water flows horizontally through a pipe of changing cross-section — no height change, so the gravitational term drops out. Where the pipe widens, the speed drops and the pressure rises. We know the speed and pressure at the narrow end A; we find the speed at the wider end B from continuity, then use Bernoulli to get the pressure at B.

   1. Find the velocity at B using the equation of continuity. For an incompressible fluid the volume flow rate is constant:

   $A_A v_A = A_B v_B$

   Given $A_A = 4\ \mathrm{cm}^2$, $A_B = 8\ \mathrm{cm}^2$, $v_A = 20\ \mathrm{m/s}$:

   $v_B = \frac{A_A}{A_B} v_A = \frac{4}{8} \times 20 = 10\ \mathrm{m/s}$

   2. Apply Bernoulli's equation for horizontal flow. Since the pipe is horizontal, $h_A = h_B$, so the height terms cancel:

   $P_A + \frac{1}{2} \rho v_A^2 = P_B + \frac{1}{2} \rho v_B^2$

   Here $\rho = 1000\ \mathrm{kg/m}^3$, $P_A = 10^5\ \mathrm{N/m}^2$, $v_A = 20\ \mathrm{m/s}$, $v_B = 10\ \mathrm{m/s}$.

   3. Solve for $P_B$. $P_B = P_A + \frac{1}{2} \rho (v_A^2 - v_B^2)$ …

   🔒 Start your 14-day free trial to see the full solution →
2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.A tank contains 140 cm height of water at the bottom and an oil of density $900\, \mathrm{kg\, m^{-3}}$ to a height of 200 cm above water. If the liquids are immiscible, then the initial velocity of efflux of water through a small opening at the bottom of the tank is (Acceleration due to gravity $= 10\, \mathrm{ms^{-2}}$) (A) $12\, \mathrm{ms^{-1}}$ (B) $6\, \mathrm{ms^{-1}}$ (C) $4\, \mathrm{ms^{-1}}$ (D) $8\, \mathrm{ms^{-1}}$

   ›Reveal solutionSolution

   The key idea is to apply Torricelli’s law using the effective pressure at the bottom due to both liquid columns. The velocity is $v = \sqrt{2 g h_{\text{eff}}}$, where $h_{\text{eff}}$ is the height of a single fluid column that would produce the same pressure. The result is $v = 8\ \mathrm{m/s}$, so the correct option is (D).

   ---

   Concept & Intuition

   Torricelli’s law says that the speed of efflux from a small hole at the bottom of a tank is $v = \sqrt{2 g h}$ — but only if the tank contains a single liquid of uniform density. Here we have two immiscible liquids: water (density $\rho_w = 1000\ \mathrm{kg/m^3}$) below oil (density $\rho_o = 900\ \mathrm{kg/m^3}$). The pressure at the bottom is the sum of the pressures from both columns. That total pressure determines the efflux speed, as if a single liquid of some equivalent height were pushing the water out. So we compute the total gauge pressure at the hole, then convert it into an equivalent height of water, and finally apply Torricelli’s law.

   ---

   Step-by-step solution

   1. Identify the given data

      • Height of water: $h_w = 140\ \text{cm} = 1.4\ \text{m}$
      • Height of oil: $h_o = 200\ \text{cm} = 2.0\ \text{m}$
      • Density of water: $\rho_w = 1000\ \mathrm{kg/m^3}$
      • Density of oil: $\rho_o = 900\ \mathrm{kg/m^3}$
      • $g = 10\ \mathrm{m/s^2}$

   2. Compute the total gauge pressure at the bottom

      The pressure due to the water column:

   $P_w = \rho_w g h_w = 1000 \times 10 \times 1.4 = 14000\ \mathrm{Pa}$

   The pressure due to the oil column:

   $P_o = \rho_o g h_o = 900 \times 10 \times 2.0 = 18000\ \mathrm{Pa}$

   Total gauge pressure at the hole:

   $P_{\text{total}} = P_w + P_o = 14000 + 18000 = 32000\ \mathrm{Pa}$

   3. Find the equivalent height of a single water column If only water were present, the pressure at depth $h_{\text{eff}}$ would be $\rho_w g h_{\text{eff}}$. Set this equal to the total pressure: $\rho_w g h_{\text{eff}} = 32000$ …

   🔒 Start your 14-day free trial to see the full solution →
3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.A point object is moving towards a concave mirror of focal length 20 cm along the principal axis with a uniform speed of $4~\mathrm{cm~s^{-1}}$. The speed of the image when the object is at a distance of 60 cm from the mirror is (A) $3~\mathrm{cm~s^{-1}}$ (B) $1~\mathrm{cm~s^{-1}}$ (C) $2~\mathrm{cm~s^{-1}}$ (D) $4~\mathrm{cm~s^{-1}}$

   ›Reveal solutionSolution

   The speed of the image is found by differentiating the mirror formula with respect to time. Using the given object speed and focal length, the image speed comes out to be $1~\mathrm{cm~s^{-1}}$, so the correct option is (B).

   Concept and intuition:

   When an object moves along the principal axis of a concave mirror, its image also moves. The relationship between object distance $u$ and image distance $v$ is given by the mirror formula:

   $\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$

   Here, $f = -20$ cm (concave mirror, focal length negative by sign convention), and $u$ is negative for real objects in front of the mirror. The speed of the image is the rate of change of $v$ with time, which we get by differentiating the mirror formula. The key is to treat $u$ and $v$ as functions of time and use the chain rule.

   Step-by-step solution:

   1. Set up sign convention and given data

      For a concave mirror, focal length $f = -20$ cm.

      Object distance $u = -60$ cm (negative because object is in front of the mirror).

      Object speed $v_o = \frac{du}{dt} = -4$ cm/s (negative because object moves toward the mirror, so $u$ becomes less negative, i.e., increases; but careful: the problem says "uniform speed of $4$ cm/s", so magnitude is $4$, direction is toward mirror, so $\frac{du}{dt} = +4$? Let's clarify:

      If object moves toward mirror, its distance from mirror decreases. Since $u$ is negative, moving toward mirror means $u$ becomes less negative, so $u$ increases. Hence $\frac{du}{dt} = +4$ cm/s. We'll use this.)

   2. Mirror formula

   $\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$

   Substitute $u = -60$, $f = -20$:

   $\frac{1}{-60} + \frac{1}{v} = \frac{1}{-20}$

   $-\frac{1}{60} + \frac{1}{v} = -\frac{1}{20}$

   $\frac{1}{v} = -\frac{1}{20} + \frac{1}{60} = -\frac{3}{60} + \frac{1}{60} = -\frac{2}{60} = -\frac{1}{30}$

   So $v = -30$ cm (image is real, in front of mirror).

   3. Differentiate the mirror formula with respect to time

   $\frac{d}{dt}\left(\frac{1}{u}\right) + \frac{d}{dt}\left(\frac{1}{v}\right) = 0$

   Using chain rule:

   $-\frac{1}{u^2} \frac{du}{dt} - \frac{1}{v^2} \frac{dv}{dt} = 0$

   Multiply through by $-1$: …

   🔒 Start your 14-day free trial to see the full solution →
4. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.When a current of 3 A flows through a uniform copper wire of radius 0.6 mm, the average drift speed of the electrons is V. If 1.5 A current is flowing in another uniform copper wire of radius 1.2 mm, then the average drift speed of the electrons is (A) $V$ (B) $\dfrac{V}{2}$ (C) $\dfrac{V}{4}$ (D) $\dfrac{V}{8}$

   ›Reveal solutionSolution

   The drift speed of electrons is directly proportional to the current and inversely proportional to the square of the wire's radius. For the given changes, the new drift speed will be $\boxed{\dfrac{V}{8}}$.

   When an electric current flows through a conductor, it's due to the collective motion of charge carriers, typically electrons. These electrons don't move in a straight line at high speed; instead, they undergo frequent collisions with the atoms of the conductor. Under the influence of an electric field, they acquire a small average velocity in a direction opposite to the field, which is called the drift speed.

   The relationship between current ($I$) and the average drift speed ($v_d$) of electrons is fundamental. The current is essentially the amount of charge passing through a cross-section of the wire per unit time. If we consider a segment of the wire, the total charge in that segment moving past a point depends on the number of free electrons per unit volume (number density, $n$), the charge of each electron ($e$), the cross-sectional area of the wire ($A$), and how fast these electrons are drifting ($v_d$).

   The current $I$ flowing through a conductor is given by:

   $I = n A v_d e$

   where:

   • $n$ is the number density of free electrons (number of electrons per unit volume)
   • $A$ is the cross-sectional area of the wire
   • $v_d$ is the average drift speed of the electrons
   • $e$ is the magnitude of the charge of an electron ($1.6 \times 10^{-19} \text{ C}$)

   For a given material like copper, the number density of free electrons ($n$) is a constant. The charge of an electron ($e$) is also a universal constant. Therefore, for a specific material, the current $I$ is directly proportional to the product of the cross-sectional area $A$ and the drift speed $v_d$.

   $I \propto A v_d$

   Since the wire is uniform and circular, its cross-sectional area $A$ is given by $\pi r^2$, where $r$ is the radius. So, we can write:

   $I \propto (\pi r^2) v_d$

   $I \propto r^2 v_d$

   This means that the drift speed $v_d$ is directly proportional to the current $I$ and inversely proportional to the square of the radius $r^2$.

   $v_d \propto \frac{I}{r^2}$

   We can use this proportionality to compare the drift speeds in the two scenarios.

   Here's how to solve the problem step-by-step:

   1. Identify the given information for both wires.

      • Wire 1:
        • Current $I_1 = 3 \text{ A}$
        • Radius $r_1 = 0.6 \text{ mm}$
        • Drift speed $v_{d1} = V$
      • Wire 2:
        • Current $I_2 = 1.5 \text{ A}$
        • Radius $r_2 = 1.2 \text{ mm}$
        • Drift speed $v_{d2} = ?$

   2. Write the current equation for each wire.

      Since both wires are made of copper, the number density of free electrons ($n$) and the charge of an electron ($e$) are the same for both.

      • For Wire 1: $I_1 = n (\pi r_1^2) v_{d1} e$
      • For Wire 2: $I_2 = n (\pi r_2^2) v_{d2} e$ …

   🔒 Start your 14-day free trial to see the full solution →
5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Water is filled in a tank up to a height of 20 cm from the bottom of the tank. Water flows through a hole of area $1\,\text{mm}^2$ at its bottom. The mass of the water coming out from the hole in a time of 0.6 s is (Density of water $= 1000\,\text{kg m}^{-3}$ and acceleration due to gravity $= 10\,\text{ms}^{-2}$) (A) $1.8\,\text{g}$ (B) $1.2\,\text{g}$ (C) $0.6\,\text{g}$ (D) $2.4\,\text{g}$

   ›Reveal solutionSolution

   Using Torricelli’s law for efflux speed and the continuity of flow, the mass of water discharged in 0.6 s is found to be 1.2 g, corresponding to option (B).

   Concept & Intuition

   When water flows out of a small hole at the bottom of a tank, the speed of efflux is given by Torricelli’s law: $v = \sqrt{2gh}$, where $h$ is the height of the water column above the hole. This is analogous to a free-falling object acquiring speed under gravity. The volume flow rate is the product of the hole’s cross-sectional area and this speed. Multiplying by density gives the mass flow rate, and then multiplying by time yields the total mass discharged. The key assumption is that the water level remains nearly constant over the short time interval (0.6 s), so we treat $h$ as constant.

   Step-by-step solution

   1. Identify the given data

      Height of water column: $h = 20\ \text{cm} = 0.2\ \text{m}$

      Hole area: $A = 1\ \text{mm}^2 = 1 \times 10^{-6}\ \text{m}^2$

      Time: $t = 0.6\ \text{s}$

      Density: $\rho = 1000\ \text{kg/m}^3$

      Gravity: $g = 10\ \text{m/s}^2$

   2. Find the efflux speed using Torricelli’s law

   $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.2} = \sqrt{4} = 2\ \text{m/s}$

   3. Compute the volume flow rate Volume per second = area × speed:

   $\frac{dV}{dt} = A v = (1 \times 10^{-6}) \times 2 = 2 \times 10^{-6}\ \text{m}^3/\text{s}$

   4. Compute the mass flow rate Mass per second = density × volume flow rate: $$\frac{dm}{dt} = \rho \cdot \frac{dV}{dt} = 1000 \times 2 \times 10^{-6} = 2 \times 10^{-3}\ \text{kg/s} …$$

   🔒 Start your 14-day free trial to see the full solution →
6. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.A tank is filled with water to a height of 80 cm from its bottom. The speed of efflux of water through a hole on the side wall of the tank near its bottom is (Acceleration due to gravity $= 10~\mathrm{ms}^{-2}$) (A) $6~\mathrm{ms}^{-1}$ (B) $4~\mathrm{ms}^{-1}$ (C) $8~\mathrm{ms}^{-1}$ (D) $2~\mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   The speed of efflux from a hole near the bottom of a tank depends only on the height of the water column above the hole. Using Torricelli’s theorem, $v = \sqrt{2gh}$, with $h = 0.8~\mathrm{m}$ and $g = 10~\mathrm{m/s}^2$, the speed is $4~\mathrm{m/s}$.

   The concept here is Torricelli’s theorem, which is a direct application of Bernoulli’s principle. When a liquid is in an open tank, the pressure at the free surface is atmospheric, and the pressure just outside the hole is also atmospheric. If the hole is small and the tank is large, the water level falls slowly enough that the velocity of the water surface can be taken as nearly zero. Bernoulli’s equation then simplifies beautifully: the speed of efflux depends only on the height of the liquid column above the hole, just like a freely falling body.

   Let’s work through it step by step.

   1. Identify the relevant height. The hole is near the bottom, so the height of the water column above it is essentially the full depth of water in the tank. That depth is given as 80 cm. Convert to SI units:

      $h = 80~\text{cm} = 0.8~\text{m}$.

   2. Apply Torricelli’s theorem. For a hole at depth $h$ below the free surface, the speed of efflux is

   $v = \sqrt{2gh}$

   This formula comes from equating the gravitational potential energy lost by a unit volume of water (as it falls from the surface to the hole) to the kinetic energy gained.

   3. Plug in the values. $v = \sqrt{2 \times 10 \times 0.8} = \sqrt{16} = 4~\mathrm{m/s}$ …

   🔒 Start your 14-day free trial to see the full solution →
7. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.Water flows through a horizontal pipe of variable cross-section at the rate of $12\pi$ litre per minute. The velocity of the water at the point where the diameter of the pipe becomes $2\,\text{cm}$ is (A) $6\,\text{ms}^{-1}$ (B) $8\,\text{ms}^{-1}$ (C) $4\,\text{ms}^{-1}$ (D) $2\,\text{ms}^{-1}$

   ›Reveal solutionSolution

   Convert the flow rate to SI and divide by the cross-sectional area: $v = Q/A = 2\ \text{ms}^{-1}$ — option (D).

   Step-by-step solution

   The volume flow rate is constant along the pipe: $Q = A v$.

   Flow rate in SI units:

   $Q = 12\pi\ \text{litre/min} = \frac{12\pi \times 10^{-3}\ \text{m}^3}{60\ \text{s}} = 2\pi \times 10^{-4}\ \text{m}^3\,\text{s}^{-1}.$

   Cross-sectional area at diameter $2\ \text{cm}$ (radius $r = 1\ \text{cm} = 10^{-2}\ \text{m}$): …

   🔒 Start your 14-day free trial to see the full solution →
8. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.The height of water level in a tank of uniform cross-section is 5 m. The volume of the water leaked in 5 s through a hole of area $2.4\,\text{mm}^2$ made at the bottom of the tank is (Assume the level of the water in the tank remains constant and acceleration due to gravity $= 10\,\text{ms}^{-2}$) (A) $90 \times 10^{-6}\ \text{m}^3$ (B) $120 \times 10^{-6}\ \text{m}^3$ (C) $80 \times 10^{-6}\ \text{m}^3$ (D) $40 \times 10^{-6}\ \text{m}^3$

   ›Reveal solutionSolution

   Using Torricelli’s law, the efflux speed is constant because the water level is kept fixed. The volume leaked in 5 s is area × speed × time, giving $120 \times 10^{-6}\ \text{m}^3$, which corresponds to option (B).

   The key idea is Torricelli’s law: for a tank with a small hole at the bottom, if the water level is kept constant, the speed of efflux is the same as the speed a drop would have if it fell freely from the water surface to the hole. That speed is $v = \sqrt{2gh}$, where $h$ is the height of the water column above the hole. Since the level doesn’t change, the flow rate is steady, and the volume leaked is simply the product of the hole’s area, the efflux speed, and the time.

   1. Identify the given data

      • Height of water column: $h = 5\ \text{m}$
      • Hole area: $A = 2.4\ \text{mm}^2 = 2.4 \times 10^{-6}\ \text{m}^2$ (since $1\ \text{mm}^2 = 10^{-6}\ \text{m}^2$)
      • Time: $t = 5\ \text{s}$
      • Acceleration due to gravity: $g = 10\ \text{m/s}^2$

   2. Apply Torricelli’s law for efflux speed

      The speed of water leaving the hole is

   $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10\ \text{m/s}.$

   This is constant because the water level is maintained at 5 m.

   3. Compute the volume flow rate Volume per second (discharge) = area × speed: $Q = A \cdot v = (2.4 \times 10^{-6}) \times 10 = 2.4 \times 10^{-5}\ \text{m}^3/\text{s}.$ …

   🔒 Start your 14-day free trial to see the full solution →
9. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.A liquid drop of diameter 2 mm breaks into 125 identical drops, then the change in the surface energy is (Surface tension of the liquid = $70 \times 10^{-3}$ Nm$^{-1}$) (A) $3.52 \times 10^{-6}$ J (B) $1.76 \times 10^{-6}$ J (C) $2.48 \times 10^{-6}$ J (D) $5.27 \times 10^{-6}$ J

   ›Reveal solutionSolution

   The key idea is that surface energy change equals surface tension times the increase in surface area when one large drop splits into many smaller ones. The final answer is $3.52 \times 10^{-6}$ J, which corresponds to option (A).

   Concept and Intuition

   Surface energy is the energy stored in a liquid’s surface due to surface tension. When a drop breaks into smaller drops, the total volume stays the same, but the total surface area increases because many small drops have more combined surface than one big drop. The extra surface area requires work, which comes from the surface energy change. The formula is:

   $\Delta E = T \times \Delta A$

   where $T$ is surface tension and $\Delta A$ is the increase in surface area.

   Step-by-step solution

   1. Find the radius of the original drop

      Diameter = 2 mm, so radius $R = 1$ mm = $1 \times 10^{-3}$ m.

   2. Volume conservation

      Volume of original drop: $V = \frac{4}{3}\pi R^3$.

      It breaks into 125 identical drops, each of radius $r$.

      Total volume: $125 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$.

      Cancel $\frac{4}{3}\pi$:

   $125 r^3 = R^3 \quad \Rightarrow \quad r = \frac{R}{\sqrt[3]{125}} = \frac{R}{5}.$

   So $r = \frac{1 \times 10^{-3}}{5} = 2 \times 10^{-4}$ m.

   3. Calculate surface areas Original surface area: $A_1 = 4\pi R^2 = 4\pi (1 \times 10^{-3})^2 = 4\pi \times 10^{-6}$ m². Surface area of one small drop: $4\pi r^2 = 4\pi (2 \times 10^{-4})^2 = 4\pi \times 4 \times 10^{-8} = 16\pi \times 10^{-8}$ m². Total surface area of 125 drops:

   $A_2 = 125 \times 16\pi \times 10^{-8} = 2000\pi \times 10^{-8} = 2\pi \times 10^{-5} \text{ m}^2.$

   4. Find the increase in surface area

   $\Delta A = A_2 - A_1 = 2\pi \times 10^{-5} - 4\pi \times 10^{-6}.$

   Write both with same exponent: $4\pi \times 10^{-6} = 0.4\pi \times 10^{-5}$.

   So

   $\Delta A = (2\pi - 0.4\pi) \times 10^{-5} = 1.6\pi \times 10^{-5} \text{ m}^2.$

   5. Compute the change in surface energy Surface tension $T = 70 \times 10^{-3}$ N/m. $\Delta E = T \times \Delta A = (70 \times 10^{-3}) \times (1.6\pi \times 10^{-5}).$ …

   🔒 Start your 14-day free trial to see the full solution →
10. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.A vessel having small hole in the bottom has to hold water without leakage, if water is poured to a height of 7 cm. Then the radius of the hole is [surface tension of water is $0.07\, \mathrm{Nm}^{-1}$, angle of contact is $0^\circ$ and $g = 10\, \mathrm{ms}^{-2}$] (A) $1\, \mathrm{mm}$ (B) $2\, \mathrm{mm}$ (C) $4\, \mathrm{mm}$ (D) $0.2\, \mathrm{mm}$

    ›Reveal solutionSolution

    The water's hydrostatic pressure $\rho g h$ is held back by the maximum surface-tension pressure $2T/r$; solving gives $r = 0.2\ \mathrm{mm}$ — option (D).

    Water does not leak while the upward pressure the surface film can support at the hole equals the hydrostatic pressure of the column. For a hole of radius $r$ with contact angle $0^\circ$, the maximum supporting pressure is

    $\Delta P_{\max} = \frac{2T}{r}.$

    Setting this equal to the hydrostatic pressure $\rho g h$ at the bottom:

    $\rho g h = \frac{2T}{r} \quad\Rightarrow\quad r = \frac{2T}{\rho g h}.$ …

    🔒 Start your 14-day free trial to see the full solution →
11. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
    Q.Two capillary tubes of radii 0.2 cm and 0.4 cm are dipped vertically in the same liquid. The ratio of the heights through which the liquid will rise in the tubes is (A) $2 : 3$ (B) $1 : 4$ (C) $2 : 1$ (D) $4 : 1$

    ›Reveal solutionSolution

    Capillary rise is inversely proportional to the radius of the tube. For tubes with radii 0.2 cm and 0.4 cm, the ratio of heights will be $2:1$.

    Capillary action is the phenomenon where a liquid spontaneously rises or falls in a narrow tube (a capillary) due to the interplay of surface tension, adhesive forces (between liquid and tube), and cohesive forces (within the liquid). When a liquid wets the tube (like water in glass), adhesive forces are stronger, causing the liquid to climb the walls and form a concave meniscus. Surface tension then pulls this meniscus upwards, lifting the entire liquid column until the upward force is balanced by the weight of the risen liquid.

    The height $h$ to which a liquid rises in a capillary tube is given by the formula:

    $h = \frac{2T \cos\theta}{\rho g r}$

    Where:

    • $T$ is the surface tension of the liquid.
    • $\theta$ is the angle of contact between the liquid and the tube material.
    • $\rho$ is the density of the liquid.
    • $g$ is the acceleration due to gravity.
    • $r$ is the radius of the capillary tube.

    For the same liquid and the same tube material (which implies the same angle of contact $\theta$) and under the same gravitational conditions, $T$, $\cos\theta$, $\rho$, and $g$ are all constant. Therefore, the height $h$ is inversely proportional to the radius $r$ of the capillary tube.

    $h \propto \frac{1}{r}$

    This means that if the radius of the tube is smaller, the liquid will rise to a greater height, and vice-versa.

    1. Identify the given radii: …

    🔒 Start your 14-day free trial to see the full solution →
12. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.Consider an increase of $1\%$ in each of radius of artery, viscosity of blood and density of blood respectively. The percentage change in flow rate of blood in artery is (A) $0.25\%$ (B) $0.50\%$ (C) $1.0\%$ (D) $3.0\%$

    ›Reveal solutionSolution

    Poiseuille: $Q \propto r^4/\eta$ (density irrelevant) → $\%\Delta Q = 4(1\%) - 1\% = 3\%$.

    Concept — Poiseuille's law. For steady laminar flow of a viscous fluid through a tube (here, blood through an artery):

    $Q = \frac{\pi\,\Delta P\, r^4}{8\,\eta\, L}$

    The volume flow rate depends on the fourth power of the radius, inversely on the viscosity $\eta$, and not at all on the fluid's density.

    Step 1 — percentage-change (small-error) formula. Taking logarithms and differentiating,

    $\frac{\Delta Q}{Q} = 4\,\frac{\Delta r}{r} - \frac{\Delta \eta}{\eta}$ …

    🔒 Start your 14-day free trial to see the full solution →