Q.The terminal velocity of a copper ball of radius $2.0\ \text{mm}$ falling through a tank of oil at $20\,^{\circ}\text{C}$ is $6.5\ \text{cm s}^{-1}$. Compute the viscosity of the oil at $20\,^{\circ}\text{C}$. Density of oil is $1.5 \times 10^{3}\ \text{kg m}^{-3}$, density of copper is $8.9 \times 10^{3}\ \text{kg m}^{-3}$.

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.A spherical rain drop of mass ‘m’ falls vertically through air with a terminal velocity of $0.2~\mathrm{ms}^{-1}$. If 27 such identical rain drops combine to form a bigger spherical drop, then the terminal velocity of the bigger drop if it falls vertically through air is (A) $0.9~\mathrm{ms}^{-1}$ (B) $3.6~\mathrm{ms}^{-1}$ (C) $5.4~\mathrm{ms}^{-1}$ (D) $1.8~\mathrm{ms}^{-1}$

   ›Reveal solutionSolution

   Terminal velocity scales with the square of the radius for spherical drops under Stokes’ law. Combining 27 identical drops conserves volume, so the big drop’s radius is 3 times that of a small drop, making its terminal velocity $3^2 = 9$ times larger: $0.2 \times 9 = 1.8~\mathrm{ms}^{-1}$.

   Concept & Intuition

   When a spherical drop falls at terminal velocity, the downward gravitational force is exactly balanced by the upward drag force. For small, slow-moving spheres in air, the drag is given by Stokes’ law: $F_{\text{drag}} = 6\pi \eta r v$, where $\eta$ is the viscosity of air, $r$ is the radius, and $v$ is the speed. The weight is $mg = \rho \cdot \frac{4}{3}\pi r^3 \cdot g$. Setting them equal gives $v \propto r^2$. So if the radius changes, the terminal velocity changes as the square of the radius ratio. When 27 identical drops merge, the total volume is conserved, so the big drop’s radius is $\sqrt[3]{27} = 3$ times the small drop’s radius. Hence the terminal velocity multiplies by $3^2 = 9$.

   Step-by-step reasoning

   1. Force balance for a single small drop At terminal velocity $v_t$, weight = drag:

   $mg = 6\pi \eta r v_t.$

   The mass $m = \rho \cdot \frac{4}{3}\pi r^3$, so

   $\rho \frac{4}{3}\pi r^3 g = 6\pi \eta r v_t.$

   Cancel common factors ($\pi$, $r$):

   $\frac{4}{3} \rho r^2 g = 6 \eta v_t.$

   Solve for $v_t$:

   $v_t = \frac{2\rho g}{9\eta} \, r^2.$

   Thus $v_t \propto r^2$ — the key scaling.

   2. Volume conservation when 27 drops combine

      Each small drop has volume $V_{\text{small}} = \frac{4}{3}\pi r^3$.

      Total volume of 27 drops: $27 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (3r)^3$.

      So the big drop’s radius $R = 3r$.

   3. Terminal velocity of the big drop

      Using the proportionality $v_t \propto r^2$:

      $$$$

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2. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.A metal sphere of density $9300~\mathrm{kg\,m^{-3}}$ and radius $1.5~\mathrm{mm}$ is falling with a terminal velocity $V$ through a vertical column of liquid of density $1300~\mathrm{kg\,m^{-3}}$ and coefficient of viscosity $\eta$. The terminal velocity of another sphere of same metal but of radius $3~\mathrm{mm}$ falling through another vertical column, of liquid of density $800~\mathrm{kg\,m^{-3}}$ and coefficient of viscosity $1.7\eta$ is (A) $4.5V$ (B) $1.5V$ (C) $2.5V$ (D) $3.5V$

   ›Reveal solutionSolution

   The terminal velocity of a sphere in a viscous fluid is determined by a balance of gravitational, buoyant, and viscous drag forces. By deriving the formula for terminal velocity and comparing the two scenarios, we find that the new terminal velocity is $\boxed{2.5V}$.

   When an object falls through a fluid, it experiences several forces. Initially, the gravitational force pulls it down, causing it to accelerate. However, as its speed increases, the fluid exerts an upward viscous drag force that opposes the motion. Simultaneously, the fluid also exerts an upward buoyant force.

   Concept: Terminal Velocity

   Terminal velocity is the constant maximum velocity attained by an object falling through a fluid when the net force acting on it becomes zero. This occurs when the downward gravitational force is balanced by the sum of the upward buoyant force and the upward viscous drag force.

   The forces acting on a spherical object of radius $r$ and density $\rho_s$ falling through a liquid of density $\rho_l$ and coefficient of viscosity $\eta$ are:

   1. Gravitational Force ($F_g$): This acts downwards. $F_g = m g = \rho_s \left(\frac{4}{3}\pi r^3\right) g$
   2. Buoyant Force ($F_b$): This acts upwards, according to Archimedes' principle. $F_b = \rho_l \left(\frac{4}{3}\pi r^3\right) g$
   3. Viscous Drag Force ($F_v$): This acts upwards, opposing the motion. For a sphere moving at velocity $v$ in a fluid, Stokes' Law gives: $F_v = 6\pi \eta r v$

   At terminal velocity ($v_t$), the forces are balanced:

   $F_g = F_b + F_v$

   $\rho_s \frac{4}{3}\pi r^3 g = \rho_l \frac{4}{3}\pi r^3 g + 6\pi \eta r v_t$

   Rearranging the terms to solve for $v_t$:

   $6\pi \eta r v_t = \left(\rho_s - \rho_l\right) \frac{4}{3}\pi r^3 g$

   $v_t = \frac{\left(\rho_s - \rho_l\right) \frac{4}{3}\pi r^3 g}{6\pi \eta r}$

   The terminal velocity $v_t$ of a sphere is given by:

   $v_t = \frac{2 \left(\rho_s - \rho_l\right) r^2 g}{9 \eta}$

   Now, let's apply this formula to the given problem.

   1. Identify the relationship for terminal velocity:

      From the derived formula, we can see that terminal velocity $v_t$ depends on the difference in densities $(\rho_s - \rho_l)$, the square of the radius $r^2$, and inversely on the coefficient of viscosity $\eta$. The terms $2$, $g$, and $9$ are constants.

      So, we can write the proportionality:

      $v_t \propto \frac{(\rho_s - \rho_l) r^2}{\eta}$

   2. Apply to the first scenario:

      Let the given values for the first sphere and liquid be:

      • Density of sphere, $\rho_s = 9300~\mathrm{kg\,m^{-3}}$
      • Radius, $r_1 = 1.5~\mathrm{mm}$
      • Density of liquid, $\rho_{l1} = 1300~\mathrm{kg\,m^{-3}}$
      • Coefficient of viscosity, $\eta_1 = \eta$
      • Terminal velocity, $V_1 = V$

      The density difference is $\rho_s - \rho_{l1} = 9300 - 1300 = 8000~\mathrm{kg\,m^{-3}}$.

      So, for the first case:

      $V \propto \frac{(8000) (1.5)^2}{\eta}$

   3. Apply to the second scenario:

      Let the given values for the second sphere and liquid be:

      • Density of sphere, $\rho_s = 9300~\mathrm{kg\,m^{-3}}$ (same metal)
      • Radius, $r_2 = 3~\mathrm{mm}$
      • Density of liquid, $\rho_{l2} = 800~\mathrm{kg\,m^{-3}}$
      • Coefficient of viscosity, $\eta_2 = 1.7\eta$
      • Terminal velocity, $V_2 = ?$

      The density difference is $\rho_s - \rho_{l2} = 9300 - 800 = 8500~\mathrm{kg\,m^{-3}}$.

      So, for the second case:

      $V_2 \propto \frac{(8500) (3)^2}{1.7\eta}$ …

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3. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.An air bubble is formed in a liquid of surface tension $5 \times 10^{-2} \ \mathrm{Nm}^{-1}$. The decrease in the pressure inside the air bubble when its radius increases from $0.2 \ \mathrm{mm}$ to $0.5 \ \mathrm{mm}$ is (A) $450 \ \mathrm{Nm}^{-2}$ (B) $150 \ \mathrm{Nm}^{-2}$ (C) $600 \ \mathrm{Nm}^{-2}$ (D) $300 \ \mathrm{Nm}^{-2}$

   ›Reveal solutionSolution

   The pressure inside a spherical air bubble is higher than the outside by $2T/r$ (for a single surface). The decrease in pressure when the radius grows from $0.2$ mm to $0.5$ mm is $300 \ \mathrm{Nm}^{-2}$.

   The key idea is that an air bubble in a liquid has one liquid-air interface (unlike a soap bubble which has two). The excess pressure inside a spherical bubble is given by $\Delta P = \frac{2T}{r}$, where $T$ is the surface tension and $r$ is the radius. As the bubble expands, the excess pressure drops. We want the difference between the excess pressures at the two radii.

   1. Recall the formula for excess pressure in a spherical bubble in a liquid.

      For a single curved surface (like a liquid drop or an air bubble in a liquid), the Laplace pressure is $\Delta P = \frac{2T}{r}$.

      Watch out

      A common mistake is to use $\frac{4T}{r}$, which applies to a soap bubble (two surfaces). Here there is only one surface, so it's $\frac{2T}{r}$.

   2. Write the excess pressure at the smaller radius.

      $r_1 = 0.2 \ \mathrm{mm} = 0.2 \times 10^{-3} \ \mathrm{m} = 2 \times 10^{-4} \ \mathrm{m}$

      $\Delta P_1 = \frac{2T}{r_1} = \frac{2 \times 5 \times 10^{-2}}{2 \times 10^{-4}} = \frac{10^{-1}}{2 \times 10^{-4}} = 5 \times 10^{2} = 500 \ \mathrm{Nm}^{-2}$

   3. Write the excess pressure at the larger radius.

      $r_2 = 0.5 \ \mathrm{mm} = 0.5 \times 10^{-3} \ \mathrm{m} = 5 \times 10^{-4} \ \mathrm{m}$ …

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4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the terminal velocity of a metal sphere of mass 8 g falling through a liquid is $3\,\mathrm{cms}^{-1}$, then the terminal velocity of another sphere of mass 64 g made of the same metal falling through same liquid is (A) $6\,\mathrm{cms}^{-1}$ (B) $3\,\mathrm{cms}^{-1}$ (C) $12\,\mathrm{cms}^{-1}$ (D) $18\,\mathrm{cms}^{-1}$

   ›Reveal solutionSolution

   Terminal velocity in a viscous fluid depends on the square of the radius (or mass^{2/3}) for spheres of the same material. Since mass scales as radius³, the terminal velocity scales as mass^{2/3}. Here, the 64 g sphere has 8 times the mass of the 8 g sphere, so its terminal velocity is $8^{2/3} = 4$ times larger, giving $12\,\mathrm{cm\,s^{-1}}$. The correct option is (C).

   ---

   Concept and intuition

   When a sphere falls through a viscous liquid at terminal velocity, the net force on it is zero: the downward weight is balanced by the upward buoyant force and the viscous drag. For small, smooth spheres at low speeds, the drag is given by Stokes’ law: $F_d = 6\pi \eta r v$, where $\eta$ is the liquid’s viscosity, $r$ is the sphere’s radius, and $v$ is its speed. The weight is $mg = \rho_{\text{sphere}} V g$, and the buoyant force is $\rho_{\text{liquid}} V g$. The terminal velocity $v_t$ satisfies:

   $mg - \rho_{\text{liquid}} V g = 6\pi \eta r v_t.$

   Since the spheres are made of the same metal, their density $\rho$ is constant. The volume $V = \frac{4}{3}\pi r^3$ and mass $m = \rho V$. The key insight: for a given liquid and same material, the terminal velocity depends only on the sphere’s size — specifically, on its radius squared. Because mass is proportional to $r^3$, we can relate terminal velocity directly to mass.

   ---

   Step-by-step solution

   1. Write the terminal velocity formula At terminal velocity, the net downward force (weight minus buoyancy) equals the viscous drag:

   $(\rho_{\text{sphere}} - \rho_{\text{liquid}}) V g = 6\pi \eta r v_t.$

   Let $\Delta\rho = \rho_{\text{sphere}} - \rho_{\text{liquid}}$. Then:

   $v_t = \frac{\Delta\rho \, V g}{6\pi \eta r}.$

   2. Express in terms of radius Volume $V = \frac{4}{3}\pi r^3$, so:

   $v_t = \frac{\Delta\rho \cdot \frac{4}{3}\pi r^3 g}{6\pi \eta r} = \frac{2\Delta\rho \, g}{9\eta} \, r^2.$

   All factors except $r^2$ are constant for the same liquid and same metal. Hence:

   $v_t \propto r^2.$

   3. Relate radius to mass Since the spheres are made of the same metal, density $\rho$ is constant: $m = \rho V = \rho \cdot \frac{4}{3}\pi r^3 \quad \Rightarrow \quad r \propto m^{1/3}.$ …

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5. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.A rain drop of diameter 1 mm falls with a terminal velocity of $0.7\,\mathrm{ms}^{-1}$ in air. If the coefficient of viscosity of air is $2 \times 10^{-5}\,\mathrm{Pas}$, the viscous force on the rain drop is (A) $13.2 \times 10^{-8}\,\mathrm{N}$ (B) $6.6 \times 10^{-8}\,\mathrm{N}$ (C) $26.4 \times 10^{-8}\,\mathrm{N}$ (D) $10.4 \times 10^{-8}\,\mathrm{N}$

   ›Reveal solutionSolution

   By Stokes' law $F = 6\pi\eta r v$, with $r=0.5\ \mathrm{mm}$, giving $F = 13.2\times10^{-8}\ \mathrm{N}$.

   Concept

   For a small sphere moving through a viscous fluid in the laminar regime, the viscous drag is given by Stokes' law:

   $F = 6\pi\eta r v$

   where $\eta$ is the viscosity, $r$ the radius and $v$ the speed.

   Substituting the values

   The diameter is $1\ \mathrm{mm}$, so the radius is $r = 0.5\times10^{-3}\ \mathrm{m}$. With $\eta = 2\times10^{-5}\ \mathrm{Pa\,s}$ and $v = 0.7\ \mathrm{m/s}$: …

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6. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.A rain drop of diameter 1 mm falls with a terminal velocity of $0.7\,\mathrm{ms}^{-1}$ in air. If the coefficient of viscosity of air is $2 \times 10^{-5}\,\mathrm{Pas}$, the viscous force on the rain drop is (A) $26.4 \times 10^{-8}\,\mathrm{N}$ (B) $13.2 \times 10^{-8}\,\mathrm{N}$ (C) $6.6 \times 10^{-8}\,\mathrm{N}$ (D) $10.4 \times 10^{-8}\,\mathrm{N}$

   ›Reveal solutionSolution

   Using Stokes’ law for viscous drag on a sphere at terminal velocity, the viscous force equals the weight of the drop. The calculated force is $6.6 \times 10^{-8}\,\mathrm{N}$, matching option (C).

   The key concept here is Stokes’ law, which gives the viscous drag force on a small sphere moving slowly through a fluid:

   $F = 6 \pi \eta r v$

   where $\eta$ is the coefficient of viscosity, $r$ is the radius, and $v$ is the velocity.

   At terminal velocity, the net force on the drop is zero — the downward weight is exactly balanced by the upward viscous drag (buoyancy is negligible for a tiny drop in air). So the viscous force equals the weight of the drop. That gives us a direct way to compute the force without needing the density of water explicitly — but we do need it.

   Let’s work it through.

   1. Find the radius Diameter = 1 mm = $1 \times 10^{-3}\,\mathrm{m}$, so radius

   $r = 0.5 \times 10^{-3}\,\mathrm{m} = 5 \times 10^{-4}\,\mathrm{m}.$

   2. Compute the volume and mass Volume of sphere:

   $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (5 \times 10^{-4})^3$

   $= \frac{4}{3} \pi (125 \times 10^{-12}) = \frac{500}{3} \pi \times 10^{-12} \,\mathrm{m}^3.$

   Density of water $\rho = 1000\,\mathrm{kg/m^3}$, so mass:

   $m = \rho V = 1000 \times \frac{500}{3} \pi \times 10^{-12} = \frac{5}{3} \pi \times 10^{-7} \,\mathrm{kg}.$

   3. Weight = viscous force Weight:

   $mg = \left( \frac{5}{3} \pi \times 10^{-7} \right) \times 9.8$

   $= \frac{5 \times 9.8}{3} \pi \times 10^{-7} = \frac{49}{3} \pi \times 10^{-7}.$

   Numerically: $\pi \approx 3.14$, so

   $\frac{49}{3} \times 3.14 \approx 16.33 \times 3.14? \text{ Wait — let's do carefully:}$

   $\frac{49}{3} \approx 16.3333, \quad 16.3333 \times 3.14 \approx 51.2867.$

   Then multiply by $10^{-7}$:

   $F \approx 5.12867 \times 10^{-6} \,\mathrm{N}?$

   That seems too large — check: $10^{-7} \times 51.3 = 5.13 \times 10^{-6}$, but the options are around $10^{-8}$. Something’s off.

   Watch out

   The above weight calculation gives a force ~$5 \times 10^{-6}\,\mathrm{N}$, which is 100 times larger than the options. This suggests the drop is not water? Or perhaps the terminal velocity given is not the one from weight balance? Actually, the problem gives terminal velocity and viscosity — we should use Stokes’ law directly with the given data, not compute weight. …

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7. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.Two spherical rain drops of radii in the ratio $4:5$ are falling vertically through air. The ratio of the terminal velocities of the rain drops is (A) $64:125$ (B) $16:25$ (C) $4:5$ (D) $1:1$

   ›Reveal solutionSolution

   Terminal velocity in a viscous medium (Stokes’ law regime) is proportional to the square of the radius. For radii in ratio 4:5, the terminal‑velocity ratio is $4^2 : 5^2 = 16:25$, so the answer is (B).

   The key concept is Stokes’ law for terminal velocity in a fluid. When a small sphere falls through a viscous fluid at low speeds, the drag force is proportional to its radius and its speed. At terminal velocity, the net force (weight minus buoyancy) is exactly balanced by this drag. Since weight depends on volume (hence $r^3$) and drag depends on $r$, the terminal speed ends up proportional to $r^2$. That’s the intuition: bigger drops fall faster, but not linearly — the square relationship makes the ratio of speeds the square of the ratio of radii.

   Let’s work it through:

   1. Forces on a falling drop

      The downward force is weight: $W = \frac{4}{3}\pi r^3 \rho g$ (where $\rho$ is the density of water).

      The upward buoyant force is $B = \frac{4}{3}\pi r^3 \sigma g$ (where $\sigma$ is the density of air).

      The net downward force (weight minus buoyancy) is $F_{\text{net}} = \frac{4}{3}\pi r^3 (\rho - \sigma) g$.

   2. Drag force at low speeds

      For a sphere moving slowly in a viscous fluid, Stokes’ law gives drag $F_d = 6\pi \eta r v$, where $\eta$ is the viscosity of air and $v$ is the speed.

   3. Terminal velocity condition

      At terminal velocity, the net downward force equals the drag:

   $\frac{4}{3}\pi r^3 (\rho - \sigma) g = 6\pi \eta r v_t$

   Cancel $\pi$ and one factor of $r$:

   $\frac{4}{3} r^2 (\rho - \sigma) g = 6 \eta v_t$

   Solve for $v_t$:

   $v_t = \frac{2}{9} \frac{(\rho - \sigma) g}{\eta} \, r^2$

   4. Ratio of terminal velocities …

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8. TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQ
   Q.Two solid steel spheres having radii in the ratio 1:2 are falling freely through vertical column of a liquid, then the ratio of their terminal velocities is (A) $\dfrac{1}{2}$ (B) $\dfrac{1}{4}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{1}{5}$

   ›Reveal solutionSolution

   Terminal velocity in a viscous liquid depends on the square of the radius for a given material and fluid. For radii in the ratio 1:2, the terminal velocities are in the ratio 1:4, so the answer is $\frac{1}{4}$.

   When a solid sphere falls through a viscous liquid, it accelerates until the net force on it becomes zero. At that point, it moves with a constant speed called the terminal velocity. The key is to understand what forces balance at terminal speed: weight pulls it down, buoyancy pushes it up, and viscous drag opposes motion. The drag force for a sphere in a viscous fluid is given by Stokes' law, which is valid for small Reynolds numbers (slow, streamlined flow).

   The beauty of this problem is that the ratio of terminal velocities depends only on the radii, because both spheres are made of the same material and fall through the same liquid. Let’s see why.

   1. Write the force balance at terminal velocity.

      For a sphere of radius $r$, density $\rho_s$, falling in a liquid of density $\rho_l$ and viscosity $\eta$, the forces are:

      • Weight: $W = \frac{4}{3}\pi r^3 \rho_s g$ (downward)
      • Buoyancy: $F_b = \frac{4}{3}\pi r^3 \rho_l g$ (upward)
      • Viscous drag (Stokes' law): $F_d = 6\pi \eta r v_t$ (upward, opposing motion)

      At terminal velocity $v_t$, net force is zero:

   $W - F_b - F_d = 0$

   $\frac{4}{3}\pi r^3 (\rho_s - \rho_l) g = 6\pi \eta r v_t$

   2. Solve for terminal velocity. Cancel common factors ($\pi$, $r$):

   $\frac{4}{3} r^2 (\rho_s - \rho_l) g = 6 \eta v_t$

   $v_t = \frac{2}{9} \frac{r^2 (\rho_s - \rho_l) g}{\eta}$

   Terminal velocity of a sphere in a viscous fluid:

   $v_t = \frac{2}{9} \frac{r^2 (\rho_s - \rho_l) g}{\eta}$

   3. Apply to the two spheres. …

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9. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
   Q.Two rain drops of same radii $'r'$, falling with terminal velocity $'v'$ merge and form a bigger drop of radius $'R'$. The terminal velocity of the bigger drop is (A) $\dfrac{vR}{r}$ (B) $v\dfrac{R^2}{r^2}$ (C) $v$ (D) $2v$

   ›Reveal solutionSolution

   When two identical drops merge, volume conservation gives the new radius; terminal velocity scales as $r^2$, so the bigger drop falls at $v \cdot 2^{2/3}$. None of the printed options is correct.

   Why terminal velocity depends on drop size

   A falling raindrop accelerates until drag balances weight. At terminal velocity the net force is zero. For a sphere moving slowly through a viscous fluid, Stokes's law tells us the drag force is $F_{\text{drag}} = 6\pi\eta r v$, where $\eta$ is the fluid's viscosity. The weight is $mg = \frac{4}{3}\pi r^3 \rho g$, where $\rho$ is the density of water.

   Setting drag equal to weight:

   $6\pi\eta r v = \frac{4}{3}\pi r^3 \rho g$

   Solve for $v$:

   $v = \frac{2r^2 \rho g}{9\eta}$

   The key insight: terminal velocity is proportional to the square of the radius, $v \propto r^2$.

   $v_{\text{terminal}} = \frac{2\rho g}{9\eta} \, r^2$

   Step-by-step solution

   1. Find the radius of the merged drop. Two drops each of radius $r$ have total volume $2 \times \frac{4}{3}\pi r^3 = \frac{8}{3}\pi r^3$. The merged drop has volume $\frac{4}{3}\pi R^3$. Equating:

   $\frac{4}{3}\pi R^3 = \frac{8}{3}\pi r^3$

   $R^3 = 2r^3 \implies R = 2^{1/3} r$

   2. Relate the new terminal velocity to the old. The original drop has terminal velocity $v \propto r^2$. The merged drop has terminal velocity $V \propto R^2$. Taking the ratio:

   $\frac{V}{v} = \frac{R^2}{r^2}$

   Substitute $R = 2^{1/3} r$:

   $\frac{V}{v} = \frac{(2^{1/3} r)^2}{r^2} = \frac{2^{2/3} r^2}{r^2} = 2^{2/3}$

   Therefore:

   $V = v \cdot 2^{2/3} = v \cdot \sqrt[3]{4}$

   3. Check against the options. …

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10. TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
    Q.Two copper spheres of masses M and 8M are falling through a column of glycerine. If the terminal velocity of the sphere of mass 8M is $V$, then the terminal velocity of the sphere of mass M is (A) $2V$ (B) $\dfrac{V}{4}$ (C) $4V$ (D) $\dfrac{V}{2}$

    ›Reveal solutionSolution

    Terminal velocity in a viscous fluid depends on the square of the radius, and mass scales with the cube of the radius. For spheres of the same material, terminal velocity is proportional to mass^(2/3). Thus if the 8M sphere has terminal velocity V, the M sphere has terminal velocity V/4.

    Concept & Intuition

    When a sphere falls through a viscous fluid like glycerine, it accelerates until the net force is zero — that’s terminal velocity. The forces are: weight downward, buoyant force upward, and viscous drag upward. For small Reynolds numbers, Stokes’ law gives drag = $6\pi\eta r v$. At terminal velocity, weight – buoyancy = drag. Since both spheres are copper (same density), the buoyant force is proportional to volume, and weight is also proportional to volume. The key is that mass scales as $r^3$, while the drag force scales as $r v$. So terminal velocity ends up scaling as $r^2$, and therefore as $m^{2/3}$.

    Step-by-step reasoning

    1. Write the terminal velocity condition For a sphere of radius $r$ and density $\rho_s$ falling in a fluid of density $\rho_f$ and viscosity $\eta$, the forces balance:

    $\frac{4}{3}\pi r^3 \rho_s g - \frac{4}{3}\pi r^3 \rho_f g = 6\pi\eta r v_t$

    The left side is the net gravitational force (weight minus buoyancy). Simplify:

    $\frac{4}{3}\pi r^3 (\rho_s - \rho_f) g = 6\pi\eta r v_t$

    2. Solve for terminal velocity Cancel common factors:

    $v_t = \frac{2}{9} \frac{(\rho_s - \rho_f) g}{\eta} \, r^2$

    So $v_t \propto r^2$ for a given fluid and material.

    3. Relate mass to radius Mass $m = \frac{4}{3}\pi r^3 \rho_s$, so $r \propto m^{1/3}$. …

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11. TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
    Q.The work done in increasing the size of a soap film from $10\,\text{cm} \times 10\,\text{cm}$ to $15\,\text{cm} \times 15\,\text{cm}$ is $750\,\mu\text{J}$. The surface tension of the soap film is (A) $30 \times 10^{-3}\ \text{Nm}^{-1}$ (B) $3 \times 10^{-3}\ \text{Nm}^{-1}$ (C) $20 \times 10^{-4}\ \text{Nm}^{-1}$ (D) $20 \times 10^{-3}\ \text{Nm}^{-1}$

    ›Reveal solutionSolution

    A soap film has two surfaces, so $W=T\,(2\Delta A)$. With $\Delta A=125\text{ cm}^2=1.25\times10^{-2}\text{ m}^2$ and $W=750\ \mu\text{J}$, $T=30\times10^{-3}\text{ N/m}$ — option (A).

    Area change.

    $A_i=10\times10=100\text{ cm}^2,\qquad A_f=15\times15=225\text{ cm}^2,$

    $\Delta A=225-100=125\text{ cm}^2=125\times10^{-4}=1.25\times10^{-2}\text{ m}^2.$

    Two surfaces. A soap film exposes both faces to air, so the new area created is $2\Delta A$:

    $W=T\,(2\Delta A).$

    Surface tension. …

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12. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.A solid metal sphere released in a vertical liquid column has attained terminal velocity in the downward direction. The magnitudes of viscous force, buoyant force and gravitational force acting on it are $F_v$, $F_B$ and $F_w$ respectively. Then the correct relation between them is (A) $F_B > F_v = F_w$ (B) $F_w = F_v + F_B$ (C) $F_B = F_w + F_v$ (D) $F_v = F_B + F_w$

    ›Reveal solutionSolution

    At terminal velocity, the net force on the sphere is zero. Since it moves downward, weight must be balanced by the sum of buoyant and viscous forces, so $F_w = F_v + F_B$.

    Concept & Intuition

    When an object moves through a fluid at constant speed (terminal velocity), Newton’s first law tells us the net force is zero. For a sphere sinking in a liquid, three vertical forces act:

    • Weight ($F_w$) downward,
    • Buoyant force ($F_B$) upward,
    • Viscous drag ($F_v$) upward (opposing motion).

    Because the sphere moves downward, the drag force points upward. At terminal velocity, the downward force must exactly equal the sum of the upward forces. This is a straightforward force balance — no acceleration, so the vector sum is zero.

    Step-by-step reasoning

    1. Identify directions

       • Weight $F_w$ acts downward (positive direction, say).
       • Buoyant force $F_B$ acts upward (negative direction).
       • Viscous force $F_v$ opposes motion; since motion is downward, $F_v$ acts upward (negative direction).

    2. Write the net force equation

       At terminal velocity, acceleration $a = 0$, so $\sum F = 0$.

       Taking downward as positive:

    $F_w - F_B - F_v = 0$

    3. Rearrange for the relation $F_w = F_B + F_v$ …

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