Welcome to Lecture 83 of our comprehensive JEE Physics Master Series. In this lecture, we explore the dynamics of bodies falling through viscous fluids: the rigorous mathematical derivation of terminal velocity, time-dependent $v$-$t$ and $y$-$t$ differential dynamics, droplet coalescence scaling laws, the $r^5$ heat dissipation theorem, and non-inertial buoyant corrections.
1. Detailed Concept Card
1.1 Mechanics of a Falling Sphere in a Viscous Medium
Consider a small, smooth, rigid sphere of radius $r$ and uniform density $\rho$ released from rest inside an infinite, homogeneous fluid of density $\sigma$ and dynamic viscosity $\eta$. At any instantaneous speed $v$ directed vertically downward, the sphere is acted upon by three concurrent forces:
- 1. Downward Gravitational Force (True Weight): $$W = m g = \left(\frac{4}{3}\pi r^3\, \rho\right) g$$
- 2. Upward Buoyant Force (Archimedes’ Upthrust): $$F_B = \left(\frac{4}{3}\pi r^3\, \sigma\right) g$$
- 3. Upward Viscous Drag Force (Stoke’s Resistance): $$F_v = 6\pi\, \eta\, r\, v$$
According to Newton’s second law, the instantaneous downward acceleration $a = \frac{dv}{dt}$ is governed by:
$$m \frac{dv}{dt} = W – F_B – F_v = \frac{4}{3}\pi r^3 (\rho – \sigma) g – 6\pi \eta r v$$
1.2 Derivation of Terminal Velocity ($v_t$)
As the sphere accelerates downward from rest, its speed $v$ increases. Consequently, the retarding viscous force $F_v \propto v$ increases continuously. Eventually, the retarding forces ($F_B + F_v$) grow to completely balance the downward weight $W$, causing the net force and acceleration to become zero ($a = \frac{dv}{dt} = 0$).
Beyond this instant, the sphere continues to move downward with a constant, asymptotic limiting velocity known as the Terminal Velocity ($v_t$):
$$W – F_B – F_v = 0 \implies 6\pi \eta r v_t = \frac{4}{3}\pi r^3 (\rho – \sigma) g$$
Solving explicitly for $v_t$:
$$v_t = \frac{2}{9}\, \frac{r^2 (\rho – \sigma) g}{\eta}$$
Physical Regimes:
- Case I ($\rho > \sigma$): $v_t > 0$. The body sinks downward to a steady terminal downward speed (e.g., steel ball in glycerine, raindrop in air).
- Case II ($\rho < \sigma$): $v_t < 0$. The net driving force is upward. The body accelerates upward to a steady terminal ascent speed (e.g., air bubble rising in water): $$v_{\text{ascent}} = \frac{2}{9}\, \frac{r^2 (\sigma – \rho) g}{\eta}$$
- Case III ($\rho = \sigma$): $v_t = 0$. The body is neutrally buoyant and remains suspended without moving.
1.3 Time-Dependent Kinematic Equations ($v$-$t$ Dynamics)
Dividing the equation of motion by mass $m = \frac{4}{3}\pi r^3 \rho$:
$$\frac{dv}{dt} = \left(1 – \frac{\sigma}{\rho}\right)g – \frac{6\pi \eta r}{\frac{4}{3}\pi r^3 \rho}\, v = g’ – \frac{9\eta}{2\rho r^2}\, v$$
Define the Relaxation Time Constant ($\tau$):
$$\tau = \frac{m}{6\pi \eta r} = \frac{2\rho r^2}{9\eta}$$
Note that $v_t = g’\tau = g\left(1 – \frac{\sigma}{\rho}\right)\tau$. The differential equation becomes:
$$\frac{dv}{dt} = \frac{v_t – v}{\tau} \implies \frac{dv}{v_t – v} = \frac{dt}{\tau}$$
Integrating from initial release at $t = 0$ ($v = 0$) to time $t$:
$$\int_0^v \frac{dv’}{v_t – v’} = \int_0^t \frac{dt’}{\tau} \implies -\ln\left(1 – \frac{v}{v_t}\right) = \frac{t}{\tau}$$
$$v(t) = v_t \left(1 – e^{-t/\tau}\right)$$
Kinematic Characteristics:
- At $t = 0$: $v = 0$, acceleration is maximum: $a(0) = g\left(1 – \frac{\sigma}{\rho}\right)$.
- At $t = \tau$: $v = v_t(1 – e^{-1}) \approx 0.632\, v_t$ ($63.2\%$ of terminal speed).
- At $t = 3\tau$: $v = v_t(1 – e^{-3}) \approx 0.950\, v_t$ ($95.0\%$ of terminal speed).
- At $t = 5\tau$: $v \approx 0.993\, v_t$. The body has effectively reached terminal velocity.
- Distance Fallen as a Function of Time: $$y(t) = \int_0^t v(t’)\, dt’ = v_t \int_0^t \left(1 – e^{-t’/\tau}\right) dt’ = v_t t – v_t \tau \left(1 – e^{-t/\tau}\right)$$ For $t \gg \tau$, $y(t) \approx v_t t – v_t \tau$, representing uniform motion with an apparent time lag $\tau$.
1.4 Proportionality & Scaling Laws
A. Scaling with Radius and Mass
$$v_t \propto r^2$$
Since mass $m = \frac{4}{3}\pi r^3 \rho \propto r^3$, we have $r \propto m^{1/3}$. Therefore:
$$v_t \propto (m^{1/3})^2 \implies v_t \propto m^{2/3}$$
B. Droplet Coalescence Law
If $n$ identical spherical droplets each of radius $r$ and terminal speed $v_t$ coalesce into a single large droplet of radius $R$:
- Volume conservation: $\frac{4}{3}\pi R^3 = n \left(\frac{4}{3}\pi r^3\right) \implies R = n^{1/3} r$.
- Terminal velocity of the merged drop: $$v_t’ = \left(\frac{R}{r}\right)^2 v_t = (n^{1/3})^2 v_t = n^{2/3}\, v_t$$
C. Viscous Heat Dissipation Theorem ($r^5$ Law)
When the sphere falls at its terminal velocity $v_t$, the mechanical potential energy lost per second is dissipated entirely into thermal energy via viscous shear. The rate of heat production is:
$$\frac{dQ}{dt} = P_{\text{dissipated}} = F_v \cdot v_t = (6\pi \eta r v_t) v_t = 6\pi \eta r\, v_t^2$$
Substitute $v_t = \frac{2}{9}\frac{r^2(\rho – \sigma)g}{\eta}$:
$$\frac{dQ}{dt} = 6\pi \eta r \left[ \frac{2}{9}\frac{r^2(\rho – \sigma)g}{\eta} \right]^2 = \frac{8\pi (\rho – \sigma)^2 g^2}{27\, \eta}\, r^5$$
$$\frac{dQ}{dt} \propto r^5$$
The rate of viscous heat dissipation scales with the fifth power of the radius! If two droplets coalesce ($n = 2$), the rate of heat dissipation increases by $(2^{1/3})^5 = 2^{5/3} \approx 3.17$ times.
1.5 Important Insights & JEE Pitfalls
- Pitfall 1 (Neglecting Buoyancy in Liquids): While $\sigma \ll \rho$ holds for raindrops in air ($\sigma_{\text{air}} \approx 1.2\text{ kg/m}^3$ vs $\rho_{\text{water}} = 1000\text{ kg/m}^3$), for steel balls in glycerine ($\rho = 7800\text{ kg/m}^3, \sigma = 1260\text{ kg/m}^3$), neglecting buoyancy introduces an error of $\approx 16\%$. Always evaluate $(\rho – \sigma)$.
- Pitfall 2 (Initial Velocity Greater than Terminal Velocity): If a projectile is fired downward into a fluid with an initial speed $v_0 > v_t$, the viscous drag exceeds the net downward gravitational force ($F_v > W – F_B$). The body decelerates asymptotically to $v_t$: $$v(t) = v_t + (v_0 – v_t)e^{-t/\tau}$$
- Pitfall 3 (Terminal Velocity in Accelerated Containers): In an elevator accelerating upward with acceleration $a_0$, the effective gravity is $g_{\text{eff}} = g + a_0$. Both weight and buoyant force scale with $g_{\text{eff}}$, so terminal velocity becomes: $$v_t’ = \frac{2}{9}\, \frac{r^2 (\rho – \sigma)(g + a_0)}{\eta} = v_t \left(1 + \frac{a_0}{g}\right)$$
2. Four Graded Solved Examples
Example 1: Steel Sphere Falling in a Viscous Oil Column (JEE Main Level)
Problem Statement: A small steel ball of radius $r = 2.0\text{ mm}$ and density $\rho = 7.8 \times 10^3\text{ kg/m}^3$ is released into a tall jar filled with castor oil of density $\sigma = 0.95 \times 10^3\text{ kg/m}^3$ and dynamic viscosity $\eta = 0.98\text{ Pa}\cdot\text{s}$. Take $g = 9.8\text{ m/s}^2$.
(a) Determine the terminal velocity $v_t$ of the ball.
(b) Calculate the magnitude of the viscous drag force acting on the ball at terminal velocity.
(c) Verify the validity of Stoke’s law by evaluating the Reynolds number $\text{Re} = \frac{\sigma v_t (2r)}{\eta}$.
Solution:
Part (a): Terminal Velocity
Parameters in standard SI units:
$r = 2.0 \times 10^{-3}\text{ m}$, $\rho = 7800\text{ kg/m}^3$, $\sigma = 950\text{ kg/m}^3$, $\eta = 0.98\text{ Pa}\cdot\text{s}$, $g = 9.8\text{ m/s}^2$.
Density difference: $\rho – \sigma = 7800 – 950 = 6850\text{ kg/m}^3$.
Applying the terminal velocity formula:
$$v_t = \frac{2}{9}\, \frac{r^2 (\rho – \sigma) g}{\eta} = \frac{2}{9} \times \frac{(2.0 \times 10^{-3})^2 \times 6850 \times 9.8}{0.98}$$
$$v_t = \frac{2}{9} \times \frac{4.0 \times 10^{-6} \times 6850 \times 9.8}{0.98} = \frac{2}{9} \times (4.0 \times 10^{-6} \times 6850 \times 10) = \frac{2}{9} \times 0.274 = 0.06089\text{ m/s} \approx 6.09\text{ cm/s}$$
Part (b): Viscous Drag Force at Terminal Velocity
At terminal velocity, the net acceleration is zero, so $F_v = W – F_B$:
$$F_v = \frac{4}{3}\pi r^3 (\rho – \sigma) g = \frac{4}{3}\pi \times (2.0 \times 10^{-3})^3 \times 6850 \times 9.8$$
$$F_v = \frac{4}{3}\pi \times (8.0 \times 10^{-9}) \times 6850 \times 9.8 = 4.1888 \times 5.3704 \times 10^{-4} \approx 2.25 \times 10^{-3}\text{ N} = 2.25\text{ mN}$$
Alternatively, via Stoke’s formula: $F_v = 6\pi \eta r v_t = 6\pi \times 0.98 \times (2.0 \times 10^{-3}) \times 0.06089 \approx 2.25 \times 10^{-3}\text{ N}$.
Part (c): Reynolds Number Check
$$\text{Re} = \frac{\sigma v_t (2r)}{\eta} = \frac{950 \times 0.06089 \times (4.0 \times 10^{-3})}{0.98} = \frac{0.2314}{0.98} \approx 0.236$$
Since $\text{Re} = 0.236 < 1$, the flow is strictly in the creeping laminar regime, validating the use of Stoke's law.
Example 2: Coalescence of 64 Identical Mist Droplets (JEE Main Level)
Problem Statement: Sixty-four ($64$) identical tiny spherical water droplets, each falling through still air with a terminal velocity of $v_t = 1.5\text{ cm/s}$, coalesce during rainfall to form a single larger spherical drop.
(a) Determine the terminal velocity $v_t’$ of the coalesced drop.
(b) Find the ratio of the rate of heat dissipation in the air by the single large drop to the total rate of heat dissipation by all $64$ droplets before coalescence.
Solution:
Part (a): Terminal Velocity Ratio
Let each small droplet have radius $r$ and the merged drop have radius $R$. By conservation of volume:
$$\frac{4}{3}\pi R^3 = 64 \times \left(\frac{4}{3}\pi r^3\right) \implies R^3 = 64 r^3 \implies R = 4 r$$
Since terminal velocity scales as $v_t \propto r^2$:
$$v_t’ = \left(\frac{R}{r}\right)^2 v_t = (4)^2 v_t = 16\, v_t$$
$$v_t’ = 16 \times 1.5\text{ cm/s} = 24.0\text{ cm/s}$$
Part (b): Rate of Heat Dissipation Ratio
The rate of heat dissipation for a falling sphere at terminal velocity is:
$$\frac{dQ}{dt} = 6\pi \eta r v_t^2 \propto r\, v_t^2 \propto r (r^2)^2 = r^5$$
For a single small droplet: $P_1 = k\, r^5$.
Total rate of heat dissipation for all $64$ separate droplets:
$$P_{\text{total, initial}} = 64 \times P_1 = 64\, k\, r^5$$
For the single coalesced drop of radius $R = 4r$:
$$P_{\text{final}} = k\, R^5 = k\, (4r)^5 = 1024\, k\, r^5$$
The ratio of the rate of heat dissipation of the single large drop to the initial total rate is:
$$\frac{P_{\text{final}}}{P_{\text{total, initial}}} = \frac{1024\, k\, r^5}{64\, k\, r^5} = \frac{1024}{64} = 16$$
Key Takeaway: Upon coalescence of $n$ droplets, the terminal speed increases by $n^{2/3}$, individual heat dissipation increases by $n^{5/3}$, and total system dissipation increases by $n^{2/3}$.
Example 3: Sphere Projected Downward at Twice Its Terminal Speed (JEE Advanced Level)
Problem Statement: A small sphere of mass $m$ and radius $r$ is projected vertically downward into a deep container of viscous oil with an initial velocity $v_0 = 2 v_t$, where $v_t$ is its terminal velocity in the oil. The relaxation time constant of the system is $\tau = \frac{m}{6\pi\eta r}$.
(a) Find the velocity $v(t)$ of the sphere as a function of time.
(b) Determine the time $t_1$ required for the speed to reduce to $1.25\, v_t$.
(c) Derive an expression for the total distance $S$ traveled by the sphere until its speed drops to $1.25\, v_t$.
Solution:
Part (a): Velocity Function $v(t)$
Taking the vertically downward direction as positive, the equation of motion is:
$$m \frac{dv}{dt} = m g’ – 6\pi \eta r v = \frac{m}{\tau}(v_t – v) \implies \frac{dv}{v – v_t} = -\frac{dt}{\tau}$$
Integrating with initial condition $v(0) = v_0 = 2v_t$:
$$\int_{2v_t}^v \frac{dv’}{v’ – v_t} = -\int_0^t \frac{dt’}{\tau} \implies \ln\left(\frac{v – v_t}{2v_t – v_t}\right) = -\frac{t}{\tau}$$
$$\frac{v – v_t}{v_t} = e^{-t/\tau} \implies v(t) = v_t + v_t e^{-t/\tau} = v_t\left(1 + e^{-t/\tau}\right)$$
Part (b): Time to Reach $v = 1.25 v_t$
Substitute $v(t_1) = 1.25 v_t$:
$$1.25 v_t = v_t(1 + e^{-t_1/\tau}) \implies e^{-t_1/\tau} = 0.25 = \frac{1}{4}$$
$$-\frac{t_1}{\tau} = -\ln 4 \implies t_1 = \tau \ln 4 = 2\tau \ln 2 \approx 1.386\, \tau$$
Part (c): Distance Fallen $S$
Distance is obtained by integrating $v(t)$ from $t = 0$ to $t = t_1$:
$$S = \int_0^{t_1} v(t)\, dt = \int_0^{t_1} v_t\left(1 + e^{-t/\tau}\right) dt = v_t \left[ t – \tau e^{-t/\tau} \right]_0^{t_1}$$
$$S = v_t \left[ (t_1 – \tau e^{-t_1/\tau}) – (0 – \tau) \right] = v_t \left[ t_1 + \tau(1 – e^{-t_1/\tau}) \right]$$
Substitute $t_1 = \tau \ln 4$ and $e^{-t_1/\tau} = \frac{1}{4}$:
$$S = v_t \left[ \tau \ln 4 + \tau\left(1 – \frac{1}{4}\right) \right] = v_t \tau \left( \ln 4 + \frac{3}{4} \right) = v_t \tau \left( 2\ln 2 + 0.75 \right) \approx 2.136\, v_t \tau$$
Key Takeaway: When a projectile enters a viscous medium with $v_0 > v_t$, the drag force produces exponential retardation towards the terminal velocity with the exact same relaxation time constant $\tau$.
Example 4: Air Bubble Rising in a Tall Water Container in an Elevator (JEE Advanced Level)
Problem Statement: A small spherical air bubble of radius $r = 1.0\text{ mm}$ rises through a tall column of water of density $\sigma = 1000\text{ kg/m}^3$ and dynamic viscosity $\eta = 1.0 \times 10^{-3}\text{ Pa}\cdot\text{s}$. Density of air is $\rho = 1.2\text{ kg/m}^3$. Take $g = 9.8\text{ m/s}^2$.
(a) Find the terminal velocity of the bubble when the container is at rest.
(b) If the entire container is placed inside an elevator accelerating upward with acceleration $a_0 = 4.9\text{ m/s}^2$, find the new terminal velocity $v_t’$.
Solution:
Part (a): Bubble Ascent Speed at Rest
Since $\rho_{\text{air}} \ll \sigma_{\text{water}}$, the buoyant force far exceeds the gravitational weight ($F_B \gg W$), driving the bubble upward. The net upward buoyant force is counteracted by downward viscous drag:
$$F_B – W = 6\pi \eta r v_{\text{ascent}} \implies \frac{4}{3}\pi r^3 (\sigma – \rho) g = 6\pi \eta r v_{\text{ascent}}$$
$$v_{\text{ascent}} = \frac{2}{9}\, \frac{r^2 (\sigma – \rho) g}{\eta}$$
Since $\sigma – \rho = 1000 – 1.2 = 998.8\text{ kg/m}^3 \approx 1000\text{ kg/m}^3$ and $r = 1.0 \times 10^{-3}\text{ m}$:
$$v_{\text{ascent}} = \frac{2}{9} \times \frac{(1.0 \times 10^{-3})^2 \times 1000 \times 9.8}{1.0 \times 10^{-3}} = \frac{2}{9} \times \frac{10^{-6} \times 1000 \times 9.8}{10^{-3}} = \frac{2}{9} \times 9.8 = \frac{19.6}{9} \approx 2.178\text{ m/s}$$
Part (b): Container Inside Upward Accelerating Elevator
In the non-inertial reference frame of the elevator accelerating upward with $a_0 = 4.9\text{ m/s}^2$, the effective gravity is:
$$g_{\text{eff}} = g + a_0 = 9.8 + 4.9 = 14.7\text{ m/s}^2 = 1.5\, g$$
Both the buoyant upthrust and gravitational weight scale with $g_{\text{eff}}$:
$$F_B’ – W’ = \frac{4}{3}\pi r^3 (\sigma – \rho) g_{\text{eff}} = 6\pi \eta r v_t’$$
$$v_t’ = \frac{2}{9}\, \frac{r^2 (\sigma – \rho) g_{\text{eff}}}{\eta} = v_{\text{ascent}} \left(\frac{g_{\text{eff}}}{g}\right) = v_{\text{ascent}} \times 1.5$$
$$v_t’ = 2.178 \times 1.5 = 3.267\text{ m/s}$$
Key Takeaway: Terminal velocity is directly proportional to effective gravity ($v_t \propto g_{\text{eff}}$). In free fall ($a_0 = -g$), $g_{\text{eff}} = 0$, and the bubble does not rise at all!
3. Practice Worksheet (10 Questions)
Questions 1 to 5: Single Correct (JEE Main level). Questions 6 to 8: One or More than One Correct (JEE Advanced level). Questions 9 and 10: Numerical Value (JEE Advanced level).
Question 1 (JEE Main Single Correct)
Two spherical raindrops of radii in the ratio $1 : 2$ fall through air with terminal velocities $v_1$ and $v_2$. The ratio $v_1 / v_2$ is:
(A) $1 : 2$
(B) $1 : 4$
(C) $1 : 8$
(D) $1 : 16$
Question 2 (JEE Main Single Correct)
A spherical ball of density $\rho$ falls through a liquid of density $\sigma$. If $\rho < \sigma$, the ball will:
(A) Move downward with constantly increasing speed.
(B) Move downward and achieve a downward terminal velocity.
(C) Move upward and achieve an upward terminal velocity.
(D) Oscillate about a mean position in the liquid.
Question 3 (JEE Main Single Correct)
Twenty-seven ($27$) identical small water droplets, each falling with terminal velocity $v_t$, merge to form a single larger drop. The terminal velocity of the larger drop is:
(A) $3\, v_t$
(B) $9\, v_t$
(C) $27\, v_t$
(D) $81\, v_t$
Question 4 (JEE Main Single Correct)
The rate of heat dissipation $\frac{dQ}{dt}$ due to viscous drag when a spherical droplet of radius $r$ falls with its terminal velocity in still air depends on $r$ as:
(A) $r^2$
(B) $r^3$
(C) $r^4$
(D) $r^5$
Question 5 (JEE Main Single Correct)
A sphere of mass $m$ and radius $r$ is released from rest in a viscous fluid. If $v_t$ is its terminal velocity, the instantaneous acceleration when its velocity reaches $0.6\, v_t$ is:
(A) $0.4\, g\left(1 – \frac{\sigma}{\rho}\right)$
(B) $0.6\, g\left(1 – \frac{\sigma}{\rho}\right)$
(C) $0.4\, g$
(D) Zero
Question 6 (JEE Advanced One or More Correct)
A small lead sphere falls from rest through a deep column of glycerine. Which of the following statements is/are correct?
(A) The acceleration of the sphere decreases monotonically with time and tends to zero.
(B) The net force acting on the sphere decreases exponentially with time.
(C) The velocity-time curve has a positive slope everywhere for $t > 0$ that approaches zero asymptotically.
(D) The viscous drag force at any instant $t$ is directly proportional to the distance fallen $y(t)$.
Question 7 (JEE Advanced One or More Correct)
Two spherical droplets $A$ and $B$ have radii in the ratio $r_A / r_B = 2$. Both are allowed to fall through air from rest until they attain their terminal velocities. Which of the following ratios is/are correct?
(A) Ratio of terminal speeds $v_{tA} / v_{tB} = 4$.
(B) Ratio of viscous drag forces at terminal speeds $F_{vA} / F_{vB} = 8$.
(C) Ratio of relaxation time constants $\tau_A / \tau_B = 4$.
(D) Ratio of rates of heat dissipation at terminal speed $P_A / P_B = 32$.
Question 8 (JEE Advanced One or More Correct)
A jar filled with viscous liquid containing a falling sphere is placed inside an elevator. Under which of the following elevator motions will the terminal velocity of the sphere be strictly zero?
(A) The elevator is moving upward with a constant velocity $v = 5\text{ m/s}$.
(B) The elevator cable snaps and the elevator falls freely under gravity.
(C) The elevator accelerates downward with acceleration $a = g$.
(D) The elevator accelerates upward with acceleration $a = g$.
Question 9 (JEE Advanced Numerical Value)
Eight ($8$) spherical raindrops each of radius $r = 1.0\text{ mm}$ falling at terminal speed $v_t = 2.5\text{ m/s}$ coalesce into a single large spherical drop. Calculate the terminal speed of the newly formed drop in $\text{m/s}$.
Question 10 (JEE Advanced Numerical Value)
A solid spherical ball of density $\rho = 8000\text{ kg/m}^3$ and radius $r = 3.0\text{ mm}$ falls through a fluid of density $\sigma = 1000\text{ kg/m}^3$. It attains a terminal velocity of $v_t = 0.70\text{ m/s}$. Find the dynamic viscosity $\eta$ of the fluid in $\text{Pa}\cdot\text{s}$. (Take $g = 9.8\text{ m/s}^2$).
4. Detailed Solutions & Answer Key
Quick Answer Key
| Question | Type | Correct Answer |
|---|---|---|
| 1 | Single Correct | (B) |
| 2 | Single Correct | (C) |
| 3 | Single Correct | (B) |
| 4 | Single Correct | (D) |
| 5 | Single Correct | (A) |
| 6 | One or More Correct | (A, B, C) |
| 7 | One or More Correct | (A, B, C, D) |
| 8 | One or More Correct | (B, C) |
| 9 | Numerical | 10 |
| 10 | Numerical | 0.20 (or $0.2$) |
Step-by-Step Solutions
Solution 1:
Terminal velocity is directly proportional to the square of the radius: $v_t \propto r^2$.
$$\frac{v_1}{v_2} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$
Correct Answer: (B)
Solution 2:
When $\rho < \sigma$, the buoyant force exceeds the true weight ($F_B > W$). The net effective force acts upward, driving the ball upward until the downward viscous drag force balances the net upward buoyancy: $F_v = F_B – W$. The ball ascends with a constant terminal upward speed.
Correct Answer: (C)
Solution 3:
For $n = 27$ identical droplets coalescing into a single droplet of radius $R$:
$$\frac{4}{3}\pi R^3 = 27 \left(\frac{4}{3}\pi r^3\right) \implies R = 3r$$
The new terminal velocity is:
$$v_t’ = \left(\frac{R}{r}\right)^2 v_t = (3)^2 v_t = 9\, v_t$$
Correct Answer: (B)
Solution 4:
Rate of heat dissipation equals the power dissipated by the viscous drag force at terminal speed:
$$\frac{dQ}{dt} = F_v \cdot v_t = (6\pi \eta r v_t) v_t = 6\pi \eta r v_t^2$$
Since $v_t \propto r^2$, we substitute $v_t^2 \propto r^4$:
$$\frac{dQ}{dt} \propto r \cdot r^4 = r^5$$
Correct Answer: (D)
Solution 5:
The equation of motion is $a = \frac{dv}{dt} = \frac{v_t – v}{\tau}$.
Initial acceleration at $v = 0$ is $a_0 = \frac{v_t}{\tau} = g\left(1 – \frac{\sigma}{\rho}\right)$.
When velocity $v = 0.6\, v_t$:
$$a = \frac{v_t – 0.6\, v_t}{\tau} = 0.4 \frac{v_t}{\tau} = 0.4\, g\left(1 – \frac{\sigma}{\rho}\right)$$
Correct Answer: (A)
Solution 6:
(A) Acceleration $a(t) = a_0 e^{-t/\tau}$ decreases exponentially and monotonically to zero. Correct.
(B) Net force $F_{\text{net}}(t) = m a(t) = m a_0 e^{-t/\tau}$ decreases exponentially. Correct.
(C) $v(t) = v_t(1 – e^{-t/\tau})$. The slope $\frac{dv}{dt} = a_0 e^{-t/\tau} > 0$ for all finite $t$, approaching $0$ as $t \to \infty$. Correct.
(D) $F_v(t) = 6\pi \eta r v(t)$ is proportional to velocity $v(t)$, not displacement $y(t)$. Incorrect.
Correct Answer: (A, B, C)
Solution 7:
Given $r_A / r_B = 2$:
(A) $v_{tA} / v_{tB} = (r_A / r_B)^2 = 2^2 = 4$. Correct.
(B) $F_{vA} / F_{vB} = \frac{6\pi \eta r_A v_{tA}}{6\pi \eta r_B v_{tB}} = \left(\frac{r_A}{r_B}\right)\left(\frac{v_{tA}}{v_{tB}}\right) = 2 \times 4 = 8$. Correct.
(C) $\tau = \frac{2\rho r^2}{9\eta} \propto r^2 \implies \tau_A / \tau_B = (r_A / r_B)^2 = 4$. Correct.
(D) $P = \frac{dQ}{dt} \propto r^5 \implies P_A / P_B = (r_A / r_B)^5 = 2^5 = 32$. Correct.
Correct Answer: (A, B, C, D)
Solution 8:
Terminal velocity is directly proportional to effective gravity: $v_t = \frac{2}{9}\frac{r^2(\rho – \sigma)g_{\text{eff}}}{\eta}$.
When the elevator falls freely ($a = g$ downward) or cables snap, $g_{\text{eff}} = g – g = 0$, so $v_t = 0$.
Constant velocity gives $a = 0 \implies g_{\text{eff}} = g \neq 0$. Upward acceleration gives $g_{\text{eff}} = g + a = 2g \neq 0$.
Correct Answer: (B, C)
Solution 9:
For $n = 8$ droplets coalescing into a single droplet of radius $R$:
$$R^3 = 8 r^3 \implies R = 2 r$$
The new terminal velocity is:
$$v_t’ = \left(\frac{R}{r}\right)^2 v_t = 2^2 \times 2.5 = 4 \times 2.5 = 10.0\text{ m/s}$$
Correct Answer: 10
Solution 10:
Parameters:
$r = 3.0 \times 10^{-3}\text{ m}$, $\rho = 8000\text{ kg/m}^3$, $\sigma = 1000\text{ kg/m}^3$, $v_t = 0.70\text{ m/s}$, $g = 9.8\text{ m/s}^2$.
$$\rho – \sigma = 8000 – 1000 = 7000\text{ kg/m}^3$$
From the terminal velocity formula:
$$v_t = \frac{2}{9}\, \frac{r^2 (\rho – \sigma) g}{\eta} \implies \eta = \frac{2}{9}\, \frac{r^2 (\rho – \sigma) g}{v_t}$$
$$\eta = \frac{2}{9} \times \frac{(3.0 \times 10^{-3})^2 \times 7000 \times 9.8}{0.70}$$
$$\eta = \frac{2}{9} \times \frac{9.0 \times 10^{-6} \times 7000 \times 9.8}{0.70} = 2 \times 10^{-6} \times 7000 \times \left(\frac{9.8}{0.70}\right)$$
Notice that $\frac{9.8}{0.70} = 14$:
$$\eta = 2 \times 10^{-6} \times 7000 \times 14 = 0.014 \times 14 = 0.196\text{ Pa}\cdot\text{s} \approx 0.20\text{ Pa}\cdot\text{s}$$
Correct Answer: 0.20