201. For spheres of the same material falling through the same liquid under Stokes-law conditions, a graph of terminal speed \(v_t\) against \(r^2\) is a straight line through the origin with slope
\[
\frac{v_t}{r^2}=4.0\times10^{4}\,\text{m}^{-1}\text{s}^{-1}.
\]
The sphere-fluid density difference is \(900\,\text{kg m}^{-3}\) and \(g=10\,\text{m s}^{-2}\). What is the liquid viscosity, and what terminal speed would a sphere have if its diameter were doubled from that of a sphere of radius \(1.0\,\text{mm}\)?
ⓐ. \(0.050\,\text{Pa s}\) and \(0.16\,\text{m s}^{-1}\)
ⓑ. \(0.10\,\text{Pa s}\) and \(0.08\,\text{m s}^{-1}\)
ⓒ. \(0.050\,\text{Pa s}\) and \(0.08\,\text{m s}^{-1}\)
ⓓ. \(0.20\,\text{Pa s}\) and \(0.16\,\text{m s}^{-1}\)
Correct Answer: \(0.050\,\text{Pa s}\) and \(0.16\,\text{m s}^{-1}\)
Explanation: The Stokes terminal-speed relation is
\[
v_t=
\frac{2r^2(\rho_s-\rho_f)g}{9\eta}
\]
Therefore the graph slope is
\[
\frac{v_t}{r^2}
=
\frac{2(\rho_s-\rho_f)g}{9\eta}
\]
\(\textbf{Infer the viscosity from the slope:}\)
\[
4.0\times10^{4}
=
\frac{2(900)(10)}{9\eta}
\]
Hence,
\[
\eta=
\frac{18000}
{9(4.0\times10^{4})}
\]
Therefore,
\[
\eta=0.050\,\text{Pa s}
\]
\(\textbf{Initial terminal speed for r=1.0\,\text{mm}:}\)
\[
v_{t1}
=
(4.0\times10^{4})(1.0\times10^{-3})^2
=
0.040\,\text{m s}^{-1}
\]
Doubling the diameter doubles the radius.
Since
\[
v_t\propto r^2,
\]
the terminal speed increases by a factor of four:
\[
v_{t2}=4(0.040)=0.16\,\text{m s}^{-1}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\eta=0.050\,\text{Pa s}\quad\text{and}\quad v_{t2}=0.16\,\text{m s}^{-1}}
\]
202. A small sphere has exactly the same density as the surrounding stationary liquid and is released from rest well away from boundaries. Assume the standard buoyancy and Stokes-drag treatment.
Statement I: Initially, its weight and buoyant force are equal in magnitude.
Statement II: If released from rest, it has no gravitational settling or rising tendency in this idealised situation.
Statement III: If it is forced to move relative to the liquid, it can still experience a Stokes viscous drag.
Statement IV: Decreasing the viscosity necessarily gives it a nonzero gravitational terminal speed even though the sphere-fluid density difference remains zero.
ⓐ. I and II only
ⓑ. II, III and IV only
ⓒ. I, II and III only
ⓓ. I, III and IV only
Correct Answer: I, II and III only
Explanation: Statement I is correct because equal sphere and fluid densities give
\[
W=\rho_sVg
\]
and
\[
F_b=\rho_fVg
\]
with
\[
\rho_s=\rho_f.
\]
Therefore,
\[
W=F_b.
\]
Statement II is correct because the gravitational driving term
\[
W-F_b
\]
is zero.
If the sphere is also initially at rest relative to the liquid, its Stokes drag is zero, so there is no net force producing settling or rising.
Statement III is correct because Stokes drag depends on relative motion:
\[
F_d=6\pi\eta rv.
\]
An externally forced relative velocity can therefore produce drag even for a neutrally buoyant sphere.
Statement IV is incorrect because the terminal-speed expression contains the density difference:
\[
v_t\propto\frac{\rho_s-\rho_f}{\eta}.
\]
If that density difference is zero, merely reducing \(\eta\) does not create a gravitational settling speed.
Hence Statements I, II and III only are correct.
203. Use the flow arrangement described below.
An incompressible liquid flows steadily through a tube that narrows from section P of area \(6.0\,\text{cm}^2\) to section Q of area \(3.0\,\text{cm}^2\). The speed at P is \(2.0\,\text{m s}^{-1}\). In a proposed streamline sketch, two different curves labelled S1 and S2 intersect at a point X and have different tangent directions at X.
Which conclusion is correct?
ⓐ. \(v_Q=1.0\,\text{m s}^{-1}\), and crossing streamlines are allowed because the flow is steady
ⓑ. \(v_Q=4.0\,\text{m s}^{-1}\); S1 and S2 cannot intersect with different tangents, as velocity at X has one direction
ⓒ. \(v_Q=4.0\,\text{m s}^{-1}\), but the intersection proves that the liquid is incompressible
ⓓ. \(v_Q=2.0\,\text{m s}^{-1}\), and streamline direction is unrelated to local velocity
Correct Answer: \(v_Q=4.0\,\text{m s}^{-1}\); S1 and S2 cannot intersect with different tangents, as velocity at X has one direction
Explanation: For steady incompressible flow through the tube,
\[
A_Pv_P=A_Qv_Q
\]
Therefore,
\[
(6.0)(2.0)=(3.0)v_Q
\]
Hence,
\[
v_Q=4.0\,\text{m s}^{-1}
\]
The speed increases because the same volume flow rate passes through the smaller area.
A streamline is tangent to the local fluid-velocity direction at every point.
At any specified point in the flow, the local velocity has one definite direction.
If two purported streamlines crossed at X with different tangents, they would assign two different velocity directions to the same point.
That is impossible for the stated streamline description.
Thus S1 and S2 cannot both be genuine streamlines.
The continuity result and the non-crossing property arise from different physical ideas: conservation of flow and uniqueness of the local velocity direction.
\(\textbf{Final answer:}\) \(v_Q=4.0\,\text{m s}^{-1}\), and the two differently directed curves cannot both be streamlines.
204. Four sections are claimed to belong to the same steady flow tube carrying a fluid whose density may vary from section to section.
| Section | Density \((\text{kg m}^{-3})\) | Area \((\text{cm}^2)\) | Speed \((\text{m s}^{-1})\) |
|---|
| I | 1.2 | 5.0 | 4.0 |
| II | 0.8 | 6.0 | 5.0 |
| III | 1.5 | 4.0 | 4.0 |
| IV | 1.0 | 3.0 | 6.0 |
Which conclusion is correct?
ⓐ. Section I is inconsistent because volume flow rate rather than mass flow rate must be constant
ⓑ. Section II is inconsistent; its speed should be \(4.0\,\text{m s}^{-1}\)
ⓒ. Section III is inconsistent; its density should be \(1.0\,\text{kg m}^{-3}\)
ⓓ. Section IV is inconsistent; its speed should be \(8.0\,\text{m s}^{-1}\)
Correct Answer: Section IV is inconsistent; its speed should be \(8.0\,\text{m s}^{-1}\)
Explanation: For steady flow when density changes,
\[
\rho Av=\text{constant}.
\]
The common conversion factor from \(\text{cm}^2\) to \(\text{m}^2\) is the same for every row, so the products can first be compared directly.
For Section I,
\[
\rho Av=(1.2)(5.0)(4.0)=24.
\]
For Section II,
\[
\rho Av=(0.8)(6.0)(5.0)=24.
\]
For Section III,
\[
\rho Av=(1.5)(4.0)(4.0)=24.
\]
Thus the first three sections have the same mass-flow factor.
For Section IV, the listed values give
\[
\rho Av=(1.0)(3.0)(6.0)=18.
\]
To obtain the common value,
\[
(1.0)(3.0)v=24.
\]
Therefore,
\[
v=8.0\,\text{m s}^{-1}.
\]
Using \(Av=\text{constant}\) here would be unjustified because density is explicitly changing.
\(\textbf{Final answer:}\) Section IV is inconsistent; its speed should be \(8.0\,\text{m s}^{-1}\).
205. An incompressible liquid flows through a main pipe of cross-sectional area \(12\,\text{cm}^2\) with speed \(u\), then divides into branches P and Q of areas \(4\,\text{cm}^2\) and \(6\,\text{cm}^2\). Valves maintain the volume-flow-rate ratio
\[
Q_P:Q_Q=2:1.
\]
For the operating conditions, the supplied critical speeds are \(6.0\,\text{m s}^{-1}\) in P and \(3.0\,\text{m s}^{-1}\) in Q. What is the greatest possible main-pipe speed that does not exceed either critical speed, and what is the corresponding speed in Q?
ⓐ. \(u_{\max}=4.5\,\text{m s}^{-1}\), \(v_Q=3.0\,\text{m s}^{-1}\)
ⓑ. \(u_{\max}=3.0\,\text{m s}^{-1}\), \(v_Q=2.0\,\text{m s}^{-1}\)
ⓒ. \(u_{\max}=2.0\,\text{m s}^{-1}\), \(v_Q=3.0\,\text{m s}^{-1}\)
ⓓ. \(u_{\max}=6.0\,\text{m s}^{-1}\), \(v_Q=2.0\,\text{m s}^{-1}\)
Correct Answer: \(u_{\max}=3.0\,\text{m s}^{-1}\), \(v_Q=2.0\,\text{m s}^{-1}\)
Explanation: The total volume flow rate in the main pipe is
\[
Q=12u
\]
when the common area unit is retained for ratio work.
The branch-flow ratio is
\[
Q_P:Q_Q=2:1
\]
Therefore,
\[
Q_P=\frac23Q=8u,
\qquad
Q_Q=\frac13Q=4u
\]
\(\textbf{Branch speeds:}\)
For P,
\[
v_P=\frac{Q_P}{A_P}
=
\frac{8u}{4}
=
2u
\]
For Q,
\[
v_Q=\frac{Q_Q}{A_Q}
=
\frac{4u}{6}
=
\frac{2u}{3}
\]
\(\textbf{Apply the critical-speed restriction in P:}\)
\[
2u\le6.0,
\]
so
\[
u\le3.0\,\text{m s}^{-1}
\]
\(\textbf{Apply the restriction in Q:}\)
\[
\frac{2u}{3}\le3.0,
\]
so
\[
u\le4.5\,\text{m s}^{-1}
\]
The stricter condition is therefore set by branch P.
At
\[
u_{\max}=3.0\,\text{m s}^{-1},
\]
\[
v_Q=\frac{2}{3}(3.0)=2.0\,\text{m s}^{-1}
\]
\(\textbf{Final answer:}\)
\[
\boxed{u_{\max}=3.0\,\text{m s}^{-1}\quad\text{and}\quad v_Q=2.0\,\text{m s}^{-1}}
\]
206. For a family of circular tubes, suppose the experimentally supplied transition relation is
\[
v_c=K\frac{\eta}{\rho D},
\]
where the same constant \(K\) applies. Define the corresponding critical volume flow rate by
\[
Q_c=Av_c.
\]
Tube 2 has viscosity \(\eta_2=\eta_1/2\), fluid density \(\rho_2=1.5\rho_1\), and diameter \(D_2=3D_1\). Which pair of ratios is correct?
ⓐ. \(v_{c2}/v_{c1}=1/3,\ Q_{c2}/Q_{c1}=3\)
ⓑ. \(v_{c2}/v_{c1}=1/9,\ Q_{c2}/Q_{c1}=1/9\)
ⓒ. \(v_{c2}/v_{c1}=1/6,\ Q_{c2}/Q_{c1}=3/2\)
ⓓ. \(v_{c2}/v_{c1}=1/9,\ Q_{c2}/Q_{c1}=1\)
Correct Answer: \(v_{c2}/v_{c1}=1/9,\ Q_{c2}/Q_{c1}=1\)
Explanation: From the supplied relation,
\[
v_c\propto\frac{\eta}{\rho D}
\]
Therefore,
\[
\frac{v_{c2}}{v_{c1}}
=
\frac{\eta_2}{\eta_1}
\frac{\rho_1}{\rho_2}
\frac{D_1}{D_2}
\]
Substitution gives
\[
\frac{v_{c2}}{v_{c1}}
=
\left(\frac12\right)
\left(\frac{1}{1.5}\right)
\left(\frac13\right)
\]
Hence,
\[
\frac{v_{c2}}{v_{c1}}=\frac19
\]
\(\textbf{Now include the change of tube area:}\)
For a circular tube,
\[
A\propto D^2
\]
Thus,
\[
\frac{A_2}{A_1}=3^2=9
\]
Since
\[
Q_c=Av_c,
\]
we obtain
\[
\frac{Q_{c2}}{Q_{c1}}
=
9\left(\frac19\right)
=
1
\]
The second tube has a much lower critical speed but a ninefold larger area, so its critical volume flow rate is unchanged.
\(\textbf{Final answer:}\) \(v_{c2}/v_{c1}=1/9\) and \(Q_{c2}/Q_{c1}=1\).
207. A gas flows through a fixed tube. For this particular system, an experimentally supplied relation states that the critical speed is directly proportional to gas viscosity when density and tube dimensions are held fixed. The gas is heated so that its viscosity increases by \(25\%\), while the density is treated as unchanged. The pump maintains the same actual mean flow speed, which before heating was \(0.90\) of the original critical speed. Which statement is correct?
ⓐ. The new operating ratio is \(1.125\), so the gas necessarily crosses above the new critical speed
ⓑ. The new critical speed decreases to \(0.75\) of its original value
ⓒ. The operating ratio remains \(0.90\) because actual speed is unchanged
ⓓ. The new operating ratio \(v/v_c=0.72\), so the flow moves farther below the critical threshold
Correct Answer: The new operating ratio \(v/v_c=0.72\), so the flow moves farther below the critical threshold
Explanation: The usual viscosity trend for a gas is an increase with increasing temperature.
The problem states quantitatively that
\[
\eta_2=1.25\eta_1
\]
The supplied relation gives
\[
v_c\propto\eta
\]
Therefore,
\[
v_{c2}=1.25v_{c1}
\]
The actual mean speed is maintained by the pump at
\[
v=0.90v_{c1}
\]
Relative to the new critical speed,
\[
\frac{v}{v_{c2}}
=
\frac{0.90v_{c1}}{1.25v_{c1}}
\]
Hence,
\[
\frac{v}{v_{c2}}=0.72
\]
Thus the operating point moves farther below the supplied transition threshold.
The fact that the actual speed itself is unchanged does not mean its ratio to the critical speed is unchanged.
\(\textbf{Final answer:}\) The new operating ratio is \(0.72\), below the new critical speed.
208. Use the fixed-point velocity observations described below.
At point P in a flowing fluid, a velocity-versus-time graph is a horizontal line at \(3.0\,\text{m s}^{-1}\). At point Q in another flow, the instantaneous velocity fluctuates irregularly with time between approximately \(2.2\,\text{m s}^{-1}\) and \(3.8\,\text{m s}^{-1}\), although its long-time average is \(3.0\,\text{m s}^{-1}\).
Which interpretation is correct?
ⓐ. Both flows are steady because their average speeds are the same
ⓑ. P is consistent with steady flow; Q is unsteady because its local instantaneous velocity varies with time
ⓒ. Q is steady because its velocity never becomes zero
ⓓ. P must be turbulent because a nonzero constant velocity cannot occur in streamline flow
Correct Answer: P is consistent with steady flow; Q is unsteady because its local instantaneous velocity varies with time
Explanation: Steady flow is judged by the behaviour of flow variables at a fixed spatial point.
At P, the measured velocity remains
\[
3.0\,\text{m s}^{-1}
\]
with time.
That observation is therefore consistent with steady flow at P.
At Q, the instantaneous velocity repeatedly changes with time at the same fixed location.
A constant long-time average does not remove those instantaneous fluctuations.
Such irregular local time variation is inconsistent with the simple steady streamline description and is characteristic of turbulent behaviour.
The criterion is not whether the velocity is nonzero.
A fluid can have a nonzero, time-independent local velocity in perfectly steady flow.
Likewise, two flows having the same average speed need not have the same flow character.
Hence option B correctly distinguishes the fixed-point observations.
209. A fluid whose density changes along a tube flows steadily through two sections. At section 1,
\[
\rho_1=1.2\,\text{kg m}^{-3},\qquad
A_1=30\,\text{cm}^2,
\]
and the supplied critical speed is \(4.0\,\text{m s}^{-1}\). At section 2,
\[
\rho_2=0.80\,\text{kg m}^{-3},\qquad
A_2=20\,\text{cm}^2,
\]
and the supplied critical speed is \(6.0\,\text{m s}^{-1}\). What is the greatest steady mass flow rate that does not exceed either critical speed, and what is the speed at section 1 at that limiting flow?
ⓐ. \(1.44\times10^{-2}\,\text{kg s}^{-1}\) and \(4.0\,\text{m s}^{-1}\)
ⓑ. \(9.6\times10^{-3}\,\text{kg s}^{-1}\) and \(2.67\,\text{m s}^{-1}\)
ⓒ. \(9.6\times10^{-3}\,\text{kg s}^{-1}\) and \(4.0\,\text{m s}^{-1}\)
ⓓ. \(1.20\times10^{-2}\,\text{kg s}^{-1}\) and \(3.33\,\text{m s}^{-1}\)
Correct Answer: \(9.6\times10^{-3}\,\text{kg s}^{-1}\) and \(2.67\,\text{m s}^{-1}\)
Explanation: For steady flow,
\[
\dot m=\rho Av
\]
Each section therefore places its own upper bound on the mass flow rate.
\(\textbf{Maximum allowed by section 1:}\)
\[
A_1=30\,\text{cm}^2=3.0\times10^{-3}\,\text{m}^2
\]
Thus,
\[
\dot m_{1,\max}
=
(1.2)(3.0\times10^{-3})(4.0)
\]
Hence,
\[
\dot m_{1,\max}=1.44\times10^{-2}\,\text{kg s}^{-1}
\]
\(\textbf{Maximum allowed by section 2:}\)
\[
A_2=20\,\text{cm}^2=2.0\times10^{-3}\,\text{m}^2
\]
Therefore,
\[
\dot m_{2,\max}
=
(0.80)(2.0\times10^{-3})(6.0)
=
9.6\times10^{-3}\,\text{kg s}^{-1}
\]
Section 2 gives the smaller allowable value and therefore sets the system limit.
\(\textbf{Speed at section 1 at this limiting mass flow:}\)
\[
v_1=
\frac{9.6\times10^{-3}}
{(1.2)(3.0\times10^{-3})}
\]
Thus,
\[
v_1\approx2.67\,\text{m s}^{-1}
\]
This is safely below section 1's own \(4.0\,\text{m s}^{-1}\) critical value.
\(\textbf{Final answer:}\)
\[
\boxed{\dot m_{\max}=9.6\times10^{-3}\,\text{kg s}^{-1}\quad\text{and}\quad v_1\approx2.67\,\text{m s}^{-1}}
\]
210. A sphere of radius \(1.0\,\text{mm}\) and density \(2800\,\text{kg m}^{-3}\) falls at terminal speed \(1.0\times10^{-2}\,\text{m s}^{-1}\) through a liquid of density \(1000\,\text{kg m}^{-3}\). Take \(g=10\,\text{m s}^{-2}\), and assume Stokes' law applies. The same liquid is then used in a circular tube of diameter \(2.0\,\text{cm}\). For this tube, an experimentally supplied relation is
\[
v_c=50\frac{\eta}{\rho D}.
\]
The liquid volume flow rate in the tube is \(2.512\times10^{-4}\,\text{m}^3\text{s}^{-1}\). Take \(\pi=3.14\). Which conclusion is correct?
ⓐ. \(\eta=0.40\,\text{Pa s}\), \(v_c=1.0\,\text{m s}^{-1}\), and the mean tube speed is \(0.80\,\text{m s}^{-1}\), below critical speed
ⓑ. \(\eta=0.20\,\text{Pa s}\), \(v_c=0.50\,\text{m s}^{-1}\), and the tube flow exceeds critical speed
ⓒ. \(\eta=0.40\,\text{Pa s}\), \(v_c=2.0\,\text{m s}^{-1}\), and the mean tube speed equals critical speed
ⓓ. \(\eta=0.80\,\text{Pa s}\), \(v_c=2.0\,\text{m s}^{-1}\), and the mean tube speed is \(1.6\,\text{m s}^{-1}\)
Correct Answer: \(\eta=0.40\,\text{Pa s}\), \(v_c=1.0\,\text{m s}^{-1}\), and the mean tube speed is \(0.80\,\text{m s}^{-1}\), below critical speed
Explanation: \(\textbf{Infer viscosity from the falling-sphere experiment:}\)
For Stokes-law terminal motion,
\[
v_t=
\frac{2r^2(\rho_s-\rho_f)g}{9\eta}
\]
Therefore,
\[
\eta=
\frac{2r^2(\rho_s-\rho_f)g}{9v_t}
\]
Here,
\[
\rho_s-\rho_f=2800-1000=1800\,\text{kg m}^{-3}
\]
Substitution gives
\[
\eta=
\frac{2(1.0\times10^{-3})^2(1800)(10)}
{9(1.0\times10^{-2})}
\]
Hence,
\[
\eta=0.40\,\text{Pa s}
\]
\(\textbf{Use the supplied critical-speed relation:}\)
\[
v_c=
50\frac{0.40}{(1000)(0.020)}
\]
Therefore,
\[
v_c=1.0\,\text{m s}^{-1}
\]
\(\textbf{Find the actual mean tube speed:}\)
The tube cross-sectional area is
\[
A=\frac{\pi D^2}{4}
\]
Thus,
\[
A=
\frac{3.14(0.020)^2}{4}
=
3.14\times10^{-4}\,\text{m}^2
\]
Since
\[
Q=Av,
\]
\[
v=
\frac{2.512\times10^{-4}}
{3.14\times10^{-4}}
=
0.80\,\text{m s}^{-1}
\]
Finally,
\[
0.80\lt 1.0\,\text{m s}^{-1},
\]
so the operating speed remains below the supplied critical speed.
The terminal-speed measurement determines viscosity, which then enters the tube-flow calculation and transition assessment using the stated critical speed.
\(\textbf{Final answer:}\) \(\eta=0.40\,\text{Pa s}\), \(v_c=1.0\,\text{m s}^{-1}\), and the mean speed is \(0.80\,\text{m s}^{-1}\), below critical.
211. A Newtonian liquid is sheared between a stationary plate and a parallel moving plate. Initially the plate separation is \(2.0\,\text{mm}\), the moving-plate speed is \(0.40\,\text{m s}^{-1}\), and the required steady tangential force is \(6.0\,\text{N}\). The same liquid and plate area are then used with separation \(1.0\,\text{mm}\) and speed \(0.60\,\text{m s}^{-1}\). What are the new tangential force and the mechanical power required to maintain the motion?
ⓐ. \(9.0\,\text{N}\) and \(5.4\,\text{W}\)
ⓑ. \(12\,\text{N}\) and \(7.2\,\text{W}\)
ⓒ. \(18\,\text{N}\) and \(10.8\,\text{W}\)
ⓓ. \(18\,\text{N}\) and \(7.2\,\text{W}\)
Correct Answer: \(18\,\text{N}\) and \(10.8\,\text{W}\)
Explanation: For the parallel-layer arrangement,
\[
F=\eta A\frac{v}{l}
\]
The liquid and plate area are unchanged, so
\[
F\propto\frac{v}{l}
\]
\(\textbf{Changed shear condition:}\)
\[
\frac{F_2}{F_1}
=
\frac{v_2}{v_1}
\frac{l_1}{l_2}
\]
Substitution gives
\[
\frac{F_2}{6.0}
=
\frac{0.60}{0.40}
\frac{2.0}{1.0}
\]
Hence,
\[
F_2=6.0(1.5)(2)=18\,\text{N}
\]
\(\textbf{Mechanical power needed at steady speed:}\)
The applied force balances the viscous resistance, so
\[
P=Fv
\]
Therefore,
\[
P_2=(18)(0.60)=10.8\,\text{W}
\]
Halving the gap and increasing the speed both increase the velocity gradient, while power contains an additional factor of the plate speed.
\(\textbf{Final answer:}\)
\[
\boxed{F_2=18\,\text{N}\quad\text{and}\quad P_2=10.8\,\text{W}}
\]
212. A dense sphere is falling at its terminal speed \(v_t\) through a liquid of viscosity \(\eta\). It suddenly enters another liquid having the same density as the first liquid but viscosity \(2\eta\). The sphere density and \(g\) are unchanged, and Stokes' law remains applicable. Let \(a_0\) be the magnitude of the sphere's initial acceleration when it was originally released from rest in the first liquid. Immediately after entering the second liquid, and after a new terminal state is eventually reached, which statement is correct?
ⓐ. Immediate acceleration is downward with magnitude \(a_0\), and new terminal speed is \(2v_t\)
ⓑ. Immediate acceleration is zero, and terminal speed remains \(v_t\)
ⓒ. Immediate acceleration is upward with magnitude \(a_0/2\), and new terminal speed is \(v_t/2\)
ⓓ. Immediate acceleration is upward with magnitude \(a_0\), and new terminal speed is \(v_t/2\)
Correct Answer: Immediate acceleration is upward with magnitude \(a_0\), and new terminal speed is \(v_t/2\)
Explanation: Let the constant effective downward force be
\[
K=W-F_b
\]
In the first liquid, at terminal speed,
\[
F_{d1}=K
\]
At the instant of entry into the second liquid, the sphere speed has not yet changed.
Stokes drag is
\[
F_d=6\pi\eta rv
\]
\(\textbf{Immediate force after entering the more viscous liquid:}\)
Since viscosity doubles while the instantaneous speed is still \(v_t\),
\[
F_{d2}=2F_{d1}=2K
\]
The effective downward force remains \(K\), because the liquid density is unchanged.
Therefore the net force is
\[
F_{\text{net}}=K-2K=-K
\]
Its direction is upward.
Originally, at release in the first liquid, the drag was zero, so
\[
a_0=\frac{K}{m}
\]
Hence the immediate acceleration in the second liquid has magnitude
\[
\frac{K}{m}=a_0
\]
\(\textbf{New terminal condition:}\)
The terminal drag must again equal \(K\).
With twice the viscosity,
\[
6\pi(2\eta)rv'_t=6\pi\eta rv_t
\]
Thus,
\[
v'_t=\frac{v_t}{2}
\]
\(\textbf{Final answer:}\) The sphere initially accelerates upward with magnitude \(a_0\), then settles to terminal speed \(v_t/2\).
213. A sphere falls with terminal speed \(v_t\) in liquid P under Stokes-law conditions. It then enters liquid Q. For the same sphere and \(g\),
\[
(\rho_s-\rho_Q)=\frac12(\rho_s-\rho_P),
\qquad
\eta_Q=\frac12\eta_P.
\]
Assume the sphere crosses the boundary with speed \(v_t\). Which conclusion is correct immediately after entry into Q?
ⓐ. It accelerates downward because the viscosity has decreased
ⓑ. It accelerates upward because buoyancy has increased
ⓒ. The sphere enters Q at Q's terminal speed, so its acceleration is zero
ⓓ. The sphere accelerates downward because its terminal speed in Q immediately doubles to \(2v_t\)
Correct Answer: The sphere enters Q at Q's terminal speed, so its acceleration is zero
Explanation: For Stokes-law terminal motion,
\[
v_t=
\frac{2r^2(\rho_s-\rho_f)g}{9\eta}.
\]
For liquid Q,
\[
\frac{v_{tQ}}{v_{tP}}
=
\frac{(\rho_s-\rho_Q)}{(\rho_s-\rho_P)}
\frac{\eta_P}{\eta_Q}.
\]
Using the given changes,
\[
\frac{v_{tQ}}{v_{tP}}
=
\left(\frac12\right)(2)=1.
\]
Thus,
\[
v_{tQ}=v_t.
\]
The sphere enters Q with exactly this speed.
The effective downward force \(W-F_b\) in Q is half its value in P.
At the same time, the Stokes drag at speed \(v_t\) is also half because the viscosity is halved.
Therefore the forces still balance immediately:
\[
W-F_b=F_d.
\]
No transient acceleration occurs at the boundary under the idealised conditions stated.
\(\textbf{Final answer:}\) The sphere is already at Q's terminal speed and has zero acceleration immediately after entry.
214. Two spheres P and Q are made of the same material and fall through the same liquid under Stokes-law conditions. Their terminal speeds satisfy
\[
v_Q=9v_P.
\]
What are the ratios of their radii and terminal viscous-drag magnitudes?
ⓐ. \(r_Q/r_P=3,\ F_{dQ}/F_{dP}=27\)
ⓑ. \(r_Q/r_P=9,\ F_{dQ}/F_{dP}=9\)
ⓒ. \(r_Q/r_P=3,\ F_{dQ}/F_{dP}=9\)
ⓓ. \(r_Q/r_P=\sqrt3,\ F_{dQ}/F_{dP}=3\sqrt3\)
Correct Answer: \(r_Q/r_P=3,\ F_{dQ}/F_{dP}=27\)
Explanation: For the same sphere material, liquid and gravitational field,
\[
v_t\propto r^2
\]
Therefore,
\[
\frac{v_Q}{v_P}
=
\left(\frac{r_Q}{r_P}\right)^2
\]
Given
\[
\frac{v_Q}{v_P}=9,
\]
we obtain
\[
\frac{r_Q}{r_P}=3
\]
\(\textbf{Terminal drag scaling:}\)
At terminal speed,
\[
F_d=W-F_b
\]
For fixed sphere and liquid densities,
\[
F_d\propto V\propto r^3
\]
Hence,
\[
\frac{F_{dQ}}{F_{dP}}
=
\left(\frac{r_Q}{r_P}\right)^3
=
3^3
=
27
\]
Stokes drag itself is linear in \(r\) at a specified speed, but the terminal speed also rises as \(r^2\); together these give the \(r^3\) terminal-drag dependence.
\(\textbf{Final answer:}\)
\[
\boxed{r_Q/r_P=3\quad\text{and}\quad F_{dQ}/F_{dP}=27}
\]
215. The same dense sphere falls through the same liquid on two planets. On planet P, the gravitational acceleration is \(g\), while on planet Q it is \(g/2\). Before the second experiment, the liquid on Q is cooled or otherwise changed so that its viscosity becomes \(2\eta\), while the relevant densities remain unchanged. Under Stokes-law conditions, how do the terminal speed and terminal viscous-drag magnitude on Q compare with those on P?
ⓐ. \(v_{tQ}=v_{tP}/4,\ F_{dQ}=F_{dP}/2\)
ⓑ. \(v_{tQ}=v_{tP}/2,\ F_{dQ}=F_{dP}/4\)
ⓒ. \(v_{tQ}=v_{tP}/4,\ F_{dQ}=F_{dP}/4\)
ⓓ. \(v_{tQ}=v_{tP},\ F_{dQ}=F_{dP}/2\)
Correct Answer: \(v_{tQ}=v_{tP}/4,\ F_{dQ}=F_{dP}/2\)
Explanation: The terminal-speed magnitude is
\[
v_t=
\frac{2r^2(\rho_s-\rho_f)g}{9\eta}
\]
The sphere and density difference are unchanged.
Therefore,
\[
\frac{v_{tQ}}{v_{tP}}
=
\frac{g_Q}{g_P}
\frac{\eta_P}{\eta_Q}
\]
Substitution gives
\[
\frac{v_{tQ}}{v_{tP}}
=
\left(\frac12\right)\left(\frac12\right)
=
\frac14
\]
\(\textbf{Terminal drag is set by force balance:}\)
At terminal speed,
\[
F_d=(\rho_s-\rho_f)Vg
\]
Thus,
\[
\frac{F_{dQ}}{F_{dP}}
=
\frac{g_Q}{g_P}
=
\frac12
\]
The doubled viscosity changes the speed required to generate the balancing drag, but the magnitude of the required terminal drag itself is determined by the effective weight.
\(\textbf{Final answer:}\) \(v_{tQ}=v_{tP}/4\) and \(F_{dQ}=F_{dP}/2\).
216. An incompressible liquid flows through a pipe whose inlet area is \(6.0\,\text{cm}^2\) and outlet area is \(2.0\,\text{cm}^2\). The inlet speed varies with time according to
\[
v_1(t)=1.0+0.20t
\]
in SI units, where \(t\) is in seconds. Assume the instantaneous volume flow rate is the same through both sections. The experimentally supplied critical speed at the outlet is \(6.0\,\text{m s}^{-1}\). Which conclusion is correct?
ⓐ. The outlet speed is constant at \(3.0\,\text{m s}^{-1}\), so the flow is steady
ⓑ. Outlet speed reaches critical at \(t=5.0\,\text{s}\); the time-dependent flow is unsteady
ⓒ. The outlet reaches critical speed at \(t=10\,\text{s}\), but the flow remains steady because continuity is satisfied
ⓓ. The outlet reaches critical speed at \(t=5.0\,\text{s}\), proving that continuity ceases to apply thereafter
Correct Answer: Outlet speed reaches critical at \(t=5.0\,\text{s}\); the time-dependent flow is unsteady
Explanation: At each instant, incompressibility gives
\[
A_1v_1=A_2v_2
\]
Therefore,
\[
v_2=
\frac{A_1}{A_2}v_1
=
3v_1
\]
Using the time-dependent inlet speed,
\[
v_2(t)
=
3(1.0+0.20t)
\]
\(\textbf{Find when the outlet reaches its supplied critical speed:}\)
\[
3(1.0+0.20t)=6.0
\]
Thus,
\[
1.0+0.20t=2.0,
\]
so
\[
t=5.0\,\text{s}
\]
\(\textbf{Classify the flow in time:}\)
At a fixed inlet or outlet section, the velocity changes as \(t\) changes.
Therefore the flow is not steady.
Continuity expresses the instantaneous conservation requirement and does not by itself imply that the flow is steady.
Crossing the supplied critical speed also does not make mass conservation disappear.
\(\textbf{Final answer:}\) The outlet reaches \(6.0\,\text{m s}^{-1}\) at \(5.0\,\text{s}\), and the flow is unsteady.
217. In a transparent tube, a narrow coloured dye filament is injected along the flowing liquid. At a mean speed of \(0.70\,\text{m s}^{-1}\), the filament remains thin and well defined. At \(1.00\,\text{m s}^{-1}\), it remains largely intact but begins to waver slightly. At \(1.20\,\text{m s}^{-1}\), it breaks into irregular structures and mixes rapidly across the flow. Which conclusion is best supported?
ⓐ. The terminal velocity lies between \(1.00\) and \(1.20\,\text{m s}^{-1}\)
ⓑ. Transition toward turbulence occurs between about \(1.00\) and \(1.20\,\text{m s}^{-1}\)
ⓒ. Viscosity becomes zero at \(1.20\,\text{m s}^{-1}\)
ⓓ. The liquid becomes compressible at the onset of dye mixing
Correct Answer: Transition toward turbulence occurs between about \(1.00\) and \(1.20\,\text{m s}^{-1}\)
Explanation: A thin persistent dye filament is characteristic of orderly flow in which neighbouring fluid elements follow comparatively well-defined paths.
At the lower speeds, the injected dye remains confined to a narrow region, consistent with streamline or laminar behaviour.
At \(1.20\,\text{m s}^{-1}\), the filament breaks into irregular structures and mixes strongly across the tube.
That qualitative change is evidence for the onset of turbulent motion.
The last condition showing mostly orderly behaviour is near \(1.00\,\text{m s}^{-1}\), while clear irregular mixing is observed by \(1.20\,\text{m s}^{-1}\).
Therefore the transition region is indicated between these speeds.
Terminal velocity concerns force balance on a moving body and is unrelated to this dye-stream experiment.
Turbulence does not require viscosity to become zero, nor does dye mixing imply that the liquid has become compressible.
Hence option B is the supported interpretation.
218. Consider the following statements about streamline and turbulent flow.
Statement I: In steady streamline flow, the velocity at a fixed point is time-independent.
Statement II: In turbulent flow, irregular velocity fluctuations and eddy-like motion may occur.
Statement III: The onset of turbulence means that viscosity has become exactly zero.
Statement IV: A constant time-averaged speed at a point does not by itself prove that the instantaneous flow there is steady.
Statement V: Crossing an experimentally identified critical speed indicates that the ordinary streamline description may no longer be valid.
Which statements are correct?
ⓐ. I, II, IV and V only
ⓑ. I, III and V only
ⓒ. II, III and IV only
ⓓ. I, II, III, IV and V
Correct Answer: I, II, IV and V only
Explanation: Statement I is correct because steady flow requires the velocity at a fixed spatial point to remain unchanged with time.
Statement II is also correct: turbulence is associated with irregular fluctuations, mixing and eddy-like motion.
Statement III is incorrect.
A fluid can be viscous and nevertheless flow turbulently.
The transition is not produced by viscosity becoming exactly zero.
Statement IV is correct because a time average can remain constant even while the instantaneous velocity fluctuates strongly.
Such fluctuations would violate the simple steady-flow condition.
Statement V is also correct in the supplied experimental sense.
Critical velocity marks the transition region beyond which ordinary streamline behaviour should not automatically be assumed.
Thus Statements I, II, IV and V only are correct.
219. For a family of tubes carrying the same liquid, the experimentally supplied critical-speed relation is
\[
v_c=K\frac{\eta}{\rho D},
\]
with the same constant \(K\). The following data are reported:
| Trial | Viscosity | Density | Diameter | Reported \(v_c\) |
|---|
| I | \(\eta\) | \(\rho\) | \(D\) | \(v_0\) |
| II | \(2\eta\) | \(\rho\) | \(2D\) | \(v_0\) |
| III | \(\eta/2\) | \(\rho/2\) | \(D\) | \(v_0\) |
| IV | \(\eta\) | \(2\rho\) | \(D/2\) | \(v_0/2\) |
Which trial is inconsistent with the supplied relation?
ⓐ. Trial I
ⓑ. Trial II
ⓒ. Trial III
ⓓ. Trial IV
Correct Answer: Trial IV
Explanation: The supplied relation gives
\[
v_c\propto\frac{\eta}{\rho D}.
\]
Trial I is the reference condition and therefore gives \(v_0\).
For Trial II,
\[
\frac{v_{cII}}{v_0}
=
\frac{2}{2}=1,
\]
so the reported value is consistent.
For Trial III,
\[
\frac{v_{cIII}}{v_0}
=
\frac{1/2}{1/2}=1,
\]
so this is also consistent.
For Trial IV,
\[
\frac{v_{cIV}}{v_0}
=
\frac{1}{(2)(1/2)}
=
1.
\]
Thus the predicted value is
\[
v_{cIV}=v_0,
\]
not
\[
\frac{v_0}{2}.
\]
The doubled density and halved diameter cancel in the denominator.
\(\textbf{Final answer:}\) Trial IV is inconsistent.
220. A gas flows steadily through two sections of a tube. At section 2, the gas density is half its value at section 1, while the cross-sectional area is twice as large. What is \(v_2/v_1\) if mass flow rate is conserved?
ⓐ. \(1/4\)
ⓑ. \(1/2\)
ⓒ. \(1\)
ⓓ. \(2\)
Correct Answer: \(1\)
Explanation: For steady flow when the density may change,
\[
\rho Av=\text{constant}
\]
Therefore,
\[
\rho_1A_1v_1
=
\rho_2A_2v_2
\]
The given changes are
\[
\rho_2=\frac{\rho_1}{2},
\qquad
A_2=2A_1
\]
Hence,
\[
\rho_2A_2
=
\left(\frac{\rho_1}{2}\right)(2A_1)
=
\rho_1A_1
\]
Therefore,
\[
v_2=v_1
\]
The density decrease alone would tend to require a greater speed for fixed mass flow, but the simultaneous doubling of area compensates exactly.
This illustrates why \(Av=\text{constant}\) cannot automatically be used when density changes.
The conserved quantity is
\[
\rho Av
\]
