101. A stationary liquid has the following gauge-pressure data.
| Depth below its free surface | Gauge pressure |
|---|
| \(0.50\,\text{m}\) | \(6.0\,\text{kPa}\) |
| \(1.50\,\text{m}\) | \(18.0\,\text{kPa}\) |
Take \(g=10\,\text{m s}^{-2}\). A second liquid is required to produce the same gauge pressure as the first liquid at \(1.50\,\text{m}\), but at a depth of only \(1.20\,\text{m}\). What are the densities of the first and second liquids respectively?
ⓐ. \(800\,\text{kg m}^{-3}\) and \(1000\,\text{kg m}^{-3}\)
ⓑ. \(1200\,\text{kg m}^{-3}\) and \(1500\,\text{kg m}^{-3}\)
ⓒ. \(1500\,\text{kg m}^{-3}\) and \(1200\,\text{kg m}^{-3}\)
ⓓ. \(1000\,\text{kg m}^{-3}\) and \(1200\,\text{kg m}^{-3}\)
Correct Answer: \(1200\,\text{kg m}^{-3}\) and \(1500\,\text{kg m}^{-3}\)
Explanation: For a uniform stationary liquid,
\[
P_g=\rho gh
\]
\(\textbf{Determine the first density:}\)
Using the reading at \(1.50\,\text{m}\),
\[
18.0\times10^{3}
=
\rho_1(10)(1.50)
\]
Therefore,
\[
\rho_1
=
\frac{18.0\times10^{3}}{15}
=
1200\,\text{kg m}^{-3}
\]
The \(0.50\,\text{m}\) data point gives the same density and confirms the linear pressure-depth relation.
\(\textbf{Determine the second density:}\)
The required gauge pressure is again
\[
18.0\times10^{3}\,\text{Pa},
\]
but now at depth
\[
h_2=1.20\,\text{m}
\]
Thus,
\[
18.0\times10^{3}
=
\rho_2(10)(1.20)
\]
Hence,
\[
\rho_2=1500\,\text{kg m}^{-3}
\]
A shallower column can produce the same pressure only if its density is correspondingly greater.
\(\textbf{Final answer:}\)
\[
\boxed{\rho_1=1200\,\text{kg m}^{-3}\quad\text{and}\quad \rho_2=1500\,\text{kg m}^{-3}}
\]
102. An ideal mercury barometer on Earth reads \(76\,\text{cm}\), where \(g_E=10\,\text{m s}^{-2}\) and \(\rho_{\text{Hg}}=1.36\times10^{4}\,\text{kg m}^{-3}\). The same atmospheric pressure is to be balanced on another planet where \(g_P=5.0\,\text{m s}^{-2}\), using a liquid of density \(1000\,\text{kg m}^{-3}\). What column height is required?
ⓐ. \(20.67\,\text{m}\)
ⓑ. \(10.34\,\text{m}\)
ⓒ. \(6.80\,\text{m}\)
ⓓ. \(27.20\,\text{m}\)
Correct Answer: \(20.67\,\text{m}\)
Explanation: The atmospheric pressure represented by the Earth mercury column is
\[
P_a=\rho_{\text{Hg}}g_Eh_{\text{Hg}}
\]
The same pressure on the other planet must satisfy
\[
P_a=\rho_Lg_Ph_L
\]
Equating the two,
\[
\rho_{\text{Hg}}g_Eh_{\text{Hg}}
=
\rho_Lg_Ph_L
\]
Therefore,
\[
h_L
=
\frac{\rho_{\text{Hg}}g_Eh_{\text{Hg}}}
{\rho_Lg_P}
\]
Substitution gives
\[
h_L
=
\frac{(1.36\times10^{4})(10)(0.76)}
{(1000)(5.0)}
\]
Hence,
\[
h_L\approx20.67\,\text{m}
\]
Two effects increase the required height: the replacement liquid is much less dense than mercury and the local gravitational acceleration is smaller.
The pressure being balanced is specified to remain the same, so it is incorrect simply to double the \(76\,\text{cm}\) reading when \(g\) is halved.
\(\textbf{Final answer:}\)
\[
\boxed{h_L\approx20.67\,\text{m}}
\]
103. A sealed tank contains water to a depth \(4.0\,\text{m}\). The gas above the water is initially at absolute pressure \(1.10\times10^{5}\,\text{Pa}\). A small piston of area \(5.0\times10^{-3}\,\text{m}^2\) is fitted at the bottom, with atmospheric pressure \(1.00\times10^{5}\,\text{Pa}\) acting on its outer face. The gas pressure above the water is then increased by \(2.0\times10^{4}\,\text{Pa}\) without changing the water depth appreciably. Take \(\rho_w=1000\,\text{kg m}^{-3}\) and \(g=10\,\text{m s}^{-2}\). By how much does the net outward force on the piston increase, and what is its final magnitude?
ⓐ. Increase \(100\,\text{N}\); final force \(150\,\text{N}\)
ⓑ. Increase \(200\,\text{N}\); final force \(350\,\text{N}\)
ⓒ. Increase \(100\,\text{N}\); final force \(250\,\text{N}\)
ⓓ. Increase \(100\,\text{N}\); final force \(350\,\text{N}\)
Correct Answer: Increase \(100\,\text{N}\); final force \(350\,\text{N}\)
Explanation: \(\textbf{Initial pressure at the inner face:}\)
\[
P_{i1}=P_{g1}+\rho gh
\]
Thus,
\[
P_{i1}
=
1.10\times10^{5}
+
(1000)(10)(4.0)
=
1.50\times10^{5}\,\text{Pa}
\]
The initial pressure difference across the piston is
\[
\Delta P_1
=
1.50\times10^{5}-1.00\times10^{5}
=
5.0\times10^{4}\,\text{Pa}
\]
\(\textbf{Initial net force:}\)
\[
F_1=\Delta P_1A
=
(5.0\times10^{4})(5.0\times10^{-3})
=
250\,\text{N}
\]
\(\textbf{Effect of increasing the gas boundary pressure:}\)
The hydrostatic term \(\rho gh\) remains unchanged because the liquid depth, density and \(g\) are unchanged.
The entire liquid pressure therefore rises by
\[
2.0\times10^{4}\,\text{Pa}
\]
The force increase is
\[
\Delta F
=
(2.0\times10^{4})(5.0\times10^{-3})
=
100\,\text{N}
\]
Hence,
\[
F_2=250+100=350\,\text{N}
\]
This separates the externally imposed boundary-pressure contribution from the gravity-generated hydrostatic pressure difference.
\(\textbf{Final answer:}\) The force increases by \(100\,\text{N}\) and becomes \(350\,\text{N}\).
104. A closed vessel contains a uniform liquid. Two points A and B inside it are separated vertically by \(0.80\,\text{m}\), with B below A. Initially the pressure of the gas above the liquid is \(P_0\). The gas pressure is then increased by \(3.0\times10^{4}\,\text{Pa}\), while liquid density, depth and \(g\) remain unchanged. Consider the following statements.
Statement I: The absolute pressure at A increases by \(3.0\times10^{4}\,\text{Pa}\).
Statement II: The absolute pressure at B increases by \(3.0\times10^{4}\,\text{Pa}\).
Statement III: The pressure difference \(P_B-P_A\) increases by \(3.0\times10^{4}\,\text{Pa}\).
Statement IV: The pressure difference \(P_B-P_A\) remains equal to \(\rho g(0.80)\).
Which statements are correct?
ⓐ. I and III only
ⓑ. II and IV only
ⓒ. I, II and IV only
ⓓ. I, II and III only
Correct Answer: I, II and IV only
Explanation: At a depth \(h\) below a boundary having pressure \(P_s\),
\[
P=P_s+\rho gh.
\]
Increasing \(P_s\) by a fixed amount raises the absolute pressure at every point in the connected static liquid by that same amount.
Therefore Statement I is correct.
For exactly the same reason, Statement II is also correct.
However, the pressure difference between two points within the same uniform liquid is
\[
P_B-P_A=\rho g(h_B-h_A).
\]
The boundary pressure \(P_s\) cancels when this difference is formed.
Since density, \(g\), and vertical separation remain unchanged, the A-B pressure difference does not change.
Statement III is therefore incorrect.
Statement IV is correct because
\[
P_B-P_A=\rho g(0.80).
\]
This distinction separates an overall pressure offset from a hydrostatic pressure gradient.
Hence Statements I, II and IV only are correct.
105. A solid body of volume \(2.0\times10^{-3}\,\text{m}^3\) and bulk modulus \(1.5\times10^{11}\,\text{Pa}\) is lowered from the free surface of an open water tank to a depth \(30\,\text{m}\). Assume the pressure inside the body initially equals atmospheric pressure and that the external hydrostatic pressure increase produces a small uniform elastic compression. Take \(\rho_w=1000\,\text{kg m}^{-3}\) and \(g=10\,\text{m s}^{-2}\). What are the magnitude of its fractional volume decrease and the magnitude of its volume decrease at that depth?
ⓐ. \(2.0\times10^{-3}\) and \(4.0\times10^{-6}\,\text{m}^3\)
ⓑ. \(2.0\times10^{-2}\) and \(4.0\times10^{-5}\,\text{m}^3\)
ⓒ. \(5.0\times10^{-4}\) and \(1.0\times10^{-6}\,\text{m}^3\)
ⓓ. \(2.0\times10^{-6}\) and \(4.0\times10^{-9}\,\text{m}^3\)
Correct Answer: \(2.0\times10^{-6}\) and \(4.0\times10^{-9}\,\text{m}^3\)
Explanation: The body initially experiences atmospheric pressure internally and externally at the free surface.
Therefore the relevant additional compressive pressure at depth is the hydrostatic increase rather than the full absolute pressure.
\(\textbf{Hydrostatic pressure increase:}\)
\[
\Delta p=\rho gh
\]
Substituting,
\[
\Delta p
=
(1000)(10)(30)
=
3.0\times10^{5}\,\text{Pa}
\]
\(\textbf{Fractional volume compression:}\)
For bulk deformation,
\[
B=\frac{\Delta p}{|\Delta V|/V}
\]
Therefore,
\[
\left|\frac{\Delta V}{V}\right|
=
\frac{\Delta p}{B}
\]
Thus,
\[
\left|\frac{\Delta V}{V}\right|
=
\frac{3.0\times10^{5}}
{1.5\times10^{11}}
=
2.0\times10^{-6}
\]
\(\textbf{Absolute volume decrease:}\)
\[
|\Delta V|
=
V\left|\frac{\Delta V}{V}\right|
\]
Hence,
\[
|\Delta V|
=
(2.0\times10^{-3})(2.0\times10^{-6})
=
4.0\times10^{-9}\,\text{m}^3
\]
\(\textbf{Final answer:}\)
\[
\boxed{\left|\frac{\Delta V}{V}\right|=2.0\times10^{-6},\quad |\Delta V|=4.0\times10^{-9}\,\text{m}^3}
\]
106. Assertion: In an ideal enclosed hydraulic system, an additional pressure applied by one piston is transmitted throughout the fluid without diminution.
Reason: Therefore pistons of different areas must experience equal forces from that additional pressure.
ⓐ. Assertion is true, but Reason is false
ⓑ. Assertion is false, but Reason is true
ⓒ. Both Assertion and Reason are true, and Reason correctly explains Assertion
ⓓ. Both Assertion and Reason are false
Correct Answer: Assertion is true, but Reason is false
Explanation: The Assertion is a statement of Pascal's law for an enclosed fluid.
An applied pressure change is transmitted throughout the fluid and to the walls without diminution in the idealised static treatment.
Thus if the transmitted additional pressure is \(\Delta P\), different pistons experience the same pressure increment at comparable levels.
However, pressure and force are not the same quantity.
For a piston,
\[
F=\Delta PA.
\]
Therefore two pistons having different areas experience different forces even though the transmitted pressure is the same.
The larger-area piston experiences the larger force.
This is precisely the physical basis of hydraulic force multiplication.
The Reason is therefore false because it incorrectly converts equal pressure into equal force.
Hence the Assertion is true but the Reason is false.
107. An ideal hydraulic lift has a small piston of diameter \(2.0\,\text{cm}\) and a large piston of diameter \(10.0\,\text{cm}\). The pistons are initially at the same level. If the small piston is pushed downward through \(50\,\text{cm}\) while supporting an input force \(200\,\text{N}\), what are the ideal upward force on the large piston and its upward displacement?
ⓐ. \(1000\,\text{N}\) and \(10\,\text{cm}\)
ⓑ. \(2500\,\text{N}\) and \(5.0\,\text{cm}\)
ⓒ. \(5000\,\text{N}\) and \(10\,\text{cm}\)
ⓓ. \(5000\,\text{N}\) and \(2.0\,\text{cm}\)
Correct Answer: \(5000\,\text{N}\) and \(2.0\,\text{cm}\)
Explanation: \(\textbf{Area ratio from the diameters:}\)
Piston area varies as the square of diameter:
\[
\frac{A_2}{A_1}
=
\left(\frac{d_2}{d_1}\right)^2
\]
Thus,
\[
\frac{A_2}{A_1}
=
\left(\frac{10}{2}\right)^2
=
25
\]
\(\textbf{Force multiplication from Pascal transmission:}\)
At equal levels in the ideal system,
\[
\frac{F_1}{A_1}
=
\frac{F_2}{A_2}
\]
Hence,
\[
F_2=25F_1
=
25(200)
=
5000\,\text{N}
\]
\(\textbf{Displacement from incompressible-volume transfer:}\)
\[
A_1x_1=A_2x_2
\]
Therefore,
\[
x_2
=
\frac{A_1}{A_2}x_1
=
\frac{50\,\text{cm}}{25}
=
2.0\,\text{cm}
\]
The force increases by a factor of 25 while the displacement decreases by the same factor.
This is consistent with ideal work conservation:
\[
F_1x_1=F_2x_2
\]
\(\textbf{Final answer:}\) The ideal output force is \(5000\,\text{N}\) and the large piston rises \(2.0\,\text{cm}\).
108. In a hydraulic system filled with oil of density \(800\,\text{kg m}^{-3}\), a small piston of area \(2.0\,\text{cm}^2\) is \(0.50\,\text{m}\) vertically above a large piston of area \(100\,\text{cm}^2\). Both outer piston faces are exposed to the same atmospheric pressure. A downward force of \(40\,\text{N}\) is applied to the small piston. Neglect piston weights and take \(g=10\,\text{m s}^{-2}\). What net upward force acts on the large piston?
ⓐ. \(1960\,\text{N}\)
ⓑ. \(2000\,\text{N}\)
ⓒ. \(2040\,\text{N}\)
ⓓ. \(2400\,\text{N}\)
Correct Answer: \(2040\,\text{N}\)
Explanation: Because both outer faces are exposed to the same atmospheric pressure, it is convenient to work with gauge pressures.
\(\textbf{Pressure added by the small piston:}\)
The small-piston area is
\[
A_1=2.0\,\text{cm}^2
=
2.0\times10^{-4}\,\text{m}^2
\]
Therefore,
\[
P_1=\frac{F_1}{A_1}
\]
Substituting,
\[
P_1
=
\frac{40}{2.0\times10^{-4}}
=
2.0\times10^{5}\,\text{Pa}
\]
\(\textbf{Hydrostatic increase at the lower piston:}\)
The large piston is \(0.50\,\text{m}\) lower, so its fluid pressure is additionally larger by
\[
\rho gh
=
(800)(10)(0.50)
=
4.0\times10^{3}\,\text{Pa}
\]
Thus,
\[
P_2
=
2.04\times10^{5}\,\text{Pa}
\]
The large-piston area is
\[
A_2=100\,\text{cm}^2
=
1.0\times10^{-2}\,\text{m}^2
\]
Hence,
\[
F_2=P_2A_2
=
(2.04\times10^{5})(1.0\times10^{-2})
=
2040\,\text{N}
\]
At unequal levels, Pascal transmission of an imposed pressure change must be combined with the hydrostatic pressure difference.
\(\textbf{Final answer:}\) The large piston experiences a net upward force of \(2040\,\text{N}\).
109. An ideal hydraulic machine has piston areas \(5.0\,\text{cm}^2\) and \(200\,\text{cm}^2\). A force of \(100\,\text{N}\) pushes the small piston through \(0.40\,\text{m}\). What are the large-piston force, large-piston displacement, and ideal work done on the load?
ⓐ. \(2000\,\text{N},\ 0.020\,\text{m},\ 40\,\text{J}\)
ⓑ. \(4000\,\text{N},\ 0.010\,\text{m},\ 40\,\text{J}\)
ⓒ. \(4000\,\text{N},\ 0.040\,\text{m},\ 160\,\text{J}\)
ⓓ. \(8000\,\text{N},\ 0.005\,\text{m},\ 40\,\text{J}\)
Correct Answer: \(4000\,\text{N},\ 0.010\,\text{m},\ 40\,\text{J}\)
Explanation: \(\textbf{Hydraulic force ratio:}\)
The area ratio is
\[
\frac{A_2}{A_1}
=
\frac{200}{5.0}
=
40
\]
Therefore,
\[
F_2=40F_1
\]
Hence,
\[
F_2=40(100)=4000\,\text{N}
\]
\(\textbf{Volume conservation:}\)
For incompressible fluid displacement,
\[
A_1x_1=A_2x_2
\]
Thus,
\[
x_2
=
\frac{A_1}{A_2}x_1
=
\frac{1}{40}(0.40)
=
0.010\,\text{m}
\]
\(\textbf{Ideal output work:}\)
\[
W_{\text{out}}
=
F_2x_2
\]
Therefore,
\[
W_{\text{out}}
=
(4000)(0.010)
=
40\,\text{J}
\]
The input work is
\[
W_{\text{in}}=(100)(0.40)=40\,\text{J},
\]
consistent with the ideal machine.
A hydraulic machine multiplies force by sacrificing displacement; it does not multiply ideal mechanical work.
\(\textbf{Final answer:}\)
\[
\boxed{4000\,\text{N},\quad 0.010\,\text{m},\quad\text{and}\quad 40\,\text{J}}
\]
110. An ideal hydraulic lift must raise a load of \(1.20\times10^{4}\,\text{N}\) using an input force no greater than \(300\,\text{N}\). The small piston has diameter \(2.0\,\text{cm}\). If the design uses the minimum large-piston area capable of meeting this requirement, what large-piston diameter is needed, and how far will it rise when the small piston moves \(0.50\,\text{m}\)?
ⓐ. \(12.6\,\text{cm}\) and \(1.25\,\text{cm}\)
ⓑ. \(8.0\,\text{cm}\) and \(3.13\,\text{cm}\)
ⓒ. \(40\,\text{cm}\) and \(1.25\,\text{cm}\)
ⓓ. \(12.6\,\text{cm}\) and \(5.0\,\text{cm}\)
Correct Answer: \(12.6\,\text{cm}\) and \(1.25\,\text{cm}\)
Explanation: \(\textbf{Required force multiplication:}\)
For the limiting input force,
\[
\frac{F_2}{F_1}
=
\frac{1.20\times10^{4}}{300}
=
40
\]
Thus the minimum required area ratio is
\[
\frac{A_2}{A_1}=40
\]
\(\textbf{Convert area ratio to diameter ratio:}\)
Because
\[
A\propto d^2,
\]
we have
\[
\left(\frac{d_2}{d_1}\right)^2=40
\]
Therefore,
\[
d_2=d_1\sqrt{40}
\]
With
\[
d_1=2.0\,\text{cm},
\]
\[
d_2
=
2.0\sqrt{40}
\approx12.6\,\text{cm}
\]
\(\textbf{Output displacement:}\)
Volume conservation gives
\[
A_1x_1=A_2x_2
\]
Hence,
\[
x_2
=
\frac{x_1}{40}
=
\frac{0.50}{40}
=
0.0125\,\text{m}
\]
Therefore,
\[
x_2=1.25\,\text{cm}
\]
The same area ratio that gives a fortyfold force multiplication gives a fortyfold reduction in piston displacement.
\(\textbf{Final answer:}\) The large piston requires diameter about \(12.6\,\text{cm}\) and rises \(1.25\,\text{cm}\).
111. In an ideal hydraulic braking arrangement, a force of \(200\,\text{N}\) acts on a master-cylinder piston of area \(2.0\,\text{cm}^2\). The transmitted pressure acts on four identical wheel-cylinder pistons, each of area \(6.0\,\text{cm}^2\). Neglect height differences. What are the force on each wheel-cylinder piston and the sum of the four such forces?
ⓐ. \(200\,\text{N}\) and \(800\,\text{N}\)
ⓑ. \(300\,\text{N}\) and \(1200\,\text{N}\)
ⓒ. \(1200\,\text{N}\) and \(4800\,\text{N}\)
ⓓ. \(600\,\text{N}\) and \(2400\,\text{N}\)
Correct Answer: \(600\,\text{N}\) and \(2400\,\text{N}\)
Explanation: \(\textbf{Pressure produced at the master cylinder:}\)
The master-cylinder area is
\[
A_m=2.0\,\text{cm}^2
=
2.0\times10^{-4}\,\text{m}^2
\]
Therefore,
\[
P=\frac{F_m}{A_m}
\]
Substituting,
\[
P
=
\frac{200}{2.0\times10^{-4}}
=
1.0\times10^{6}\,\text{Pa}
\]
Pascal's law transmits this pressure increase through the brake fluid to the wheel cylinders.
\(\textbf{Force at one wheel cylinder:}\)
Each wheel-piston area is
\[
A_w=6.0\times10^{-4}\,\text{m}^2
\]
Thus,
\[
F_w=PA_w
=
(1.0\times10^{6})(6.0\times10^{-4})
=
600\,\text{N}
\]
There are four identical wheel cylinders, so
\[
F_{\text{sum}}=4(600)=2400\,\text{N}
\]
The fluid transmits pressure; the larger wheel-piston area converts that pressure into a greater force.
\(\textbf{Final answer:}\) Each wheel-cylinder piston receives \(600\,\text{N}\), and their force sum is \(2400\,\text{N}\).
112. An ideal hydraulic brake system has a master piston and four identical wheel pistons. The master-piston force and stroke are kept unchanged. Each wheel piston is replaced by another piston having twice the original area, while the number of wheel pistons is unchanged. Neglect height differences. Compared with the original system, which statement is correct?
ⓐ. The transmitted pressure doubles, each wheel force doubles, and each wheel-piston displacement is unchanged
ⓑ. The transmitted pressure is unchanged, each wheel force doubles, and each wheel-piston displacement is halved
ⓒ. The transmitted pressure is halved, each wheel force is unchanged, and each wheel-piston displacement doubles
ⓓ. The transmitted pressure is unchanged, each wheel force is halved, and each wheel-piston displacement doubles
Correct Answer: The transmitted pressure is unchanged, each wheel force doubles, and each wheel-piston displacement is halved
Explanation: The master-piston force and master-piston area are unchanged.
Therefore the pressure generated by the master cylinder remains
\[
P=\frac{F_m}{A_m},
\]
so the transmitted pressure is unchanged.
For one wheel piston,
\[
F_w=PA_w
\]
Doubling \(A_w\) while keeping \(P\) unchanged therefore doubles the force exerted by each wheel piston.
The displacement consequence follows from incompressible-volume conservation.
If the master piston displaces volume
\[
\Delta V=A_mx_m,
\]
that same total fluid volume is distributed among the four identical wheel cylinders.
Originally,
\[
A_mx_m=4A_wx_w
\]
After each wheel area becomes \(2A_w\),
\[
A_mx_m=4(2A_w)x'_w
\]
Comparing the two equations gives
\[
x'_w=\frac{x_w}{2}
\]
Thus greater force at the wheel pistons is accompanied by smaller wheel-piston displacement for the same master stroke.
\(\textbf{Final answer:}\) Pressure is unchanged, wheel force doubles, and wheel-piston displacement becomes half.
113. Consider an enclosed liquid at rest connected to pistons of different areas and to rigid containing walls. A small additional pressure is applied through one piston.
Statement I: The additional pressure is transmitted throughout the enclosed liquid without diminution in the ideal static treatment.
Statement II: If two pistons are at the same level, the additional forces produced on them are proportional to their areas.
Statement III: Pistons at different vertical levels must have equal total pressure merely because Pascal's law operates.
Statement IV: The fluid pressure force on a small surface element acts normally to that surface.
Which statements are correct?
ⓐ. I and III only
ⓑ. II, III and IV only
ⓒ. I, II and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: Statement I is correct and expresses the pressure-transmission content of Pascal's law.
An imposed pressure increment in an enclosed fluid is transmitted throughout the fluid in the ideal static description.
Statement II is also correct because force and pressure are related by
\[
F=\Delta PA.
\]
If the transmitted pressure increment is the same, a larger piston area produces a proportionally larger additional force.
Statement III is incorrect.
Pascal's law concerns transmission of an imposed pressure change; it does not remove the ordinary hydrostatic variation of total pressure with height.
For two levels separated vertically by \(h\),
\[
\Delta P_{\text{hydrostatic}}=\rho gh
\]
still applies in a gravitational field.
Statement IV is correct because pressure force exerted by a static fluid is normal to the surface on which it acts.
The central distinction is therefore equal transmitted pressure increment versus possibly unequal total pressures and unequal piston forces.
Hence Statements I, II and IV only are correct.
114. An ideal hydraulic system contains oil of density \(800\,\text{kg m}^{-3}\). A small piston of area \(4.0\,\text{cm}^2\) is \(0.75\,\text{m}\) vertically below a large piston of area \(100\,\text{cm}^2\). Both outer piston faces are exposed to the same atmospheric pressure. At a certain instant, a downward force of \(100\,\text{N}\) is applied to the small piston. If the small piston is then displaced downward through \(0.25\,\text{m}\), neglect the small change in piston height difference during the stroke. Take \(g=10\,\text{m s}^{-2}\). What are the net upward force on the large piston and its displacement?
ⓐ. \(2500\,\text{N}\) and \(1.0\,\text{cm}\)
ⓑ. \(2560\,\text{N}\) and \(1.0\,\text{cm}\)
ⓒ. \(2440\,\text{N}\) and \(2.5\,\text{cm}\)
ⓓ. \(2440\,\text{N}\) and \(1.0\,\text{cm}\)
Correct Answer: \(2440\,\text{N}\) and \(1.0\,\text{cm}\)
Explanation: \(\textbf{Pressure beneath the small piston:}\)
The small-piston area is
\[
A_1=4.0\times10^{-4}\,\text{m}^2
\]
Its gauge pressure contribution is
\[
P_1=\frac{F_1}{A_1}
\]
Thus,
\[
P_1=
\frac{100}{4.0\times10^{-4}}
=
2.50\times10^{5}\,\text{Pa}
\]
\(\textbf{Correct for the higher large-piston level:}\)
The large piston is \(0.75\,\text{m}\) above the small piston, so its fluid pressure is lower by
\[
\rho gh=(800)(10)(0.75)=6.0\times10^{3}\,\text{Pa}
\]
Hence,
\[
P_2=
2.50\times10^{5}-6.0\times10^{3}
=
2.44\times10^{5}\,\text{Pa}
\]
The large-piston area is
\[
A_2=100\,\text{cm}^2=1.0\times10^{-2}\,\text{m}^2
\]
Therefore,
\[
F_2=P_2A_2
=
(2.44\times10^{5})(1.0\times10^{-2})
=
2440\,\text{N}
\]
\(\textbf{Volume transferred by the piston motion:}\)
For an incompressible fluid,
\[
A_1x_1=A_2x_2
\]
Thus,
\[
x_2=
\frac{A_1}{A_2}x_1
=
\frac{4.0}{100}(0.25)
=
0.010\,\text{m}
\]
Hence,
\[
x_2=1.0\,\text{cm}
\]
The height correction affects the force calculation, whereas the ideal displacement ratio follows from volume conservation.
\(\textbf{Final answer:}\)
\[
\boxed{F_2=2440\,\text{N}\quad\text{and}\quad x_2=1.0\,\text{cm}}
\]
115. A hydraulic lift contains oil of density \(900\,\text{kg m}^{-3}\). The small piston of area \(5.0\,\text{cm}^2\) is \(0.50\,\text{m}\) above the large piston of area \(250\,\text{cm}^2\). Both exposed piston faces are under the same atmospheric pressure. A load of \(1.20\times10^{4}\,\text{N}\) is just supported by the large piston. Take \(g=10\,\text{m s}^{-2}\). If the small piston is subsequently pushed through \(0.30\,\text{m}\), neglect the small change in level difference during this short stroke. Which pair is correct?
ⓐ. Input force \(\approx238\,\text{N}\); large-piston displacement \(=6.0\,\text{mm}\)
ⓑ. Input force \(=240\,\text{N}\); large-piston displacement \(=30\,\text{mm}\)
ⓒ. Input force \(\approx242\,\text{N}\); large-piston displacement \(=6.0\,\text{mm}\)
ⓓ. Input force \(\approx238\,\text{N}\); large-piston displacement \(=15\,\text{mm}\)
Correct Answer: Input force \(\approx238\,\text{N}\); large-piston displacement \(=6.0\,\text{mm}\)
Explanation: \(\textbf{Gauge pressure required beneath the load:}\)
The large-piston area is
\[
A_2=250\,\text{cm}^2
=
2.50\times10^{-2}\,\text{m}^2
\]
To support the load,
\[
P_2=\frac{F_2}{A_2}
\]
Thus,
\[
P_2=
\frac{1.20\times10^{4}}
{2.50\times10^{-2}}
=
4.80\times10^{5}\,\text{Pa}
\]
\(\textbf{Move upward to the small piston:}\)
The small piston is \(0.50\,\text{m}\) higher, so
\[
P_1=P_2-\rho gh
\]
Therefore,
\[
P_1=
4.80\times10^{5}
-
(900)(10)(0.50)
\]
Hence,
\[
P_1=4.755\times10^{5}\,\text{Pa}
\]
The small-piston area is
\[
A_1=5.0\times10^{-4}\,\text{m}^2
\]
Thus,
\[
F_1=P_1A_1
=
(4.755\times10^{5})(5.0\times10^{-4})
=
237.75\,\text{N}
\]
So,
\[
F_1\approx238\,\text{N}
\]
\(\textbf{Use incompressible-volume transfer:}\)
\[
A_1x_1=A_2x_2
\]
Hence,
\[
x_2=
\frac{5.0}{250}(0.30)
=
0.0060\,\text{m}
\]
Therefore,
\[
x_2=6.0\,\text{mm}
\]
The pressure required at the upper small piston is slightly less than that under the lower large piston because hydrostatic pressure increases downward.
\(\textbf{Final answer:}\) Input force \(\approx238\,\text{N}\) and large-piston displacement \(=6.0\,\text{mm}\).
116. In an ideal hydraulic braking system, a master piston of area \(2.0\,\text{cm}^2\) is acted on by a force of \(180\,\text{N}\). A front wheel-cylinder piston of area \(6.0\,\text{cm}^2\) is at the same level as the master piston. A rear wheel-cylinder piston of area \(8.0\,\text{cm}^2\) is \(0.50\,\text{m}\) lower. The brake fluid has density \(900\,\text{kg m}^{-3}\). All exposed outer piston faces experience the same atmospheric pressure. Take \(g=10\,\text{m s}^{-2}\). What forces act on the front and rear wheel-cylinder pistons respectively?
ⓐ. \(540\,\text{N}\) and \(720\,\text{N}\)
ⓑ. \(600\,\text{N}\) and \(720\,\text{N}\)
ⓒ. \(540\,\text{N}\) and \(684\,\text{N}\)
ⓓ. \(540\,\text{N}\) and \(723.6\,\text{N}\)
Correct Answer: \(540\,\text{N}\) and \(723.6\,\text{N}\)
Explanation: \(\textbf{Pressure generated at the master cylinder:}\)
\[
A_m=2.0\times10^{-4}\,\text{m}^2
\]
Therefore,
\[
P_m=\frac{180}{2.0\times10^{-4}}
=
9.0\times10^{5}\,\text{Pa}
\]
\(\textbf{Front wheel at the same level:}\)
The transmitted gauge pressure is
\[
P_f=9.0\times10^{5}\,\text{Pa}
\]
With
\[
A_f=6.0\times10^{-4}\,\text{m}^2,
\]
the front force is
\[
F_f=P_fA_f
\]
Hence,
\[
F_f=
(9.0\times10^{5})(6.0\times10^{-4})
=
540\,\text{N}
\]
\(\textbf{Rear wheel at the lower level:}\)
Hydrostatic pressure adds
\[
\rho gh=(900)(10)(0.50)
=
4.5\times10^{3}\,\text{Pa}
\]
Thus,
\[
P_r=
9.0\times10^{5}
+
4.5\times10^{3}
=
9.045\times10^{5}\,\text{Pa}
\]
Since
\[
A_r=8.0\times10^{-4}\,\text{m}^2,
\]
\[
F_r=P_rA_r
=
(9.045\times10^{5})(8.0\times10^{-4})
=
723.6\,\text{N}
\]
Equal transmitted pressure increment does not imply equal wheel forces when piston areas differ, and unequal heights additionally introduce a hydrostatic correction.
\(\textbf{Final answer:}\) \(F_f=540\,\text{N}\) and \(F_r=723.6\,\text{N}\).
117. In the idealised operation of a hydraulic braking system, which sequence is physically correct?
ⓐ. Wheel-cylinder piston moves \(\rightarrow\) pressure is created in the master cylinder \(\rightarrow\) pedal force is applied \(\rightarrow\) brake pads act
ⓑ. Pedal force is applied \(\rightarrow\) brake pads act directly \(\rightarrow\) fluid pressure decreases \(\rightarrow\) wheel pistons move
ⓒ. Pedal force acts on the master piston \(\rightarrow\) fluid pressure increases \(\rightarrow\) pressure is transmitted through the brake fluid \(\rightarrow\) wheel-cylinder pistons exert braking force
ⓓ. Pedal force increases fluid volume \(\rightarrow\) equal forces appear at all pistons \(\rightarrow\) wheel pistons move \(\rightarrow\) pressure is transmitted
Correct Answer: Pedal force acts on the master piston \(\rightarrow\) fluid pressure increases \(\rightarrow\) pressure is transmitted through the brake fluid \(\rightarrow\) wheel-cylinder pistons exert braking force
Explanation: The process begins when the driver applies force to the pedal and therefore to the master-cylinder piston.
The master piston acts on the brake fluid and creates a pressure increase.
Pascal's law describes the transmission of this pressure increase through the enclosed brake fluid.
The pressure then acts over the areas of the wheel-cylinder pistons.
Each wheel-cylinder force is determined by
\[
F=PA.
\]
Those piston forces are used to press the braking elements into contact.
The fluid does not need to generate identical forces at every piston.
Instead, the transmitted pressure is the common intermediate physical quantity in the idealised description.
The piston areas determine the corresponding forces.
Thus the physically correct sequence is force on master piston, pressure generation, pressure transmission and wheel-cylinder braking action.
118. A sealed tank contains a liquid of density \(900\,\text{kg m}^{-3}\). Points A and B lie \(1.0\,\text{m}\) and \(4.0\,\text{m}\) below the liquid surface respectively. Initially the gas pressure above the liquid is \(1.10\times10^{5}\,\text{Pa}\). The gas pressure is then increased by \(2.5\times10^{4}\,\text{Pa}\). Take \(g=10\,\text{m s}^{-2}\). What are the final absolute pressures at A and B, and what happens to \(P_B-P_A\)?
ⓐ. \(1.44\times10^{5}\,\text{Pa},\ 1.71\times10^{5}\,\text{Pa}\); the difference increases by \(2.5\times10^{4}\,\text{Pa}\)
ⓑ. \(1.35\times10^{5}\,\text{Pa},\ 1.62\times10^{5}\,\text{Pa}\); the difference remains \(2.7\times10^{4}\,\text{Pa}\)
ⓒ. \(1.44\times10^{5}\,\text{Pa},\ 1.71\times10^{5}\,\text{Pa}\); the difference remains \(2.7\times10^{4}\,\text{Pa}\)
ⓓ. \(1.44\times10^{5}\,\text{Pa},\ 1.69\times10^{5}\,\text{Pa}\); the difference becomes \(2.5\times10^{4}\,\text{Pa}\)
Correct Answer: \(1.44\times10^{5}\,\text{Pa},\ 1.71\times10^{5}\,\text{Pa}\); the difference remains \(2.7\times10^{4}\,\text{Pa}\)
Explanation: \(\textbf{New boundary pressure:}\)
The gas pressure becomes
\[
P_s=
1.10\times10^{5}
+
2.5\times10^{4}
=
1.35\times10^{5}\,\text{Pa}
\]
\(\textbf{Final pressure at A:}\)
\[
P_A=P_s+\rho gh_A
\]
Thus,
\[
P_A=
1.35\times10^{5}
+
(900)(10)(1.0)
\]
Hence,
\[
P_A=1.44\times10^{5}\,\text{Pa}
\]
\(\textbf{Final pressure at B:}\)
\[
P_B=
1.35\times10^{5}
+
(900)(10)(4.0)
\]
Therefore,
\[
P_B=
1.71\times10^{5}\,\text{Pa}
\]
\(\textbf{Pressure difference between the two depths:}\)
\[
P_B-P_A
=
\rho g(h_B-h_A)
\]
Thus,
\[
P_B-P_A
=
(900)(10)(3.0)
=
2.7\times10^{4}\,\text{Pa}
\]
Changing the boundary pressure shifts the absolute pressure at both points by the same amount and therefore does not change the gravity-generated pressure difference between them.
\(\textbf{Final answer:}\) \(P_A=1.44\times10^{5}\,\text{Pa}\), \(P_B=1.71\times10^{5}\,\text{Pa}\), and \(P_B-P_A\) remains \(2.7\times10^{4}\,\text{Pa}\).
119. The same liquid of density \(1000\,\text{kg m}^{-3}\) is placed in open tanks on two planets. On planet P, \(g_P=10\,\text{m s}^{-2}\), atmospheric pressure is \(1.0\times10^{5}\,\text{Pa}\), and a point lies \(2.0\,\text{m}\) below the surface. On planet Q, \(g_Q=4.0\,\text{m s}^{-2}\), atmospheric pressure is \(8.0\times10^{4}\,\text{Pa}\), and a point lies \(5.0\,\text{m}\) below the surface. Which comparison is correct?
ⓐ. Both gauge and absolute pressures are equal
ⓑ. Planet P has twice the gauge pressure of planet Q, but their absolute pressures are equal
ⓒ. Planet Q has the larger gauge pressure and the smaller absolute pressure
ⓓ. The gauge pressures are equal, but the absolute pressure is larger on planet P
Correct Answer: The gauge pressures are equal, but the absolute pressure is larger on planet P
Explanation: \(\textbf{Gauge pressure on planet P:}\)
\[
P_{gP}=\rho g_Ph_P
\]
Thus,
\[
P_{gP}
=
(1000)(10)(2.0)
=
2.0\times10^{4}\,\text{Pa}
\]
Its absolute pressure is
\[
P_{\text{abs},P}
=
1.0\times10^{5}
+
2.0\times10^{4}
=
1.20\times10^{5}\,\text{Pa}
\]
\(\textbf{Gauge pressure on planet Q:}\)
\[
P_{gQ}
=
(1000)(4.0)(5.0)
=
2.0\times10^{4}\,\text{Pa}
\]
Thus the gravity-generated pressure increments are equal.
However,
\[
P_{\text{abs},Q}
=
8.0\times10^{4}
+
2.0\times10^{4}
=
1.00\times10^{5}\,\text{Pa}
\]
The product \(gh\) controls the hydrostatic contribution, whereas absolute pressure also contains the atmospheric boundary pressure.
\(\textbf{Final answer:}\) The gauge pressures are equal, but planet P has the larger absolute pressure.
120. A sealed water tank has depth \(5.0\,\text{m}\). On planet P, \(g=10\,\text{m s}^{-2}\) and the gas above the water is maintained at absolute pressure \(1.10\times10^{5}\,\text{Pa}\). The tank is transferred to a planet where \(g=4.0\,\text{m s}^{-2}\). Assume the depth and water density \(1000\,\text{kg m}^{-3}\) remain unchanged. To keep the absolute pressure at the bottom equal to its original value, what gas pressure is now required above the water?
ⓐ. \(1.40\times10^{5}\,\text{Pa}\), an increase of \(3.0\times10^{4}\,\text{Pa}\)
ⓑ. \(1.70\times10^{5}\,\text{Pa}\), an increase of \(6.0\times10^{4}\,\text{Pa}\)
ⓒ. \(1.10\times10^{5}\,\text{Pa}\), because Pascal's law keeps bottom pressure unchanged
ⓓ. \(8.0\times10^{4}\,\text{Pa}\), a decrease of \(3.0\times10^{4}\,\text{Pa}\)
Correct Answer: \(1.40\times10^{5}\,\text{Pa}\), an increase of \(3.0\times10^{4}\,\text{Pa}\)
Explanation: \(\textbf{Original bottom pressure:}\)
\[
P_{b1}=P_{s1}+\rho g_1h
\]
Therefore,
\[
P_{b1}
=
1.10\times10^{5}
+
(1000)(10)(5.0)
\]
Hence,
\[
P_{b1}
=
1.60\times10^{5}\,\text{Pa}
\]
\(\textbf{Hydrostatic contribution on the second planet:}\)
\[
\rho g_2h
=
(1000)(4.0)(5.0)
=
2.0\times10^{4}\,\text{Pa}
\]
To preserve the original bottom pressure,
\[
P_{s2}+2.0\times10^{4}
=
1.60\times10^{5}
\]
Therefore,
\[
P_{s2}
=
1.40\times10^{5}\,\text{Pa}
\]
The required increase in boundary pressure is
\[
1.40\times10^{5}
-
1.10\times10^{5}
=
3.0\times10^{4}\,\text{Pa}
\]
Reducing \(g\) weakens only the hydrostatic contribution; the missing pressure can be supplied by increasing the imposed gas pressure.
\(\textbf{Final answer:}\) \(P_{s2}=1.40\times10^{5}\,\text{Pa}\), requiring an increase of \(3.0\times10^{4}\,\text{Pa}\).
