101. In a hydraulic lift, a pressure of \(3.0\times10^{5}\,\text{Pa}\) acts on a large piston of area \(0.20\,\text{m}^2\). The load supported by this piston is
ⓐ. \(1.5\times10^{5}\,\text{N}\)
ⓑ. \(1.5\times10^{6}\,\text{N}\)
ⓒ. \(6.0\times10^{4}\,\text{N}\)
ⓓ. \(6.0\times10^{5}\,\text{N}\)
Correct Answer: \(6.0\times10^{4}\,\text{N}\)
Explanation: \( \textbf{Given:} \) Pressure \(P=3.0\times10^{5}\,\text{Pa}\) and piston area \(A=0.20\,\text{m}^2\).
\( \textbf{Required:} \) Force \(F\) supported by the piston.
Pressure and force are related by:
\[
P=\frac{F}{A}
\]
Rearranging:
\[
F=PA
\]
Substituting:
\[
F=(3.0\times10^{5})(0.20)
\]
Calculation:
\[
F=6.0\times10^{4}\,\text{N}
\]
The large force comes from applying the same pressure over a comparatively large area.
\( \textbf{Final answer:} \) The load supported is \(6.0\times10^{4}\,\text{N}\).
102. A student says, “A hydraulic press multiplies force because it creates energy inside the liquid.” The correct correction is that the press
ⓐ. creates extra work by increasing both force and displacement
ⓑ. works only because pressure disappears at the large piston
ⓒ. violates conservation of energy in an ideal case
ⓓ. gives a larger force with a smaller displacement
Correct Answer: gives a larger force with a smaller displacement
Explanation: An ideal hydraulic press does not create energy. It transmits pressure through an enclosed fluid and uses different piston areas to change the force. A larger output piston gives a larger force because \(F=PA\). However, the output piston moves through a smaller distance than the input piston. In an ideal machine, \(F_1d_1=F_2d_2\), so the gain in force is balanced by a loss in displacement.
103. A hydraulic machine has \(A_2=25A_1\). If the small piston moves downward by \(50\,\text{cm}\), the upward displacement of the large piston is
ⓐ. \(2.0\,\text{cm}\)
ⓑ. \(25\,\text{cm}\)
ⓒ. \(50\,\text{cm}\)
ⓓ. \(1250\,\text{cm}\)
Correct Answer: \(2.0\,\text{cm}\)
Explanation: \( \textbf{Given:} \) \(A_2=25A_1\) and \(d_1=50\,\text{cm}\).
\( \textbf{Required:} \) Displacement \(d_2\) of the large piston.
For an ideal incompressible fluid, the volume pushed in equals the volume lifted out:
\[
A_1d_1=A_2d_2
\]
Substitute \(A_2=25A_1\):
\[
A_1(50)=25A_1d_2
\]
Cancel \(A_1\):
\[
50=25d_2
\]
Therefore:
\[
d_2=\frac{50}{25}=2.0\,\text{cm}
\]
This shows the displacement trade-off that accompanies force multiplication.
\( \textbf{Final answer:} \) The large piston rises by \(2.0\,\text{cm}\).
104. Match the hydraulic-system quantities with their correct roles.
| Column I | Column II |
|---|
| P. \(\frac{F_1}{A_1}=\frac{F_2}{A_2}\) | 1. Work relation for an ideal hydraulic machine |
| Q. \(F_1d_1=F_2d_2\) | 2. Equal pressure condition |
| R. Larger output piston area | 3. Smaller output displacement for same fluid volume |
| S. \(A_1d_1=A_2d_2\) | 4. Larger output force for same pressure |
ⓐ. P-1, Q-2, R-3, S-4
ⓑ. P-2, Q-4, R-1, S-3
ⓒ. P-2, Q-1, R-4, S-3
ⓓ. P-3, Q-1, R-4, S-2
Correct Answer: P-2, Q-1, R-4, S-3
Explanation: The relation \(\frac{F_1}{A_1}=\frac{F_2}{A_2}\) expresses equal pressure in an ideal hydraulic system. The relation \(F_1d_1=F_2d_2\) expresses work conservation when losses are neglected. A larger output piston gives a larger force for the same pressure because \(F=PA\). The relation \(A_1d_1=A_2d_2\) shows equal displaced volumes of an incompressible fluid. The force gain and displacement loss are two parts of the same hydraulic action.
105. A graph is plotted for an ideal hydraulic press with output force \(F_2\) on the vertical axis and output piston area \(A_2\) on the horizontal axis. The input pressure is kept constant. The graph should be
ⓐ. straight line through origin; slope is pressure
ⓑ. a curve showing \(F_2\propto\frac{1}{A_2}\)
ⓒ. a horizontal line because force is independent of area
ⓓ. a vertical line because area cannot change
Correct Answer: straight line through origin; slope is pressure
Explanation: \( \textbf{Relation used:} \)
\[
F_2=PA_2
\]
Here the pressure \(P\) transmitted by the fluid is constant.
This can be compared with:
\[
y=mx
\]
The vertical-axis quantity is \(F_2\), and the horizontal-axis quantity is \(A_2\).
Therefore, the slope is:
\[
P
\]
Since \(F_2\) is directly proportional to \(A_2\), the graph is a straight line through the origin.
Increasing output area increases output force only because the same pressure acts over a larger surface.
\( \textbf{Final answer:} \) The graph is a straight line through the origin with slope equal to pressure.
106. A hydraulic jack has a small piston of area \(2.0\times10^{-4}\,\text{m}^2\) and a large piston of area \(4.0\times10^{-2}\,\text{m}^2\). What input force is needed to lift a load of \(2.0\times10^{4}\,\text{N}\), ignoring losses?
ⓐ. \(100\,\text{N}\)
ⓑ. \(50\,\text{N}\)
ⓒ. \(200\,\text{N}\)
ⓓ. \(400\,\text{N}\)
Correct Answer: \(100\,\text{N}\)
Explanation: \( \textbf{Given:} \) \(A_1=2.0\times10^{-4}\,\text{m}^2\), \(A_2=4.0\times10^{-2}\,\text{m}^2\), and \(F_2=2.0\times10^{4}\,\text{N}\).
\( \textbf{Required:} \) Input force \(F_1\).
In an ideal hydraulic jack:
\[
\frac{F_1}{A_1}=\frac{F_2}{A_2}
\]
Rearranging:
\[
F_1=F_2\frac{A_1}{A_2}
\]
Substitute:
\[
F_1=(2.0\times10^{4})\left(\frac{2.0\times10^{-4}}{4.0\times10^{-2}}\right)
\]
Area ratio:
\[
\frac{2.0\times10^{-4}}{4.0\times10^{-2}}=0.5\times10^{-2}=5.0\times10^{-3}
\]
Hence:
\[
F_1=(2.0\times10^{4})(5.0\times10^{-3})
\]
\[
F_1=100\,\text{N}
\]
\( \textbf{Final answer:} \) The required input force is \(100\,\text{N}\).
107. Consider the following statements about an ideal hydraulic lift.
Statement I: The pressure is the same at both pistons if height differences are ignored.
Statement II: The larger piston can exert a larger force.
Statement III: The larger piston must move through a larger distance than the smaller piston.
ⓐ. II and III only
ⓑ. I and III only
ⓒ. I, II and III
ⓓ. I and II only
Correct Answer: I and II only
Explanation: In an ideal hydraulic lift, Pascal’s law gives equal transmitted pressure at both pistons when height differences and losses are ignored. Since \(F=PA\), the larger piston can exert a larger force under the same pressure. Statement III is false because the larger piston moves through a smaller distance for the same volume of fluid displaced. This follows from \(A_1d_1=A_2d_2\). The machine trades displacement for force rather than increasing both together.
108. Use the case below and answer the question.
A small piston and a large piston are connected by an enclosed liquid. The small piston has area \(A\), and the large piston has area \(10A\). A worker applies a force \(F\) on the small piston and observes that the load on the large piston rises slowly.
For an ideal system, the load force on the large piston and its displacement compared with the small piston are
ⓐ. \(F\) and ten times as large
ⓑ. \(10F\) and one-tenth as large
ⓒ. \(\frac{F}{10}\) and ten times as large
ⓓ. \(10F\) and ten times as large
Correct Answer: \(10F\) and one-tenth as large
Explanation: The same pressure is transmitted through the enclosed liquid. At the small piston, the pressure is \(P=\frac{F}{A}\). At the large piston of area \(10A\), the force is \(F_2=P(10A)=\frac{F}{A}(10A)=10F\). For displacement, the displaced volumes must match: \(Ad_1=(10A)d_2\). Hence \(d_2=\frac{d_1}{10}\). A hydraulic lift multiplies force but reduces output displacement in the same area ratio.
109. A hydraulic press has \(A_1=10\,\text{cm}^2\) and \(A_2=500\,\text{cm}^2\). A force \(F_1=200\,\text{N}\) is applied on the small piston. If the small piston moves \(25\,\text{cm}\), the output force and output displacement are
ⓐ. \(1.0\times10^{4}\,\text{N}\) and \(25\,\text{cm}\)
ⓑ. \(4.0\times10^{3}\,\text{N}\) and \(2.0\,\text{cm}\)
ⓒ. \(200\,\text{N}\) and \(1250\,\text{cm}\)
ⓓ. \(1.0\times10^{4}\,\text{N}\) and \(0.50\,\text{cm}\)
Correct Answer: \(1.0\times10^{4}\,\text{N}\) and \(0.50\,\text{cm}\)
Explanation: \( \textbf{Given:} \) \(A_1=10\,\text{cm}^2\), \(A_2=500\,\text{cm}^2\), \(F_1=200\,\text{N}\), and \(d_1=25\,\text{cm}\).
\( \textbf{Area ratio:} \)
\[
\frac{A_2}{A_1}=\frac{500}{10}=50
\]
Equal pressure gives:
\[
\frac{F_1}{A_1}=\frac{F_2}{A_2}
\]
Therefore:
\[
F_2=F_1\frac{A_2}{A_1}
\]
Substitute:
\[
F_2=200\times50=1.0\times10^{4}\,\text{N}
\]
Equal displaced volume gives:
\[
A_1d_1=A_2d_2
\]
So:
\[
d_2=d_1\frac{A_1}{A_2}
\]
Substitute:
\[
d_2=25\left(\frac{10}{500}\right)=0.50\,\text{cm}
\]
\( \textbf{Final answer:} \) The output force is \(1.0\times10^{4}\,\text{N}\), and the output displacement is \(0.50\,\text{cm}\).
110. In a hydraulic system, the small piston is pressed down slowly. The large piston lifts a heavy object. The reason a small input force can lift a heavy load is best described as
ⓐ. the applied pressure acts on a much larger output area
ⓑ. the liquid reduces the weight of the object to zero
ⓒ. the large piston moves faster than the small piston
ⓓ. atmospheric pressure is removed from both pistons
Correct Answer: the applied pressure acts on a much larger output area
Explanation: The small input force produces pressure in the enclosed liquid. Pascal’s law transmits this pressure through the fluid. When the same pressure acts on a large output piston, the output force becomes large because \(F=PA\). The object’s weight is not reduced to zero; it is balanced or lifted by the upward hydraulic force. The large piston usually moves more slowly and through a smaller distance than the small piston.
111. A table compares pressure and force in a hydraulic press.
| Row | Small piston | Large piston | Correct conclusion? |
|---|
| P | Pressure \(P\) | Pressure \(P\) | Same pressure may act on both pistons. |
| Q | Force \(F\) | Force \(F\) | Force must remain equal on both pistons. |
| R | Area \(A\) | Area \(4A\) | Output force can be \(4F\) for input force \(F\). |
| S | Displacement \(d\) | Displacement \(d\) | Both pistons must move through equal distances. |
Which set of rows contains correct conclusions for an ideal hydraulic press?
ⓐ. P and R only
ⓑ. Q and S only
ⓒ. P, Q and R only
ⓓ. P, R and S only
Correct Answer: P and R only
Explanation: Row P is correct because Pascal’s law allows the same pressure to be transmitted through the enclosed fluid. Row R is also correct because if the large piston has area \(4A\), the same pressure can give four times the force. Row Q is not correct because equal pressure does not require equal force when areas differ. Row S is not correct because equal fluid volume displacement gives \(A_1d_1=A_2d_2\), so different piston areas imply different displacements. The correct comparison is equal pressure, not equal force or equal distance.
112. A hydraulic lift raises a car of weight \(1.5\times10^{4}\,\text{N}\) using a large piston of area \(0.30\,\text{m}^2\). The small piston has area \(6.0\times10^{-3}\,\text{m}^2\). Ignoring losses, the minimum force on the small piston is
ⓐ. \(150\,\text{N}\)
ⓑ. \(450\,\text{N}\)
ⓒ. \(900\,\text{N}\)
ⓓ. \(300\,\text{N}\)
Correct Answer: \(300\,\text{N}\)
Explanation: \( \textbf{Given:} \) \(F_2=1.5\times10^{4}\,\text{N}\), \(A_2=0.30\,\text{m}^2\), and \(A_1=6.0\times10^{-3}\,\text{m}^2\).
\( \textbf{Required:} \) Input force \(F_1\).
Equal pressure in an ideal hydraulic lift gives:
\[
\frac{F_1}{A_1}=\frac{F_2}{A_2}
\]
Rearranging:
\[
F_1=F_2\frac{A_1}{A_2}
\]
Substitute values:
\[
F_1=(1.5\times10^{4})\left(\frac{6.0\times10^{-3}}{0.30}\right)
\]
Evaluate the area ratio:
\[
\frac{6.0\times10^{-3}}{0.30}=2.0\times10^{-2}
\]
Therefore:
\[
F_1=(1.5\times10^{4})(2.0\times10^{-2})
\]
\[
F_1=3.0\times10^{2}\,\text{N}
\]
\( \textbf{Final answer:} \) The required input force is \(300\,\text{N}\).
113. Two vessels of different shapes contain the same liquid up to the same vertical height. If their bottom points are at the same depth, the pressure at those bottom points is
ⓐ. the same in both vessels
ⓑ. greater in the vessel containing larger total volume
ⓒ. greater in the vessel with sloping walls
ⓓ. zero if both vessels are open to air
Correct Answer: the same in both vessels
Explanation: In a liquid at rest, gauge pressure at a depth is given by \(P=\rho gh\). For the same liquid, \(\rho\) is the same, and at the same place \(g\) is also the same. If the vertical height of the liquid above the bottom points is the same, then \(h\) is the same. Therefore, the bottom pressure is the same even when vessel shapes are different. The total amount of liquid may differ, but pressure at a point depends on depth, not on total volume.
114. The hydrostatic paradox refers to the idea that the pressure at the base of a liquid-filled vessel depends on
ⓐ. total curved surface area of the vessel only
ⓑ. total mass of liquid only, even when height changes
ⓒ. \(\rho\), \(g\), and \(h\), not directly on vessel shape
ⓓ. whether the vessel wall is transparent or opaque
Correct Answer: \(\rho\), \(g\), and \(h\), not directly on vessel shape
Explanation: Hydrostatic pressure at a depth in a liquid at rest is \(P=\rho gh\). This relation contains density \(\rho\), gravitational acceleration \(g\), and vertical depth \(h\). It does not contain the shape of the vessel or the total volume of liquid. This leads to the hydrostatic paradox, where vessels of different shapes can have the same pressure at equal depth. The paradox feels surprising only if pressure is confused with total weight of the liquid.
115. Use the arrangement described below.
Three open vessels have equal base areas and contain the same liquid up to the same vertical height \(h\). One vessel is straight-sided, one widens upward, and one narrows upward.
Ignoring atmospheric pressure, the force on the base of each vessel is
ⓐ. greatest for the vessel that widens upward
ⓑ. greatest for the vessel that narrows upward
ⓒ. zero because the vessels are open
ⓓ. equal for all three vessels
Correct Answer: equal for all three vessels
Explanation: The gauge pressure at the base of each vessel is \(P=\rho gh\). Since the liquid, height, and gravitational acceleration are the same, the base pressure is the same for all three vessels. The base areas are also stated to be equal. The force on a horizontal base is \(F=PA\), so equal \(P\) and equal \(A\) give equal base force. Different vessel shapes can contain different amounts of liquid, but that does not change the base pressure for the same height.
116. Two vessels contain water to the same height. Vessel P has base area \(A\), while vessel Q has base area \(3A\). The gauge pressure at the base and the force on the base satisfy
ⓐ. pressure ratio \(1:3\), force ratio \(1:1\)
ⓑ. pressure ratio \(1:1\), force ratio \(1:3\)
ⓒ. pressure ratio \(3:1\), force ratio \(1:1\)
ⓓ. pressure ratio \(1:1\), force ratio \(3:1\)
Correct Answer: pressure ratio \(1:1\), force ratio \(1:3\)
Explanation: \( \textbf{Pressure comparison:} \)
\[
P=\rho gh
\]
Both vessels contain the same liquid to the same height, so the base pressure is the same.
Therefore:
\[
P_P:P_Q=1:1
\]
\( \textbf{Force relation:} \)
\[
F=PA
\]
Vessel P has base area \(A\), and vessel Q has base area \(3A\).
Since \(P\) is the same:
\[
F_P:F_Q=PA:P(3A)=1:3
\]
\( \textbf{Final answer:} \) The pressure ratio is \(1:1\), and the base-force ratio is \(1:3\).
117. Consider the following statements about the hydrostatic paradox.
Statement I: Same liquid height can give the same base pressure in vessels of different shapes.
Statement II: The total force on a horizontal base depends on the base area.
Statement III: Base pressure is decided by the total volume of liquid in the vessel.
ⓐ. II and III only
ⓑ. I and III only
ⓒ. I and II only
ⓓ. I, II and III
Correct Answer: I and II only
Explanation: Statement I is true because \(P=\rho gh\), so equal liquid height in the same liquid gives equal base pressure. Statement II is also true because the force on a horizontal base is \(F=PA\). If pressure is the same but base area changes, the base force changes. Statement III is not true because hydrostatic pressure at a point is not decided by the total volume of liquid. The pressure-volume confusion is the main reason the hydrostatic paradox seems surprising.
118. Study the table and identify the row that gives the best conclusion.
| Row | Situation | Conclusion |
|---|
| P | Same liquid, same height, different vessel shapes | Base pressure is the same. |
| Q | Same liquid, same height, larger base area | Base pressure must be larger. |
| R | Same pressure, larger horizontal area | Force on the area must be smaller. |
| S | Same depth, same liquid | Pressure depends mainly on total liquid volume. |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row P
Explanation: Row P correctly states the hydrostatic pressure idea: for the same liquid and same height, base pressure is the same even if vessel shapes differ. Row Q is not suitable because base area does not directly change pressure at a given depth. Row R is not suitable because for the same pressure, force is \(F=PA\), so a larger area gives a larger force. Row S repeats the volume-based misunderstanding. Depth, density, and gravity decide pressure at a point in a static liquid.
119. A wide tank and a narrow tube are connected at the bottom and contain the same liquid at rest. The liquid levels are the same in both arms. The pressure at two bottom points at the same horizontal level is
ⓐ. larger in the wide tank because it contains more liquid
ⓑ. larger in the narrow tube because its liquid column is thinner
ⓒ. the same in both parts
ⓓ. zero in the narrow tube because its area is small
Correct Answer: the same in both parts
Explanation: The connected liquid is at rest, so pressure at the same horizontal level must be the same. If it were not the same, the liquid would move until the pressure difference disappeared. The wide tank may contain more total liquid, but the pressure at a point is governed by vertical depth. A narrow tube can still have the same pressure at the same level as a wide container. The cross-sectional area affects volume and possible force on a surface, not the pressure equality at the same level.
120. A horizontal base of area \(0.50\,\text{m}^2\) is under a uniform gauge pressure of \(4.0\times10^{4}\,\text{Pa}\). The force exerted by the liquid on the base is
ⓐ. \(8.0\times10^{4}\,\text{N}\)
ⓑ. \(4.0\times10^{4}\,\text{N}\)
ⓒ. \(2.0\times10^{4}\,\text{N}\)
ⓓ. \(2.0\times10^{5}\,\text{N}\)
Correct Answer: \(2.0\times10^{4}\,\text{N}\)
Explanation: \( \textbf{Given:} \) Gauge pressure \(P=4.0\times10^{4}\,\text{Pa}\) and base area \(A=0.50\,\text{m}^2\).
\( \textbf{Required:} \) Force on the horizontal base.
For uniform pressure on a surface:
\[
F=PA
\]
Substituting:
\[
F=(4.0\times10^{4})(0.50)
\]
Calculation:
\[
F=2.0\times10^{4}\,\text{N}
\]
The force is found from pressure and area; it is not found by using the container shape.
\( \textbf{Final answer:} \) The force on the base is \(2.0\times10^{4}\,\text{N}\).
