Pascal's Law Questions in English

(C) According to Pascal's law,which was stated by Blaise Pascal in $1652$,a change in pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of its container.
This pressure change is independent of the orientation or height within the fluid.
Therefore,when air is blown into a closed pipe containing liquid,the increase in pressure is transmitted equally throughout the fluid,meaning the pressure will increase in all directions.

(D) The statement describes $Pascal's$ $Law$.
$Pascal's$ $Law$ states that a change in pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel.
Therefore,the correct option is $D$.

(A) Pascal's law states that the external static pressure applied on a confined fluid is distributed or transmitted equally throughout the fluid in all directions.

(C) According to Pascal's Law,the pressure exerted on the fluid is transmitted equally in all directions. Therefore,$P_1 = P_2$.
This implies $\frac{F_1}{A_1} = \frac{F_2}{A_2}$.
Given: $A_1 = 100 \, cm^2 = 10^2 \, cm^2$,$F_1 = 10^7 \, dynes$,$m = 2000 \, kg = 2 \times 10^6 \, g$,and $g \approx 980 \, cm/s^2 \approx 10^3 \, cm/s^2$.
Force $F_2 = m \times g = 2000 \times 10^3 \times 10^3 \, dynes = 2 \times 10^9 \, dynes$.
Substituting the values: $\frac{10^7}{10^2} = \frac{2 \times 10^9}{A_2}$.
$10^5 = \frac{2 \times 10^9}{A_2}$.
$A_2 = \frac{2 \times 10^9}{10^5} = 2 \times 10^4 \, cm^2$.

(C) The pressure exerted by the mass on the piston is balanced by the hydrostatic pressure due to the height difference $h$ of the water column.
Pressure $P = \frac{F}{A} = \frac{mg}{A}$.
Given: $m = 12 \ kg$,$g = 10 \ m/s^2$,$A = 800 \ cm^2 = 800 \times 10^{-4} \ m^2 = 0.08 \ m^2$.
Pressure due to mass = $\frac{12 \times 10}{0.08} = \frac{120}{0.08} = 1500 \ Pa$.
Hydrostatic pressure $P = \rho g h$,where $\rho = 1000 \ kg/m^3$ for water.
$1500 = 1000 \times 10 \times h$.
$1500 = 10000 \times h$.
$h = \frac{1500}{10000} = 0.15 \ m$.
Converting to centimeters: $h = 0.15 \times 100 = 15 \ cm$.

(B) According to Pascal's law,the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel.
For a hydraulic lift,the pressure at both sides must be equal:
$P_1 = P_2$
Since pressure $P = \frac{F}{A}$,we have:
$\frac{F_1}{A_1} = \frac{F_2}{A_2}$
From the diagram,it is clear that the area at side $1$ $(A_1)$ is smaller than the area at side $2$ $(A_2)$,i.e.,$A_1 < A_2$.
Rearranging the equation for force $F_2$:
$F_2 = F_1 \times \frac{A_2}{A_1}$
Since $\frac{A_2}{A_1} > 1$,it follows that $F_2 > F_1$,or $F_1 < F_2$.
Therefore,the force at side $1$ is less than the force at side $2$.

(B) The mechanical advantage $(MA)$ of a hydraulic press is defined as the ratio of the area of the larger piston $(A_1)$ to the area of the smaller piston $(A_2)$.
$MA = \frac{A_1}{A_2} = \frac{\pi (D_1/2)^2}{\pi (D_2/2)^2} = \frac{D_1^2}{D_2^2}$
Given the diameters $D_1 = 0.6\, m$ and $D_2 = 0.1\, m$:
$MA = \frac{(0.6)^2}{(0.1)^2} = \frac{0.36}{0.01} = 36$
Therefore,the mechanical advantage is $36$.

(A) The pressure exerted by the mass on the piston is given by $P = \frac{mg}{A}$.
This pressure is balanced by the hydrostatic pressure of the water column of height $h$,which is $P = \rho gh$.
Equating the two,we get $\frac{mg}{A} = \rho gh$.
Therefore,$h = \frac{m}{\rho A}$.
Given: $m = 12\,kg$,$A = 800\,cm^2 = 800 \times 10^{-4}\,m^2 = 0.08\,m^2$,and density of water $\rho = 1000\,kg/m^3$.
Substituting the values: $h = \frac{12}{1000 \times 0.08} = \frac{12}{80} = 0.15\,m$.

(A) According to Pascal's Law,pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
$(a)$ Let $F_1 = 10 \; N$ be the force on the smaller piston with diameter $d_1 = 1.0 \; cm$ and $F_2$ be the force on the larger piston with diameter $d_2 = 3.0 \; cm$.
The areas are $A_1 = \pi (d_1/2)^2$ and $A_2 = \pi (d_2/2)^2$.
Since pressure $P = F_1/A_1 = F_2/A_2$,we have $F_2 = F_1 \times (A_2/A_1) = F_1 \times (d_2/d_1)^2$.
$F_2 = 10 \; N \times (3.0 \; cm / 1.0 \; cm)^2 = 10 \times 9 = 90 \; N$.
$(b)$ Since water is incompressible,the volume of water displaced by the smaller piston must equal the volume displaced by the larger piston.
$V = A_1 L_1 = A_2 L_2$,where $L_1 = 6.0 \; cm$ is the displacement of the smaller piston.
$L_2 = L_1 \times (A_1/A_2) = L_1 \times (d_1/d_2)^2$.
$L_2 = 6.0 \; cm \times (1.0 \; cm / 3.0 \; cm)^2 = 6.0 \times (1/9) = 0.67 \; cm$.

(N/A) According to Pascal's Law,pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
Given:
Radius of small piston,$r_{1} = 5.0 \; cm = 5.0 \times 10^{-2} \; m$
Radius of large piston,$r_{2} = 15 \; cm = 15 \times 10^{-2} \; m$
Mass of the car,$m = 1350 \; kg$
Acceleration due to gravity,$g = 9.8 \; m s^{-2}$
The force exerted by the car on the large piston is $F_{2} = m \times g = 1350 \times 9.8 = 13230 \; N$.
Using the principle of hydraulic lift,$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$,where $A_{1} = \pi r_{1}^{2}$ and $A_{2} = \pi r_{2}^{2}$.
$F_{1} = F_{2} \times \frac{A_{1}}{A_{2}} = F_{2} \times \left(\frac{r_{1}}{r_{2}}\right)^{2}$
$F_{1} = 13230 \times \left(\frac{5}{15}\right)^{2} = 13230 \times \left(\frac{1}{3}\right)^{2} = 13230 \times \frac{1}{9} = 1470 \; N$.
Rounding to two significant figures,$F_{1} \approx 1.5 \times 10^{3} \; N$.
The pressure $P$ required is $P = \frac{F_{1}}{A_{1}} = \frac{1470}{\pi \times (5.0 \times 10^{-2})^{2}} = \frac{1470}{3.14159 \times 25 \times 10^{-4}} \approx 1.87 \times 10^{5} \; Pa \approx 1.9 \times 10^{5} \; Pa$.

(C) The maximum mass of the car to be lifted is $m = 3000 \; kg$.
The area of the cross-section of the load-carrying piston is $A = 425 \; cm^{2} = 425 \times 10^{-4} \; m^{2}$.
The maximum force exerted by the load is $F = mg = 3000 \times 9.8 = 29400 \; N$.
The pressure exerted on the load-carrying piston is $P = \frac{F}{A}$.
Substituting the values,$P = \frac{29400}{425 \times 10^{-4}} = 6.917 \times 10^{5} \; Pa$.
According to Pascal's Law,pressure is transmitted equally in all directions in an enclosed fluid. Therefore,the maximum pressure that the smaller piston must bear is $6.917 \times 10^{5} \; Pa$.

(A) fluid at rest exerts a force on the surface of a vessel in contact with it.
According to Pascal's law and the properties of static fluids,this force is always directed perpendicular to the surface of the vessel at every point.
If the force had a component parallel to the surface,the fluid would experience a tangential stress,causing it to flow along the surface.
Since the fluid is in a state of rest (static),there can be no tangential component of the force.
Therefore,the force exerted by the fluid is always normal (perpendicular) to the surface.

(N/A) Law: "The pressure in a fluid at rest is the same at all points in all directions if gravity is neglected."
Proof: Consider a small element $ABC-DEF$ in the interior of a fluid at rest, in the form of a right-angled prism. Since the element is very small, the effect of gravity can be neglected.
Let the areas of the surfaces be $A_a$ (bottom surface $BEFC$), $A_c$ (vertical surface $ABED$), and $A_b$ (inclined surface $ADFC$).
Let the forces acting perpendicular to these surfaces be $F_a$, $F_c$, and $F_b$ respectively.
From the geometry of the prism:
$A_a = A_b \cos \theta$
$A_c = A_b \sin \theta$
For the element to be in equilibrium, the net force in any direction must be zero:
Horizontal direction: $F_c = F_b \sin \theta$
Vertical direction: $F_a = F_b \cos \theta$
Since pressure $P = F/A$, we have:
$P_c = F_c / A_c = (F_b \sin \theta) / (A_b \sin \theta) = F_b / A_b = P_b$
$P_a = F_a / A_a = (F_b \cos \theta) / (A_b \cos \theta) = F_b / A_b = P_b$
Thus, $P_a = P_c = P_b$. This proves that the pressure at all points in a fluid at rest is the same.

(N/A) Pascal's law of pressure transmission states: "The pressure in a fluid at rest is the same at all points which are at the same height. $A$ change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel."
Consider a vessel with a piston and several vertical tubes at different points. The pressure in the vessel is indicated by the height of the liquid column in the vertical tubes.
When the piston is pushed, the fluid level rises in all the tubes, reaching the same level in each of them.
This indicates that when the pressure on the fluid was increased, it was distributed uniformly throughout the entire volume.
Many devices are based on this law, such as door closers, hydraulic brakes, hydraulic lifts, and shock absorbers in vehicles.

(N/A) Pascal's Law: "$A$ change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel."
$A$ hydraulic lift is based on this principle.
As shown in the figure, consider two cylinders with cross-sectional areas $A_{1}$ and $A_{2}$, where $A_{1} < < A_{2}$.
$A$ liquid is filled in this vessel.
$A$ piston of small cross-section $A_{1}$ is used to exert a force $F_{1}$ directly on the liquid. The pressure $P_{1} = \frac{F_{1}}{A_{1}}$ is transmitted throughout the liquid to the large cylinder attached with a larger piston of area $A_{2}$.
According to Pascal's law, the pressure is transmitted undiminished, so $P_{1} = P_{2}$.
Therefore, $\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$, which implies $F_{2} = F_{1} \left( \frac{A_{2}}{A_{1}} \right)$.
Since $A_{2} > > A_{1}$, the force $F_{2}$ exerted on the large piston is much larger than the applied force $F_{1}$, allowing the lift to support heavy loads like a car.

(N/A) Hydraulic brakes operate on the principle of Pascal's law.
When a driver applies a force $F_1$ on the brake pedal,the master piston,which has a small cross-sectional area $A_1$,moves into the master cylinder. This action pushes the brake oil through the tubes into the wheel cylinder.
The wheel cylinder contains two pistons with a much larger cross-sectional area $A_2$. According to Pascal's law,the pressure applied is transmitted equally throughout the fluid. Since $A_2 \gg A_1$,a much larger force $F_2$ is exerted on the wheel pistons,where $F_2 = F_1 \times (A_2 / A_1)$.
This large force $F_2$ pushes the brake shoes outward,bringing them into contact with the rotating wheel rim,thereby applying the brakes.
When the driver releases the brake pedal,a restoring spring pulls the brake shoes back to their original position,releasing the wheel rim. The brake oil flows back into the master cylinder as the pistons return to their initial positions.

(N/A) Pascal's law states that the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the container.
Mathematically,if an external pressure $P_{ext}$ is applied to a confined fluid,the change in pressure $\Delta P$ at any point within the fluid is equal to the applied pressure,i.e.,$\Delta P = P_{ext}$.

(N/A) Pascal's law states that a change in pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel.
Significance:
$1$. It forms the fundamental principle behind hydraulic systems,such as hydraulic lifts,hydraulic brakes,and hydraulic jacks.
$2$. It allows for the multiplication of force,where a small input force applied to a small area can result in a much larger output force on a larger area,as defined by the relation $F_2 = F_1 \times (A_2 / A_1)$.

(N/A) Pascal's law states that a change in pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the container.
Two common devices that operate on this principle are:
$1$. Hydraulic lift: Used to lift heavy objects like cars in service stations.
$2$. Hydraulic brakes: Used in automobiles to stop or slow down the vehicle by applying pressure to the brake fluid.

(N/A) When the cork of area $a$ is hit with a force $F$,it exerts a pressure $P = \frac{F}{a}$ on the liquid. According to Pascal's law,this pressure is transmitted equally and undiminished throughout the entire liquid. Since the area of the bottom of the bottle $A$ is much larger than the area of the cork $a$,the force exerted on the bottom is $F_{bottom} = P \times A$. Because $A \gg a$,the resulting force $F_{bottom}$ is significantly large,which causes the bottom of the bottle to break.

(N/A) The devices working on the principle of Pascal's law are:
$(1)$ Hydraulic lift: It is used to lift heavy objects like cars by applying a small force.
$(2)$ Hydraulic brakes: It is used in vehicles to stop or slow down the motion by transmitting pressure through a fluid.

(D) According to Pascal's Law,the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
The force acting on the bigger piston due to the weight of the car is:
$F = m \times g$
$F = 3000 \, kg \times 9.8 \, m/s^{2} = 29400 \, N$
The area of the bigger piston is:
$A = 425 \, cm^{2} = 425 \times 10^{-4} \, m^{2} = 0.0425 \, m^{2}$
The pressure exerted on the bigger piston is:
$P = \frac{F}{A}$
$P = \frac{29400 \, N}{0.0425 \, m^{2}}$
$P \approx 6.92 \times 10^{5} \, Pa$
Since the pressure is transmitted equally throughout the fluid,the smaller piston must bear the same pressure of $6.92 \times 10^{5} \, Pa$.

(B) According to Pascal's law,the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. For a hydraulic lift,the pressure at both pistons is equal:
$\frac{F_1}{A_1} = \frac{F_2}{A_2}$
In the first case,let $A_1$ be the area of the smaller piston and $A_2$ be the area of the larger piston. The force on the smaller piston is $mg$ and on the larger piston is $100g$:
$\frac{mg}{A_1} = \frac{100g}{A_2} \implies \frac{100}{m} = \frac{A_2}{A_1} \quad .......(1)$
Since $A = \frac{\pi d^2}{4}$,the ratio of areas is proportional to the square of the ratio of diameters: $\frac{A_2}{A_1} = \left(\frac{d_2}{d_1}\right)^2$.
In the second case,the new diameter of the larger piston is $d_2' = 4d_2$ and the new diameter of the smaller piston is $d_1' = \frac{d_1}{4}$.
Let the new mass lifted be $M_0$. The new areas are $A_2' = (4)^2 A_2 = 16A_2$ and $A_1' = (\frac{1}{4})^2 A_1 = \frac{A_1}{16}$.
Applying Pascal's law again:
$\frac{mg}{A_1'} = \frac{M_0g}{A_2'} \implies \frac{M_0}{m} = \frac{A_2'}{A_1'} = \frac{16A_2}{A_1/16} = 256 \left(\frac{A_2}{A_1}\right)$
Substituting $\frac{A_2}{A_1} = \frac{100}{m}$ from equation $(1)$:
$\frac{M_0}{m} = 256 \left(\frac{100}{m}\right)$
$M_0 = 256 \times 100 = 25600 \, kg$.

(A) According to Pascal's Law,the pressure at any point in a static fluid depends only on the depth $h$ of that point below the free surface of the fluid. The pressure is given by $p = h \rho g$.
Since the lengths $AB$ and $CD$ are very small compared to the total height $h$ of the water column,the depth of all points on the surfaces $AB$,$BC$,and $CD$ is approximately equal to $h$.
Pressure is a scalar quantity,which means it acts equally in all directions at a given depth. Therefore,the magnitude of the pressure exerted by the water on any surface at a depth $h$ is independent of the orientation of the surface.
Thus,$p_1 = p_2 = p_3 = h \rho g$.

(B) The correct option is $B$.
Pascal's law states that whenever external pressure is applied to any part of a confined fluid at rest,the pressure is transmitted undiminished to every portion of the fluid and to the walls of the container.

(B) According to Pascal's Law,the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. Therefore,the pressure at both pistons must be equal.
$\frac{F_1}{A_1} = \frac{F_2}{A_2}$
Given:
$F_1 = F$
$A_1 = 10 \, cm^2 = 10 \times 10^{-4} \, m^2$
$A_2 = 10 \, m^2$
$F_2 = W \times g = 800 \, kg \times 10 \, m/s^2 = 8000 \, N$
Substituting the values into the equation:
$\frac{F}{10 \times 10^{-4}} = \frac{8000}{10}$
$F = 800 \times 10 \times 10^{-4}$
$F = 8000 \times 10^{-4}$
$F = 0.8 \, N$

(C) Assertion $A$ is correct because squeezing a tube of toothpaste is a practical application of Pascal's principle,where pressure applied at one point is transmitted through the fluid to the opening.
Reason $R$ is the formal definition of Pascal's principle,which states that a change in pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of the container.
Since the phenomenon described in $A$ occurs precisely because of the physical law described in $R$,$R$ is the correct explanation of $A$.

(C) The force exerted by the load is given by $F = mg$.
Given $m = 5000\,kg$ and $g = 10\,m/s^2$,we have $F = 5000 \times 10 = 50000\,N$.
The area of the cross-section is $A = 250\,cm^2 = 250 \times 10^{-4}\,m^2 = 2.5 \times 10^{-2}\,m^2$.
According to Pascal's Law,the pressure applied to the fluid is transmitted equally throughout. The pressure $P$ exerted by the load is $P = \frac{F}{A}$.
$P = \frac{50000}{250 \times 10^{-4}} = \frac{5 \times 10^4}{2.5 \times 10^{-2}} = 2 \times 10^6\,Pa$.

(B) For a static fluid,the pressure at any point at the same horizontal depth $h$ is given by $P = P_{atm} + \rho gh$. Since $P_{atm}$,$\rho$,$g$,and $h$ are constant for a given horizontal level,the pressure is the same at all points at the same level. Thus,Statement $I$ is true.
According to Pascal's Law,a change in pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel. Thus,Statement $II$ is true.

(B) According to Pascal's Law, the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
Therefore, the pressure at both arms must be equal for equilibrium:
$\frac{F_1}{A_1} = \frac{F_2}{A_2}$
Here, $F_1$ is the force on the thicker arm, $A_1$ is the area of the thicker arm, $F_2 = 10 \,N$ is the force on the thinner arm, and $A_2$ is the area of the thinner arm.
The diameter of the thicker arm is $D_1 = 14 \,cm$, so its radius is $r_1 = 7 \,cm$.
The diameter of the thinner arm is $D_2 = 1.4 \,cm$, so its radius is $r_2 = 0.7 \,cm$.
Substituting the areas $A = \pi r^2$:
$\frac{F_1}{\pi (7)^2} = \frac{10}{\pi (0.7)^2}$
$F_1 = 10 \times \frac{49}{0.49}$
$F_1 = 10 \times 100 = 1000 \,N$.

(D) According to Pascal's law,the pressure applied at the input piston is transmitted equally to the output piston.
$P_1 = P_2 \implies \frac{F_1}{A_1} = \frac{F_2}{A_2}$
Given: $F_1 = 100 \ N$,$A_1 = 6 \ cm^2$,$A_2 = 1500 \ cm^2$.
$F_2 = F_1 \times \frac{A_2}{A_1} = 100 \times \frac{1500}{6} = 100 \times 250 = 25000 \ N$.
The work done on the output piston is $W = F_2 \times d_2$,where $d_2 = 20 \ cm = 0.2 \ m$.
$W = 25000 \ N \times 0.2 \ m = 5000 \ J$.
Since $1 \ kJ = 1000 \ J$,the work done is $5 \ kJ$.

(C) The 'Hydraulic lift' operates on the principle of Pascal's Law.
Pascal's Law states that a change in pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the container.
$A$ hydraulic lift uses this principle by applying a small force on a small piston to create pressure, which is then transmitted through the fluid to a larger piston, resulting in a larger force capable of lifting heavy objects.
Mathematically, $Pressure = \frac{Force}{Area}$. Since the pressure is constant throughout the system, $P = \frac{F_1}{A_1} = \frac{F_2}{A_2}$, which allows a small input force to generate a large output force.

(D) According to Pascal's Law,the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
For a hydraulic lift,the pressure on both pistons is equal: $P_1 = P_2$.
This implies $\frac{F_1}{A_1} = \frac{F_2}{A_2}$,where $F_1$ is the force on the smaller piston,$A_1$ is its area,$F_2$ is the force (weight) on the larger piston,and $A_2$ is its area.
The area of a circular piston is given by $A = \pi r^2$.
Thus,$\frac{F_1}{\pi r_1^2} = \frac{F_2}{\pi r_2^2}$,which simplifies to $F_2 = F_1 \times (\frac{r_2}{r_1})^2$.
Given $F_1 = 250 \ N$,$r_1 = 5 \ cm$,and $r_2 = 50 \ cm$.
Substituting the values: $F_2 = 250 \times (\frac{50}{5})^2 = 250 \times (10)^2 = 250 \times 100 = 25,000 \ N$.
Converting to kiloNewtons: $F_2 = 25 \ kN$.

(C) The principle of a hydraulic lift is based on Pascal's law,which states that pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
Pressure on the small piston $(P_1)$ = Pressure on the large piston $(P_2)$.
$\frac{F_1}{A_1} = \frac{F_2}{A_2}$
Here,$F_1 = F$,$r_1 = 3 \ cm = 3 \times 10^{-2} \ m$,$r_2 = 5 \ cm = 5 \times 10^{-2} \ m$,and $F_2 = mg = 1875 \times 10 = 18750 \ N$.
Substituting the values:
$\frac{F}{\pi (3 \times 10^{-2})^2} = \frac{18750}{\pi (5 \times 10^{-2})^2}$
$\frac{F}{9 \times 10^{-4}} = \frac{18750}{25 \times 10^{-4}}$
$F = \frac{18750 \times 9}{25}$
$F = 750 \times 9 = 6750 \ N$.

(A) In a hydraulic machine, the pressure at both pistons is the same according to Pascal's Law.
Pressure at piston $P_1 = $ Pressure at piston $P_2$
$\Rightarrow \frac{F_1}{A_1} = \frac{F_2}{A_2}$
Where $F_1 = mg = 2 \,kg \times 10 \,ms^{-2} = 20 \,N$.
$A_1 = \pi R_1^2 = \pi (2)^2 = 4\pi \,m^2$.
$A_2 = \pi R_2^2 = \pi (8)^2 = 64\pi \,m^2$.
Now, $F_2 = F_1 \times \frac{A_2}{A_1} = 20 \times \frac{64\pi}{4\pi} = 20 \times 16 = 320 \,N$.

(B) Pascal's law states that a change in pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel.
$A$ hydraulic lift is a practical application of Pascal's law. In a hydraulic lift,a small force $F_1$ applied to a small area $A_1$ creates a pressure $p_1 = F_1 / A_1$. This pressure is transmitted through the fluid to a larger area $A_2$,where it exerts a larger force $F_2 = p_2 \times A_2$. Since $p_1 = p_2$,we have:
$\frac{F_1}{A_1} = \frac{F_2}{A_2}$
Thus,the hydraulic lift works on the principle of Pascal's law.

(A) According to Pascal's law, the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.
Given:
Diameter of small piston, $d_1 = 8 \text{ cm}$
Diameter of large piston, $d_2 = 24 \text{ cm}$
Mass of the vehicle, $M = 1400 \text{ kg}$
Acceleration due to gravity, $g = 10 \text{ ms}^{-2}$
Force on large piston, $F_2 = M \times g = 1400 \times 10 = 14000 \text{ N}$
Pressure on small piston = Pressure on large piston
$\frac{F_1}{A_1} = \frac{F_2}{A_2}$
Since $A = \frac{\pi d^2}{4}$, we have:
$\frac{F_1}{d_1^2} = \frac{F_2}{d_2^2}$
$F_1 = F_2 \times \left(\frac{d_1}{d_2}\right)^2$
$F_1 = 14000 \times \left(\frac{8}{24}\right)^2$
$F_1 = 14000 \times \left(\frac{1}{3}\right)^2 = \frac{14000}{9} \approx 1555.55 \text{ N}$
Rounding to the nearest option, $F \approx 1600 \text{ N}$.

(D) Given radii: $r_A = 10 \,cm = 0.1 \,m$, $r_B = 100 \,cm = 1 \,m$, $r_C = 5 \,cm = 0.05 \,m$.
Weight placed on piston $A$ is $F_A = m_A g = 2g$.
According to Pascal's law, the pressure is transmitted equally throughout the fluid: $\frac{F_A}{A_A} = \frac{F_B}{A_B} = \frac{F_C}{A_C}$.
Since $A = \pi r^2$, we have $\frac{F_A}{r_A^2} = \frac{F_B}{r_B^2} = \frac{F_C}{r_C^2}$.
For piston $B$: $F_B = F_A \times \left(\frac{r_B}{r_A}\right)^2 = 2g \times \left(\frac{1}{0.1}\right)^2 = 2g \times 100 = 200g$.
Thus, the mass that can be lifted by $B$ is $m_B = 200 \,kg$.
For piston $C$: $F_C = F_A \times \left(\frac{r_C}{r_A}\right)^2 = 2g \times \left(\frac{0.05}{0.1}\right)^2 = 2g \times (0.5)^2 = 2g \times 0.25 = 0.5g$.
Thus, the mass that can be lifted by $C$ is $m_C = 0.5 \,kg$.
The calculated masses are $200 \,kg$ and $0.5 \,kg$. Since this does not match any of the given options, the correct choice is $D$.

(B) Given radii of pistons are $r_1 = 2 \ m$ and $r_2 = 5 \ m$. Let $m_1$ be the mass placed on piston $P_1$ and $m_2 = x$ be the mass placed on piston $P_2$.
According to Pascal's law,the pressure at the same horizontal level in a continuous static fluid is the same.
$p_1 = p_2$
$\frac{F_1}{A_1} = \frac{F_2}{A_2}$
Since $F = mg$ and $A = \pi r^2$,we have:
$\frac{m_1 g}{\pi r_1^2} = \frac{m_2 g}{\pi r_2^2}$
$\frac{m_1}{r_1^2} = \frac{m_2}{r_2^2}$
Substituting the given values:
$\frac{m_1}{2^2} = \frac{x}{5^2}$
$\frac{m_1}{4} = \frac{x}{25}$
$m_1 = \frac{4}{25} x = 0.16 x$
Thus,the minimum mass that should be kept on $P_1$ is $0.16 x$.

(A) According to Pascal's law, the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel. Therefore, the pressure at both pistons $P_1$ and $P_2$ must be equal.
Let $F_1$ and $F_2$ be the forces applied on pistons $P_1$ and $P_2$ respectively, and $A_1$ and $A_2$ be their respective areas.
$F_1 = m_1 g = 2g$
$F_2 = m_2 g = 32g$
$A_1 = \pi (2)^2 = 4\pi$
$A_2 = \pi R^2$
Equating the pressures: $\frac{F_1}{A_1} = \frac{F_2}{A_2}$
$\frac{2g}{4\pi} = \frac{32g}{\pi R^2}$
$\frac{1}{2} = \frac{32}{R^2}$
$R^2 = 64$
$R = 8 \,m$

(B) According to Pascal's law,the pressure applied at both pistons must be equal for equilibrium.
$P_1 = P_2$
$\frac{F_1}{A_1} = \frac{F_2}{A_2}$
Here,$F_1 = M_1 g = 1000 \times g$ and $F_2 = M_2 g$.
Given $A_1 = 1 \ m^2$ and $A_2 = 0.01 \ m^2$.
Substituting the values:
$\frac{1000 \times g}{1} = \frac{M_2 \times g}{0.01}$
$1000 = \frac{M_2}{0.01}$
$M_2 = 1000 \times 0.01 = 10 \ kg$.
Thus,a mass of $10 \ kg$ needs to be placed on piston $(P_2)$ to lift the $1000 \ kg$ mass.