Q: In a hydraulic lift, a force of 100\text{ N} is applied on a piston of area 0.01\text{ m}^2. The force produced on the larger piston of area 0.5\text{ m}^2 is:

5000\text{ N}. Pressure is common: \dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}, so F_2 = 100 \times \dfrac{0.5}{0.01} = 100 \times 50 = 5000\text{ N}.

Q: A force of 200\text{ N} acts on the small piston of a hydraulic press of area 5\text{ cm}^2. If the large piston has an area of 500\text{ cm}^2, the force it exerts is:

20000\text{ N}. F_2 = F_1 \dfrac{A_2}{A_1} = 200 \times \dfrac{500}{5} = 200 \times 100 = 20000\text{ N}.

Q: In a hydraulic lift the pistons have areas in the ratio 1 : 10. If the smaller piston is pushed down by 20\text{ cm}, the larger piston rises by:

2\text{ cm}. The liquid is incompressible, so the volumes swept are equal: A_1 d_1 = A_2 d_2. Hence d_2 = 20 \times \dfrac{1}{10} = 2\text{ cm}.

Question 1

Pascal's law states that a pressure applied to an enclosed incompressible fluid at rest is:

Accepted Answer

Transmitted undiminished to every point of the fluid and the walls. Pascal's law is the basis of all hydraulic machines: the added pressure appears equally everywhere in the enclosed fluid, regardless of direction or distance.

Question 2

In a hydraulic lift, a force of 100	ext{ N} is applied on a piston of area 0.01	ext{ m}^2. The force produced on the larger piston of area 0.5	ext{ m}^2 is:

Accepted Answer

5000	ext{ N}. Pressure is common: \dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}, so F_2 = 100 	imes \dfrac{0.5}{0.01} = 100 	imes 50 = 5000	ext{ N}.

Question 3

A force of 200	ext{ N} acts on the small piston of a hydraulic press of area 5	ext{ cm}^2. If the large piston has an area of 500	ext{ cm}^2, the force it exerts is:

Accepted Answer

20000	ext{ N}. F_2 = F_1 \dfrac{A_2}{A_1} = 200 	imes \dfrac{500}{5} = 200 	imes 100 = 20000	ext{ N}.

Question 4

In a hydraulic lift, the ratio of the forces on the two pistons is equal to:

Accepted Answer

The ratio of their areas. Since Pascal's law makes the pressure equal, \dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}, hence \dfrac{F_1}{F_2} = \dfrac{A_1}{A_2}.

Question 5

The pistons of a hydraulic lift have diameters in the ratio 1 : 10. The ratio of the forces they can exert is:

Accepted Answer

1 : 100. Area \propto d^2, so the area ratio is 1 : 100. Since force is proportional to area at constant pressure, the force ratio is also 1 : 100.

Question 6

The mechanical advantage of a hydraulic press with input piston area A_1 and output piston area A_2 is:

Accepted Answer

\dfrac{A_2}{A_1}. Mechanical advantage is the ratio of output force to input force, which equals \dfrac{A_2}{A_1} for a hydraulic press.

Question 7

In an ideal hydraulic press, the work done by the small piston compared with the work done on the load is:

Accepted Answer

Equal, since energy is conserved. A hydraulic press multiplies force but not energy. The small piston moves through a larger distance in exactly the proportion that reduces its force, so F_1 d_1 = F_2 d_2.

Question 8

In a hydraulic lift the pistons have areas in the ratio 1 : 10. If the smaller piston is pushed down by 20	ext{ cm}, the larger piston rises by:

Accepted Answer

2	ext{ cm}. The liquid is incompressible, so the volumes swept are equal: A_1 d_1 = A_2 d_2. Hence d_2 = 20 	imes \dfrac{1}{10} = 2	ext{ cm}.

Question 9

Hydraulic brakes used in automobiles work on the principle of:

Accepted Answer

Pascal's law. Pressing the brake pedal raises the pressure in the brake fluid; that increase is transmitted undiminished to all the wheel cylinders, so every wheel is braked equally.

Question 10

Which of the following does NOT work on Pascal's law?

Accepted Answer

A mercury barometer. A barometer measures atmospheric pressure by balancing it against a mercury column — that is hydrostatics (P = \rho g h), not the transmission of pressure through an enclosed fluid.

Question 11

For Pascal's law to hold, the fluid must be:

Accepted Answer

Enclosed, incompressible and at rest. The law applies to a confined fluid in equilibrium. If the fluid were compressible or moving, part of the applied pressure would go into compression or kinetic energy instead of being transmitted intact.

Question 12

A hydraulic jack lifts a car of weight 12000	ext{ N} using an effort of 400	ext{ N}. The ratio of the piston areas A_2 : A_1 is:

Accepted Answer

30 : 1. \dfrac{A_2}{A_1} = \dfrac{F_2}{F_1} = \dfrac{12000}{400} = 30, so the ratio is 30 : 1.

Question 13

Neglecting the weight of the fluid itself, the pressure at every point in an enclosed hydraulic system is:

Accepted Answer

The same throughout. By Pascal's law the transmitted pressure is uniform. The forces differ only because the pistons have different areas, since F = PA.

Question 14

A hydraulic machine is said to multiply:

Accepted Answer

Force, but not energy. Energy conservation forbids getting more work out than is put in. The machine trades distance for force: a large output force acts over a correspondingly small displacement.

Question 15

A dentist's chair is raised using a hydraulic system because such a system provides:

Accepted Answer

A large, smooth and uniformly distributed lifting force from a small effort. The pressure is transmitted equally in all directions, so the lift is steady and jerk-free, and the large output piston converts a modest effort into a large force.

Question 16

If the radius of the larger piston of a hydraulic lift is tripled while the smaller piston is unchanged, the output force for the same effort becomes:

Accepted Answer

9 times. Output force \propto A_2 \propto r_2^2. Tripling the radius multiplies the area, and hence the force, by 9.

Question 17

A force of 150	ext{ N} is applied to a hydraulic piston of area 0.02	ext{ m}^2. The force on the output piston of area 0.6	ext{ m}^2 is:

Accepted Answer

4500	ext{ N}. F_2 = F_1\dfrac{A_2}{A_1} = 150 	imes \dfrac{0.6}{0.02} = 150 	imes 30 = 4500	ext{ N}.

Question 18

The pistons of a hydraulic press have radii in the ratio 1 : 5. The ratio of the forces they exert is:

Accepted Answer

1 : 25. Area \propto r^2, so the areas are in the ratio 1 : 25, and at equal pressure the forces follow the same ratio.

Question 19

A force of 50	ext{ N} acts on the small piston of a hydraulic press of area 4	ext{ cm}^2. If the larger piston has an area of 200	ext{ cm}^2, the output force is:

Accepted Answer

2500	ext{ N}. F_2 = 50 	imes \dfrac{200}{4} = 50 	imes 50 = 2500	ext{ N}.

Question 20

In a hydraulic lift the piston areas are in the ratio 1 : 8. If the smaller piston moves down by 40	ext{ cm}, the larger piston rises by:

Accepted Answer

5	ext{ cm}. The swept volumes are equal: A_1 d_1 = A_2 d_2 \Rightarrow d_2 = 40 	imes \dfrac{1}{8} = 5	ext{ cm}.

Pascal's Law & Applications — MH-CET Physics MCQs with Solutions

Questions with solutions