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101. Pascal’s law states that:
ⓐ. Pressure in a liquid decreases with depth
ⓑ. Pressure applied to a confined fluid is transmitted undiminished in all directions
ⓒ. Pressure in a gas is independent of volume
ⓓ. Fluids cannot exert pressure
Correct Answer: Pressure applied to a confined fluid is transmitted undiminished in all directions
Explanation: Pascal’s law states that when pressure is applied to a fluid in an enclosed container, the pressure is transmitted equally and undiminished throughout the fluid. This is the basis of hydraulic machines.
102. Which scientist first formulated Pascal’s law?
ⓐ. Robert Boyle
ⓑ. Evangelista Torricelli
ⓒ. Blaise Pascal
ⓓ. Isaac Newton
Correct Answer: Blaise Pascal
Explanation: Blaise Pascal formulated this principle in the 17th century. Boyle worked on gas laws, Torricelli invented the barometer, and Newton gave laws of motion and gravitation.
103. Which of the following is a direct application of Pascal’s law?
ⓐ. Buoyancy of ships
ⓑ. Flow of blood in veins
ⓒ. Hydraulic lift in car workshops
ⓓ. Airplane wings generating lift
Correct Answer: Hydraulic lift in car workshops
Explanation: Hydraulic lifts apply pressure at one point and transmit it through a confined fluid to lift heavy loads. This works because of Pascal’s law. Buoyancy is Archimedes’ principle, while wings use Bernoulli’s principle.
104. In a hydraulic press, a force of 200 N is applied on a piston of area $0.01 \, \text{m}^2$. The area of the larger piston is $0.5 \, \text{m}^2$. What is the output force?
ⓐ. 1000 N
ⓑ. 5000 N
ⓒ. 10,000 N
ⓓ. 20,000 N
Correct Answer: 10,000 N
Explanation: Pressure = $\frac{F}{A} = \frac{200}{0.01} = 2 \times 10^4 \, \text{Pa}$. Output force = $P \times A_2 = (2 \times 10^4)(0.5) = 1.0 \times 10^4 \, \text{N}$.
105. Which of the following medical devices works on Pascal’s law?
ⓐ. Sphygmomanometer (blood pressure monitor)
ⓑ. Stethoscope
ⓒ. X-ray machine
ⓓ. Thermometer
Correct Answer: Sphygmomanometer (blood pressure monitor)
Explanation: A sphygmomanometer uses a cuff filled with air. The applied air pressure is transmitted uniformly around the arm, demonstrating Pascal’s law.
106. The pressure in a hydraulic lift is 4000 Pa. If the area of the large piston is $0.5 \, \text{m}^2$, what is the force exerted on it?
ⓐ. 200 N
ⓑ. 1000 N
ⓒ. 2000 N
ⓓ. 4000 N
Correct Answer: 2000 N
Explanation: Force = Pressure × Area = $(4000)(0.5) = 2000 \, \text{N}$.
107. Why do hydraulic brakes work more efficiently than mechanical brakes?
ⓐ. Because fluids reduce friction
ⓑ. Because fluids cannot transmit pressure
ⓒ. Because pressure applied at one point is transmitted equally to all wheels
ⓓ. Because brakes work without pressure
Correct Answer: Because pressure applied at one point is transmitted equally to all wheels
Explanation: Pascal’s law ensures that pressure applied on brake fluid in the master cylinder is transmitted equally to all four wheels, giving efficient braking with less effort.
108. In a hydraulic system, the input piston area is $0.02 \, \text{m}^2$ and the output piston area is $0.2 \, \text{m}^2$. If an input force of 100 N is applied, calculate the output force.
ⓐ. 500 N
ⓑ. 1000 N
ⓒ. 2000 N
ⓓ. 3000 N
Correct Answer: 1000 N
Explanation: Pressure = $\frac{100}{0.02} = 5000 \, \text{Pa}$. Output force = $5000 \times 0.2 = 1000 \, \text{N}$.
109. Which principle explains why toothpaste can be squeezed out of a closed tube by pressing at any point?
ⓐ. Archimedes’ principle
ⓑ. Pascal’s law
ⓒ. Bernoulli’s theorem
ⓓ. Newton’s second law
Correct Answer: Pascal’s law
Explanation: When pressure is applied to the toothpaste tube, it is transmitted equally through the fluid paste, forcing it out of the opening.
110. Which of the following statements is NOT an application of Pascal’s law?
ⓐ. Hydraulic lifts
ⓑ. Hydraulic brakes
ⓒ. Transmission of pressure in liquids
ⓓ. Flotation of ships
Correct Answer: Flotation of ships
Explanation: Flotation of ships is explained by Archimedes’ principle. Hydraulic lifts, brakes, and pressure transmission in liquids are direct applications of Pascal’s law.
111. In a hydraulic press, the input piston has an area of $0.01 \, \text{m}^2$ and the output piston has an area of $0.5 \, \text{m}^2$. If a force of 150 N is applied on the small piston, calculate the force on the large piston.
ⓐ. 5000 N
ⓑ. 6000 N
ⓒ. 7000 N
ⓓ. 7500 N
Correct Answer: 7500 N
Explanation: Pressure = $\frac{F_1}{A_1} = \frac{150}{0.01} = 1.5 \times 10^4 \, \text{Pa}$. Force on large piston = $P \times A_2 = (1.5 \times 10^4)(0.5) = 7500 \, \text{N}$.
112. Which of the following is NOT a hydraulic system application?
ⓐ. Car brakes
ⓑ. Airplane control systems
ⓒ. Lifting cranes
ⓓ. Optical lens systems
Correct Answer: Optical lens systems
Explanation: Hydraulic systems rely on Pascal’s principle for transmitting pressure. Car brakes, cranes, and airplane control use hydraulics. Optical lenses work on refraction, not hydraulics.
113. A hydraulic lift has a mechanical advantage of 50. If an input force of 200 N is applied, the output force is:
ⓐ. 2000 N
ⓑ. 5000 N
ⓒ. 8000 N
ⓓ. 10,000 N
Correct Answer: 10,000 N
Explanation: Mechanical advantage (M.A.) = $\frac{F_{out}}{F_{in}}$. Thus, $F_{out} = M.A. \times F_{in} = 50 \times 200 = 10,000 \, \text{N}$.
114. Why are hydraulic systems more efficient than mechanical lever systems?
ⓐ. Fluids cannot leak
ⓑ. Fluids can transmit pressure equally in all directions
ⓒ. Fluids reduce friction to zero
ⓓ. Fluids have infinite density
Correct Answer: Fluids can transmit pressure equally in all directions
Explanation: According to Pascal’s law, pressure applied on a confined fluid is transmitted equally in all directions. This allows a small input force to produce a large output force, which makes hydraulic systems efficient.
115. Which of the following correctly explains the working of hydraulic jacks?
ⓐ. They work by Bernoulli’s theorem
ⓑ. They amplify input force using Pascal’s principle
ⓒ. They reduce input force using Archimedes’ principle
ⓓ. They depend on surface tension
Correct Answer: They amplify input force using Pascal’s principle
Explanation: Hydraulic jacks use Pascal’s law: pressure applied on a small piston is transmitted to a larger piston, producing a much greater lifting force.
116. In a hydraulic system, the pressure applied on the input piston is $2 \times 10^4 \, \text{Pa}$. If the output piston has an area of $0.4 \, \text{m}^2$, find the output force.
ⓐ. 500 N
ⓑ. 2000 N
ⓒ. 4000 N
ⓓ. 8000 N
Correct Answer: 8000 N
Explanation: Force = Pressure × Area = $(2 \times 10^4)(0.4) = 8000 \, \text{N}$.
117. Why are hydraulic brakes preferred in modern vehicles?
ⓐ. They require no external fluid
ⓑ. They transmit pressure equally to all four wheels with little effort
ⓒ. They increase the braking distance
ⓓ. They reduce pressure transmission
Correct Answer: They transmit pressure equally to all four wheels with little effort
Explanation: Hydraulic brakes apply Pascal’s law to distribute pressure uniformly through brake fluid, providing effective braking simultaneously at all wheels.
118. A hydraulic lift is used to raise a car of mass 1200 kg. If the area of the larger piston is $0.3 \, \text{m}^2$, calculate the pressure required in the system. (Take $g = 10 \, \text{m/s}^2$)
ⓐ. $2.0 \times 10^4 \, \text{Pa}$
ⓑ. $3.0 \times 10^4 \, \text{Pa}$
ⓒ. $4.0 \times 10^4 \, \text{Pa}$
ⓓ. $5.0 \times 10^4 \, \text{Pa}$
Correct Answer: $4.0 \times 10^4 \, \text{Pa}$
Explanation: Weight of car = $mg = 1200 \times 10 = 12000 \, \text{N}$. Pressure = $\frac{F}{A} = \frac{12000}{0.3} = 4.0 \times 10^4 \, \text{Pa}$.
119. In airplane hydraulic systems, Pascal’s principle is used to:
ⓐ. Control airflow over wings
ⓑ. Transmit pressure to control surfaces like flaps and rudders
ⓒ. Reduce drag force
ⓓ. Increase lift using surface tension
Correct Answer: Transmit pressure to control surfaces like flaps and rudders
Explanation: Hydraulic systems in airplanes transmit pressure through fluid to operate control surfaces (flaps, rudders, landing gear). This allows precise control with little pilot effort.
120. A hydraulic machine has two pistons. The small piston has an area of $0.005 \, \text{m}^2$ and the large piston has an area of $0.5 \, \text{m}^2$. If a force of 100 N is applied on the small piston, calculate the force produced on the large piston.
ⓐ. 500 N
ⓑ. 1000 N
ⓒ. 5000 N
ⓓ. 10,000 N
Correct Answer: 10,000 N
Explanation: Pressure on small piston = $\frac{100}{0.005} = 2 \times 10^4 \, \text{Pa}$. Output force = $P \times A = (2 \times 10^4)(0.5) = 1.0 \times 10^4 \, \text{N}$.
121. A hydraulic lift has a small piston of area $0.01 \, \text{m}^2$ and a large piston of area $0.5 \, \text{m}^2$. If a force of 200 N is applied on the small piston, what load can be lifted on the large piston?
ⓐ. 2000 N
ⓑ. 4000 N
ⓒ. 6000 N
ⓓ. 10,000 N
Correct Answer: 10,000 N
Explanation: Pressure = $\frac{200}{0.01} = 2.0 \times 10^4 \, \text{Pa}$. Force on large piston = $P \times A = (2.0 \times 10^4)(0.5) = 1.0 \times 10^4 \, \text{N}$.
122. Hydraulic brakes work efficiently because:
ⓐ. Liquids are highly compressible
ⓑ. Pressure applied is transmitted equally to all parts of the liquid
ⓒ. Friction is reduced in the fluid
ⓓ. Brake fluid increases force by chemical reaction
Correct Answer: Pressure applied is transmitted equally to all parts of the liquid
Explanation: Hydraulic brakes are based on Pascal’s law. Pressure applied on the brake pedal is transmitted equally through brake fluid, applying equal pressure to all wheels.
123. In a hydraulic brake system, if the master cylinder exerts a pressure of $3 \times 10^5 \, \text{Pa}$ and each brake piston has an area of $0.01 \, \text{m}^2$, what force is exerted on each brake shoe?
ⓐ. 1000 N
ⓑ. 2000 N
ⓒ. 3000 N
ⓓ. 4000 N
Correct Answer: 3000 N
Explanation: Force = Pressure × Area = $(3 \times 10^5)(0.01) = 3000 \, \text{N}$.
124. Which of the following is NOT a limitation of hydraulic brakes?
ⓐ. Leakage of brake fluid
ⓑ. Formation of air bubbles in the fluid
ⓒ. Uniform pressure transmission
ⓓ. Brake fluid heating under repeated use
Correct Answer: Uniform pressure transmission
Explanation: Uniform pressure transmission is the main advantage of hydraulic brakes. Limitations include leakage, heating, and air bubble formation which reduce effectiveness.
125. A car weighing 15,000 N is lifted using a hydraulic lift. The large piston has an area of $0.3 \, \text{m}^2$. What is the minimum pressure required in the system?
ⓐ. $2.5 \times 10^4 \, \text{Pa}$
ⓑ. $3.5 \times 10^4 \, \text{Pa}$
ⓒ. $4.5 \times 10^4 \, \text{Pa}$
ⓓ. $5.0 \times 10^4 \, \text{Pa}$
Correct Answer: $5.0 \times 10^4 \, \text{Pa}$
Explanation: Pressure = Force / Area = $\frac{15000}{0.3} = 5.0 \times 10^4 \, \text{Pa}$.
126. Why do hydraulic brakes apply equal force on all four wheels of a car?
ⓐ. Brake pads are identical in all wheels
ⓑ. Fluids transmit pressure equally in all directions (Pascal’s law)
ⓒ. Mechanical levers distribute force
ⓓ. Braking fluid evaporates uniformly
Correct Answer: Fluids transmit pressure equally in all directions (Pascal’s law)
Explanation: Pascal’s principle ensures equal pressure distribution in all directions, so all four wheels receive the same braking force.
127. In a hydraulic lift, the input piston area is $0.02 \, \text{m}^2$ and the output piston area is $0.5 \, \text{m}^2$. If an operator applies 400 N on the small piston, what load is lifted?
ⓐ. 5000 N
ⓑ. 8000 N
ⓒ. 9000 N
ⓓ. 10,000 N
Correct Answer: 10,000 N
Explanation: Pressure = $\frac{400}{0.02} = 2.0 \times 10^4 \, \text{Pa}$. Output force = $2.0 \times 10^4 \times 0.5 = 10,000 \, \text{N}$.
128. Which safety issue is commonly faced in hydraulic brakes?
ⓐ. Non-uniform pressure
ⓑ. Air bubble formation in brake fluid
ⓒ. Infinite fluid density
ⓓ. Overcooling of brake fluid
Correct Answer: Air bubble formation in brake fluid
Explanation: Air bubbles compress easily and reduce pressure transmission, leading to brake failure. This is why hydraulic systems are bled regularly to remove trapped air.
129. A hydraulic jack has an input piston of radius 2 cm and an output piston of radius 20 cm. If a force of 100 N is applied on the smaller piston, find the output force.
ⓐ. 10,000 N
ⓑ. 8000 N
ⓒ. 6000 N
ⓓ. 5000 N
Correct Answer: 10,000 N
Explanation: Area ratio = $\frac{A_2}{A_1} = \frac{\pi (0.2)^2}{\pi (0.02)^2} = 100$. Force on larger piston = $100 \times 100 = 10,000 \, N$.
130. Which real-life example demonstrates both Pascal’s principle and hydraulic system application?
ⓐ. Bicycle pump
ⓑ. Syringe
ⓒ. Water fountain
ⓓ. Hydraulic car lift
Correct Answer: Hydraulic car lift
Explanation: A hydraulic car lift works on Pascal’s principle, transmitting pressure through fluid to lift heavy vehicles with a small applied force. Syringes and pumps involve fluid flow but not hydraulic multiplication.
131. In a hydraulic press, the input piston has an area of $0.02 \, \text{m}^2$. A force of 100 N is applied on it. If the output piston has an area of $0.5 \, \text{m}^2$, calculate the load lifted.
ⓐ. 2000 N
ⓑ. 2500 N
ⓒ. 3000 N
ⓓ. 3500 N
Correct Answer: 2500 N
Explanation: Pressure = $\frac{F}{A} = \frac{100}{0.02} = 5000 \, \text{Pa}$. Load on large piston = $P \times A_2 = 5000 \times 0.5 = 2500 \, \text{N}$.
132. A hydraulic lift has a large piston of area $0.4 \, \text{m}^2$ and a small piston of area $0.01 \, \text{m}^2$. What force must be applied to lift a 2000 N load?
ⓐ. 25 N
ⓑ. 40 N
ⓒ. 50 N
ⓓ. 100 N
Correct Answer: 25 N
Explanation: $F_1 = \frac{A_1}{A_2} \times F_2 = \frac{0.01}{0.4} \times 2000 = 25 \, \text{N}$.
133. The master piston of a hydraulic brake has an area of $0.001 \, \text{m}^2$. If the driver applies a force of 50 N, what pressure is transmitted to the brake fluid?
ⓐ. $2.0 \times 10^4 \, \text{Pa}$
ⓑ. $5.0 \times 10^4 \, \text{Pa}$
ⓒ. $2.5 \times 10^5 \, \text{Pa}$
ⓓ. $5.0 \times 10^5 \, \text{Pa}$
Correct Answer: $5.0 \times 10^4 \, \text{Pa}$
Explanation: Pressure = $\frac{F}{A} = \frac{50}{0.001} = 5.0 \times 10^4 \, \text{Pa}$.
134. A hydraulic jack has an input piston of radius 5 cm and an output piston of radius 20 cm. If a force of 200 N is applied on the input piston, calculate the output force.
ⓐ. 2000 N
ⓑ. 2500 N
ⓒ. 3000 N
ⓓ. 3200 N
Correct Answer: 3200 N
Explanation: Area ratio = $\frac{A_2}{A_1} = \frac{\pi (0.2)^2}{\pi (0.05)^2} = \frac{0.1256}{0.00785} = 16$. Output force = $200 \times 16 = 3200 \, N$.
135. A hydraulic lift is used to raise a car of mass 1500 kg. If the area of the large piston is $0.5 \, \text{m}^2$, find the pressure needed in the system. (Take $g = 9.8 \, \text{m/s}^2$)
ⓐ. $2.94 \times 10^4 \, \text{Pa}$
ⓑ. $3.94 \times 10^4 \, \text{Pa}$
ⓒ. $4.94 \times 10^4 \, \text{Pa}$
ⓓ. $5.94 \times 10^4 \, \text{Pa}$
Correct Answer: $2.94 \times 10^4 \, \text{Pa}$
Explanation: Weight of car = $1500 \times 9.8 = 14,700 \, N$. Pressure = $\frac{F}{A} = \frac{14,700}{0.5} = 2.94 \times 10^4 \, \text{Pa}$.
136. A force of 120 N is applied on a piston of area $0.02 \, \text{m}^2$. The pressure transmitted to the fluid is:
ⓐ. 5000 Pa
ⓑ. 6000 Pa
ⓒ. 7000 Pa
ⓓ. 8000 Pa
Correct Answer: 6000 Pa
Explanation: Pressure = $\frac{F}{A} = \frac{120}{0.02} = 6000 \, \text{Pa}$.
137. A hydraulic brake system applies a pressure of $2 \times 10^5 \, \text{Pa}$ to the brake shoes. If the area of each brake piston is $0.005 \, \text{m}^2$, what force acts on each brake shoe?
ⓐ. 500 N
ⓑ. 750 N
ⓒ. 1000 N
ⓓ. 1200 N
Correct Answer: 1000 N
Explanation: Force = Pressure × Area = $(2 \times 10^5)(0.005) = 1000 \, N$.
138. A hydraulic lift has two pistons. The smaller piston has an area of $0.01 \, \text{m}^2$ and the larger piston has an area of $0.4 \, \text{m}^2$. If a load of 8000 N is to be lifted, calculate the force required on the smaller piston.
ⓐ. 150 N
ⓑ. 200 N
ⓒ. 250 N
ⓓ. 300 N
Correct Answer: 200 N
Explanation: $F_1 = \frac{A_1}{A_2} \times F_2 = \frac{0.01}{0.4} \times 8000 = 200 \, N$.
139. An input piston of area $0.005 \, \text{m}^2$ is used to apply a force of 60 N. Find the pressure and also the force produced on an output piston of area $0.25 \, \text{m}^2$.
ⓐ. $1.2 \times 10^4 \, \text{Pa}, 3000 \, N$
ⓑ. $1.2 \times 10^4 \, \text{Pa}, 2000 \, N$
ⓒ. $1.0 \times 10^4 \, \text{Pa}, 2500 \, N$
ⓓ. $1.2 \times 10^4 \, \text{Pa}, 4000 \, N$
Correct Answer: $1.2 \times 10^4 \, \text{Pa}, 3000 \, N$
Explanation: Pressure = $\frac{F}{A} = \frac{60}{0.005} = 1.2 \times 10^4 \, \text{Pa}$. Output force = $P \times A_2 = (1.2 \times 10^4)(0.25) = 3000 \, N$.
140. A car weighing $20,000 \, N$ is lifted using a hydraulic lift. The large piston has an area of $0.4 \, \text{m}^2$. What minimum force is required on the small piston of area $0.02 \, \text{m}^2$?
ⓐ. 800 N
ⓑ. 900 N
ⓒ. 1000 N
ⓓ. 1100 N
Correct Answer: 1000 N
Explanation: Pressure needed = $\frac{F_2}{A_2} = \frac{20,000}{0.4} = 5.0 \times 10^4 \, \text{Pa}$. Input force = $P \times A_1 = (5.0 \times 10^4)(0.02) = 1000 \, N$.
141. Hydrostatic pressure at a depth $h$ in a liquid of density $\rho$ is given by:
ⓐ. $P = \rho g h$
ⓑ. $P = \frac{\rho}{gh}$
ⓒ. $P = \rho h^2 g$
ⓓ. $P = \frac{1}{\rho g h}$
Correct Answer: $P = \rho g h$
Explanation: Hydrostatic pressure is directly proportional to depth, density of liquid, and gravitational acceleration: $P = \rho g h$. Other expressions are dimensionally incorrect.
142. A container is filled with water ($\rho = 1000 \, \text{kg/m}^3$) to a height of 10 m. Find the hydrostatic pressure at the bottom. (Take $g = 9.8 \, \text{m/s}^2$)
ⓐ. $9.8 \times 10^3 \, \text{Pa}$
ⓑ. $9.8 \times 10^4 \, \text{Pa}$
ⓒ. $1.98 \times 10^5 \, \text{Pa}$
ⓓ. $9.8 \times 10^5 \, \text{Pa}$
Correct Answer: $9.8 \times 10^4 \, \text{Pa}$
Explanation: $P = \rho g h = (1000)(9.8)(10) = 9.8 \times 10^4 \, \text{Pa}$.
143. Hydrostatic pressure does NOT depend on:
ⓐ. Depth of the fluid
ⓑ. Density of the fluid
ⓒ. Gravitational acceleration
ⓓ. Shape of the container
Correct Answer: Shape of the container
Explanation: Hydrostatic pressure is independent of container shape, depending only on depth, density, and gravity. This is the hydrostatic paradox.
144. A swimmer is 5 m below the surface of a swimming pool. What hydrostatic pressure acts on him due to water? ($\rho = 1000 \, \text{kg/m}^3, g = 10 \, \text{m/s}^2$)
ⓐ. $2.5 \times 10^3 \, \text{Pa}$
ⓑ. $3.5 \times 10^4 \, \text{Pa}$
ⓒ. $5.0 \times 10^4 \, \text{Pa}$
ⓓ. $1.0 \times 10^5 \, \text{Pa}$
Correct Answer: $5.0 \times 10^4 \, \text{Pa}$
Explanation: $P = \rho g h = (1000)(10)(5) = 5.0 \times 10^4 \, \text{Pa}$.
145. A dam is built to hold water of depth 40 m. What is the hydrostatic pressure at the base of the dam? ($\rho = 1000 \, \text{kg/m}^3, g = 9.8 \, \text{m/s}^2$)
ⓐ. $2.92 \times 10^5 \, \text{Pa}$
ⓑ. $3.92 \times 10^5 \, \text{Pa}$
ⓒ. $4.92 \times 10^5 \, \text{Pa}$
ⓓ. $5.92 \times 10^5 \, \text{Pa}$
Correct Answer: $3.92 \times 10^5 \, \text{Pa}$
Explanation: $P = (1000)(9.8)(40) = 3.92 \times 10^5 \, \text{Pa}$.
146. Which of the following explains why two containers of different shapes but same depth have the same pressure at the bottom?
ⓐ. Bernoulli’s theorem
ⓑ. Hydrostatic paradox
ⓒ. Pascal’s principle
ⓓ. Archimedes’ principle
Correct Answer: Hydrostatic paradox
Explanation: The hydrostatic paradox states that hydrostatic pressure at a depth depends only on vertical depth, not on the shape or volume of the container.
147. A cylindrical vessel is filled with oil ($\rho = 800 \, \text{kg/m}^3$) up to 6 m height. Find the hydrostatic pressure at the bottom. (Take $g = 10 \, \text{m/s}^2$)
ⓐ. $4.0 \times 10^4 \, \text{Pa}$
ⓑ. $4.5 \times 10^4 \, \text{Pa}$
ⓒ. $5.0 \times 10^4 \, \text{Pa}$
ⓓ. $4.8 \times 10^4 \, \text{Pa}$
Correct Answer: $4.8 \times 10^4 \, \text{Pa}$
Explanation: $P = \rho g h = (800)(10)(6) = 4.8 \times 10^4 \, \text{Pa}$.
148. Hydrostatic pressure inside a liquid column increases:
ⓐ. Linearly with depth
ⓑ. Exponentially with depth
ⓒ. Decreases with depth
ⓓ. Remains constant with depth
Correct Answer: Linearly with depth
Explanation: $P = \rho g h$ shows that pressure increases linearly with depth $h$. Doubling the depth doubles the pressure.
149. A tank has kerosene of density $820 \, \text{kg/m}^3$ to a height of 4 m. Calculate the hydrostatic pressure at the bottom. (Take $g = 9.8 \, \text{m/s}^2$)
ⓐ. $3.21 \times 10^4 \, \text{Pa}$
ⓑ. $3.50 \times 10^4 \, \text{Pa}$
ⓒ. $4.21 \times 10^4 \, \text{Pa}$
ⓓ. $5.00 \times 10^4 \, \text{Pa}$
Correct Answer: $3.21 \times 10^4 \, \text{Pa}$
Explanation: $P = (820)(9.8)(4) \approx 3.21 \times 10^4 \, \text{Pa}$.
150. The hydrostatic pressure at 10 m depth in mercury is: ($\rho = 13.6 \times 10^3 \, \text{kg/m}^3, g = 9.8 \, \text{m/s}^2$)
ⓐ. $1.23 \times 10^5 \, \text{Pa}$
ⓑ. $9.8 \times 10^4 \, \text{Pa}$
ⓒ. $1.33 \times 10^6 \, \text{Pa}$
ⓓ. $1.36 \times 10^6 \, \text{Pa}$
Correct Answer: $1.33 \times 10^6 \, \text{Pa}$
Explanation: $P = \rho g h = (13.6 \times 10^3)(9.8)(10) \approx 1.33 \times 10^6 \, \text{Pa}$. This very high value is due to mercury’s large density.
151. The variation of pressure with height in a fluid under gravity is expressed as:
ⓐ. $\frac{dP}{dh} = -\rho g$
ⓑ. $\frac{dP}{dh} = \rho g$
ⓒ. $\frac{dP}{dh} = -\frac{g}{\rho}$
ⓓ. $\frac{dP}{dh} = \frac{\rho}{g}$
Correct Answer: $\frac{dP}{dh} = -\rho g$
Explanation: Pressure decreases with increase in height. The negative sign indicates that as height $h$ increases, pressure $P$ decreases.
152. The pressure at height $h$ in a liquid column open to the atmosphere is given by:
ⓐ. $P = P_0 + \rho g h$
ⓑ. $P = P_0 – \rho g h$
ⓒ. $P = \rho g h$ only
ⓓ. $P = \frac{P_0}{\rho g h}$
Correct Answer: $P = P_0 + \rho g h$
Explanation: The absolute pressure at depth is equal to atmospheric pressure plus hydrostatic pressure.
153. Which statement is correct about atmospheric pressure with altitude?
ⓐ. Increases linearly with height
ⓑ. Remains constant with height
ⓒ. Decreases with height
ⓓ. Increases exponentially with height
Correct Answer: Decreases with height
Explanation: Atmospheric pressure decreases with height because the air column above reduces as altitude increases.
154. In the atmosphere, the variation of pressure with height is approximately:
ⓐ. Linear decrease
ⓑ. Exponential decrease
ⓒ. Constant everywhere
ⓓ. Proportional to height squared
Correct Answer: Exponential decrease
Explanation: For an isothermal atmosphere, $P = P_0 e^{-\frac{Mgh}{RT}}$. This shows that pressure falls exponentially with height.
155. If air pressure at sea level is $1.01 \times 10^5 \, \text{Pa}$, what is the approximate air pressure at a height of 5 km? (Assume $P = P_0 e^{-h/H}$, with scale height $H = 8 \, \text{km}$).
ⓐ. $0.80 \times 10^5 \, \text{Pa}$
ⓑ. $0.6 \times 10^5 \, \text{Pa}$
ⓒ. $0.30 \times 10^5 \, \text{Pa}$
ⓓ. $1.01 \times 10^5 \, \text{Pa}$
Correct Answer: $0.6 \times 10^5 \, \text{Pa}$
Explanation: $P = 1.01 \times 10^5 e^{-5000/8000} \approx 1.01 \times 10^5 \times 0.61 \approx 0.62 \times 10^5 \, \text{Pa}$. Closest option is B, but correct calculation gives about $0.62 \times 10^5 \, \text{Pa}$.
156. Why does it become difficult to breathe at high altitudes?
ⓐ. Oxygen content decreases at height
ⓑ. Air pressure decreases with altitude
ⓒ. Gravity becomes negligible
ⓓ. Temperature always increases with height
Correct Answer: Air pressure decreases with altitude
Explanation: Although oxygen percentage in air remains the same, reduced air pressure at high altitudes lowers oxygen partial pressure, making breathing difficult.
157. A mountain climber at 6000 m experiences reduced pressure compared to sea level. This reduction occurs because:
ⓐ. Gravitational acceleration is higher at altitude
ⓑ. Density of air decreases with height
ⓒ. Temperature increases drastically
ⓓ. Air becomes heavier at altitude
Correct Answer: Density of air decreases with height
Explanation: At higher altitudes, air density reduces due to lower pressure. The lower density decreases oxygen availability.
158. The pressure at a point in a static fluid at height $h$ above some reference level is:
ⓐ. $P = P_0 + \rho g h$
ⓑ. $P = P_0 – \rho g h$
ⓒ. $P = \rho g h$
ⓓ. $P = \frac{P_0}{\rho g h}$
Correct Answer: $P = P_0 – \rho g h$
Explanation: As we go upward (increasing height), pressure decreases relative to atmospheric reference: $P = P_0 – \rho g h$.
159. An air bubble rises from the bottom of a lake 40 m deep. If atmospheric pressure is $1.0 \times 10^5 \, \text{Pa}$, find the pressure inside the bubble at the bottom. (Take $ \rho = 1000 \, \text{kg/m}^3, g = 10 \, \text{m/s}^2$)
ⓐ. $2.0 \times 10^5 \, \text{Pa}$
ⓑ. $3.0 \times 10^5 \, \text{Pa}$
ⓒ. $4.0 \times 10^5 \, \text{Pa}$
ⓓ. $5.0 \times 10^5 \, \text{Pa}$
Correct Answer: $3.0 \times 10^5 \, \text{Pa}$
Explanation: Pressure = $P_{atm} + \rho g h = 1.0 \times 10^5 + (1000)(10)(40) = 3.0 \times 10^5 \, \text{Pa}$.
160. The pressure difference between two points at heights $h_1$ and $h_2$ in a liquid is:
ⓐ. $\Delta P = \rho g (h_1 – h_2)$
ⓑ. $\Delta P = \rho g (h_2 – h_1)$
ⓒ. $\Delta P = \frac{h_1}{h_2}$
ⓓ. $\Delta P = \rho g \frac{h_1}{h_2}$
Correct Answer: $\Delta P = \rho g (h_2 – h_1)$
Explanation: Hydrostatic law: pressure difference between two levels in a static fluid depends only on vertical height difference: $\Delta P = \rho g \Delta h$.
161. A tank is filled with water to a depth of 12 m. Calculate the pressure at the bottom of the tank. ($ \rho = 1000 \, \text{kg/m}^3, g = 9.8 \, \text{m/s}^2$)
ⓐ. $9.8 \times 10^3 \, \text{Pa}$
ⓑ. $1.18 \times 10^5 \, \text{Pa}$
ⓒ. $2.18 \times 10^5 \, \text{Pa}$
ⓓ. $3.18 \times 10^5 \, \text{Pa}$
Correct Answer: $1.18 \times 10^5 \, \text{Pa}$
Explanation: $P = \rho g h = (1000)(9.8)(12) = 1.18 \times 10^5 \, \text{Pa}$.
162. A diver descends to 30 m depth in seawater of density $1025 \, \text{kg/m}^3$. Find the gauge pressure at that depth. ($g = 9.8 \, \text{m/s}^2$)
ⓐ. $2.0 \times 10^5 \, \text{Pa}$
ⓑ. $2.5 \times 10^5 \, \text{Pa}$
ⓒ. $3.0 \times 10^5 \, \text{Pa}$
ⓓ. $3.5 \times 10^5 \, \text{Pa}$
Correct Answer: $3.0 \times 10^5 \, \text{Pa}$
Explanation: $P = \rho g h = (1025)(9.8)(30) \approx 3.0 \times 10^5 \, \text{Pa}$.
163. A cylindrical vessel is filled with oil ($\rho = 800 \, \text{kg/m}^3$) up to a height of 5 m. Find the pressure at the bottom due to oil. ($g = 10 \, \text{m/s}^2$)
ⓐ. $2.0 \times 10^4 \, \text{Pa}$
ⓑ. $3.0 \times 10^4 \, \text{Pa}$
ⓒ. $4.0 \times 10^4 \, \text{Pa}$
ⓓ. $5.0 \times 10^4 \, \text{Pa}$
Correct Answer: $4.0 \times 10^4 \, \text{Pa}$
Explanation: $P = \rho g h = (800)(10)(5) = 4.0 \times 10^4 \, \text{Pa}$.
164. A dam holds back water of depth 50 m. Calculate the hydrostatic pressure at the base. ($\rho = 1000 \, \text{kg/m}^3, g = 9.8 \, \text{m/s}^2$)
ⓐ. $4.9 \times 10^4 \, \text{Pa}$
ⓑ. $2.5 \times 10^5 \, \text{Pa}$
ⓒ. $3.5 \times 10^5 \, \text{Pa}$
ⓓ. $4.9 \times 10^5 \, \text{Pa}$
Correct Answer: $4.9 \times 10^5 \, \text{Pa}$
Explanation: $P = \rho g h = (1000)(9.8)(50) = 4.9 \times 10^5 \, \text{Pa}$.
165. If the pressure at the bottom of a lake is $2.5 \times 10^5 \, \text{Pa}$, calculate the depth of the lake. ($ \rho = 1000 \, \text{kg/m}^3, g = 9.8 \, \text{m/s}^2$)
ⓐ. 15 m
ⓑ. 20 m
ⓒ. 25 m
ⓓ. 30 m
Correct Answer: 25 m
Explanation: $h = \frac{P}{\rho g} = \frac{2.5 \times 10^5}{1000 \times 9.8} \approx 25.5 \, \text{m}$. Closest answer is 25 m.
166. A U-tube is filled with water on one side and mercury ($ \rho = 13.6 \times 10^3 \, \text{kg/m}^3$) on the other side. If water column height is 1.36 m, find the equivalent height of mercury column.
ⓐ. 0.01 m
ⓑ. 0.10 m
ⓒ. 0.136 m
ⓓ. 1.0 m
Correct Answer: 0.136 m
Explanation: Pressure balance: $\rho_{w} h_{w} = \rho_{Hg} h_{Hg}$. So, $(1000)(1.36) = (13,600) h_{Hg} \Rightarrow h_{Hg} = 0.136 \, \text{m}$.
167. A swimming pool is 2 m deep. What is the pressure at the bottom due to water? ($\rho = 1000 \, \text{kg/m}^3, g = 10 \, \text{m/s}^2$)
ⓐ. $1.0 \times 10^3 \, \text{Pa}$
ⓑ. $2.0 \times 10^3 \, \text{Pa}$
ⓒ. $1.0 \times 10^4 \, \text{Pa}$
ⓓ. $2.0 \times 10^4 \, \text{Pa}$
Correct Answer: $2.0 \times 10^4 \, \text{Pa}$
Explanation: $P = \rho g h = (1000)(10)(2) = 2.0 \times 10^4 \, \text{Pa}$.
168. The pressure at a depth of 1 m in mercury ($ \rho = 13.6 \times 10^3 \, \text{kg/m}^3$) is: (Take $g = 9.8 \, \text{m/s}^2$)
ⓐ. $9.8 \times 10^4 \, \text{Pa}$
ⓑ. $1.33 \times 10^5 \, \text{Pa}$
ⓒ. $2.0 \times 10^5 \, \text{Pa}$
ⓓ. $1.0 \times 10^6 \, \text{Pa}$
Correct Answer: $1.33 \times 10^5 \, \text{Pa}$
Explanation: $P = \rho g h = (13.6 \times 10^3)(9.8)(1) = 1.33 \times 10^5 \, \text{Pa}$.
169. The bottom of a tank filled with two liquids has water (2 m depth) and oil ($ \rho = 800 \, \text{kg/m}^3$) of 3 m depth above it. Find the total pressure at the bottom. ($g = 10 \, \text{m/s}^2$)
ⓐ. $3.0 \times 10^4 \, \text{Pa}$
ⓑ. $4.1 \times 10^4 \, \text{Pa}$
ⓒ. $4.4 \times 10^4 \, \text{Pa}$
ⓓ. $6.0 \times 10^4 \, \text{Pa}$
Correct Answer: $4.4 \times 10^4 \, \text{Pa}$
Explanation: Pressure due to oil = $(800)(10)(3) = 2.4 \times 10^4 \, \text{Pa}$. Pressure due to water = $(1000)(10)(2) = 2.0 \times 10^4 \, \text{Pa}$. Total = $4.4 \times 10^4 \, \text{Pa}$.
170. The pressure difference between two points in a liquid column 10 m apart (vertical) with density $\rho = 1200 \, \text{kg/m}^3$ is: (Take $g = 9.8 \, \text{m/s}^2$)
ⓐ. $9.8 \times 10^3 \, \text{Pa}$
ⓑ. $1.18 \times 10^5 \, \text{Pa}$
ⓒ. $2.18 \times 10^5 \, \text{Pa}$
ⓓ. $1.50 \times 10^5 \, \text{Pa}$
Correct Answer: $1.18 \times 10^5 \, \text{Pa}$
Explanation: $\Delta P = \rho g h = (1200)(9.8)(10) = 1.18 \times 10^5 \, \text{Pa}$.
171. Viscosity of a fluid is defined as:
ⓐ. Resistance offered to compressibility
ⓑ. Resistance offered to flow due to internal friction between layers
ⓒ. Resistance to changes in temperature
ⓓ. Resistance to external pressure
Correct Answer: Resistance offered to flow due to internal friction between layers
Explanation: Viscosity is a measure of the internal friction between adjacent fluid layers moving at different velocities. This internal resistance slows down flow.
172. Which of the following fluids has the highest viscosity at room temperature?
ⓐ. Water
ⓑ. Honey
ⓒ. Air
ⓓ. Petrol
Correct Answer: Honey
Explanation: Honey flows slowly because it has very high internal friction between its layers (high viscosity). Air has very low viscosity, petrol and water have moderate viscosities.
173. The SI unit of viscosity is:
ⓐ. Poise
ⓑ. Newton-second per square meter (N·s/m$^2$)
ⓒ. Pascal per second (Pa/s)
ⓓ. Both B and C
Correct Answer: Both B and C
Explanation: The SI unit of viscosity is $\text{Pa·s} = \text{N·s/m}^2$. The CGS unit is poise, where $1 \, \text{Pa·s} = 10 \, \text{poise}$.
174. The dimensional formula of viscosity is:
ⓐ. $[M^0 L^0 T^0]$
ⓑ. $[M L^{-1} T^{-1}]$
ⓒ. $[M L T^{-2}]$
ⓓ. $[M^1 L^{-2} T^{-2}]$
Correct Answer: $[M L^{-1} T^{-1}]$
Explanation: From Newton’s law of viscosity, $\eta = \frac{\tau}{dv/dy}$. Since stress = $F/A = MLT^{-2}/L^2 = M L^{-1} T^{-2}$ and velocity gradient has dimension $T^{-1}$, viscosity = $[M L^{-1} T^{-1}]$.
175. Newton’s law of viscosity is mathematically expressed as:
ⓐ. $\tau = \eta \frac{dv}{dy}$
ⓑ. $\tau = \rho g h$
ⓒ. $\tau = m a$
ⓓ. $\tau = \eta \frac{dy}{dv}$
Correct Answer: $\tau = \eta \frac{dv}{dy}$
Explanation: Newton’s law of viscosity states that shear stress ($\tau$) in a fluid is directly proportional to the velocity gradient $\frac{dv}{dy}$. The proportionality constant is the coefficient of viscosity ($\eta$).
176. If a liquid has higher viscosity, then:
ⓐ. It flows more easily
ⓑ. It resists flow more strongly
ⓒ. It has lower internal friction
ⓓ. It cannot transmit pressure
Correct Answer: It resists flow more strongly
Explanation: High viscosity means greater internal friction between fluid layers, resulting in slower flow (e.g., honey). Low viscosity fluids like air flow easily.
177. Which of the following is an example of a low-viscosity fluid?
ⓐ. Honey
ⓑ. Glycerin
ⓒ. Air
ⓓ. Tar
Correct Answer: Air
Explanation: Air has very low viscosity, so its flow resistance is negligible compared to liquids like honey, glycerin, or tar, which have high viscosity.
178. Which physical property directly causes viscosity in fluids?
ⓐ. Molecular attraction between fluid particles
ⓑ. Gravitational force
ⓒ. Elasticity of molecules
ⓓ. Capillary action
Correct Answer: Molecular attraction between fluid particles
Explanation: Viscosity arises because molecules of adjacent fluid layers exert intermolecular forces on each other, resisting relative motion.
179. In which type of fluids does viscosity remain constant irrespective of applied stress?
ⓐ. Newtonian fluids
ⓑ. Non-Newtonian fluids
ⓒ. Ideal fluids
ⓓ. Superfluids
Correct Answer: Newtonian fluids
Explanation: Newtonian fluids obey Newton’s law of viscosity, where viscosity is constant for a given temperature and pressure (e.g., water, air). Non-Newtonian fluids do not follow this law.
180. Which of the following is NOT correct about viscosity?
ⓐ. It represents internal friction of fluids
ⓑ. It is temperature-dependent
ⓒ. It decreases with rise in temperature for liquids
ⓓ. It increases with rise in temperature for liquids
Correct Answer: It increases with rise in temperature for liquids
Explanation: For liquids, viscosity decreases with increase in temperature (weaker intermolecular forces). For gases, viscosity increases with temperature due to faster molecular collisions.
181. Which of the following is a Newtonian fluid?
ⓐ. Water
ⓑ. Blood
ⓒ. Toothpaste
ⓓ. Custard
Correct Answer: Water
Explanation: Newtonian fluids, like water and air, follow Newton’s law of viscosity where shear stress is directly proportional to velocity gradient, with constant viscosity at a given temperature. Blood, toothpaste, and custard are non-Newtonian.
182. In Newtonian fluids, the relationship between shear stress $\tau$ and velocity gradient $\frac{dv}{dy}$ is:
ⓐ. Non-linear
ⓑ. Independent of viscosity
ⓒ. Linear and proportional
ⓓ. Random
Correct Answer: Linear and proportional
Explanation: For Newtonian fluids, $\tau = \eta \frac{dv}{dy}$. The constant of proportionality $\eta$ is the viscosity, which does not change with shear rate.
183. Which of the following best describes a non-Newtonian fluid?
ⓐ. A fluid with constant viscosity
ⓑ. A fluid with viscosity dependent on shear rate
ⓒ. A fluid that never flows
ⓓ. A fluid with zero viscosity
Correct Answer: A fluid with viscosity dependent on shear rate
Explanation: Non-Newtonian fluids have viscosities that vary with applied shear stress or shear rate (e.g., ketchup, blood, toothpaste).
184. Which of the following is an example of shear-thinning (pseudoplastic) non-Newtonian fluid?
ⓐ. Cornstarch paste
ⓑ. Water
ⓒ. Air
ⓓ. Ketchup
Correct Answer: Ketchup
Explanation: Ketchup flows more easily when shaken or squeezed, showing decreasing viscosity with increasing shear rate, a property of shear-thinning fluids.
185. Which of the following is an example of shear-thickening (dilatant) fluid?
ⓐ. Honey
ⓑ. Cornstarch suspension (oobleck)
ⓒ. Water
ⓓ. Glycerin
Correct Answer: Cornstarch suspension (oobleck)
Explanation: A cornstarch-water mixture (oobleck) becomes harder and more viscous when stress is applied, demonstrating shear-thickening behavior.
186. Which of the following fluids is often modeled as non-Newtonian in biological studies?
ⓐ. Water
ⓑ. Air
ⓒ. Blood
ⓓ. Mercury
Correct Answer: Blood
Explanation: Blood behaves as a non-Newtonian fluid since its viscosity changes with shear rate and presence of cells. At low shear rates, it behaves more viscous; at high shear rates, less viscous.
187. Which of the following fluids behaves like a Bingham plastic (requires a yield stress before flowing)?
ⓐ. Water
ⓑ. Toothpaste
ⓒ. Air
ⓓ. Oil
Correct Answer: Toothpaste
Explanation: Toothpaste behaves as a Bingham plastic, meaning it won’t flow until a minimum stress (yield stress) is applied, after which it flows like a viscous fluid.
188. Which statement is TRUE about Newtonian fluids?
ⓐ. They always have viscosity increasing with shear rate
ⓑ. Their viscosity is constant for a given temperature and pressure
ⓒ. Their viscosity decreases with shear rate
ⓓ. They show yield stress before flow
Correct Answer: Their viscosity is constant for a given temperature and pressure
Explanation: Newtonian fluids have constant viscosity independent of shear stress. Non-Newtonian fluids show variations like shear-thinning, shear-thickening, or yield stress.
189. Which of the following pairs correctly identifies one Newtonian and one Non-Newtonian fluid?
ⓐ. Air and Honey
ⓑ. Water and Ketchup
ⓒ. Mercury and Oxygen
ⓓ. Oil and Petrol
Correct Answer: Water and Ketchup
Explanation: Water is Newtonian with constant viscosity. Ketchup is non-Newtonian because its viscosity changes with shear rate (shear-thinning).
190. Which law of fluid mechanics applies strictly to Newtonian fluids but not to non-Newtonian fluids?
ⓐ. Archimedes’ principle
ⓑ. Newton’s law of viscosity
ⓒ. Bernoulli’s theorem
ⓓ. Pascal’s law
Correct Answer: Newton’s law of viscosity
Explanation: Newton’s law of viscosity states $\tau = \eta \frac{dv}{dy}$ with constant $\eta$, which applies only to Newtonian fluids. Non-Newtonian fluids deviate from this law.
191. The instrument commonly used to measure viscosity of liquids in laboratories is:
ⓐ. Barometer
ⓑ. Manometer
ⓒ. Ostwald viscometer
ⓓ. Calorimeter
Correct Answer: Ostwald viscometer
Explanation: The Ostwald viscometer measures viscosity by recording the time a liquid takes to flow between two marks under gravity. Barometers and manometers measure pressure, not viscosity.
192. Which principle does an Ostwald viscometer use for viscosity measurement?
ⓐ. Bernoulli’s principle
ⓑ. Hagen–Poiseuille’s law
ⓒ. Pascal’s principle
ⓓ. Archimedes’ principle
Correct Answer: Hagen–Poiseuille’s law
Explanation: Ostwald viscometer uses laminar flow through a capillary. The relation between flow time, density, and viscosity is based on Poiseuille’s law.
193. In an Ostwald viscometer, the viscosity of a liquid is determined relative to:
ⓐ. Water
ⓑ. Air
ⓒ. Mercury
ⓓ. Alcohol
Correct Answer: Water
Explanation: Ostwald viscometer generally measures viscosity of unknown liquids by comparing their flow times with that of water (reference liquid).
194. Which viscometer measures viscosity by observing the falling speed of a ball through a liquid?
ⓐ. Ostwald viscometer
ⓑ. Falling-sphere viscometer
ⓒ. Rotational viscometer
ⓓ. Redwood viscometer
Correct Answer: Falling-sphere viscometer
Explanation: Falling-sphere viscometer applies Stokes’ law. The terminal velocity of a sphere falling through liquid is used to calculate viscosity.
195. Stokes’ law for viscosity measurement using a falling-sphere method is:
ⓐ. $F = 6 \pi \eta r v$
ⓑ. $F = \rho g h$
ⓒ. $F = ma$
ⓓ. $F = \eta \frac{dv}{dy}$
Correct Answer: $F = 6 \pi \eta r v$
Explanation: According to Stokes’ law, the viscous drag force on a sphere of radius $r$ moving at velocity $v$ in a fluid of viscosity $\eta$ is $6 \pi \eta r v$.
196. In a rotational viscometer, viscosity is measured by:
ⓐ. Time taken for liquid to fall
ⓑ. Torque required to rotate one cylinder in another containing fluid
ⓒ. Pressure difference in a liquid column
ⓓ. Surface tension changes
Correct Answer: Torque required to rotate one cylinder in another containing fluid
Explanation: Rotational viscometers measure viscosity by calculating the torque needed to rotate a spindle or cylinder in a fluid at a constant angular velocity.
197. The Redwood viscometer is used to measure:
ⓐ. Viscosity of gases only
ⓑ. Absolute viscosity of oils
ⓒ. Kinematic viscosity of lubricating oils
ⓓ. Density of liquids
Correct Answer: Kinematic viscosity of lubricating oils
Explanation: The Redwood viscometer is widely used in industries to measure the kinematic viscosity of lubricating oils, based on flow through a capillary under gravity.
198. The relationship used in an Ostwald viscometer for comparing viscosities of two liquids is:
ⓐ. $\frac{\eta_1}{\eta_2} = \frac{t_1}{t_2}$
ⓑ. $\frac{\eta_1}{\eta_2} = \frac{\rho_1 t_1}{\rho_2 t_2}$
ⓒ. $\frac{\eta_1}{\eta_2} = \frac{\rho_2 t_1}{\rho_1 t_2}$
ⓓ. $\frac{\eta_1}{\eta_2} = \frac{t_2}{t_1}$
Correct Answer: $\frac{\eta_1}{\eta_2} = \frac{\rho_1 t_1}{\rho_2 t_2}$
Explanation: Viscosity ratio depends on both density ($\rho$) and flow time ($t$) of the two liquids as derived from Poiseuille’s law.
199. In a capillary tube viscometer, the volume flow rate for laminar flow is given by:
ⓐ. $Q = \frac{\pi r^4 \Delta P}{8 \eta L}$
ⓑ. $Q = \frac{\pi r^2 \Delta P}{\eta L}$
ⓒ. $Q = \rho g h$
ⓓ. $Q = \eta \frac{dv}{dy}$
Correct Answer: $Q = \frac{\pi r^4 \Delta P}{8 \eta L}$
Explanation: Poiseuille’s equation states that the volume flow rate $Q$ through a capillary depends on pressure difference $\Delta P$, tube radius $r$, length $L$, and fluid viscosity $\eta$.
200. Which of the following instruments is most suitable for measuring viscosity of highly viscous liquids like tar or bitumen?
ⓐ. Ostwald viscometer
ⓑ. Redwood viscometer
ⓒ. Falling-sphere viscometer
ⓓ. Rotational viscometer
Correct Answer: Rotational viscometer
Explanation: Rotational viscometers can measure high-viscosity fluids because they rely on torque measurements rather than gravity-driven flow, which is very slow for viscous substances like tar.
The chapter Mechanical Properties of Fluids from Class 11 Physics is a vital part of the NCERT/CBSE syllabus and is frequently tested in board exams and competitive exams like JEE, NEET, and other state-level entrance tests. This part focuses on applications of pressure in fluids, buoyancy, and Archimedes’ principle. Across all 7 parts, we provide 700 MCQs with detailed answers designed to strengthen both theoretical understanding and numerical problem-solving skills. Here in Part 2, you will find another 100 solved MCQs for effective practice and exam preparation.
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