Fluid Mechanics

01

PYQ 2022

medium

physics ID: ap-eapce

A cylindrical tank having large diameter is filled with water to a height H. A hole of cross-sectional area 5 cm in the tank allows water to drain out. If the water drains out at the rate of m s, then the value of H is} (acceleration due to gravity = ms)

1

$80 cm$

2

$120 cm$

3

$60 cm$

4

$90 cm$

Official Solution

Correct Option: (1)

$According to Torricelli's theorem, the speed of efflux () of a liquid through a hole at a depth below the surface is given by, where is the acceleration due to gravity. This assumes the diameter of the tank is large compared to the hole, so the velocity of the liquid surface falling is negligible. The volume flow rate () of the liquid draining out is given by, where is the cross-sectional area of the hole. So, . Given values: Area of the hole . Volume flow rate . Acceleration due to gravity . To convert to centimeters (cm), since : .$

02

PYQ 2023

medium

physics ID: ap-eapce

$The gauge pressure at a depth of 50 m in a sea is$

1

$1025 Pa$

2

$512500 Pa$

3

$20000 Pa$

4

$15000 Pa$

Official Solution

Correct Option: (2)

$The formula for gauge pressure at a certain depth in a fluid is given by: where: - is the gauge pressure, - is the density of the fluid (1025 kg/m³ for sea water), - is the acceleration due to gravity (10 m/s²), - is the depth (50 m). Substituting the values: Thus, the correct answer is option (2), 512500 Pa.$

03

PYQ 2023

medium

physics ID: ap-eapce

$The fractional change in the volume of a glass slab when subjected to hydraulic pressure of 14 atm is (Bulk modulus of glass = )$

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Official Solution

Correct Option: (2)

$The fractional change in volume is given by the formula: Where: - is the pressure, - is the bulk modulus. We are given: - , - . The fractional change in volume is: Thus, the fractional change in volume is .$

04

PYQ 2023

medium

physics ID: ap-eapce

$A swimming pool has depth of 3 meters. The pressure at the bottom of the pool due to water alone is (Density of water is kg/m, m/s)$

1

$Pa$

2

$Pa$

3

$Pa$

4

$Pa$

Official Solution

Correct Option: (4)

$Hydrostatic pressure at depth is . Substitute:,, m. So, Pa. That is Pa. This is pressure due to water alone (gauge pressure). Atmospheric pressure is not included in this.$

05

PYQ 2023

medium

physics ID: ap-eapce

$A liquid of density 800 kg/m is used instead of mercury in the Torricelli barometer. What is the height of the liquid column if atmospheric pressure is 76 cm of Hg? (Density of mercury = kg/m)$

1

$13.2 m$

2

$12.9 m$

3

$13.6 m$

4

$800 m$

Official Solution

Correct Option: (2)

$Pressure due to mercury column = Given m, kg/m Same pressure should be balanced by the new liquid column: So, \Rightarrow m Approximate answer = 12.9 m$

06

PYQ 2023

medium

physics ID: ap-eapce

$A spherical ball of volume 2000 cm is subjected to a hydraulic pressure of 15 atm. If the change in volume is cm, the bulk modulus of the material of the spherical ball is:$

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Official Solution

Correct Option: (1)

$The bulk modulus is given by the relation: Where: - (since) - - Substitute the values: Thus, the correct answer is:$

07

PYQ 2023

medium

physics ID: ap-eapce

$A person of height 1.65 m is standing upright. The additional external force required by blood vessel of length 1 cm, diameter 1 mm at feet to balance the pressure compared to similar blood vessel in head is (Density of blood =)$

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Official Solution

Correct Option: (1)

$The force required to balance the pressure at feet is equal to the pressure due to the column of blood at height . The pressure at the feet is given by: where is the density of the blood, is the acceleration due to gravity, and is the height of the person. Now, the volume of the blood column is given by the formula for the volume of a cylinder: where is the radius of the blood vessel (since diameter = 1 mm), and is the length of the blood vessel. The mass of the blood column is: The force required to balance the pressure at the feet is: Substituting the values: Simplifying: Hence, the additional external force required is .$

08

PYQ 2023

medium

physics ID: ap-eapce

$A tank of oil has height of 4 m and density of 850 kg m . The gauge pressure at the bottom of the tank is (1 atm = 10 Pa, Acceleration due to gravity = 10 ms)$

1

$kPa$

2

$kPa$

3

$kPa$

4

$kPa$

Official Solution

Correct Option: (1)

$The gauge pressure at a certain depth in a fluid of density is given by the formula: $ is the acceleration due to gravity. Given: Height of the oil tank m Density of the oil kg m Acceleration due to gravity ms Substitute the values into the formula: \) $ The gauge pressure at the bottom of the tank is 34 kPa.$

09

PYQ 2023

medium

physics ID: ap-eapce

$A liquid of density and coefficient of viscosity flows through a pipe with diameter with a velocity of . The Reynolds number of the fluid is$

1

$300$

2

$400$

3

$500$

4

$600$

Official Solution

Correct Option: (1)

$The Reynolds number is given by the formula: where,,, and . Substituting the values: Thus, the Reynolds number is 300.$

10

PYQ 2024

medium

physics ID: ap-eapce

$A spherical ball of radius m and density kgm falls freely under gravity before entering water. The distance before velocity change in water is:$

1

$cm$

2

$mm$

3

$m$

4

$m$

Official Solution

Correct Option: (3)

$Step 1: Using Stoke’s Law The velocity before entering water: Solving, we get: Thus, the correct answer is option (3).$

11

PYQ 2024

easy

physics ID: ap-eapce

$The radius of cross-section of the cylindrical tube of a spray pump is 2 cm. One end of the pump has 50 fine holes each of radius 0.4 mm. If the speed of flow of the liquid inside the tube is 0.04 m/s, the speed of ejection of the liquid from the holes is:$

1

$6 m/s$

2

$2 m/s$

3

$4 m/s$

4

$3 m/s$

Official Solution

Correct Option: (2)

$To find the speed of ejection of the liquid from the holes, we can use the principle of conservation of volume flow rate. The volume flow rate must be constant throughout the entire system, which means the flow rate in the cylindrical tube is equal to the combined flow rate out of the holes. Mathematically, this can be represented as: where and are the cross-sectional area and velocity in the tube, and and are the total cross-sectional area and velocity of ejection from the holes. First, calculate the cross-sectional areas: The radius of the cylindrical tube The area of the tube Next, calculate the combined area of the holes: Radius of each hole Number of holes = 50 Area of one hole Total area of holes Now use the conservation equation: Simplify and solve for : Check the calculation. A conversion error occurred previously. Let's redo the calculation where previously using a wrong constant; see the revised logical steps for final correct answer calculation below. Therefore, the speed of ejection is correctly 2 m/s .$

12

PYQ 2025

medium

physics ID: ap-eapce

$A rectangular slab consists of two cubes of copper and brass of equal sides having thermal conductivities in the ratio . If the free face of brass is at and that of copper is at, then the temperature of their interface is$

1

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4

Official Solution

Correct Option: (1)

$Step 1: Let the thermal conductivity of copper be, and that of brass be . Let be the temperature at the interface. Since both cubes have equal cross-sectional area and equal length, and the system is in steady-state, the rate of heat flow through both materials is the same. % Final Answer$

13

PYQ 2025

medium

physics ID: ap-eapce

$A mercury drop of radius cm is divided into droplets of equal size. If the surface tension of mercury is N/m, then the change in surface energy in the process is:$

1

$J$

2

$erg$

3

$J$

4

$erg$

Official Solution

Correct Option: (3)

$Step 1: Surface Energy Formula Surface energy is given by: where: - N/m (surface tension), - = Change in surface area. Step 2: Surface Area Calculations Initial surface area: where cm, Final surface area for droplets of radius : Since volume is conserved: Solving for : Final surface area: Change in surface area: Step 3: Compute Conclusion Thus, the correct answer is:$

14

PYQ 2025

medium

physics ID: ap-eapce

$A mercury drop of radius cm is divided into droplets of equal size. If the surface tension of mercury is N/m, then the change in surface energy in the process is:$

1

$J$

2

$erg$

3

$J$

4

$erg$

Official Solution

Correct Option: (3)

$Step 1: Surface Energy Formula Surface energy is given by: where: - N/m (surface tension), - = Change in surface area. Step 2: Surface Area Calculations Initial surface area: where cm, Final surface area for droplets of radius : Since volume is conserved: Solving for : Final surface area: Change in surface area: Step 3: Compute Conclusion Thus, the correct answer is:$

15

PYQ 2025

medium

physics ID: ap-eapce

$A liquid drop of diameter splits into small identical drops. If is the surface tension of the liquid, then the change in the surface energy in the process is$

1

2

3

4

Official Solution

Correct Option: (1)

$Initial radius Volume conserved: Initial surface area: Surface area of each small drop: Total surface area: Change in area: Change in energy$

16

PYQ 2025

medium

physics ID: ap-eapce

$When the temperature increases, the viscosity of$

1

$gases decreases but liquids increases.$

2

$gases increases but liquids decreases.$

3

$both gases and liquids increase.$

4

$both gases and liquids decrease.$

Official Solution

Correct Option: (1)

$As the temperature increases, the viscosity of gases generally decreases, while the viscosity of liquids increases due to the higher molecular movement and interaction in the liquid phase.$

17

PYQ 2025

medium

physics ID: ap-eapce

$An aeroplane of mass kg and total wing area of is travelling at a constant height. The pressure difference between the upper and lower surfaces of its wings is (Acceleration due to gravity)$

1

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4

Official Solution

Correct Option: (4)

$The aeroplane is travelling at a constant height. This means the vertical forces are balanced. The upward lift force generated by the wings must be equal to the weight of the aeroplane. Mass kg. Acceleration due to gravity . Weight N. So, N. The lift force is generated due to the pressure difference between the lower and upper surfaces of the wings., where is the total wing area. Given . So, . This matches option (4).$

18

PYQ 2025

medium

physics ID: ap-eapce

$In a water tank, an air bubble rises from the bottom to the top surface of the water. If the depth of the water in the tank is 7.28 m and atmospheric pressure is 10 m of water, then the ratio of the radii of the bubble at the bottom of the tank and at the top surface of the water is (Temperature of the water in the tank is constant)$

1

$2:3$

2

$5:6$

3

$3:4$

4

$4:5$

Official Solution

Correct Option: (2)

$Let and be the pressure and radius of the bubble at the bottom of the tank, and and be the pressure and radius at the top surface. At the bottom, m of water (where is the density of water and is the depth). At the top, m of water. Since the temperature is constant, we can use Boyle's law: . If m of water, and m, then and . so .$

19

PYQ 2025

medium

physics ID: ap-eapce

$If water flows with a velocity of 20 cm/s in a pipe of radius 2 cm, then the flow is (The coefficient of viscosity of water is and density of water is)$

1

$turbulent$

2

$steady flow$

3

$non-viscous$

4

$unsteady$

Official Solution

Correct Option: (1)

$Step 1: Calculate Reynolds number (Re) Reynolds number determines the type of flow and is given by: where, (density), (velocity), (diameter), (viscosity). Step 2: Substitute values Step 3: Interpret Reynolds number - For pipe flow, if, flow is laminar . - If, flow is turbulent . - Between 2000 and 4000 is transition region. Here,, so the flow is turbulent . Step 4: Conclusion Therefore, the water flow in the pipe is turbulent.$

20

PYQ 2025

medium

physics ID: ap-eapce

$An aeroplane of mass kg and total wing area of is travelling at a constant height. The pressure difference between the upper and lower surfaces of its wings is (Acceleration due to gravity)$

1

2

3

4

Official Solution

Correct Option: (4)

$The aeroplane is travelling at a constant height. This means the vertical forces are balanced. The upward lift force generated by the wings must be equal to the weight of the aeroplane. Mass kg. Acceleration due to gravity . Weight N. So, N. The lift force is generated due to the pressure difference between the lower and upper surfaces of the wings., where is the total wing area. Given . So, . This matches option (4).$

High-Yield Trend

Chapter Questions
20 MCQs

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Fluid Mechanics

High-Yield Trend

Chapter Questions 20 MCQs

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Chapter Questions
20 MCQs