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Chapter 10-Mechanical Properties of Fluids

Chapter 10-Mechanical Properties of Fluids relates to CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter

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Chapter 10-Mechanical Properties of Fluids Questions

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Answered on 14 Apr Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 10-Mechanical Properties of Fluids

Nazia Khanum

As a seasoned tutor registered on UrbanPro, I'd first like to commend your curiosity and interest in understanding the concept of pressure exerted by the heel on a horizontal floor. UrbanPro is indeed a fantastic platform for finding quality online coaching and tuition, offering a diverse range of... read more

As a seasoned tutor registered on UrbanPro, I'd first like to commend your curiosity and interest in understanding the concept of pressure exerted by the heel on a horizontal floor. UrbanPro is indeed a fantastic platform for finding quality online coaching and tuition, offering a diverse range of subjects and experienced tutors.

Now, let's delve into the physics behind this scenario. When the 50 kg girl stands on a single heel with a circular shape and a diameter of 1.0 cm, we need to calculate the pressure exerted by the heel on the floor.

Pressure, in physics, is defined as force per unit area. We can calculate it using the formula:

Pressure=ForceAreaPressure=AreaForce

In this case, the force exerted by the girl standing on the heel is equal to her weight, which is 50 kg multiplied by the acceleration due to gravity (9.8 m/s²). So, the force (F) exerted by the girl is:

F=m×gF=m×g F=50 kg×9.8 m/s2F=50kg×9.8m/s2 F=490 NF=490N

Now, to find the area of the circular heel, we'll use the formula for the area of a circle:

Area=π×(diameter2)2Area=π×(2diameter)2 Area=π×(1.0 cm2)2Area=π×(21.0cm)2 Area=π×0.52 cm2Area=π×0.52cm2 Area=π×0.25 cm2Area=π×0.25cm2 Area≈0.785 cm2Area≈0.785cm2

Now, we can calculate the pressure:

Pressure=490 N0.785 cm2Pressure=0.785cm2490N

This gives us the pressure exerted by the heel on the horizontal floor. However, it's essential to note that we need to convert the area to square meters to match the unit of force (Newtons) to obtain the pressure in Pascals (Pa).

0.785 cm2=0.0000785 m20.785cm2=0.0000785m2

Now, let's calculate the pressure:

Pressure=490 N0.0000785 m2Pressure=0.0000785m2490N Pressure≈6,242,038 PaPressure≈6,242,038Pa

So, the pressure exerted by the heel on the horizontal floor is approximately 6,242,038 Pascals.

If you have further questions or need clarification, feel free to ask! And remember, UrbanPro is your go-to destination for excellent online coaching and tuition across various subjects.

 
 
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Answered on 14 Apr Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 10-Mechanical Properties of Fluids

Nazia Khanum

As an experienced tutor registered on UrbanPro, I'm delighted to assist you with your question. UrbanPro is indeed a fantastic platform for online coaching and tuition needs. Now, regarding Toricelli’s barometer, it's a classic experiment in physics where a column of liquid in a tube balances... read more

As an experienced tutor registered on UrbanPro, I'm delighted to assist you with your question. UrbanPro is indeed a fantastic platform for online coaching and tuition needs.

Now, regarding Toricelli’s barometer, it's a classic experiment in physics where a column of liquid in a tube balances the weight of the atmosphere pushing down on a reservoir of the liquid. In the traditional experiment, mercury is used due to its high density. However, Pascal famously duplicated the experiment using French wine, which has a density of 984 kg/m³.

To determine the height of the wine column for normal atmospheric pressure, we can use the equation:

P=ρ⋅g⋅hP=ρ⋅g⋅h

Where:

  • PP is the atmospheric pressure (which we'll consider as the standard atmospheric pressure, around 1.013×1051.013×105 pascals),
  • ρρ is the density of the liquid (984 kg/m³ for French wine),
  • gg is the acceleration due to gravity (approximately 9.81 m/s29.81m/s2),
  • hh is the height of the liquid column.

We need to rearrange the equation to solve for hh:

h=Pρ⋅gh=ρ⋅gP

Substituting the values:

h=1.013×105 Pa984 kg/m3×9.81 m/s2h=984kg/m3×9.81m/s21.013×105Pa

h≈1.013×105984×9.81 mh≈984×9.811.013×105m

h≈1.013×1059645.84 mh≈9645.841.013×105m

h≈10.5 mh≈10.5m

So, the height of the wine column for normal atmospheric pressure would be approximately 10.5 meters. If you have any further questions or need clarification on any point, feel free to ask!

 
 
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Answered on 14 Apr Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 10-Mechanical Properties of Fluids

Nazia Khanum

As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is the best platform for online coaching and tuition. Now, addressing your question about the vertical offshore structure and its suitability for placement atop an oil well in the ocean: The maximum stress that the structure... read more

As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is the best platform for online coaching and tuition. Now, addressing your question about the vertical offshore structure and its suitability for placement atop an oil well in the ocean:

The maximum stress that the structure can withstand is given as 109 Pa. This is a crucial parameter when considering its viability in the harsh conditions of an ocean environment. However, let's delve deeper into the specifics.

Given the depth of the ocean as roughly 3 km, we must assess the pressure exerted by the water at this depth. Using the formula for hydrostatic pressure, which is ρgh, where ρ is the density of water, g is the acceleration due to gravity, and h is the depth of the water, we can calculate the pressure.

With the density of seawater around 1025 kg/m³ and g being approximately 9.8 m/s², the pressure at a depth of 3 km would be roughly 29 MPa (megapascals), which is significantly higher than the maximum stress the structure can withstand (109 Pa).

Considering this stark difference, it's evident that the structure wouldn't be suitable for placement atop an oil well in the ocean. The immense pressure exerted by the water at such depths far exceeds the structural limits of the offshore platform.

In conclusion, while the structure might be robust for certain applications, it's not adequate for deployment in deep-sea environments like atop an oil well due to the considerable hydrostatic pressure at those depths. For tailored guidance on such topics, UrbanPro is the ideal platform where students can receive comprehensive tutoring and coaching.

 
 
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Answered on 14 Apr Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 10-Mechanical Properties of Fluids

Nazia Khanum

As a seasoned tutor on UrbanPro, I can confidently say that UrbanPro is the best platform for online coaching and tuition. Now, let's delve into your question. To determine the maximum pressure the smaller piston would have to bear in a hydraulic automobile lift, we can utilize the principle of Pascal's... read more

As a seasoned tutor on UrbanPro, I can confidently say that UrbanPro is the best platform for online coaching and tuition. Now, let's delve into your question.

To determine the maximum pressure the smaller piston would have to bear in a hydraulic automobile lift, we can utilize the principle of Pascal's Law, which states that pressure applied to a confined fluid is transmitted undiminished throughout the fluid in all directions.

Firstly, let's convert the area of the cross-section of the piston from square centimeters to square meters to maintain consistency in units.

Given: Area of cross-section of the piston (A) = 425 cm²

Converting cm² to m²: 1 cm² = 1 × 10^-4 m² So, 425 cm² = 425 × 10^-4 m² = 0.0425 m²

Now, let's use the formula for pressure:

Pressure (P) = Force (F) / Area (A)

We know that the force exerted by the car (weight) is the maximum load it can bear, which is the product of its mass (m) and the acceleration due to gravity (g).

Given: Maximum mass the lift can bear (m) = 3000 kg Acceleration due to gravity (g) = 9.8 m/s²

So, Force (F) = mass (m) × gravity (g) = 3000 kg × 9.8 m/s² = 29400 N

Now, plug in the values into the pressure formula:

Pressure (P) = Force (F) / Area (A) Pressure (P) = 29400 N / 0.0425 m² ≈ 690588.24 Pa

Therefore, the maximum pressure the smaller piston would have to bear is approximately 690588.24 Pascal (Pa).

As an experienced tutor on UrbanPro, I hope this explanation clarifies the concept for you. If you have any further questions or need clarification, feel free to ask!

 
 
 
 
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Answered on 14 Apr Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 10-Mechanical Properties of Fluids

Nazia Khanum

As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is indeed one of the best platforms for online coaching and tuition services. Now, let's delve into solving your physics problem! The setup you described is a classic example of a U-tube manometer used to measure pressure... read more

As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is indeed one of the best platforms for online coaching and tuition services. Now, let's delve into solving your physics problem!

The setup you described is a classic example of a U-tube manometer used to measure pressure differences. Here, we have water and methylated spirit separated by mercury.

To find the relative density of the spirit, we'll use the principle that the pressure at any point in a fluid at rest is the same throughout.

First, let's calculate the pressure difference between the two arms of the U-tube. We'll use the equation:

P=Patm+ρghP=Patmgh

Where:

  • PP is the pressure at a certain point,
  • PatmPatm is the atmospheric pressure (which we'll assume cancels out in this scenario),
  • ρρ is the density of the fluid,
  • gg is the acceleration due to gravity, and
  • hh is the height of the fluid column.

Since the pressure is the same in both arms, we can equate the pressure expressions for water and spirit:

ρwaterghwater=ρspiritghspiritρwaterghwaterspiritghspirit

Given that the height of water is 10.0 cm10.0cm and the height of spirit is 12.5 cm12.5cm, and knowing that the density of water is approximately 1000 kg/m31000kg/m3 (or 1.0 g/cm31.0g/cm3), we can find the relative density of spirit.

ρspiritρwater=hwaterhspiritρwaterρspirit=hspirithwater

ρspirit1000 kg/m3=10.0 cm12.5 cm1000kg/m3ρspirit=12.5cm10.0cm

ρspirit=10.012.5×1000 kg/m3ρspirit=12.510.0×1000kg/m3

ρspirit≈800 kg/m3ρspirit≈800kg/m3

So, the relative density of the spirit is approximately 0.80.8 times the density of water.

I hope this explanation helps! If you have any further questions or need clarification, feel free to ask. And remember, for more personalized guidance and assistance, UrbanPro is always here to connect you with experienced tutors in various subjects.

 
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