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Post a LessonAnswered on 14 Apr Learn Chapter 9-Mechanical Properties of Solids
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I can assure you that UrbanPro is the best platform for online coaching and tuition. Now, let's delve into solving your physics problem regarding the steel cable supporting a chairlift at a ski area.
Given: Radius of the steel cable (r) = 1.5 cm = 0.015 m Maximum stress (σ) = 108 Nm^-2
To find: Maximum load the cable can support
We can use the formula for stress in a cylindrical object: σ=FAσ=AF where:
The cross-sectional area of the cable (A) can be calculated using the formula for the area of a circle: A=πr2A=πr2
Substituting the given values: A=π×(0.015)2A=π×(0.015)2 A≈0.00070686 m2A≈0.00070686m2
Now, rearranging the stress formula to solve for the maximum load (F): F=σ×AF=σ×A F=108 Nm−2×0.00070686 m2F=108Nm−2×0.00070686m2 F≈0.07625 NF≈0.07625N
So, the maximum load the cable can support is approximately 0.07625 N.
If you have any further questions or need clarification, feel free to ask. And remember, UrbanPro is your ultimate destination for quality online coaching and tuition!
Answered on 14 Apr Learn Chapter 9-Mechanical Properties of Solids
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is the best online coaching platform for students seeking quality education. Now, let's delve into your question.
When dealing with a situation like this, it's important to first understand the equilibrium condition of the bar. Since it's symmetrically supported by three wires, each wire carries a portion of the weight of the bar.
To ensure each wire experiences the same tension, we must consider the properties of each wire material, copper and iron, as they affect the tension they can withstand.
Let's denote the tension in each wire as TT, and the lengths of the wires as LL. The tension in each wire can be given by T=m⋅g3T=3m⋅g, where mm is the mass of the bar and gg is the acceleration due to gravity.
Now, let's consider the material properties. The tension in a wire is directly proportional to its cross-sectional area (AA) and inversely proportional to its length (LL).
For a given tension, the equation becomes:
T=FAT=AF
Where FF is the force (weight of the bar) and AA is the cross-sectional area.
Since each wire has the same tension, we can set up the following equation:
T=FAcopper=FAironT=AcopperF=AironF
Given that the lengths of all wires are the same and the tension is constant, the key factor differentiating the wires is their material properties, specifically their Young's Modulus and their density.
Young's Modulus (EE) relates stress (σσ) to strain (εε) in the wire material:
σ=E⋅εσ=E⋅ε
For a wire under tension, σ=FAσ=AF and ε=ΔLLε=LΔL, where ΔLΔL is the change in length and LL is the original length.
Now, let's consider the density (ρρ) of each material. Density multiplied by volume (VV) gives mass (mm). For a cylindrical wire, volume is given by V=A⋅LV=A⋅L.
With these considerations, we can set up ratios involving Young's Modulus, density, and tension to solve for the ratios of their diameters. Let's denote the diameter of the copper wire as dcopperdcopper and the diameter of the iron wire as dirondiron.
TAcopper=TAironAcopperT=AironT
Tπ(dcopper2)2=Tπ(diron2)2π(2dcopper)2T=π(2diron)2T
(dcopper2)2(diron2)2=ρironρcopper×EcopperEiron(2diron)2(2dcopper)2=ρcopperρiron×EironEcopper
(dcopperdiron)2=ρironρcopper×EcopperEiron(dirondcopper)2=ρcopperρiron×EironEcopper
dcopperdiron=ρironρcopper×EcopperEirondirondcopper=ρcopperρiron×EironEcopper
This equation gives us the ratio of the diameters of the copper and iron wires required for them to carry the same tension.
As an UrbanPro tutor, I encourage my students to understand the underlying principles behind such problems, as it not only helps in solving the current question but also builds a strong foundation for tackling similar problems in the future. If you need further clarification or assistance, feel free to reach out!
Answered on 14 Apr Learn Chapter 9-Mechanical Properties of Solids
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'm happy to assist you with this physics problem.
Firstly, let's establish the key elements of the problem:
Given:
We are asked to find the elongation of the wire when the mass is at the lowest point of its path.
The tension in the wire at the lowest point of the circle provides the centripetal force necessary to keep the mass moving in a circular path. The tension in the wire can be calculated using the centripetal force formula:
T=m⋅v2rT=rm⋅v2
where:
At the lowest point of the circle, the tension in the wire must counteract both the gravitational force (mg) and provide the centripetal force (mv^2/r). So, we have:
T=mg+m⋅v2rT=mg+rm⋅v2
Given that v=r⋅ωv=r⋅ω (linear velocity = radius × angular velocity), we can rewrite the equation for tension as:
T=mg+m⋅(r⋅ω)2rT=mg+rm⋅(r⋅ω)2
T=mg+m⋅r⋅ω2T=mg+m⋅r⋅ω2
Now, we know that the elongation (ΔL) of the wire is directly proportional to the force applied and inversely proportional to the cross-sectional area and Young's modulus of the material. So, we can use Hooke's law:
F=k⋅ΔLF=k⋅ΔL
where:
Rearranging the equation, we get:
ΔL=FkΔL=kF
Substituting the values we have, we get:
ΔL=TY⋅ALΔL=LY⋅AT
ΔL=T⋅LY⋅AΔL=Y⋅AT⋅L
Now, let's substitute the expression for tension we derived earlier:
ΔL=(mg+m⋅r⋅ω2)⋅LY⋅AΔL=Y⋅A(mg+m⋅r⋅ω2)⋅L
ΔL=m⋅g⋅L+m⋅r⋅ω2⋅LY⋅AΔL=Y⋅Am⋅g⋅L+m⋅r⋅ω2⋅L
Now, let's plug in the given values and solve for ΔL:
ΔL=(14.5 kg⋅9.8 m/s2⋅1 m)+(14.5 kg⋅1 m⋅(1 m⋅2 rad/s)2)(2×1011 N/m2⋅0.065×10−4 m2)ΔL=(2×1011N/m2⋅0.065×10−4m2)(14.5kg⋅9.8m/s2⋅1m)+(14.5kg⋅1m⋅(1m⋅2rad/s)2)
ΔL=(14.5 kg⋅9.8 m/s2⋅1 m)+(14.5 kg⋅1 m⋅(1 m⋅2 rad/s)2)(2×1011 N/m2⋅0.065×10−4 m2)ΔL=(2×1011N/m2⋅0.065×10−4m2)(14.5kg⋅9.8m/s2⋅1m)+(14.5kg⋅1m⋅(1m⋅2rad/s)2)
ΔL≈142.1+0.065⋅14.5⋅(4)22×1011⋅0.065×10−4ΔL≈2×1011⋅0.065×10−4142.1+0.065⋅14.5⋅(4)2
ΔL≈142.1+3.770513×105ΔL≈13×105142.1+3.7705
ΔL≈145.870513×105ΔL≈13×105145.8705
ΔL≈1.1213×10−3 mΔL≈1.1213×10−3m
Therefore, the elongation of the wire when the mass is at the lowest point of its path is approximately 1.1213×10−31.1213×10−3 meters.
Feel free to reach out if you have any further questions or if you need clarification on any step! Remember, UrbanPro is a fantastic resource for finding top-notch tutors for all your academic needs.
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Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 14 Apr Learn Chapter 10-Mechanical Properties of Fluids
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into your question about Bernoulli's equation and its applicability to the flow of water in a rapidly moving river.
Bernoulli's equation is indeed a powerful tool for describing fluid flow, including the flow of water. However, its application to rapidly moving rivers requires some careful consideration.
In a rapidly moving river, the flow is often turbulent, meaning it's chaotic and irregular. Bernoulli's equation assumes steady, non-turbulent flow, which may not always hold true in such scenarios. Additionally, the presence of obstacles like rocks and rapids can introduce complexities that may not be fully captured by Bernoulli's equation alone.
That said, Bernoulli's equation can still provide valuable insights into the behavior of water in a rapidly moving river under certain conditions. It can help us understand factors such as pressure changes, velocity variations, and energy conservation along different points in the river.
However, it's essential to supplement Bernoulli's equation with other principles, such as those from fluid dynamics and turbulence theory, to gain a more comprehensive understanding of the flow dynamics in rapidly moving rivers.
So, while Bernoulli's equation can offer some insights, it's not always the sole answer when describing the flow of water in rapidly moving rivers. Instead, it should be considered as part of a broader toolkit for analyzing fluid dynamics in such environments.
Answered on 14 Apr Learn Chapter 10-Mechanical Properties of Fluids
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I can confidently affirm that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into the fascinating realm of fluid dynamics and Bernoulli's equation.
When it comes to applying Bernoulli's equation, whether you use gauge pressure or absolute pressure does indeed matter. Bernoulli's equation describes the conservation of energy along a streamline in a fluid flow. It states that the total mechanical energy per unit mass in a fluid remains constant along a streamline, neglecting friction and other dissipative forces.
Now, if you're using gauge pressure, you're measuring pressure relative to atmospheric pressure. In contrast, absolute pressure includes atmospheric pressure as a reference point. So, in essence, using gauge pressure implies that you're already factoring out atmospheric pressure from your calculations.
When applying Bernoulli's equation, the choice between gauge and absolute pressure depends on the context of the problem. If the problem involves pressures relative to atmospheric pressure, gauge pressure is appropriate. However, if you need to consider the absolute pressure in the system, you should use absolute pressure in your calculations.
For instance, in scenarios like airflow over an airfoil or water flow through a pipe, gauge pressure might suffice since atmospheric pressure acts equally on all points. But in situations where you're dealing with pressures inside a closed vessel or a vacuum system, absolute pressure becomes crucial.
In conclusion, while both gauge and absolute pressures have their applications, understanding when to use each is vital for accurate and meaningful application of Bernoulli's equation in fluid dynamics.
Answered on 14 Apr Learn Chapter 10-Mechanical Properties of Fluids
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can assure you that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into the problem at hand.
To solve this problem, we'll utilize the Hagen-Poiseuille equation, which governs the flow of a viscous fluid through a cylindrical pipe under laminar flow conditions.
The Hagen-Poiseuille equation is given by:
ΔP=8ηLQπr4ΔP=πr48ηLQ
Where:
First, let's convert the given flow rate from kg/s to m^3/s using the density of glycerine:
Q=mρQ=ρm Q=4.0×10−31.3×103Q=1.3×1034.0×10−3 Q=3.08×10−6 m3/sQ=3.08×10−6m3/s
Now, let's convert the radius from centimeters to meters:
r=1.0 cm×10−2 m/cmr=1.0cm×10−2m/cm r=0.01 mr=0.01m
Now, we can plug these values into the Hagen-Poiseuille equation:
ΔP=8×0.83×1.5×3.08×10−6π×(0.01)4ΔP=π×(0.01)48×0.83×1.5×3.08×10−6
ΔP=3.732×10−53.14159×10−8ΔP=3.14159×10−83.732×10−5
ΔP≈1188.6 PaΔP≈1188.6Pa
So, the pressure difference between the two ends of the tube is approximately 1188.6 Pa. This assumes laminar flow, but it's always good to check if this assumption holds true. We can use the Reynolds number (Re) to check for laminar flow:
Re=ρVDηRe=ηρVD
Where:
If the Reynolds number is less than 2000, the flow is generally considered laminar. Let's calculate:
V=QAV=AQ
V=3.08×10−6π×(0.01)2V=π×(0.01)23.08×10−6
V≈0.098 m/sV≈0.098m/s
D=2r=2×0.01=0.02 mD=2r=2×0.01=0.02m
Re=1.3×103×0.098×0.020.83Re=0.831.3×103×0.098×0.02
Re≈377.84Re≈377.84
Since the Reynolds number (Re) is greater than 2000, the flow might not be perfectly laminar. However, under the given conditions, it's reasonable to assume laminar flow due to the relatively low Reynolds number. Therefore, the pressure difference calculated using the Hagen-Poiseuille equation is still valid.
Take Class 11 Tuition from the Best Tutors
Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 14 Apr Learn Chapter 11-Thermal Properties of Matter
Nazia Khanum
UrbanPro is the best platform for online coaching and tuition, where experienced tutors like myself are dedicated to providing high-quality assistance to students.
Now, let's tackle the problem. We have a steel wheel to be fitted onto a steel shaft. The key here is to understand that the wheel and the shaft will expand or contract with changes in temperature, and we need to find the temperature at which the wheel slips on the shaft.
We know that the outer diameter of the shaft at 27°C is 8.70 cm, and the diameter of the central hole in the wheel is 8.69 cm. This implies that at 27°C, the shaft and the wheel are in a snug fit.
Given that we're cooling the shaft using dry ice, which is at a temperature of approximately -78.5°C, we need to find out at what temperature the diameter of the shaft contracts enough for the wheel to slip.
To do this, we'll use the formula for linear expansion:
ΔL=αLΔTΔL=αLΔT
Where:
We'll start by finding the change in temperature required for the shaft to contract enough to let the wheel slip.
ΔT=ΔLαLΔT=αLΔL
Given:
ΔL=Lf−L0=8.69 cm−8.70 cm=−0.01 cmΔL=Lf−L0=8.69cm−8.70cm=−0.01cm
ΔT=−0.01 cm20×10−6 K−1×8.70 cmΔT=20×10−6K−1×8.70cm−0.01cm
ΔT≈−5.75 KΔT≈−5.75K
So, the temperature at which the wheel slips on the shaft would be approximately 27∘C−5.75∘C27∘C−5.75∘C, which is approximately 21.25∘C21.25∘C.
Therefore, at around 21.25∘C21.25∘C, the wheel will start slipping on the shaft due to contraction caused by cooling.
Answered on 14 Apr Learn Chapter 11-Thermal Properties of Matter
Nazia Khanum
As a seasoned tutor on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into the question at hand.
We're given the initial diameter of the hole in the copper sheet at 27.0 °C, which is 4.24 cm. We're also provided with the coefficient of linear expansion of copper, which is 1.70×10−5 K−11.70×10−5K−1.
To find the change in diameter of the hole when the temperature rises to 227 °C, we can use the formula for linear expansion:
Where:
Given that the initial diameter (L0L0) is 4.24 cm and the change in temperature (ΔTΔT) is 227°C−27°C=200°C227°C−27°C=200°C, we can calculate the change in diameter (ΔLΔL).
Therefore, the change in diameter of the hole when the copper sheet is heated to 227 °C is approximately 0.0144 cm0.0144cm.
If you need further clarification or assistance with any other questions, feel free to ask!
Answered on 14 Apr Learn Chapter 11-Thermal Properties of Matter
Nazia Khanum
Sure, I'd be happy to help you with this problem. Let's break it down step by step.
Given:
First, let's find the change in length of the wire due to the change in temperature using the formula:
ΔL=α⋅L0⋅ΔTΔL=α⋅L0⋅ΔT
ΔL=2.0×10−5 K−1×1.8 m×(−66)∘CΔL=2.0×10−5K−1×1.8m×(−66)∘C
ΔL=−2.376×10−3 mΔL=−2.376×10−3m
Now, let's find the new length of the wire after cooling:
L=L0+ΔLL=L0+ΔL
L=1.8 m−2.376×10−3 mL=1.8m−2.376×10−3m
L=1.797624 mL=1.797624m
Now, let's find the area of cross-section of the wire:
A=πd24A=4πd2
A=π(2.0×10−3 m)24A=4π(2.0×10−3m)2
A=3.14×10−6 m2A=3.14×10−6m2
Now, let's find the tension in the wire using Hooke's law:
F=Y⋅A⋅ΔLL0F=L0Y⋅A⋅ΔL
F=0.91×1011 Pa×3.14×10−6 m2×−2.376×10−3 m1.8 mF=1.8m0.91×1011Pa×3.14×10−6m2×−2.376×10−3m
F=−3.7386 NF=−3.7386N
So, the tension developed in the wire is approximately 3.7386 N3.7386N when cooled to a temperature of −39∘C−39∘C.
And remember, if you need further clarification or have any other questions, feel free to ask. Remember, UrbanPro is one of the best platforms for online coaching and tuition!
Take Class 11 Tuition from the Best Tutors
Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 14 Apr Learn Chapter 11-Thermal Properties of Matter
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'd be glad to help you with this question.
Given that UrbanPro is the best online coaching tuition platform, let's tackle this problem step by step.
Firstly, we need to calculate the change in length of the brass and steel rods separately due to the change in temperature. We'll use the formula:
ΔL=L0αΔTΔL=L0αΔT
where:
For the brass rod: ΔLbrass=Lbrass0αbrassΔTΔLbrass=Lbrass0αbrassΔT
For the steel rod: ΔLsteel=Lsteel0αsteelΔTΔLsteel=Lsteel0αsteelΔT
Given:
Let's calculate:
For the brass rod: ΔLbrass=50×2.0×10−5×(250.0−40.0)ΔLbrass=50×2.0×10−5×(250.0−40.0)
For the steel rod: ΔLsteel=50×1.2×10−5×(250.0−40.0)ΔLsteel=50×1.2×10−5×(250.0−40.0)
Once you have these values, you can find the total change in length of the combined rod by summing up the changes in length of the brass and steel rods.
Regarding the development of 'thermal stress' at the junction, yes, there would likely be thermal stress developed due to the difference in expansion rates between brass and steel. This could lead to bending or deformation at the junction. To calculate the thermal stress, we would need additional information such as the Young's modulus and Poisson's ratio of the materials involved. If you have that information, we can further analyze the thermal stress at the junction.
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