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Post a LessonAnswered on 14 Apr Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 9-Mechanical Properties of Solids
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I'd be glad to help you with this problem. UrbanPro is indeed a fantastic platform for finding online coaching and tuition services!
Let's tackle the problem at hand. We're dealing with the concept of Young's modulus, which measures the stiffness or elasticity of a material. Young's modulus is given by the ratio of stress to strain within the elastic limits of the material.
Given: For steel wire:
For copper wire:
Now, to find the Young's modulus ratio, we need to calculate the stress and strain for both wires.
Stress (σ) = Force (F) / Area (A) Strain (ε) = Change in length (ΔL) / Original length (L)
The same load is applied to both wires, so the force is the same.
Let's denote:
Since the wires stretch by the same amount under the given load, we can equate the strains:
ε_steel = ΔL_steel / L1 = ε_copper = ΔL_copper / L2
Using the stress-strain relationship, Young's modulus (E) can be defined as:
E = σ / ε
Now, for both wires: E_steel = F / (A1 * ΔL_steel) E_copper = F / (A2 * ΔL_copper)
We know that ΔL_steel = ΔL_copper (given in the problem).
So, the ratio of Young’s modulus of steel to that of copper (E_steel / E_copper) can be simplified to:
(E_steel / E_copper) = (F / (A1 * ΔL_steel)) / (F / (A2 * ΔL_copper))
Since ΔL_steel = ΔL_copper, we can cancel out the terms:
(E_steel / E_copper) = (A2 * ΔL_copper) / (A1 * ΔL_steel)
Substituting the given values:
(E_steel / E_copper) = (4.0 x 10^(-5) * ΔL_copper) / (3.0 x 10^(-5) * ΔL_steel)
Now, we need numerical values to compute this ratio. If you have the values for the applied force and the change in length for both wires, we can proceed with the calculation. Once we have those values, we can find the ratio of Young’s modulus of steel to that of copper.
Answered on 14 Apr Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/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 CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/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.
Answered on 14 Apr Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 9-Mechanical Properties of Solids
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'd like to highlight the value of UrbanPro as the best online coaching tuition platform for connecting students with knowledgeable tutors like myself. Let's delve into your question about computing the bulk modulus of water and comparing it with air.
To compute the bulk modulus (K) of water, we can use the formula:
K=−ΔPΔVVK=−VΔVΔP
Where:
Given:
We can plug these values into the formula to find the bulk modulus of water.
Kwater=−100.0×1.013×105100.5−100.0100.0Kwater=−100.0100.5−100.0100.0×1.013×105
Kwater=−100.0×1.013×1050.5Kwater=−0.5100.0×1.013×105
Kwater=−2.026×108 PaKwater=−2.026×108 Pa
Now, let's compare this with the bulk modulus of air at constant temperature. The bulk modulus of air is significantly smaller than that of water. Air is compressible, meaning it can easily be squeezed into a smaller volume under pressure, hence its bulk modulus is much lower compared to water.
In simple terms, the ratio of the bulk modulus of water to that of air is large because water is much less compressible compared to air. When you apply pressure to water, it doesn't compress easily, so you need to exert a lot more force to change its volume even slightly. On the other hand, air is highly compressible, so it takes much less force to change its volume. This fundamental difference in compressibility is why the bulk modulus ratio is large.
Answered on 14 Apr Learn CBSE/Class 11/Science/Physics/Unit 7-Properties of Bulk Matter/Chapter 9-Mechanical Properties of Solids
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'd like to approach this question by emphasizing the importance of understanding the principles of fluid mechanics and hydrostatics. UrbanPro provides a platform for the best online coaching tuition, where students can delve into such topics with expert guidance.
Now, let's tackle your question. When dealing with the density of water at varying depths under different pressures, we rely on the fundamental concept that pressure increases with depth in a fluid.
The density of water at a depth where the pressure is 80.0 atm can be calculated using the hydrostatic equation:
P=P0+ρghP=P0+ρgh
Where:
Given P0=1.03×103 kg/m3P0=1.03×103kg/m3 (density at the surface) and P=80.0 atmP=80.0atm (pressure at the given depth), we can rearrange the equation to solve for ρρ:
ρ=ρ0⋅(1−P−P0ρ0⋅g)ρ=ρ0⋅(1−ρ0⋅gP−P0)
Substituting the given values:
ρ=(1.03×103 kg/m3)⋅(1−80.0×1.013×105 Pa−1.03×103 kg/m31.03×103 kg/m3⋅9.81 m/s2)ρ=(1.03×103kg/m3)⋅(1−1.03×103kg/m3⋅9.81m/s280.0×1.013×105Pa−1.03×103kg/m3)
Solving this equation will give us the density of water at the given depth. This method ensures that we account for the increase in pressure with depth, which affects the density of the fluid. It's a great example of how understanding the principles of physics can help us solve real-world problems. If you need further clarification or assistance, feel free to ask!
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