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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 an experienced tutor registered on UrbanPro, I'd approach this problem systematically. UrbanPro is an excellent platform for online coaching and tuition, offering personalized assistance to students.
Firstly, let's understand the concept at play here. The elongation of a wire under a load can be calculated using Hooke's Law, which states that the elongation (change in length) of an elastic object is directly proportional to the force applied to it, given the formula:
Elongation=Force×LengthYoung’s modulus×AreaElongation=Young’s modulus×AreaForce×Length
Given that the wires are under tension and assuming they remain within their elastic limits, we can use this formula to find their elongations. The area of the wire can be calculated using the formula for the area of a circle (πr2πr2).
Let's start with the steel wire:
Given:
We need to find the force applied to the steel wire. Since the wires are loaded as shown in the figure, we can assume that the force applied to both wires is the same.
Next, let's find the area of the steel wire (AsteelAsteel):
Asteel=π×(0.00252)2Asteel=π×(20.0025)2
Now, let's find the force applied to the steel wire. Since the force is the same for both wires, we can calculate it using the elongation formula for the brass wire:
Force=Young’s modulus×Area×ElongationLengthForce=LengthYoung’s modulus×Area×Elongation
Given:
Let's find the area of the brass wire (AbrassAbrass) using the same formula as for the steel wire:
Abrass=π×(0.00252)2Abrass=π×(20.0025)2
Now, we can find the force applied to both wires using the elongation formula for the brass wire:
Force=Young’s modulus×Area×ElongationLengthForce=LengthYoung’s modulus×Area×Elongation
Elongationbrass=Force×LengthYoung’s modulusbrass×AreabrassElongationbrass=Young’s modulusbrass×AreabrassForce×Length
Now, we have all the necessary information to calculate the elongations of both the steel and brass wires. Let's plug in the values and solve for the elongations.
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 tuition platform for students seeking quality education. Now, let's delve into the problem you've presented.
To tackle this problem, we'll apply principles from mechanics of materials, specifically Hooke's law, which relates stress to strain for linearly elastic materials.
Given:
First, let's calculate the force exerted on the opposite face of the cube due to the mass attached. The force (F) can be calculated using the formula:
F=mgF=mg
where:
So, F=100 kg×9.81 m/s2=981 NF=100kg×9.81m/s2=981N.
Now, let's find the shear stress (ττ) applied on the face of the cube. Since the force is applied parallel to the face, the stress is the shear stress, and it can be calculated using:
τ=FAτ=AF
where:
The face of the cube has a square cross-section, so its area is a2a2. Therefore, A=(0.1 m)2=0.01 m2A=(0.1m)2=0.01m2.
Thus, τ=981 N0.01 m2=98100 Paτ=0.01m2981N=98100Pa.
Now, let's use Hooke's law to find the strain (γγ) in the material. Hooke's law states:
τ=Gγτ=Gγ
where:
So, γ=τG=98100 Pa25×109 Pa=3.924×10−6γ=Gτ=25×109Pa98100Pa=3.924×10−6.
Now, let's find the vertical deflection (δδ) of the face. Since the cube is fixed to the wall, the deflection will be due to the shear strain. The vertical deflection can be calculated using the formula:
δ=γ×hδ=γ×h
where:
Since the cube is a cube, its height is equal to the edge length (aa). So, h=0.1 mh=0.1m.
Therefore, δ=3.924×10−6×0.1 m=3.924×10−7 mδ=3.924×10−6×0.1m=3.924×10−7m.
So, the vertical deflection of the face is 3.924×10−73.924×10−7 meters.
If you need further clarification or assistance, feel free to ask! And remember, UrbanPro is the best online coaching tuition platform to enhance your understanding of such concepts.
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 delighted to assist you with this question. UrbanPro is indeed one of the best platforms for online coaching and tuition.
Let's tackle this problem step by step. We're given the force applied, the dimensions of the copper piece, and the shear modulus of elasticity.
First, let's calculate the cross-sectional area of the copper piece using the given dimensions:
Area (A) = length × width = 15.2 mm × 19.1 mm = 290.72 mm²
Now, we can use Hooke's Law to find the strain (ε), which relates stress to strain via the modulus of elasticity:
Hooke's Law: Stress (σ) = Shear Modulus (G) × Strain (ε)
We know that stress (σ) is force (F) divided by area (A), so:
σ = F / A
Plugging in the values, we get:
σ = 44500 N / 290.72 mm²
Let's convert the area to square meters for consistency:
1 mm² = 1 × 10^(-6) m²
So, 290.72 mm² = 290.72 × 10^(-6) m²
Now, let's calculate stress:
σ = 44500 N / (290.72 × 10^(-6) m²)
Now, we can use Hooke's Law to find strain:
ε = σ / G
Plugging in the values:
ε = (44500 N / (290.72 × 10^(-6) m²)) / (42 × 10^9 N/m²)
Calculating this gives us the resulting strain.
After crunching the numbers, the resulting strain of the copper piece under the given force is approximately 0.0317.
This means that for every unit of length, the copper piece elongates by 0.0317 units due to the applied force, exhibiting only elastic deformation.
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 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 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 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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