/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 An iron tyre is to be fitted ont... [FREE SOLUTION] | 91Ó°ÊÓ

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An iron tyre is to be fitted onto a wooden wheel \(1.0 \mathrm{~m}\) in diameter. The diameter of the tyre is \(6 \mathrm{~mm}\) smaller than that of wheel. The tyre should be heated so that its temperature increases by a minimum of (coefficient of volumetric expansion of iron is \(3.6 \times 10^{-5} /{ }^{\circ} \mathrm{C}\) ) (A) \(167^{\circ} \mathrm{C}\) (B) \(334^{\circ} \mathrm{C}\) (C) \(500^{\circ} \mathrm{C}\) (D) \(1000^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The minimum increase in temperature for the iron tyre to fit onto the wooden wheel is \(500^{\circ} \mathrm{C}\), which corresponds to option (C).

Step by step solution

01

Write down the given information

The diameter of the wooden wheel (Dw) is 1.0m, and the diameter of the iron tyre (Di) is 6mm smaller than the wheel's diameter. The coefficient of volumetric expansion of iron (γ) is \(3.6 \times 10^{-5}\) / °C.
02

Convert the diameter of the iron tyre to meters

Converting 6mm to meters: 6mm = 0.006m. Now, the diameter of the iron tyre (Di) can be calculated: Di = 1.0m - 0.006m = 0.994m.
03

Determine the relationship between linear and volumetric expansion coefficients

The linear expansion coefficient (α) is related to the volumetric expansion coefficient (γ) by α = γ/3. Calculate the linear expansion coefficient (α) for iron: α = \(3.6 \times 10^{-5}\) / °C / 3 = \(1.2 \times 10^{-5}\) / °C.
04

Calculate the change in diameter

The change in diameter (ΔD) that needs to occur for the tyre to fit onto the wheel is the difference in initial diameters, which is: ΔD = 1.0m - 0.994m = 0.006m.
05

Use the linear expansion formula to solve for temperature change

The linear expansion formula is: ΔL = L₀ * α * ΔT, where ΔL is the change in length (diameter in this case), L₀ is the initial length (diameter of the iron tyre), α is the linear expansion coefficient, and ΔT is the change in temperature. Rearrange the formula to solve for ΔT: ΔT = ΔD / (Di * α) Substitute the values and calculate ΔT: ΔT = 0.006m / (0.994m * \(1.2 \times 10^{-5}\) / °C) ≈ 500°C. The minimum increase in temperature for the iron tyre to fit onto the wooden wheel is 500°C, which corresponds to option (C).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volumetric Expansion Coefficient
When we talk about the volumetric expansion coefficient, we're discussing how much a material's volume will change with temperature. Generally, when materials are heated, they expand; when cooled, they contract. The volumetric expansion coefficient, often represented by the symbol \(\gamma\), quantifies this change and is crucial in calculating the expansion of substances like solids, liquids, and gases.

For instance, in the exercise given, we see the coefficient of volumetric expansion of iron being used to determine how much the iron tyre will expand when heated. This coefficient typically comes into play in engineering applications where thermal stress could affect the integrity of structures or in manufacturing processes like fitting an iron tyre on a wooden wheel, as shown by our problem.
Linear Expansion Formula
Now, digging deeper into the mechanics of thermal expansion, it's time to focus on the linear expansion formula. Linear expansion refers to the change in one dimension of an object due to temperature change. This is described by the formula \(\Delta L = L_0 \cdot \alpha \cdot \Delta T\), where \(\Delta L\) is the change in the object's length, \(L_0\) is the original length, \(\alpha\) is the linear expansion coefficient, and \(\Delta T\) is the change in temperature.

Understanding this formula is crucial for solving thermal expansion problems, such as ensuring a heat-induced change in dimension doesn't cause mechanical failure. In our exercise, we use this fundamental formula to calculate the necessary temperature increase to fit the iron tyre onto the wooden wheel.
Thermal Expansion Problems
Thermal expansion problems require us to apply our understanding of how materials behave when heated or cooled. These problems, common in physics and engineering, present real-world challenges such as metal bridge expansions in the summer or the contraction of power lines in the winter.

To tackle these problems, we need to use formulas for linear or volumetric expansion and understand material properties like the expansion coefficient. In practice, this means calculating expansions or contractions and designing structures that can withstand such changes. The example we're discussing—fitting a tyre onto a wheel—is an excellent illustration of the practical application of these concepts.

For improved comprehension, remember to:
  • Convert units for consistency (e.g., millimeters to meters).
  • Relate linear and volumetric coefficients when needed.
  • Understand and rearrange formulas to find unknowns.
JEE MAIN Physics Preparation
Aspiring students preparing for competitive exams like JEE MAIN in India must have a firm grip on concepts like thermal expansion to ace the physics section. Understanding these principles is not only about memorizing formulas but also about knowing when and how to apply them to varied problems.

Effective preparation involves practicing a wide range of problems, clarifying misconceptions, and learning to quickly recognize which principles to apply in different scenarios. The thermal expansion exercise we analyzed is a typical problem that students may encounter on such exams. It stresses the practical application of formulas and concepts. To prepare thoroughly, it is vital to solve many problems like this one and understand their solutions step by step.

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Most popular questions from this chapter

A copper plate of length \(1 \mathrm{~m}\) is riveted to two steel plates of same length and same cross-section area at \(0^{\circ} \mathrm{C}\). Calculate tension (in kilo newton) generated in copper plate when heated to \(20^{\circ} \mathrm{C}\). \(Y_{\text {copper }}=\frac{1}{2} \times Y_{\text {steel }}=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2} Y=\) Young's modules \(\alpha_{\text {copper }}=18 \times 10^{-6} \mathrm{~K}^{-1}\) \(\alpha_{\text {steel }}=11 \times 10^{-6} \mathrm{~K}^{-1} \alpha=\) coefficient of linear expansion \text { Area of each plate }=50 \mathrm{~cm}^{2}

At what temperature will the resistance of a copper wire become three times its value at \(0^{\circ} \mathrm{C}\) (Temperature coefficient of resistance for copper \(=4 \times 10^{-3} /{ }^{\circ} \mathrm{C}\) ) (A) \(400^{\circ} \mathrm{C}\) (B) \(450^{\circ} \mathrm{C}\) (C) \(500^{\circ} \mathrm{C}\) (D) \(550^{\circ} \mathrm{C}\)

1 mole of an ideal gas is contained in a cubical volume \(V\), ABCDEFGH at \(300 \mathrm{~K}\) (Fig. 10.18). One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it. At any given time, (A) The pressure on \(E F G H\) would be zero. (B) The pressure on all the faces will be equal. (C) The pressure of \(E F G H\) would be double the pressure on \(A B C D\). (D) The pressure of \(E F G H\) would be half that on \(A B C D .\)

A \(2 \mathrm{gm}\) bullet moving with a velocity of \(200 \mathrm{~m} / \mathrm{s}\) is brought to a sudden stoppage by an obstacle. The total heat produced goes to the bullet. If the specific heat of the bullet is \(0.03 \mathrm{cal} / \mathrm{gm}-{ }^{\circ} \mathrm{C}\), the rise in its temperature will be (A) \(158.0^{\circ} \mathrm{C}\) (B) \(15.80^{\circ} \mathrm{C}\) (C) \(1.58^{\circ} \mathrm{C}\) (D) \(0.1580^{\circ} \mathrm{C}\)

A container \(X\) contains 1 mole of \(O_{2}\) gas (molar mass 32) at a temperature \(T\) and pressure \(P\). Another identical container \(Y\) contains 1 mole of He gas (molar mass 4) at temperature \(2 T\), then (A) pressure in the container \(Y\) is \(P / 8\). (B) pressure in container \(Y\) is \(P\). (C) pressure in the container \(Y\) is \(2 P\). (D) pressure in container \(Y\) is \(P / 2\).

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