/*! 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] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 24

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

Calculate the change in diameter

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

Use the linear expansion formula to solve for temperature change

The linear expansion formula is: 螖L = L鈧� * 伪 * 螖T, where 螖L is the change in length (diameter in this case), L鈧� is the initial length (diameter of the iron tyre), 伪 is the linear expansion coefficient, and 螖T is the change in temperature. Rearrange the formula to solve for 螖T: 螖T = 螖D / (Di * 伪) Substitute the values and calculate 螖T: 螖T = 0.006m / (0.994m * \(1.2 \times 10^{-5}\) / 掳C) 鈮� 500掳C. The minimum increase in temperature for the iron tyre to fit onto the wooden wheel is 500掳C, which corresponds to option (C).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If spring is disconnected and top part of cylinder is removed, then find the angular frequency for small oscillation. (Assuming pressure of gas at equilibrium position is \(P_{1}\) and length of gas column is \(l_{1}^{-}\)) (A) \(\sqrt{\frac{\gamma P_{1} S_{0}}{m l_{1}}}\) (B) \(\sqrt{\frac{2 \gamma P S_{0}}{m l_{1}}}\) (C) \(\sqrt{\frac{\gamma P_{1} S_{0}}{4 m l_{1}}}\) (D) \(\sqrt{\frac{\gamma P_{1} S_{0}}{2 m l_{1}}}\)

A wire fixed at the upper end stretches by length \(l\) by applying a force \(F .\) The work done in stretching is [2004] (A) \(F / 2 l\) (B) \(F l\) (C) \(2 F l\) (D) \(F l / 2\)

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) \(0.1580^{\circ} \mathrm{C}\)

Cooking gas containers are loaded on to a truck moving with uniform speed. The temperature of the gas molecules inside the containers will (A) increase. (B) decrease. (C) remain same. (D) decrease for some, whereas increase for others.

Water flows at the rate of \(0.1500 \mathrm{~kg} / \mathrm{min}\) through a tube and is heated by a heater dissipating \(25.2 \mathrm{~W}\). The inflow and outflow water temperatures are \(15.2^{\circ} \mathrm{C}\) and \(17.4^{\circ} \mathrm{C}\), respectively. When the rate of flow is increased to \(0.2318 \mathrm{~kg} / \mathrm{min}\) and the rate of heating to \(37.8 \mathrm{~W}\), the inflow and outflow temperatures are unaltered. Find the rate of loss of heat from the tube
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Astrophysics

Read Explanation

Thermodynamics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Classical Mechanics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.