/*! 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 93 A pendulum clock loses 12 s a da... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 93

A pendulum clock loses 12 s a day if the temperature is \(40^{\circ} \mathrm{C}\) and gains \(4 \mathrm{~s}\) day if the temperature is \(20^{\circ} \mathrm{C}\). The temperature at which the clock will show correct time, and the co-efficient of linear expansion \((\alpha)\) of the metal of the pendulum shaft are respectively. [2016] (A) \(60^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-4} /{ }^{\circ} \mathrm{C}\) (B) \(30^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-3} /{ }^{\circ} \mathrm{C}\) (C) \(55^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-2} /{ }^{\circ} \mathrm{C}\) (D) \(25^{\circ} \mathrm{C} ; \alpha=1.85 \times 10^{-5} /{ }^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The correct temperature at which the pendulum will show the correct time is 20\(^{\circ}\) Celsius, and the coefficient of linear expansion (\(\alpha\)) of the metal of the pendulum shaft is \(0.02/{}^{\circ}C\). Therefore, none of the options (A), (B), (C), or (D) are correct.

Step by step solution

01

Determine the rate of change in time per degree change in temperature

According to the problem, the clock loses 12 seconds for a 20 degree increase in temperature, from 20 degrees Celsius to 40 degrees Celsius, and gains 4 seconds for a 20 degree decrease in temperature, from 40 degrees Celsius to 20 degrees Celsius. We calculate the rate of change in time per degree increase in temperature by subtracting the two times and dividing by the change in temperature: rate = \((-12 - 4) / 20 = -0.8\) s/掳C.
02

Establish the equation for change in time

Given the rate, we can establish the equation for time change: \(\Delta T = -0.8(T-20)\). Setting \(\Delta T = 0\) gives the temperature T such that the time displayed by the clock is correct. Solving the equation, \(0 = -0.8(T - 20)\) gives us \(T = 20\)\(^\circ\)C.
03

Determine the coefficient alpha

The change in time is also given by the formula: \(\Delta T = -2 \alpha x \Delta t\), where \(\Delta t\) is the change in temperature and \(\alpha\) is the co-efficient of linear expansion we are looking for. We can rearrange this formula to solve for alpha: \(\alpha = -\Delta T / (2 \Delta t)\). Substituting \(\Delta T = -0.8(T - 20)\) and \(\Delta t = 20\), we get \(\alpha = 0.8 / (2 x 20) = 0.02\).

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Most popular questions from this chapter

A long metallic bar is carrying heat from one end to the other under steady state. The variation of temperature \(\theta\) along the length \(x\) of the bar from its hot end is best described by which of the following.

Two rods of length \(L_{1}\) and \(L_{2}\) are made of materials whose coefficients of linear expansion are \(\alpha_{1}\) and \(\alpha_{2}\). If the difference between the two lengths is independent of temperature. (A) \(\left(L_{1} / L_{2}\right)=\left(\alpha_{1} / \alpha_{2}\right)\) (B) \(\left(L_{1} / L_{2}\right)=\left(\alpha_{2} / \alpha_{1}\right)\) (C) \(L_{1}^{2} \alpha_{1}=L_{2}^{2} \alpha_{2}\) (D) \(\alpha_{1}^{2} L_{1}=\alpha_{2}^{2} L_{2}\)

If the degree of freedom of a gas molecule is \(f\), then the ratio of two specific heats \(C_{p} / C_{v}\) is given by (A) \(\frac{2}{f}+1\) (B) \(1-\frac{2}{f}\) (C) \(1+\frac{1}{f}\) (D) \(1-\frac{1}{f^{2}}\)

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) \(\sqrt{\frac{\gamma P_{1} S_{0}}{2 m l_{1}}}\)

Two rods of length \(L_{1}\) and \(L_{2}\) are made of materials whose coefficients of linear expansion are \(\alpha_{1}\) and \(\alpha_{2}\) If the difference between the two lengths is independent of temperature (A) \(\left(L_{1} / L_{2}\right)=\left(\alpha_{1} / \alpha_{2}\right)\) (B) \(\left(L_{1} / L_{2}\right)=\left(\alpha_{2} / \alpha_{1}\right)\) (C) \(L_{1}^{2} \alpha_{1}=L_{2}^{2} \alpha_{2}\) (D) \(\alpha_{1}^{2} L_{1}=\alpha_{2}^{2} L_{2}\)
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