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Chapter 8: Problem 1116

A Small spherical body of radius \(\mathrm{r}\) is falling under gravity in a viscous medium. Due to friction the medium gets heated. How does the rate of heating depend on radius of body when it attains terminal velocity! (A) \(r^{2}\) (B) \(r^{3}\) (C) \(\mathrm{r}^{4}\) (D) \(\mathrm{r}^{5}\)

Short Answer

Expert verified
The rate of heating depends on the radius of the spherical body as \(r^{2}\). Thus, the correct answer is (A) \(r^{2}\).

Step by step solution

01

Find Terminal Velocity of Sphere

When the sphere reaches terminal velocity, the gravitational force equals the viscous force acting on it. Let's denote the terminal velocity as \(v_t\), mass of the sphere as \(m\), and the coefficient of viscosity of the medium as \(\eta\). The gravitational force \(\mathrm{F_g}\) is given by \(mg\), where \(g\) is the acceleration due to gravity. For the viscous force \(\mathrm{F_v}\), we have to use Stoke's law, which states that the viscous force acting on a sphere falling through a viscous medium is given by: \[F_v = 6\pi \eta r v_t\] Now, equate the gravitational force and viscous force: \[mg = 6\pi \eta r v_t\] Solve for terminal velocity, \(v_t\): \[v_t = \dfrac{mg}{6\pi \eta r}\]
02

Obtain the Rate of Energy Dissipation

As the sphere falls at terminal velocity, the work done by viscous force gets converted into heat. The rate at which this energy gets dissipated (the rate of heating), is given by the product of viscous force and terminal velocity: \[\text{Rate of heating} = F_v \cdot v_t\] Using the expressions we obtained for \(F_v\) and \(v_t\) in Step 1: \[\text{Rate of heating} = (6\pi \eta r v_t) \cdot \dfrac{mg}{6\pi \eta r}\]
03

Express Rate of Heating in Terms of Radius

Now, we will simplify the expression for the rate of heating to find its dependence on the radius \(r\): \[\text{Rate of heating} = 6\pi \eta r \cdot \dfrac{mg}{6\pi \eta r}\] Cancelling out common terms, we get: \[\text{Rate of heating} \propto m g r^2\] Therefore, the rate of heating is proportional to \(r^2\).
04

Choose the Correct Answer

From our analysis, we have found that the rate of heating depends on the radius \(r\) as \(r^2\). Comparing this result with the given choices (A) \(r^{2}\), (B) \(r^{3}\), (C) \(\mathrm{r}^{4}\), and (D) \(\mathrm{r}^{5}\), the correct answer is (A) \(r^{2}\).

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Most popular questions from this chapter

Two cylinders \(\mathrm{A}\) and \(\mathrm{B}\) fitted with piston contain equal amounts of an ideal diatomic gas at \(300 \mathrm{k}\). The piston of \(\mathrm{A}\) is free to move, While that of \(B\) is held fixed. The same amount of heat is given to the gas in each cylinders. If the rise in temperature of the gas in \(\mathrm{A}\) is \(30 \mathrm{~K}\), then the rise in temperature of the gas in \(\mathrm{B}\) is. (A) \(30 \mathrm{~K}\) (B) \(42 \mathrm{~K}\) (C) \(18 \mathrm{~K}\) (D) \(50 \mathrm{~K}\)

Starting with the same initial Conditions, an ideal gas expands from Volume \(\mathrm{V}_{1}\) to \(\mathrm{V}_{2}\) in three different ways. The Work done by the gas is \(\mathrm{W}_{1}\) if the process is purely isothermal, \(\mathrm{W}_{2}\) if purely isobaric and \(\mathrm{W}_{3}\) if purely adiabatic Then (A) \(\mathrm{W}_{2}>\mathrm{W}_{1}>\mathrm{W}_{3}\) (B) \(\mathrm{W}_{2}>\mathrm{W}_{3}>\mathrm{W}_{1}\) (C) \(\mathrm{W}_{1}>\mathrm{W}_{2}>\mathrm{W}_{3}\) (D) \(\mathrm{W}_{1}>\mathrm{W}_{3}>\mathrm{W}_{2}\)

Air is filled in a motor tube at \(27^{\circ} \mathrm{C}\) and at a Pressure of 8 atmosphere. The tube suddenly bursts. Then what is the temperature of air. given \(\gamma\) of air \(=1.5\) (A) \(150 \mathrm{~K}\) (B) \(150^{\circ} \mathrm{C}\) (C) \(75 \mathrm{~K}\) (D) \(27.5^{\circ} \mathrm{C}\)

An ideal gas at \(27 \mathrm{C}\) is Compressed adiabatically, to \(\\{8 / 27\\}\) of its original Volume. If \(\mathrm{v}=(5 / 3)\), then the rise in temperature is (A) \(225 \mathrm{k}\) (B) \(450 \mathrm{~K}\) (C) \(375 \mathrm{~K}\) (D) \(405 \mathrm{~K}\)

A diatomic gas initially at \(18^{\circ} \mathrm{C}\) is Compressed adiabatically to one eighth of its original volume. The temperature after Compression will be (A) \(10^{\circ} \mathrm{C}\) (B) \(668 \mathrm{~K}\) (C) \(887^{\circ} \mathrm{C}\) (D) \(144^{\circ} \mathrm{C}\)
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