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Chapter 12: Problem 94

If the terminal speed of a sphere of gold (density \(=9.5 \mathrm{~kg} / \mathrm{m}^{3}\) ) is \(0.2 \mathrm{~m} / \mathrm{s}\) in a viscous liquid (density \(=1.5 \mathrm{~kg} / \mathrm{m}^{3}\) ). Find the terminal speed of sphere of silver (density \(=10.5 \mathrm{~kg} / \mathrm{m}^{3}\) ) of the same size in the same liquid [WB JEE 2010] (a) \(0.133 \mathrm{~m} / \mathrm{s}\) (b) \(0.1 \mathrm{~m} / \mathrm{s}\) (c) \(0.2 \mathrm{~m} / \mathrm{s}\) (d) \(0.4 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The given problem and options may contain errors; computed speed is 0.225 m/s.

Step by step solution

01

Understand Terminal Speed Formula

To find the terminal speed of the silver sphere, we need to use the formula for terminal speed in a viscous medium: \[v_t = \frac{2}{9} \frac{(\rho_s - \rho_l) g R^2}{\eta}\]where \( \rho_s \) is the density of the sphere, \( \rho_l \) is the density of the liquid, \( g \) is the acceleration due to gravity, \( R \) is the radius of the sphere, and \( \eta \) is the viscosity of the liquid. However, since the viscosity and radius are not specified but are constant for both spheres, we can relate their terminal speeds directly using their densities.
02

Write Ratio of Terminal Speeds

For the spheres to be compared, from the terminal velocity formula, the dependence on density can be expressed as:\[v_{t1} \propto (\rho_{s1} - \rho_l)\]\[v_{t2} \propto (\rho_{s2} - \rho_l)\]where \( v_{t1} \) and \( v_{t2} \) are the terminal speeds of the gold and silver spheres, respectively. From this, the ratio of terminal speeds is:\[\frac{v_{t2}}{v_{t1}} = \frac{(\rho_{s2} - \rho_l)}{(\rho_{s1} - \rho_l)}\]
03

Substitute Known Values

Substitute the known values into the ratio formula:\[\frac{v_{t2}}{0.2} = \frac{(10.5 - 1.5)}{(9.5 - 1.5)}\]Simplifying gives:\[\frac{v_{t2}}{0.2} = \frac{9}{8}\]
04

Solve for Terminal Speed of Silver Sphere

Solve the above equation for \( v_{t2} \):\[v_{t2} = 0.2 \times \frac{9}{8} = 0.225 \, \text{m/s}\]This computed speed does not match the provided options, which suggests an initial approximation or option mismatch in the given question.

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Key Concepts

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

Viscous Medium
In physics, a viscous medium refers to a fluid that resists motion due to internal friction. This friction is caused by the fluid's viscosity, which is essentially a measure of how "thick" or "sticky" the fluid is. Viscosity is an important factor in determining how objects move through the fluid.

When an object moves through a viscous medium, like a sphere through honey or oil, it experiences a drag force. This force opposes the motion, slowing down the object until it reaches a constant speed. This constant speed is known as the terminal velocity. At terminal velocity, the drag force exactly balances the forces of gravity and buoyancy acting on the object.

Important points about a viscous medium:
  • It opposes the motion of objects moving through it.
  • The greater the viscosity, the greater the drag force experienced by a moving object.
  • Terminal velocity is achieved when the forces on the object balance out.
Understanding how different variables, like the viscosity of the medium, affect terminal velocity aids in solving problems involving movement through a viscous liquid.
Density of Solids
The density of a solid is a key factor that determines how it interacts with its surroundings, particularly when it is submerged in a fluid. Density is defined as mass per unit volume (\[ \rho = \frac{m}{V} \]). It tells us how much mass is contained in a specific volume of the material.

When discussing terminal velocity, the density of the solid plays a critical role. A denser object will have a greater gravitational force acting on it compared to a less dense one of the same volume. This influences how quickly the object will reach terminal velocity when dropped in a fluid.

Key takeaways regarding density:
  • It is mass per unit volume, a fundamental property of materials.
  • Higher density means more mass in a given space, affecting gravitational pull.
  • In the context of terminal velocity, a denser object typically reaches terminal velocity faster.
Examining density helps us understand how different materials behave in the same fluid, as shown in solving for the silver and gold sphere problem.
Buoyancy in Fluids
Buoyancy is a force that occurs when an object is submerged in a fluid, counteracting the force of gravity. It results from pressure differences in the fluid surrounding the object. The buoyant force acts upward, providing a lifting force on the object.

Archimedes' principle governs buoyancy. It states that the buoyant force on an object is equal to the weight of the fluid it displaces. This principle helps explain why objects float or sink.

Understanding buoyancy includes:
  • Recognizing that it acts in opposition to gravity.
  • The buoyant force is proportional to the volume of fluid displaced.
  • It plays a significant role in determining the net force acting on an object.
In terminal velocity problems, buoyancy is considered to calculate the effective weight of the object in the liquid. Balancing buoyancy with gravity and viscous drag explains why an object settles at a terminal velocity.

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Most popular questions from this chapter

The rate of steady volume flow of water through a capillary tube of length \(l\) and radius \(r\) under a pressure difference of \(p\), is \(V\). This tube is connected with another tube of the same length but \(\underline{\text { half the }}\) radius in series. Then the rate of steady volume flow through them is (The pressure difference across the combination is \(p\) ) (a) \(\frac{v}{16}\) (b) \(\frac{v}{17}\) (c) \(\frac{16 \mathrm{~V}}{17}\) (d) \(\frac{17 \mathrm{~V}}{16}\)

A beaker containing water is balanced on the pan of a common balance. A solid of specific gravity 1 and mass \(5 \mathrm{~g}\) is tied to the arm of the balance and immersed in water contained in the beaker. The scale pan with the beaker (a) goes down (b) goes up (c) remains unchanged (d) None of these

The U-tube has a uniform cross-section as shown in figure. A liquid is filled in the two arms upto heights \(h_{1}\) and \(h_{2}\) and then the liquid is allowed to move. Neglect viscosity and surface tension. When the level equalize in the two arms, the liquid will (a) be at rest (b) be moving with an acceleration of \(g\left(\frac{h_{1}-h_{2}}{h_{1}+h_{2}+2}\right)\) (c) be moving with a velocity of \(\left(h_{1}-h_{2}\right) \sqrt{\frac{g}{2\left(h_{1}+h+h\right)}}\) (d) exert a net force to the right on the cube

Two soap bubbles \(A\) and \(B\) are formed at the two open ends of a tube. The bubble \(A\) is smaller than bubble \(B\). Valve and air can flow freely between the bubbles, then (a] there is no change in the size of the bubbles (b) the two bubbles will become of equal size (c) \(A\) will become smaller and \(B\) will become larget (d) \(B\) will become smaller and \(A\) will become larger

The area of cross-section of one limb of an U-tube is twice that of the other. Both the limbs contains mercury at the same level. Water is poured in the wider tube so that mercury level in it goes down by \(1 \mathrm{~cm}\). The height of water column is (Density of water \(=10^{3} \mathrm{kgm}^{-\text {I }}\), density of mercury \(=13.6 \times 10^{3} \mathrm{kgm}^{-3}\) ) \(\quad\) ?erala CET 200s] (a) \(13.6 \mathrm{~cm}\) (b) \(40.8 \mathrm{~cm}\) (c) \(6.8 \mathrm{~cm}\) (d) \(54.4 \mathrm{~cm}\)
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