Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 27: Problem 2811

The velocity of a small ball of mass \(\mathrm{m}\) and density \(\mathrm{d}_{1}\) when dropped in a container filled with glycerine becomes constants after sometime. What is the viscous force acting on the ball ? (density of glycerine is \(\mathrm{d}_{2}\) ) (A) \(m g\left\\{1-\left(d_{1} / d_{2}\right)\right\\}\) (B) \(m g\left\\{1-\left(d_{2} / d_{1}\right)\right\\}\) (C) \(m g\left\\{d_{1} /\left(d_{1}-d_{2}\right)\right\\}\) (D) $\operatorname{mg}\left\\{\mathrm{d}_{2} /\left(\mathrm{d}_{1}-\mathrm{d}_{2}\right)\right\\}$

Short Answer

Expert verified
(A) \(mg\\ \{1 - (d_1 / d_2)\\}\)

Step by step solution

01

Identifying the forces acting on the ball

When the ball is moving through the glycerine, it experiences three forces: 1. Gravitational force (weight) = mg, acting downward 2. Buoyant force = the upward force due to the displacement of glycerine, given by \(V \times d_2 \times g\), where V is the volume of the ball 3. Viscous force = the force due to the glycerine acting upward and opposite to the motion of the ball
02

Finding the terminal velocity

Terminal velocity occurs when the net force acting on the ball becomes zero, and the ball moves with a constant velocity. At terminal velocity, the sum of the buoyant force and the viscous force equals the gravitational force acting on the ball. So, we have: Buoyant force + Viscous force = Gravitational force
03

Expressing the forces in terms of mass and density

The volume V of the ball can be expressed as \(V = \frac{m}{d_1}\), where m is the mass of the ball and \(d_1\) is its density. Using this, we can rewrite the equation from Step 2 as: \(\frac{m}{d_1} \times d_2 \times g\) + Viscous force = mg
04

Solving for the viscous force

Rearrange the equation in Step 3 to find the expression for the viscous force: Viscous force = mg - \(\frac{m}{d_1}\) × \(d_2g\) Now, factor out the mg term to simplify: Viscous force = mg\(\\ \{1 - (d_1 / d_2)\\}\) Comparing this expression with the given options, we see that it matches option (A). Therefore, the viscous force acting on the ball is
05

(A) \(mg\\ \{1 - (d_1 / d_2)\\}\)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A steel ball of diameter \(3 \mathrm{~mm}\) falls through glycerine and covers a distance of \(25 \mathrm{~cm}\) in \(10 \mathrm{~S} .\) The specific gravity of steel and glycerine are \(7.8\) and \(1.26\) respectively. The viscosity of glycerine is about........ pascal-sec (A) \(1.3\) (B) \(1.5\) (C) \(0.8\) (D) \(1.0\)

A steel ball of diameter \(3.2 \mathrm{~mm}\) falls under gravity through an oil of density \(920 \mathrm{~kg} \mathrm{~m}^{-3}\) and viscosity $1.64 \mathrm{Ns} \mathrm{m}^{-2}\(. The density of steel. may be taken as \)7820 \mathrm{~kg} \mathrm{~m}^{-3}$. What is the terminal velocity of the ball ? (A) \(3.45 \mathrm{cms}^{-1}\) (B) \(4.25 \mathrm{cms}^{-1}\) (C) \(1.85 \mathrm{~cm}^{-1}\) (D) \(2.39 \mathrm{cms}^{-1}\)

The terminal velocity of fall of lead shots in the cylindrical vessel is influenced by the nearness of the walls of the vessel. Hence the velocity in formula is divided by a factor \([1+\\{(2.4 r) / R\\}] .\) Where \(r\) is the radius of the lead shots and \(\mathrm{R}\) the radius of the vessel. This is called.......... (A) Ladenburg correction (B) Velocity correction (C) End-correction (D) Rydberg correction

The C.G.S. unit of coefficient of viscosity is poise and the SI unit is \(\mathrm{Pa}-\mathrm{S}\). What is the relation between the two ? (A) \(1 \mathrm{~Pa}-\mathrm{s}=10^{3}\) poise (B) \(1 \mathrm{~Pa}-\mathrm{s}=10^{4}\) poise (C) \(1 \mathrm{~Pa}-\mathrm{s}=10\) poise (D) \(1 \mathrm{~Pa}-\mathrm{s}=100\) poise

64 equal drops of water are falling through air with a steady velocity \(\mathrm{V}_{0}\), if the drops coalesce, what is their new velocity? (A) \(4 \mathrm{~V}_{0}\) (B) \(2^{\\{(-1) / 3\\}} \mathrm{V}_{0}\) (C) \(2 \mathrm{~V}_{0}\) (D) \(2^{(1 / 3)} \mathrm{V}_{0}\)
See all solutions

Recommended explanations on English Textbooks

Rhetoric

Read Explanation

Essay Plan

Read Explanation

Language Acquisition

Read Explanation

Listening and Speaking

Read Explanation

Phonetics

Read Explanation

Language Analysis

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.

Sign-up for free