/*! 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 5 A wooden cylinder floats in wate... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 5

A wooden cylinder floats in water with two-third of its volume inside the water. The density of wood is (A) \(\frac{1000}{3} \mathrm{~kg} / \mathrm{m}^{3}\) (B) \(\frac{2000}{3} \mathrm{~kg} / \mathrm{m}^{3}\) (C) \(\frac{500}{3} \mathrm{~kg} / \mathrm{m}^{3}\) (D) \(250 \mathrm{~kg} / \mathrm{m}^{3}\)

Short Answer

Expert verified
The density of wood is (B) \(\frac{2000}{3} \mathrm{~kg/m^3}\).

Step by step solution

01

Write down the buoyant force equation

Archimedes' principle states that the buoyant force experienced by an object submerged in fluid is equal to the weight of the fluid displaced by the object. The buoyant force equation is given as: \( F_b = V_{submerged} \cdot \rho_{fluid} \cdot g \) Where: - \(F_b\) is the buoyant force - \(V_{submerged}\) is the volume of the submerged part of the object - \(\rho_{fluid}\) is the density of the fluid - \(g\) is the acceleration due to gravity (approximately \(9.8 \mathrm{~m/s^2}\))
02

Write down the weight of the wooden cylinder

The weight of the wooden cylinder can be calculated as: \( W_{wood} = V_{wood} \cdot \rho_{wood} \cdot g \) Since the cylinder is floating, the buoyant force is equal to the weight of the cylinder. This implies: \( V_{submerged} \cdot \rho_{fluid} \cdot g = V_{wood} \cdot \rho_{wood} \cdot g \)
03

Substitute the given values

It is given that two-third of the wooden cylinder's volume is submerged in water. Therefore, we can write the submerged volume as: \( V_{submerged} = \frac{2}{3}V_{wood} \) Substitute this into the buoyant force equation along with the density of water (\(\rho_{fluid} = 1000 \mathrm{~kg/m^3}\)): \( \frac{2}{3}V_{wood} \cdot 1000 \mathrm{~kg/m^3} \cdot g = V_{wood} \cdot \rho_{wood} \cdot g \)
04

Solve for the density of wood

Divide both sides of the equation by \(V_{wood} \cdot g\), then simplify the equation to get the density of wood: \( \frac{\frac{2}{3} \cdot 1000 \mathrm{~kg/m^3}}{1} = \rho_{wood} \) \( \rho_{wood} = \frac{2000}{3} \mathrm{~kg/m^3} \) Therefore, the correct answer is (B) \(\frac{2000}{3} \mathrm{~kg/m^3}\).

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

A gas having density \(\rho\) flows with a velocity \(v\) along a pipe of cross- sectional area \(s\) and bent at an angle of \(90^{\circ}\) at a point \(A\). The force exerted by the gas on the pipe at \(A\) is (A) \(\frac{\sqrt{2} s v}{\rho}\) (B) \(s v^{2} \rho\) (C) \(\frac{\sqrt{3} s v^{2} \rho}{2}\) (D) \(s v^{2} \rho\)

The pressure just below the meniscus of water (A) is greater than just above it. (B) is lesser than just above it. (C) is same as just above it. (D) is always equal to atmospheric pressure.

A conical block floats in water with \(90 \%\) height immersed in it. Height \(h\) of the block is equal to the diameter of the block, i.e., \(20 \mathrm{~cm}\). The mass to be kept on the block, so that the block just floats at the surface of water, is (A) \(568 \mathrm{~g}\) (B) \(980 \mathrm{~g}\) (C) \(112 \mathrm{~g}\) (D) \(196 \mathrm{~g}\)

A vessel of \(1 \times 10^{-3} \mathrm{~m}^{3}\) volume contains oil, when a pressure of \(1.2 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\) is applied on it, then volume decreases by \(0.3 \times 10^{-6} \mathrm{~m}^{3}\). The bulk modulus of oil is (A) \(1 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\) (B) \(2 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}\) (C) \(4 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}\) (D) \(6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\)

A cylindrical vessel of \(90 \mathrm{~cm}\) height is kept filled up to its brim. It has four holes \(1,2,3,4\) which are, respectively, at heights of \(20 \mathrm{~cm}, 30 \mathrm{~cm}, 45 \mathrm{~cm}\), and \(50 \mathrm{~cm}\) from the horizontal floor \(P Q\). The water falling at the maximum horizontal distance from the vessel comes from (A) Hole number 4 (B) Hole number 3 (C) Hole number 2 (D) Hole number 1
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Energy Physics

Read Explanation

Medical Physics

Read Explanation

Radiation

Read Explanation

Wave Optics

Read Explanation

Oscillations

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.