/*! 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 81 A ball is dropped into a well in... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 81

A ball is dropped into a well in which the water level is at a depth \(h\) below the top \((t=0)\). If the speed of sound be \(c\), then the time after which the splash is heard will be given by (A) \(h\left[\sqrt{\frac{2}{g h}}+\frac{1}{c}\right]\) (B) \(h\left[\sqrt{\frac{2}{g h}}-\frac{1}{c}\right]\) (C) \(h\left[\frac{2}{g}+\frac{1}{c}\right]\) (D) \(h\left[\frac{2}{g}-\frac{1}{c}\right]\)

Short Answer

Expert verified
The correct option is (A) \(h\left[\sqrt{\frac{2}{g h}}+\frac{1}{c}\right]\)

Step by step solution

01

Calculate the time taken by the ball to drop

For the first part of the journey, the ball is falling freely under gravity \(g\). The time \(t1\) it takes to fall a distance \(h\) under gravity is given by the formula derived from the second equation of motion: \[t1 = \sqrt{\frac{2h}{g}}\] We're using this because we know the initial velocity of the ball is 0 (as it's being dropped), the distance it's falling is \(h\) and it's undergoing acceleration due to gravity.
02

Calculate the time taken by the sound to reach the top

For the second part of the journey, the sound of the splash travels back up to the surface at the speed of sound \(c\). The time \(t2\) it takes for the sound to travel a distance \(h\) at speed \(c\) is: \[t2 = \frac{h}{c}\] This uses the basic formula of speed = distance/time.
03

Add up the two times

The total time \(t\) after which the splash is heard is the sum of the above two times (the splash is heard after the ball falls and makes the splash, and the sound travels back up): \[t = t1 + t2 = \sqrt{\frac{2h}{g}} + \frac{h}{c}\]
04

Identify the correct option

Substitute \(t1\) and \(t2\) in the options given and see which matches. Doing this we can see that option (A) is correct.

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

If a simple harmonic motion is represented by \(\frac{d^{2} x}{d t^{2}}+\alpha x=0\), its time period is (A) \(\frac{2 \pi}{\sqrt{\alpha}}\) (B) \(\frac{2 \pi}{\alpha}\) (C) \(2 \pi \sqrt{\alpha}\) (D) \(2 \pi \alpha\)

A traveling wave pulse is given by \(y=\frac{6}{2+(x+3 t)^{2}}\), where symbols have their usual meanings, \(x, y\) are in metre and \(\mathrm{t}\) is in second. Then (A) The pulse is traveling along +ve \(x\)-axis with velocity \(3 \mathrm{~m} / \mathrm{s}\) (B) The pulse is traveling along -ve \(x\)-axis with velocity \(3 \mathrm{~m} / \mathrm{s}\). (C) The amplitude of the wave pulse is \(3 \mathrm{~m}\). (D) The pulse is a symmetric pulse.

Let a disturbance \(y\) be propagated as a plane wave along the \(x\)-axis. The wave profiles at the instants \(t=t_{1}\) and \(t=t_{2}\) are represented, respectively, as \(y_{1}=f\left(x_{1}-v t_{1}\right)\) and \(y_{2}=f\left(x_{2}-v t_{2}\right)\). The wave is propagating without change of shape. (A) The velocity of the wave is \(2 v\). (B) The velocity of the wave is \(v=\frac{x_{2}-x_{1}}{t_{2}}\). (C) The particle velocity is \(v_{p}=v\). (D) None of these.

A particle of mass \(m\) is attached to a spring (of spring constant \(k\) ) and has a natural angular frequency \(\omega_{0}\). An external force \(F(t)\) proportional to \(\cos \omega t\left(\omega \neq \omega_{0}\right)\) is applied to the oscillator. The time displacement of the oscillator will be proportional to \(\quad\) (A) \(\frac{1}{m\left(\omega_{0}^{2}+\omega^{2}\right)}\) (B) \(\frac{1}{m\left(\omega_{0}^{2}-\omega^{2}\right)}\) (C) \(\frac{m}{\omega_{0}^{2}-\omega^{2}}\) (D) \(\frac{m}{\left(\omega_{0}^{2}+\omega^{2}\right)}\)

In a simple harmonic oscillator, at the mean position (A) kinetic energy is minimum, potential energy is maximum. (B) both kinetic and potential energies are maximum. (C) kinetic energy is maximum, potential energy is minimum. (D) both kinetic and potential energies are minimum.
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Medical Physics

Read Explanation

Force

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Particle Model of Matter

Read Explanation

Geometrical and Physical Optics

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.