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Chapter 7: Problem 926

When a pressure of 100 atmosphere is applied on a spherical ball then its volume reduces to \(0.01 \%\). What is the bulk modulus of the material of the rubber in \(\left(\right.\) dyne \(\left./ \mathrm{cm}^{2}\right)\) (A) \(10 \times 10^{12}\) (B) \(1 \times 10^{12}\) (C) \(100 \times 10^{12}\) (D) \(20 \times 10^{12}\)

Short Answer

Expert verified
The Bulk Modulus (B) of the rubber is approximately \(1 \times 10^{12}\) dyne/cm^2. (Option B)

Step by step solution

01

Determine the given values from the problem statement

The given values in the problem are the applied pressure (100 atm) and the reduced volume (0.01%). We need to calculate the Bulk Modulus (B) in dyne/cm^2.
02

Convert the pressure to dyne/cm^2

We are given the pressure in atmospheres, but we need it in dyne/cm^2. To convert from atmospheres to dyne/cm^2, use the conversion factor: \(1 atm = 1.01325 \times 10^6 dyne/cm^2\). Therefore, the pressure in dyne/cm^2 is: \(100 atm \times \frac{1.01325 \times 10^6 dyne/cm^2}{1 atm} = 1.01325 \times 10^8 dyne/cm^2 \)
03

Calculate the relative volume change

The given volume reduction is 0.01% or 0.0001 in decimal form. Since the volume reduces, we must include a negative sign, so the relative volume change is -0.0001.
04

Apply the Bulk Modulus formula

The formula for Bulk Modulus (B) is: \(B = - \frac{P}{\Delta V / V}\) Where P is the applied pressure and \(\Delta V / V\) is the relative volume change. Now plug in the values we found in steps 2 and 3: \(B = \frac{1.01325 \times 10^8 dyne/cm^2}{0.0001} = 1.01325 \times 10^{12} dyne/cm^2 \)
05

Identify the correct answer in the multiple-choice options

Comparing our calculated Bulk Modulus to the given options, we see that our result is closest to the option (B). Hence, the correct answer is: (B) \(1 \times 10^{12}\) dyne/cm^2

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Most popular questions from this chapter

The compressibility of water \(4 \times 10^{-5}\) per unit atmospheric pressure. The decrease in volume of 100 cubic centimeter of water under a pressure of 100 atmosphere will be.......... (A) \(4 \times 10^{-5} \mathrm{CC}\) (B) \(4 \times 10^{-5} \mathrm{CC}\) (C) \(0.025 \mathrm{CC}\) (D) \(0.004 \mathrm{CC}\)

At what temperature the centigrade (celsius) and Fahrenheit readings at the same. \((\mathrm{A})-40^{\circ}\) (B) \(+40^{\circ} \mathrm{C}\) (C) \(36.6^{\circ}\) (D) \(-37^{\circ} \mathrm{C}\)

A large number of water drops each of radius \(r\) combine to have a drop of radius \(\mathrm{R}\). If the surface tension is \(\mathrm{T}\) and the mechanical equivalent at heat is \(\mathrm{J}\) then the rise in temperature will be (A) \((2 \mathrm{~T} / \mathrm{rJ})\) (B) \((3 \mathrm{~T} / \mathrm{RJ})\) (C) \((3 \mathrm{~T} / \mathrm{J})\\{(1 / \mathrm{r})-(1 / \mathrm{R})\\}\) (D) \((2 \mathrm{~T} / \mathrm{J})\\{(1 / \mathrm{r})-(1 / \mathrm{R})\\}\)

Radius of a soap bubble is \(\mathrm{r}^{\prime}\), surface tension of soap solution is \(\mathrm{T}\). Then without increasing the temperature how much energy will be needed to double its radius. (A) \(4 \pi r^{2} T\) (B) \(2 \pi r^{2} T\) (C) \(12 \pi r^{2} T\) (D) \(24 \pi r^{2} T\)

Shearing stress causes change in (A) length (B) breadth (C) shape (D) volume
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