FlipIt for College Physics (Algebra Version - Six Months Access)
FlipIt for College Physics (Algebra Version - Six Months Access)
17th Edition
ISBN: 9781319032432
Author: Todd Ruskell
Publisher: W.H. Freeman & Co
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Chapter 9, Problem 68QAP
To determine

(a)

The increase volume of diamond once it is removed from the chamber.

Expert Solution
Check Mark

Answer to Problem 68QAP

The increase volume of diamond once it is removed from the chamber is 2.1×108m3.

Explanation of Solution

Given:

Initial Pressure will act like volume stress P=FA=58×109N/m2

Bulk modulus of diamond B=4.43×1011N/m2

Let ΔV be the increase volume

Density of diamond ρ=3.52g/cm3

One carat mass m=0.200g

Volume of 2.8-carat diamond V=2.8×mρ=2.8×0.200g3.52g/cm3=0.16cm3=0.16×106m3

Formula used:

Bulk Modulus B=volumestressvolumestrain=F/AΔV/V

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

  B=F/AΔV/Vor,ΔV=F/AB/Vor,ΔV=58× 109N/m2( 4.43× 10 11 N/m2 )×(0.16× 10 6m3)or,ΔV=2.1×108m3

Hence, the increase volume of diamond once it is removed from the chamber is 2.1×108m3.

Conclusion:

Thus, the increase volume of diamond once it is removed from the chamber is 2.1×108m3.

To determine

(b)

The increase diamond radius.

Expert Solution
Check Mark

Answer to Problem 68QAP

The increase diamond radius is 1.71mm.

Explanation of Solution

Given:

The increase volume of diamond once it is removed from the chamber is 2.1×108m3.

Let Δrbe the increase radius of spherical diamond ball.

Formula used:

  Increase volume of spherical ball(ΔV)=43π(Δr)3

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

  (ΔV)=43π(Δr)3or,Δr= 3( ΔV ) 4×π3

  or,Δr= 3×( 2.1× 10 8 m3 ) 4×π3or,Δr=1.71×103m=1.71mm

Hence, the increase diamond radius is 1.71mm.

Conclusion:

Thus, the increase diamond radius is 1.71mm.

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