COLLEGE PHYSICS,VOLUME 1
COLLEGE PHYSICS,VOLUME 1
2nd Edition
ISBN: 9781319115104
Author: Freedman
Publisher: MAC HIGHER
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Chapter 9, Problem 69QAP
To determine

(a)

The volume of marble shrinks in the cylinder.

Expert Solution
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Answer to Problem 69QAP

The volume of marble shrinks in the cylinder is 0.133×108m3.

Explanation of Solution

Given:

Radius of glass marble r=0.500×102m

Level of mercury in cylinder h=20.0×102m

Atmospheric Pressure at the surface P0=1.0×105N/m2

Acting Pressure at the bottom (volume stress) P=FA=2.7×104N/m2+1.0×105N/m2=1.27×105N/m2

Bulk modulus of glass B=50×106N/m2

Let ΔV be the shrink volume

Original volume of glass marble V=43πr3=43π(0.500× 10 2m)3=0.524×106m3

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/V

  or,ΔV=1.27× 105N/m2( 50× 106 N/m2 )×(0.524× 10 6m3)or,ΔV=0.133×108m3

Hence, the volume of marble shrinks in the cylinder is 0.133×108m3.

Conclusion:

Thus, the volume of marble shrinks in the cylinder is 0.133×108m3.

To determine

(b)

The change in the radius of glass marble

Expert Solution
Check Mark

Answer to Problem 69QAP

The change in the radius of glass marble is 0.68mm.

Explanation of Solution

Given:

The volume of marble shrinks in the cylinder is ΔV=0.133×108m3.

Let Δrbe the change in the radius of glass marble.

Formula used:

  shrink volume of glassmarble(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×( 0.133× 10 8 m3 ) 4×π3or,Δr=0.32× 10 9m3or,Δr=0.68×103m=0.68mm

Hence, the change in the radius of glass marble is 0.68mm.

Conclusion:

Thus, the change in the radius of glass marble is 0.68mm.

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COLLEGE PHYSICS,VOLUME 1

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