COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
COLLEGE PHYSICS-ACHIEVE AC (1-TERM)
3rd Edition
ISBN: 9781319453916
Author: Freedman
Publisher: MAC HIGHER
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Chapter 9, Problem 52QAP
To determine

(a)

The change in volume of the sphere

Expert Solution
Check Mark

Answer to Problem 52QAP

The change in volume of the sphere is 2.38×105m3.

Explanation of Solution

Given:

Original radius of sphere r0=5.00cm=5.00×102m

Original volume of sphere V0=43πr03=43π(5.00× 10 2m)3=5.24×104m3

Bulk Modulus of copper B=140×109Nm2

Surface Area of sphere A=4πr02=4π×(5.00× 10 2m)2=3.14×102m2

Let the change in volume be ΔV

Applied force F=2.00×108N

Formula used:

Bulk Modulus, B=F/AΔV/V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

  ΔV=FB×V0AΔV=2.00× 108N140× 109Nm 2×5.24× 10 4m33.14× 10 2m2ΔV=2.38×105m3

Hence, the change in volume of the sphere is 2.38×105m3.

Conclusion:

Thus, the change in volume of the sphere is 2.38×105m3.

To determine

(b)

The radius of the sphere.

Expert Solution
Check Mark

Answer to Problem 52QAP

The final radius of the sphere is 5.08cm.

Explanation of Solution

Given:

Original radius of sphere r0=5.00cm=5.00×102m

Original volume of sphere V0=43πr03=43π(5.00× 10 2m)3=5.24×104m3

Let the final radius of the sphere be rf

Let the final volume of the sphere be Vf

The change in volume of the sphere is ΔV=2.38×105m3.

Formula used:

Final volume of the sphere Vf=ΔV+V0

Here, all alphabets are in their usual meanings.

Calculation:

Substituting the given values in the formula,

  Vf=ΔV+V0Vf=2.38×105m3+5.24×104m3Vf=0.238×104m3+5.24×104m3Vf=5.48×104m3

But final volume of sphere Vf=43πrf3=5.48×104m3

So, final radius of the sphere

  rf=3 4π×5.48× 10 4m33rf=3 4π×5.48× 10 4m33rf=130.8× 10 6m33rf=5.08×102mrf=5.08cm

Hence, the final radius of the sphere is 5.08cm.

Conclusion:

Thus, the final radius of the sphere is 5.08cm.

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Chapter 9 Solutions

COLLEGE PHYSICS-ACHIEVE AC (1-TERM)

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