Gravitation Question 216
Question: A rubber of volume 2000 cc is alternately subjected to tension and released. The The stress-strain curve of rubber. Each curve is a quadrant of an ellipse. The. amount of energy lost as heat per cycle per unit volume will be
Options:
A)$ [\left( \frac{\pi }{2}-1 \right)\times 16\times 10^{2}J] $
B) $ [\left( \frac{\pi }{4}-1 \right)\times 8\times 10^{2}J] $
C)$ [\left( \frac{\pi }{4}-1 \right)\times 32\times 10^{2}J] $
D)$ [\left( \frac{\pi }{2}-1 \right)\times 32\times 10^{2}J] $
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Answer:
Correct Answer: D
Solution:
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Hystersis loss corresponding to elasticity per unit volume of a substance is given by the area of hysteresis loop, i.e., stress-strain curve corresponding to one complete loading and deloading. Area of an ellipse = $ [\pi \times ] $ semi-major axis x semi-minor axis
$ [A_{1}=\frac{1}{4}(\pi \times 8\times 4\times 10^{2})] $ And$ [A_{2}=\frac{1}{4}(\pi \times 8\times 4\times 10^{2})] $
Also, $ [A_{3}=8\times 4\times 10^{2}] $ Area of hysteresis loop is $ [A=A_{1}+A_{2}-A_{3}] $
$ [A=2\left[ \frac{\pi }{4}\times 8\times 4\times 10^{2} \right]-[8\times 4\times 10^{2}]] $
$ [=\left[ \frac{\pi }{2}-1 \right]\times 32\times 10^{2}J] $ = work done per cycle = energy lost per cycle per unit volume
BETA
