question_answer

A rubber of volume 2000 cc is alternately subjected to tension and released. The figure shows 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

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)  \[\left( \frac{\pi }{2}-1 \right)\times 32\times {{10}^{2}}J\]

Correct Answer: D

Solution :

[d] 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