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 volumeYou need to login to perform this action.
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