T = 3av1−v
T = 1+v2a2v3
T = 2v(1−v)3−a4v32a3v3+(1−v)2
2v(1−v)3−a4v32a3v3−(1−v)3
Correct answer is D
av1−v = √2v+Ta+2T
(av)3(1−v)3 = 2v+Ta+2T
a3v3(13−v)3 = 2v+Ta+2T
= 2v(1−v)3−a4v32a3v3−(1−v)3
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