A man slides down a light rope whose breaking strength is η times his weight (η < 1). What should be his maximum acceleration so that the rope just breaks?(a) ηg(b) g(1 - η) (c) g / 1 + η(d) g / 2 - η

A man slides down a light rope whose breaking strength is η times hi

1.	A man slides down a light rope whose breaking strength is η times his weight (η < 1). What should be his maximum acceleration so that the rope just breaks?(a) ηg(b) g(1 - η) (c) g / 1 + η(d) g / 2 - η
Answer» Correct answer is: (b) g(1 - η)

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A man slides down a light rope whose breaking strength is η times his weight (η &lt; 1). What should be his maximum acceleration so that the rope just breaks?(a) ηg(b) g(1 - η) (c) g / 1 + η(d) g / 2 - η

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A man slides down a light rope whose breaking strength is η times his weight (η < 1). What should be his maximum acceleration so that the rope just breaks?(a) ηg(b) g(1 - η) (c) g / 1 + η(d) g / 2 - η