A man slides down a light rope the breaking strength of which is β times his weight (β < 1). The maximum acceleration of man so that the rope just breaks is(A) βg(B) (1 - β)g(C) g/(1 + β)(D) g/(2 + β)

A man slides down a light rope the breaking strength of which is β

1.	A man slides down a light rope the breaking strength of which is β times his weight (β < 1). The maximum acceleration of man so that the rope just breaks is(A) βg(B) (1 - β)g(C) g/(1 + β)(D) g/(2 + β)
Answer» (B) The maximum acceleration of man so that the rope just breaks is (1 - β)g.

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A man slides down a light rope the breaking strength of which is β times his weight (β &lt; 1). The maximum acceleration of man so that the rope just breaks is(A) βg(B) (1 - β)g(C) g/(1 + β)(D) g/(2 + β)

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A man slides down a light rope the breaking strength of which is β times his weight (β < 1). The maximum acceleration of man so that the rope just breaks is(A) βg(B) (1 - β)g(C) g/(1 + β)(D) g/(2 + β)