A 70 kg man in sea is being lifted by a helicopter with the help of a rope , which can bear a maximum tension of 100 kg wt . With what maximum acceleration the helicopter should rise so that the rope does not break ? Take `g = 9.8 m//s^(2)` .

A 50 kg man stuck in flood is being lifted vertically by an army helicopter with the help of light rope which can bear a maximum tension of 70 kg-wt. The maximum acceleration with which helicopter can rise so that rope does not breaks is (g = 9.8 ms^(-2))

A 50 kg man stuck in flood is being lifted vertically by an army helicopter with the help of light rope which can bear a maximum tension of 70 kg-wt. The maximum acceleration with which helicopter can rise so that rope does not breaks is (g = 9.8 ms^-2)

A string tied on a roof can bear a maximum tension of 50 kg wt. The minimum acceleration that can be acquired by a man of 98 kg to descend will be [Take g=9.8 m//s^(2) ]

A string tied on a roof can bear a maximum tension of 50 kg wt. The minimum acceleration that can be acquired by a man of 98 kg to descend will be [take g = 9*8 m/s^(2) ]

A string tied on a roof can bear a maximum tension of 50 kg wt. The minimum acceleration that can be acquired by a man of 98 kg to descend will be [take g = 9*8 m/s^(2) ]

The strings of a parachute can bear a maximum tension of 72 kg wt . By what minimum acceleration can a person of 90 kg descend by means of this parachute ?

The strings of a parachute can bear a maximum tension of 72 kg wt. By what minimum acceleration can a person of 96 kg descend by means of this parachute?

A monkey of mass 30 kg climbs a rope which can withstand a maximum tension of 360 N. The maximum acceleration which this rope can tolerate for the climbing of monkey is: (g=10m//s^(2))

A fireman is sliding down with the help of a rope. If the breaking strength of the rope is ( 2mg)/( 3) where m is the mass of the man, the minimum acceleration with which the fireman can slide, so that the rope does not break, is :