Tension in the rope at the rigid support is `(g=10m//s^(2))`

Two persons are holding a rope of negligible mass horizontally. A 20 kg mass is attached to the rope at the midpoint, as a result the rope deviates from the horizontal direction. The tension required to completely straighten the rope is (g=10m//s^(2))

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 lift of mass 1000 kg is going up. The variation in the speed of lift is shown in figure. Then the minimum tension in the rope will be (g=10 m//s^(2))

A lift of mass 1000 kg is going up. The variation in the speed of lift is shown in figure. Then the minimum tension in the rope will be (g=10 m//s^(2))

In order to raise a mass of 100kg a of mass 60kg fastens a rope to it and passes the rope over a smooth pulley. He climbs the rope with acceleration 5g//4 relative to the rope. The tension in the rope is: Take g=10m//s^(2)

A chain of length 'L' and mass 'M' is hanging by fixing its upper end to a rigid support. The tension in the chain at a distance 'x' from the rigid support is

A chain of mass M and length L is held vertical by fixing its upper end to a rigid support. The tension in the chain at a distance y from the rigid support is:

A chain of length 'L' and mass 'M' is hanging by fixing its upper end to a rigid support. The tension in the chain at a distance 'x' from the rigid support is