In order to raise a mass of 100kg a man 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)`

Options

A. `1432N`
B. `928N`
C. `1218N`
D. `642N`

Answer

C

Explanation

`m_(m) =60kg, m_(B) = 100kg`
'a' be acceleration of rope
`a_(rel) = (5g)/(4),a_(rel) =a_(m) + a, a_(m) = a_(rel) -a`
`a_(m) = (5g)/(4) -a`
`T -m_(B) g =m_(B) a rarr (1)`
`T -m_(m) g =m_(2) a_(m) rarr (2)`
solving (1) & (2) we get
`T -100g =100a----(3)`
`T -60g = 60((5g)/(4) -a)----(4)`
`(3) -(4), -40 g = 100a -75g +60a`
`160a =350 rArr a = (35)/(16) ms^(-2)`
`T =100g + 100a =1000 + 1000 +100xx (35)/(16) =1218N`
.