A monkey of mass `m` clings a rope to a slung over a fixed pulley .The opposite end of the rope is tried to a weight of mass `M` tying on a horizontal table is `mu` Find the acceleration of weight and the tension of the rope for two cases .The monkey move downward with respect to the rope with an acceleration b.

The tension of rope is

Options

A. `(Mm(mu g+g +b))/(M +m)`
B. `(Mm(mu g-g +b))/(M +m)`
C. `(Mm(mu g+g -b))/(M +m)`
D. `(Mm(mu g-g -b))/(M +m)`

Answer

c

Explanation

As the monkey move downward with respect to rope with an acceleration b. absulute acceleration is `a+ b`, where a is the acceleration of rope, Therefore equation of motion are
`mg - T = m(a+b)`…(i)
`T - mu Mg = Ma`…(ii)
Putting the value of T from Eqs(ii) and (i) we get
`(m - mu M)g = (M+m)a +mb`
`rArr (m(g - b) - mu Mg)/((M +m))= a`
`T = mu Mg+ (Mmg - Mmb - mu M^(2)g)/(M +m)`
`= (mu M^(2)g + mu Mmg + Mmg - Mmb- mu M^(2)g)/(M +m)`
`= (Mm(mu g +g - b))/(M +m)`