A system consists of three masses `m_(1),m_(2) " and " m_(3)` connected by a string passing over a pulley P. The mass `m_(1)` hangs freely and `m_(2) " and " m_(3)` are on a rough horizontal table (the coefficient of friction=`mu`). The pulley is frictionless and of negligible mass. The downward acceleration of mass `m_(1)` is (assume, `m_(1)=m_(2)=m_(3)=m`)

Options

A. `(g(1-gmu))/(9)`
B. `(2gmu)/(3)`
C. `(g(1-2 mu))/(3)`
D. `(g(1-2 mu))/(2)`

Answer

C

Explanation

First of all consider the forces on the blocks

For the Ist block, `(because m_(1)=m_(2)=m_(3))`
`mg-T_(1)=m xx a` ….(i)
`rArr` Let us consider 2nd and 3rd blocks as a system.
So, `T_(1)-2 mu mg=2m xx a` ....(ii)
On solving Eqs. (i) and (ii), we get
`rArr mg-T_(1)=m xx a`
`T_(1)-2mu mg = 2m xx a`
`mg(1-2mu)=3m xx a rArr a=(g)/(3)(1-2mu)`