The correct option is
A g6Let the acceleration of pulley, block of mass
m and
2m be
ap,
a and
a2 respectively.
So, from the method of intercept we have
l1+l2=constant Differentiating w.r.t time, we get equation in the form of velocity i.e
.l1+.l2=0 Again differentiating w.r.t time, we get equation in the form of acceleration i.e
⇒ ap+ap−a=0 ⇒a=2ap ........... (1)
Also,
l3+l4=constant Double differentiating w.r.t time, we get equation in the form of acceleration i.e
.l3+.l4=0 ⇒−ap+a2=0 ⇒a2=ap...............(2)
From (1) & (2), we can say that
a=2a2 Now, from the FBD of
m, Pulley and
2m we have,
From the FBD of block of mass
m,
T−mgsinθ=m(2a2) As
θ=30∘ ⇒ T−mg2=2ma2..........(3)
From FBD of pulley,
T′=2T From the FBD of block of mass
2m,
2mg−T′=2ma2⇒ 2mg−2T=2ma2 ⇒ mg−T=ma2...........(4)
From (3) & (4), we have
mg−mg2=3ma2⇒a2=g6 Thus, acceleration of the mass
2m is
g6