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On block

$Mg−T=ma_{1}$..........(1)

On cylinder

$a_{2}=∝R$

We so at $P$

Cylinder is climbing up and string is moving down

$→a_{1}=∝R−a_{2}sinθ$

$=a_{2}(1−sinθ)$

$a_{2}=a_{1}/(1−sinθ)$

from $τ=I∝$ at $Q$

$(Tcosθ−mgsinθ)×R=23mR_{2} ∝$

$Tcosθ−mgsinθ=23ma_{2} $

From (1) and (2)

$Mgcosθ−mgsinθ=Ma_{1}cosθ+23ma_{2} $

$(Mgcosθ−mgsinθ)=Ma_{1}cosθ+23m (1−sinθ)a_{1} $

$a_{1}=2Mcosθ(1−sinθ)+3m2(Mgcosθ−mgsinθ)(1−sinθ) $

$a_{1}=2Mcosθ(1−sinθ)+3m2g(Mcosθ−msinθ)(1−sinθ) $

$a_{1}>0 $

i.e

$Mcosθ>msinθ$

$M>mtanθ=2(43 )$

$M>23 kg=1.5kg $

Solve any question of Laws of Motion with:-

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