In the figure shown, A & B are free to mo | vedclass

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Question

$\mathrm{ma}_{0} \sin 	heta+\mathrm{N}=\operatorname{mg} \cos 	heta$

$\Rightarrow \mathrm{N}=\operatorname{mg} \cos 	heta-\mathrm{ma}_{0} \sin \

Vedclass · Accepted Answer

the acceleration of $A$ will be more than $g\ sin\ 	heta $

Vedclass · Answer

$\mathrm{ma}_{0} \sin 	heta+\mathrm{N}=\operatorname{mg} \cos 	heta$

$\Rightarrow \mathrm{N}=\operatorname{mg} \cos 	heta-\mathrm{ma}_{0} \sin 	heta$

$\Rightarrow \mathrm{N}<\mathrm{mg} \cos 	heta$

Hence, $(D)$ is true.

$m a_{0} \cos 	heta+m g \sin 	heta=m a$

$\Rightarrow \mathrm{a}=\mathrm{g} \sin 	heta+\mathrm{a}_{0} \cos 	heta$

Hence acceleration of $A$

$=\sqrt{\left(a-a_{0} \cos 	hetaight)^{2}+\left(a_{0} \sin 	hetaight)^{2}}>g \sin 	heta$

In the figure shown, $A$ & $B$ are free to move. All the surfaces are smooth. then $(0 < \theta < 90^o)$

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