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4-1.Newton's Laws of Motion
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In the figure shown, $A$ & $B$ are free to move. All the surfaces are smooth. then $(0 < \theta < 90^o)$
Athe acceleration of $A$ will be more than $g\ sin\ \theta $
Bthe acceleration of $A$ will be less than $g\ sin\ \theta $
Cnormal force on $A$ due to $B$ will be more than $mg\ cos\ \theta $
Dnormal force on $A$ due to $B$ will be equal to $mg\ cos\ \theta $
Solution
$\mathrm{ma}_{0} \sin \theta+\mathrm{N}=\operatorname{mg} \cos \theta$
$\Rightarrow \mathrm{N}=\operatorname{mg} \cos \theta-\mathrm{ma}_{0} \sin \theta$
$\Rightarrow \mathrm{N}<\mathrm{mg} \cos \theta$
Hence, $(D)$ is true.
$m a_{0} \cos \theta+m g \sin \theta=m a$
$\Rightarrow \mathrm{a}=\mathrm{g} \sin \theta+\mathrm{a}_{0} \cos \theta$
Hence acceleration of $A$
$=\sqrt{\left(a-a_{0} \cos \theta\right)^{2}+\left(a_{0} \sin \theta\right)^{2}}>g \sin \theta$
$\Rightarrow \mathrm{N}=\operatorname{mg} \cos \theta-\mathrm{ma}_{0} \sin \theta$
$\Rightarrow \mathrm{N}<\mathrm{mg} \cos \theta$
Hence, $(D)$ is true.
$m a_{0} \cos \theta+m g \sin \theta=m a$
$\Rightarrow \mathrm{a}=\mathrm{g} \sin \theta+\mathrm{a}_{0} \cos \theta$
Hence acceleration of $A$
$=\sqrt{\left(a-a_{0} \cos \theta\right)^{2}+\left(a_{0} \sin \theta\right)^{2}}>g \sin \theta$
Standard 11
Physics
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