Speed $v$ of a particle moving along a straight line, when it is at a distance $x$ from a fixed point on the line is given by $v^2 = 108 - 9x^2$ (all quantities in $S.I.$ unit). Then
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$V =3 \sqrt{12- x ^2}$
Thus, $a =\frac{ dv }{ dt }=\frac{ dv }{ dx } \cdot \frac{ dx }{ dt }$ by chain rule
$\text { or, } a=v \frac{d v}{d x}=\left(3 \sqrt{12-x^2}\right) \cdot(-6 x) \cdot 1 / 2 \cdot\left(12-x^2\right)^{-1 / 2}$
At $x =3, a =3 \cdot \sqrt{3} \cdot(-18) / 2 \cdot \frac{1}{\sqrt{3}}=-27 m / s ^2$
Thus magnitude $=27$, hence $B$ is correct.
$A$ is wrong because acceleration is not uniform, but dependent on $x$. D is incorrect because maximum displacement means $v =0$, which gives $O =$ $12-x^2$ or $x=\sqrt{12}$
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