A) \[\sqrt{\frac{2b}{a}}\]
B) \[\sqrt{\frac{a}{2b}}\]
C) \[\sqrt{\frac{a}{b}}\]
D) \[\sqrt{\frac{b}{a}}\]
Correct Answer: B
Solution :
\[y=b{{x}^{2}}\] \[\frac{dy}{dt}=2bx\frac{dx}{dt}\] \[\frac{{{d}^{2}}y}{d{{t}^{2}}}=2b\left[ x\frac{{{d}^{2}}x}{d{{t}^{2}}}+{{\left( \frac{dx}{dt} \right)}^{2}} \right]\] The acceleration of the particle is along y-axis so, its acceleration along the z-axis is 0, therefore, \[\frac{{{d}^{2}}x}{d{{t}^{2}}}=0\] \[\therefore \] \[\frac{{{d}^{2}}y}{d{{t}^{2}}}=2b{{\left( \frac{dx}{dx} \right)}^{2}}\] or, \[a=2v\,v_{x}^{2}\,\,\Rightarrow \,\,{{v}_{x}}=\sqrt{\frac{a}{2b}}\] Hence, the correction option is [b].
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