A) \[\sqrt{\frac{2\text{b}}{\text{a}}}\]
B) \[\sqrt{\frac{\text{a}}{2\text{b}}}\]
C) \[\sqrt{\frac{\text{a}}{\text{b}}}\]
D) \[\sqrt{\frac{\text{b}}{\text{a}}}\]
Correct Answer: B
Solution :
| [b] \[\text{y = b}{{\text{x}}^{\text{2}}}\]. Differentiating w.r.t to t an both \[\frac{\text{dy}}{\text{dx}}=\text{b2x}\frac{\text{dx}}{\text{dt}}\Rightarrow {{\text{v}}_{y}}=2\text{bx}{{\text{v}}_{x}}\] |
| Again differentiating w.r.t to t on both sides we get \[\frac{\text{d}{{\text{v}}_{y}}}{\text{dt}}=2\text{b}{{\text{v}}_{\text{x}}}\frac{\text{dx}}{\text{dt}}+2\text{bx}\frac{\text{d}{{\text{v}}_{x}}}{\text{dt}}=2\text{bv}_{\text{x}}^{2}+0\] |
| [\[\frac{\text{d}{{\text{v}}_{x}}}{\text{dt}}=0,\] because the particle had constant acceleration along y-direction] |
| \[\text{Now, }\frac{\text{d}{{\text{v}}_{y}}}{\text{dt}}=\text{a=2bv}_{\text{x}}^{2};\] \[\text{v}_{\text{x}}^{2}=\frac{\text{a}}{2\text{b}}\Rightarrow {{\text{v}}_{x}}=\sqrt{\frac{\text{a}}{\text{2b}}}\] |
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