A) \[f\propto v\]
B) \[f\propto {{v}^{4}}\]
C) \[f\propto {{v}^{2}}\]
D) \[f\propto {{v}^{3}}\]
Correct Answer: C
Solution :
On differentiating gives equation, we get \[dt=(2px+q)dx\] \[\therefore \] \[v=\frac{dx}{dt}=\frac{1}{2px+q}\] ?(i) \[\Rightarrow \] \[\frac{dv}{dt}=f=\frac{-2p}{{{(2px+q)}^{2}}}=-2p{{v}^{2}}\] [from Eq.(i)] So, \[f\propto {{v}^{2}}\]You need to login to perform this action.
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