A) \[{{t}^{-1}}\]
B) \[{{t}^{-1/2}}\]
C) \[{{t}^{1/2}}\]
D) t
Correct Answer: B
Solution :
Here.KE\[=\frac{1}{2}m{{v}^{2}}=kt\]where, k is a constant Differentiating w.r.t. t, we get \[\frac{1}{2}m2v\frac{dv}{dt}=k\] or \[m\frac{dv}{dt}=\frac{k}{v}\] or \[F\frac{k}{v}\] but \[v=\sqrt{\frac{2kt}{m}}\] \[\therefore \] \[F=\frac{k}{\sqrt{\frac{2kt}{m}}}\] or \[F\propto {{t}^{-1/2}}\]
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