A) \[\sqrt{\frac{mk}{2}}\,{{t}^{{}^{1}/{}_{2}}}\]
B) \[\sqrt{mk}\,{{t}^{{}^{-1}/{}_{2}}}\]
C) \[\sqrt{2mk}\,{{t}^{{}^{-1}/{}_{2}}}\]
D) \[\frac{1}{2}\sqrt{mk}\,{{t}^{{}^{-1}/{}_{2}}}\]
Correct Answer: D
Solution :
| As the machine delivers a constant power |
| So F. v = constant \[=k\] (watts) |
| \[\Rightarrow \] \[m\frac{dv}{dt}.v=k\] |
| \[\Rightarrow \] \[\int{vdv}=\frac{k}{m}\int{dt}\] |
| \[\Rightarrow \] \[\frac{{{v}^{2}}}{2}=\frac{k}{m}t\Rightarrow c=\sqrt{\frac{2k}{m}t}\] |
| Now force on the particle is given by |
| \[F=m\frac{dv}{dt}=m\frac{d}{dt}{{\left( \frac{2kt}{m} \right)}^{\frac{1}{2}}}\] |
| \[=\sqrt{2km}.\,\left( \frac{1}{2}{{t}^{-\frac{1}{2}}} \right)\] |
| \[=\sqrt{\frac{mk}{2}}.{{t}^{-\frac{1}{2}}}\] |
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