A) \[2s\left( \frac{1}{f}+\frac{1}{r} \right)\]
B) \[\frac{2s}{\frac{1}{f}+\frac{1}{r}}\]
C) \[\sqrt{2s\,(f+r)}\]
D) \[\sqrt{2s\,\left( \frac{1}{f}+\frac{1}{r} \right)}\]
Correct Answer: D
Solution :
Here, \[{{x}_{1}}+{{x}_{2}}=s\] and \[{{t}_{1}}+{{t}_{2}}=t\] When a particle moves from A to C, \[{{v}^{2}}={{u}^{2}}+2f{{x}_{1}}\] \[{{v}^{2}}=0+2f{{x}_{1}}\] \[\Rightarrow \] \[{{x}_{1}}=\frac{{{v}^{2}}}{2f}\] ... (i) and \[v=u+ft\] \[\Rightarrow \] \[v=0+f{{t}_{1}}\] \[\Rightarrow \] \[{{t}_{1}}=\frac{v}{f}\] ... (ii) From C to \[{{v}^{2}}={{u}^{2}}+2fs\] \[v{{'}^{2}}={{v}^{2}}-2r{{x}_{2}}\] \[0={{v}^{2}}-2r{{x}_{2}}\] \[\Rightarrow \] \[{{x}_{2}}=\frac{{{v}^{2}}}{2r}\] ?. (iii) and \[v'=v-r{{t}_{2}}\] \[\Rightarrow \] \[0=v-r{{t}_{2}}\] \[\Rightarrow \] \[{{t}_{2}}=\frac{v}{r}\] ... (iv) On adding Eqs. (i) and (iii), we get \[{{x}_{1}}+{{x}_{2}}=\frac{{{v}^{2}}}{2}\left( \frac{1}{f}+\frac{1}{r} \right)\] \[\Rightarrow \] \[2s={{v}^{2}}\left( \frac{1}{f}+\frac{1}{f} \right)\] ... (v) On adding Eqs. (ii) and (iv), we get \[{{t}_{1}}+{{t}_{2}}=v\left( \frac{1}{f}+\frac{1}{r} \right)\] \[\Rightarrow \] \[t=v\left( \frac{1}{f}+\frac{1}{r} \right)\] \[\Rightarrow \] \[{{t}^{2}}={{v}^{2}}{{\left( \frac{1}{f}+\frac{1}{r} \right)}^{2}}\] ?. (vi) On dividing Eq. (vi) by Eq. (v), we get \[\frac{{{t}^{2}}}{2\,s}=\frac{{{v}^{2}}{{\left( \frac{1}{f}+\frac{1}{r} \right)}^{2}}}{{{v}^{2}}\left( \frac{1}{f}+\frac{1}{r} \right)}\] \[\Rightarrow \] \[\frac{{{t}^{2}}}{2\,s}=\left( \frac{1}{f}+\frac{1}{r} \right)\] \[\Rightarrow \] \[t=\sqrt{2\,s\left( \frac{1}{f}+\frac{1}{r} \right)}\]You need to login to perform this action.
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