A) \[v\propto {{x}^{\frac{1}{2}}}\]
B) \[v\propto x\]
C) \[v\propto {{x}^{-\frac{1}{2}}}\]
D) \[v\propto {{x}^{-1}}\]
Correct Answer: C
Solution :
| [c] \[\tan \theta =\frac{{{F}_{e}}}{mg}\simeq \theta \] |
| \[\frac{K{{q}^{2}}}{{{x}^{2}}mg}=\frac{x}{2\ell }\] |
| or \[\] ..(1) |
| or \[{{x}^{3/2}}\propto q\] ..(2) |
| differentiate eq.(i) w. r .t. time \[3{{x}^{2}}\frac{dx}{dt}\propto 2q\frac{dq}{dt}\] but \[\frac{dq}{dt}\] is constant so \[{{x}^{2}}(v)\propto q\]replace q from eq. (2) |
| \[{{x}^{2}}(v)\propto {{x}^{3/2}}\] |
| or \[\] |
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