A) \[k=\pm 1\]
B) \[k=\pm \,a\]
C) \[k=\pm \,\,\frac{1}{a}\]
D) \[k=\pm \,\,2\]
Correct Answer: A
Solution :
Since, given triangle is right angled isosceles triangle. \[{{F}_{net}}=\frac{{{q}^{2}}\sqrt{2}}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}+\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{(\sqrt{2}a)}^{2}}}\] Angle between two given lines = \[45{}^\circ \] \[{{F}_{net}}=\left( \frac{1+2\sqrt{2}}{2} \right)\,\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\] \[l=\frac{V}{R}=\frac{9}{9}\,=1\,A\] Where, \[R=9\,\Omega \] and \[mg=kRv=\frac{4}{3}\,\pi {{R}^{3}}\rho g=kR{{v}_{T}}\] From Eq. (i). \[{{v}_{T}}=\frac{4}{3}\,\pi \frac{{{R}^{2}}\rho g}{k}\] \[{{C}_{4}}{{H}_{10}}O\] \[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}OH\] \[C{{H}_{3}}-C{{H}_{2}}-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{3}}\] \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,H-C{{H}_{2}}-OH\]
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