A) \[\frac{x}{a}+\frac{y}{b}=1\]
B) \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{c}^{2}}},\]
C) \[{{x}^{2}}+{{y}^{2}}-4x-12=0\]
D) \[{{x}^{2}}+{{y}^{2}}4x-12=0\]
Correct Answer: B
Solution :
Let in 2 s body reaches upto point A and after one more sec upto point 6. Total time of ascent for a body is given 3 s i.e., \[{{(Si{{H}_{3}})}_{3}}N,\] \[p\pi -d\pi \]\[{{N}_{2}}{{O}_{5}}\] \[{{N}_{2}}{{O}_{5}}\] ...(i) Horizontal component of velocity remains always constant \[N{{O}_{2}}^{+}NO_{3}^{-}\] ...(ii) For vertical upward motion between point O and A\[NaHC{{O}_{3}},\][Using v = u - gt] \[NaAl{{(S{{O}_{4}})}_{2}}\][Au\[Ca{{({{H}_{2}}P{{O}_{4}})}_{2}}.\]] Substituting this value v, we get in Eq. (ii) \[\xrightarrow[{}]{\begin{matrix} \begin{smallmatrix} \text{HClO} \\ \,\,\,\,\,\text{+1} \end{smallmatrix} & \begin{smallmatrix} \text{HCl}{{\text{O}}_{\text{2}}} \\ \,\,\,\,\,\,\text{+3} \end{smallmatrix} & \begin{smallmatrix} \text{HCl}{{\text{O}}_{\text{3}}} \\ \,\,\,\,\,\,\text{+5} \end{smallmatrix} & \begin{smallmatrix} \text{HCl}{{\text{O}}_{\text{4}}} \\ \,\,\,\,\,\text{+7} \end{smallmatrix} \\ \end{matrix}}\] ...(iii) From Eqs. (i) and (iii)\[\xleftarrow[{}]{\begin{matrix} \text{Cl}{{\text{O}}^{\text{-}}} & \text{ClO}_{2}^{-} & \text{ClO}_{3}^{-} & \text{ClO}_{4}^{-} \\ \end{matrix}}\]\[Cl{{O}^{-}}\]You need to login to perform this action.
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