A) \[{{\text{t}}_{\text{2}}}\text{=-}{{\text{t}}_{\text{1}}}\text{+}\frac{\text{2}}{{{\text{t}}_{\text{1}}}}\]
B) \[{{\text{t}}_{\text{2}}}\text{=-}{{\text{t}}_{\text{1}}}-\frac{\text{2}}{{{\text{t}}_{\text{1}}}}\]
C) \[{{\text{t}}_{\text{2}}}\text{=}{{\text{t}}_{\text{1}}}-\frac{\text{2}}{{{\text{t}}_{\text{1}}}}\]
D) \[{{\text{t}}_{\text{2}}}\text{=}{{\text{t}}_{\text{1}}}+\frac{\text{2}}{{{\text{t}}_{\text{1}}}}\]
Correct Answer: B
Solution :
Equation of the normal at point \[{{C}_{6}}{{H}_{6}}\] on parbola is \[=\frac{10}{60}=\frac{1}{6}\] It also passes through \[=8\times \frac{1}{8}+6\times \frac{1}{2}=4\]then \[=12\times \frac{1}{4}=3\] \[\text{C}{{\text{u}}_{\text{4}}}\text{A}{{\text{g}}_{\text{3}}}\text{Au}\text{.}\] \[-nF{{E}^{\circ }}_{cell}=-RT\] \[{{E}^{\circ }}_{cell}=\frac{RT}{nF}\] \[{{E}^{\circ }}_{cell}\] \[=\frac{1}{2}\frac{RT}{F}\] \[\begin{align} & N{{H}_{4}}Cl\xrightarrow{\Delta }N{{H}_{3}}+HCl \\ & ABC \\ \end{align}\] \[\begin{align} & N{{H}_{3}}+HCl\to N{{H}_{4}}Cl \\ & \text{Bwhite fumes} \\ \end{align}\] \[\begin{align} & N{{H}_{3}}+2{{K}_{2}}\left[ Hg{{I}_{4}} \right]+3KOH\to \\ & B\text{Nessler }\!\!'\!\!\text{ sreagent} \\ \end{align}\] \[\begin{align} & {{H}_{2}}NHgO.HgI+7KI+{{H}_{2}}O \\ & \text{brown ppt}\text{.} \\ & \text{iodineofMillion }\!\!'\!\!\text{ sbase} \\ \end{align}\] \[\begin{align} & HCl+MN{{O}_{3}}\to MCl+HN{{O}_{3}} \\ & \text{Cwhite ppt}\text{.} \\ \end{align}\] \[\left( M=A{{g}^{+}},p{{b}^{+}},H{{g}^{+}} \right)\] \[194u=\frac{28.9}{100}\times 194=56u\]
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