question_answer

The kinetic energy & of a particle moving along a circle of radius R depends on the distance covered. It is given as \[8\sqrt{3}\]where a is a constant. The force acting on the particle is

A) \[4\sqrt{2}\]                      

B) \[16\sqrt{3}\]

C)  2as                                       

D) \[16\sqrt{2}\]

Correct Answer: B

Solution :

In non-uniform circular motion two forces will work on a particle \[\underset{-3}{\mathop{NH_{4}^{+}}}\,\underset{+5}{\mathop{NO_{3}^{-}}}\,\xrightarrow[{}]{{}}\underset{+1}{\mathop{{{N}_{2}}O}}\,+2{{H}_{2}}O\] and \[\underset{-3}{\mathop{NH_{4}^{+}}}\,\underset{+3}{\mathop{NO_{2}^{-}}}\,\xrightarrow[{}]{{}}\underset{0}{\mathop{{{N}_{2}}}}\,\] So, the net force\[\underset{+5-1}{\mathop{PC{{l}_{5}}}}\,\xrightarrow[{}]{{}}\underset{+3-1}{\mathop{PC{{l}_{3}}}}\,+\underset{0}{\mathop{C{{l}_{2}}}}\,\]           ?(i) Centripetal force \[F{{e}^{2+}}+{{e}^{-}}\xrightarrow[{}]{{}}F{{e}^{+}}\]                ... (ii) [given \[NO\xrightarrow[{}]{{}}N{{O}^{+}}+{{e}^{-}}\] given] Again from \[F{{e}^{+}}\]\[F{{e}^{2+}}\]\[N{{O}^{+}}\] \[Li-Mg\to \]\[Ca-Mg\to \] Tangential acceleration \[{{r}_{n}}\] \[{{A}^{3+}}\]\[{{B}^{2-}}\] \[{{A}_{2}}{{B}_{3}}.\]and \[\left( p+\frac{a}{{{V}^{2}}} \right)(V-b)=\]                                   ...(iii) Now, on substituting value of \[\times c{{m}^{5}}\] and \[\times c{{m}^{4}}\] in Eq.(i), we get \[/c{{m}^{3}}\]\[/c{{m}^{2}}\]