A) Zero
B) \[\text{1}0\text{m}{{\text{s}}^{\text{-1}}}\]
C) \[{{\cos }^{-1}}\left( \frac{3}{4} \right)\]
D) \[{{\sin }^{-1}}\left( \frac{3}{4} \right)\]
Correct Answer: A
Solution :
Redrawing the diagram. For block B of mass 2m \[\text{31}%\text{M}{{\text{g}}^{+}}+\text{69}%\text{M}{{\text{g}}^{\text{2}+}}\] \[C{{H}_{3}}CHO+N{{H}_{2}}.N{{H}_{2}}\to A\]\[\xrightarrow{B}C{{H}_{3}}C{{H}_{3}}+{{N}_{2}}\] \[C{{H}_{3}}CH=NN{{H}_{2}}\,and\,{{C}_{2}}{{H}_{5}}ONa\]\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{N}{{\text{H}}_{\text{2}}}\text{ and}\,{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{ONa}\] As \[C{{H}_{3}}-NH-NH-C{{H}_{3}}\,and\,{{C}_{2}}{{H}_{5}}OH\] then \[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{N}{{\text{H}}_{\text{2}}}\text{ and }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH}\] \[N{{a}^{+}}\] the masses will not move. So, acceleration of the system will be zero.You need to login to perform this action.
You will be redirected in
3 sec