question_answer

consider the diagram shown below in which two masses of m and 2m are placed on a fixed triangular wedge. The coefficient of friction between block A and the wedge is 2/3, while that for block B and the wedge is 1/3. If the whole system is released from rest, then acceleration of block A is

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.