question_answer

Two masses \[{{M}_{1}}\] and \[{{M}_{2}}\] are attached to the ends of a string which pass over a pulley attached to the top of a double inclined plane of angles of inclination \[\alpha \] and \[\beta \]. If \[{{M}_{2}}>{{M}_{1}},\] the acceleration a of the system is given by

A)  \[\frac{{{M}_{2}}(\sin \beta )g}{{{M}_{1}}+{{M}_{2}}}\]

B)                  \[\frac{{{M}_{1}}(\sin \alpha )g}{{{M}_{1}}+{{M}_{2}}}\]

C)                  \[\left( \frac{{{M}_{2}}\sin \beta -{{M}_{1}}\sin \alpha }{{{M}_{1}}+{{M}_{2}}} \right)g\]

D)  zero

Correct Answer: C

Solution :

    Let \[{{M}_{2}}\] moves downwards with an acceleration a and the tension in the string is T.             \[\therefore \]     \[{{M}_{2}}g\,\sin \beta -T={{M}_{2}}a\]                              ?.(i)                                 \[T-{{M}_{1}}g\,\sin \alpha ={{M}_{1}}a\]                   ?...(ii) Adding (i) and (ii), we get \[{{M}_{2}}g\,\sin \beta -{{M}_{1}}g\,\sin \alpha =({{M}_{2}}+{{M}_{1}})\,a\] \[a=g\frac{({{M}_{2}}\,\sin \beta -{{M}_{1}}\sin \alpha )}{({{M}_{2}}+{{M}_{1}})}\].