question_answer

A light string passing over a smooth light pulley connects two blocks of masses \[{{m}_{1}}\] and \[{{m}_{2}}\] (vertically). If the acceleration of the system is g/8, then the ratio of the masses is   AIEEE  Solved  Paper-2002

A) \[8:1\]     

B)                           \[9:7\]  

C) \[4:3\]                     

D)           \[5:3\]

Correct Answer: B

Solution :

As the string is inextensible, both masses have the same acceleration a. Also, the pulley is massless and frictionless, hence the tension at both ends of the string is the same. Suppose, the     mass, \[{{m}_{2}}\] is greater than mass \[{{m}_{1}}\], so the heavier mass is accelerated downward and the lighter mass, \[{{m}_{2}}\] is accelerated upwards,                              Therefore, by Newton's 2nd law,              \[T-{{m}_{1}}g={{m}_{1}}a\]                                           ... (i)              \[{{m}_{2}}g-T={{m}_{2}}a\]                                           ... (ii)              After solving Eqs. (i) and (ii),              \[a=\frac{({{m}_{2}}(1-{{m}_{1}}/{{m}_{2\grave{\ }}})}{({{m}_{1}}+{{m}_{2}})}.g=\frac{g}{8}\]        (given)              So         \[\frac{g}{8}=\frac{{{m}_{2}}(1-{{m}_{1}}/{{m}_{2}})}{{{m}_{2}}(1-{{m}_{1}}/{{m}_{2}})}.g\]              .... (iii)              Let        \[\frac{{{m}_{1}}}{{{m}_{2}}}=x\]              Thus, Eq. (iii) becomes              \[\frac{1-x}{1+x}=\frac{1}{8}\]                      or           \[x=\frac{7}{9}\] or  \[\frac{{{m}_{2}}}{{{m}_{1}}}=\frac{9}{7}\] So, the ratio of the masses is \[9:7\].