question_answer

                Two masses \[{{M}_{1}}=5\,\,kg,\,\,{{M}_{2}}=10\,kg\] are connected at the ends of an inextensible string passing over a frictionless pulley as shown. When masses are released, then acceleration of masses will be:                                                                                                                                                                                                

A)                 g                             

B)                 \[\frac{g}{2}\]                   

C)                 \[\frac{g}{3}\]   

D)                 \[\frac{g}{4}\]

Correct Answer: C

Solution :

                                In the case of masses hanging from a pulled a string, the tension in whole string is same say equal to T.                 As \[{{M}_{2}}>{{M}_{1}},\] so mass \[{{M}_{2}}\] moves down and mass \[{{M}_{1}}\] moves up with the same acceleration a (say). The arrangement of the motion is represented in the figure.                 Equation of motion of mass \[{{M}_{2}},\] is                 \[{{M}_{2}}g-T={{M}_{2}}a\]                 Equation of motion of mass M, is,                 \[T-{{M}_{1}}g={{M}_{1}}a\]                 Adding Eqs. (i) and (ii), we get                 \[({{M}_{2}}g-T)\,+(T-{{M}_{1}}g)=({{M}_{1}}+{{M}_{2}})a\]                 \[({{M}_{2}}-{{M}_{1}})g\,=({{M}_{1}}\,+{{M}_{2}})a\]                 \[\Rightarrow a=\left( \frac{{{M}_{2}}-{{M}_{1}}}{{{M}_{1}}+{{M}_{2}}} \right)g\]                 Given, \[{{M}_{1}}=5\,kg,\,{{M}_{2}}=10\,kg\]                 Hence,   \[a=\left( \frac{10-5}{5+10} \right)g=\frac{5}{15}g=\frac{g}{3}\]                 Note:    In a mass-pulley system, the tension in the string is always towards the pulley.