question_answer

Two masses \[{{M}_{1}}=5\text{ }kg,\text{ }{{M}_{2}}=10\text{ }kg\] 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: [AIPMT 2000]

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

\[\left( {{M}_{2}}g-T \right)+(T-\left( {{M}_{1}}g \right)=\left( {{M}_{1}}+{{M}_{2}} \right)a\]

\[\left( {{M}_{2}}-{{M}_{1}}g \right)=\left( {{M}_{1}}+{{M}_{2}} \right)a\]

\[\Rightarrow \]   \[a=\left( \frac{{{M}_{2}}-{{M}_{1}}}{{{M}_{1}}+{{M}_{2}}} \right)g\]

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.