Question

Two particles, having masses in the ratio $n:1$, are interconnected by a light inextensible string that passes over a smooth pulley. If the system is released from rest, the acceleration of the centre of mass of the system is,

• A. $\left(\frac{n-1}{n+1}\right)g$
• B. ${\left(\frac{n+1}{n-1}\right)}^{2}g$
• C. ${\left(\frac{n-1}{n+1}\right)}^{2}g$
• D. $\left(\frac{n+1}{n-1}\right)g$

Answer 1

The correct option is C ${\left(\frac{n-1}{n+1}\right)}^{2}g$


According to $\text{NLM},$

$a=\frac{\text{Net pulling force}}{\text{Total mass}}$

$\Rightarrow a=\frac{nmg-mg}{nm+m}$

$a=\frac{(n-1)}{(n+1)}g-\left(1\right)$

Also,
$a_{com}=\frac{m_1{a}_1+m_2{a}_2}{m_1+m_2}$

$\Rightarrow a_{com}=\frac{nma-ma}{nm+m}$
$=\frac{(n-1)}{(n+1)}a-\left(2\right)$
from $\left(1\right)$ & $\left(2\right)$, we get,
$a_{com}=\frac{(n-1{)}^2}{(n+1{)}^2}g={\left(\frac{n-1}{n+1}\right)}^{2}g$

Hence, $\left(C\right)$ is the correct answer.