Question

An inextensible string passing over a smooth pulley connects two blocks of masses \(m_1\) and \(m_2\) (\(m_2 > m_1\)) vertically. If the acceleration of the system is \((\frac{g}{n})\), then the ratio of masses \((\frac{m_2}{m_1})\) is

• A. \(\frac{n}{n+1}\)
• B. \(\frac{n}{n-1}\)
• C. \(\frac{n+1}{n-1}\)
• D. \(\frac{n+1}{n}\)

Hint:

Atwood acceleration is a = (m2 - m1) g / (m1 + m2). Set a = g/n and solve for m2/m1.

Answer 1

The Correct Option is C

Step 1: Use the net force over the total mass:
The driving force is the weight difference, $(m_2 - m_1)g$, and the whole system of mass $m_1 + m_2$ accelerates. So $a = \frac{m_2-m_1}{m_2+m_1}g$.

Step 2: Introduce the ratio:
Let $r = m_2/m_1$. Then $a = \frac{r-1}{r+1}g$. Set equal to $\frac gn$: $n(r-1) = r+1$.

Step 3: Solve:
$nr - n = r + 1$, so $r(n-1) = n+1$ and $r = \frac{n+1}{n-1}$.

Step 4: Validity:
Since $m_2 > m_1$ we need $r > 1$, which requires $n > 1$, as expected for an acceleration less than $g$.

Final Answer:
Option (C). \[ \boxed{\frac{n+1}{n-1} \text{ (C)}} \]