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: Understanding the Concept:
Two masses hang on either side of a smooth pulley. The heavier mass \(m_2\) goes down and the lighter mass \(m_1\) goes up with the same acceleration \(a\).

Step 2: Key Formula or Approach:
For \(m_2\): \(m_2g - T = m_2a\). For \(m_1\): \(T - m_1g = m_1a\). Add them to remove \(T\):
\[ a = \frac{(m_2 - m_1)\,g}{m_1 + m_2} \]

Step 3: Detailed Explanation:
Set \(a = \dfrac gn\):
\[ \frac{m_2 - m_1}{m_2 + m_1} = \frac1n \Rightarrow n(m_2 - m_1) = m_2 + m_1 \]
\[ nm_2 - m_2 = nm_1 + m_1 \Rightarrow m_2(n-1) = m_1(n+1) \]
\[ \frac{m_2}{m_1} = \frac{n+1}{n-1} \]
Check with \(n = 3\): ratio 2. Then \(a = \dfrac{(2-1)g}{3} = \dfrac g3\), which fits. Options (A) and (D) give ratios below 1 for \(n>1\) or not matching, and (B) gives \(\tfrac{3}{2}\) for \(n=3\), which would make \(a = g/5\).

Final Answer:
The ratio \(m_2/m_1 = \dfrac{n+1}{n-1}\), option (C). \[ \boxed{\frac{n+1}{n-1} \text{ (C)}} \]