An inextensible string passing over a smooth pulley connects two blocks of masses m₁ and m₂ ( m₂ > m₁) vertically. If the acceleration of the system is (g/n), then the ratio of masses ((m₂)/(m₁)) is (a) n/n+1 (b) n/n-1 (c) (n+1)/(n-1) (d) n+1/n

Answer: (c). Atwood machine: a = ((m₂ - m₁)g)/(m₂ + m₁) = g/n. Put r = (m₂)/(m₁): (r - 1)/(r + 1) = 1/n ⇒ nr - n = r + 1. r = (n + 1)/(n - 1), so the answer is (c).

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Connected Bodies & Pulleys

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 $\left(\frac{g}{n}\right)$ , then the ratio of masses $\left(\frac{m_2}{m_1}\right)$ is

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Answer: (c)

Atwood machine: $a = \dfrac{(m_2 - m_1)g}{m_2 + m_1} = \dfrac{g}{n}$ .
Put $r = \dfrac{m_2}{m_1}$ : $\dfrac{r - 1}{r + 1} = \dfrac{1}{n} \Rightarrow nr - n = r + 1$ .
$r = \dfrac{n + 1}{n - 1}$ , so the answer is (c).

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