Q: Two blocks of masses m_{1} and m_{2} connected by a string and placed on an rough inclined plane having coefficient of friction μ as shown in figure. The ratio of masses **m _{1}/m_{2}** so that the block m

_{1}starts moving downward.

(a) $\displaystyle \frac{m_1}{m_2} > (\mu cos\theta – sin\theta ) $

(b) $ \displaystyle \frac{m_1}{m_2} > ( sin\theta +\mu cos\theta ) $

(c) $ \displaystyle \frac{m_1}{m_2} > ( cos\theta – \mu sin\theta ) $

(b) $ \displaystyle \frac{m_1}{m_2} > ( sin\theta -\mu cos\theta ) $

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Ans: (a)

Sol: As block m_{1} starts moving downwards , so

$ \displaystyle m_2 g sin\theta > \mu N + m_1 g $

$ \displaystyle m_2 g sin\theta > \mu m_2 g cos\theta + m_1 g $

$ \displaystyle m_2 g (sin\theta – \mu cos\theta ) > m_1 g $

$ \displaystyle m_2 (sin\theta – \mu cos\theta ) > m_1 $

$ \displaystyle (sin\theta – \mu cos\theta ) > \frac{m_1}{m_2} $

$ \displaystyle – ( \mu cos\theta – sin\theta ) > \frac{m_1}{m_2} $

$ \displaystyle ( \mu cos\theta – sin\theta ) < \frac{m_1}{m_2} $