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(B) The acceleration $a$ of the system is given by $a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) g$.Using the equation of motion $S = ut + \frac{1}{2}

Vedclass · Accepted Answer

$\frac{8}{7}$

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(B) The acceleration $a$ of the system is given by $a = \left(\frac{m_1 - m_2}{m_1 + m_2}ight) g$.Using the equation of motion $S = ut + \frac{1}{2}at^2$,where $S = 3 \ m$,$u = 0$,and $t = 3 \ s$:$3 = 0 + \frac{1}{2} 	imes a 	imes (3)^2$$3 = \frac{1}{2} 	imes a 	imes 9$$a = \frac{6}{9} = \frac{2}{3} \ ms^{-2}$.Now,substitute $a$ into the acceleration formula:$\frac{2}{3} = \left(\frac{m_1 - m_2}{m_1 + m_2}ight) 	imes 10$$\frac{2}{30} = \frac{m_1 - m_2}{m_1 + m_2}$$\frac{1}{15} = \frac{m_1 - m_2}{m_1 + m_2}$$m_1 + m_2 = 15m_1 - 15m_2$$16m_2 = 14m_1$$\frac{m_1}{m_2} = \frac{16}{14} = \frac{8}{7}$.

Two masses $m_1$ and $m_2$ are connected by a light string passing over a smooth pulley. When set free,$m_1$ moves downwards by $3 \ m$ in $3 \ s$. The ratio of $\frac{m_1}{m_2}$ is $(g = 10 \ ms^{-2})$.

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