JEE MAIN - Physics (2002 - No. 60)
A light string passing over a smooth light pulley connects two blocks of masses $${m_1}$$ and $${m_2}$$ (vertically). If the acceleration of the system is $$g/8$$, then the ratio of the masses is
$$8:1$$
$$9:7$$
$$4:3$$
$$5:3$$
Selitys
Assume that, mass m1 is greater than mass m2, so the heavier mass m1 is accelerating downward and the lighter mass m2 is accelerating upwards.
For mass $${m_1}$$ the equation will be
$${m_1}$$$$g-T=$$$${m_1}$$$$a$$
For mass $${m_2}$$ the equation will be
$$T-$$$${m_2}$$$$g=$$$${m_2}$$$$a$$
Adding those equations we get
$$a = {{\left( {{m_1} - {m_2}} \right)g} \over {{m_1} + {m_2}}}$$
$$\therefore$$ $${g \over 8} = {{\left( {{m_1} - {m_2}} \right)g} \over {{m_1} + {m_2}}}$$
$$ \Rightarrow {1 \over 8} = {{{{{m_1}} \over {{m_2}}} - 1} \over {{{{m_1}} \over {{m_2}}} + 1}}$$
$$ \Rightarrow$$ $${{{m_1}} \over {{m_2}}} + 1 =$$ $${8\left( {{{{m_1}} \over {{m_2}}} - 1} \right)}$$
$$ \Rightarrow$$ $${{{m_1}} \over {{m_2}}} = {9 \over 7}$$