Two bodies with masses `m_1` and `m_2 (m_2 gt m) `connected by a thread lie on a smooth table. A force Q is first applied to the larger mass and then to the smaller mass. The ratio of tensions in the thread in the two cases will be

Options

A. `(m_1)/(m_2)`
B. `(m_1)/(m_2)`
C. `(m_2)/(m_1 + m_2)`
D. `(m_1)/(m_2-m_2)`

Answer

A

Explanation

The equations and T_1 is tension
similary , when the force is applied to `m_1 ` and if the tension is `T_2 ` then
` Q - T_2 =m_1 a`
`T_2 = m_2 a`

from the above equations , we get
` T_1 (Qm_1)/( m_1 +m_2) and T_2 =(Qm_2)/(m_1 +m_2) therefore (T_1)/(T_2) =(m_1)/(m_2)`