-\mathsf{a = \dfrac{g}{7}}a= 7g \mathsf{T_2 = \dfrac{6g}{7}}T 2 = 76g Option → CGiven :-\begin{lgathered}m_3 = 1 kg \\ m_2 = 3kg \\ m_1 = 3 kg\end{lgathered} m 3 =1kgm 2 =3kgm 1 =3kg Where,m_3 = a, m_2 = b, m_1 = cm 3 =a,m 2 =b,m 1 =cTo find :-The ACCELERATION of the block of MASS 1 kg and Tension FORCE between A and B.Solution:-Let T_1T 1 be the Tension force between Block B and Block CAND T_2T 2 be the tension force between B and A.Now,Consider Block C as system.The force acting on it :-Mg force downward.Tension force upward.Acceleration is upward.\mathsf{m_1 g - T_1 = m_1 a}m 1 g−T 1 =m 1 a\mathsf{T_1 = m_1g - m_1 a}----1)T 1 =m 1 g−m 1 a−−−−1)For Block B.\mathsf{m_2g + T_2 - T_1 = m_2a}----2)m 2 g+T 2 −T 1 =m 2 a−−−−2)Take Block A as system.\mathsf{ m_3g - T_2 = m_3a}m 3 g−T 2 =m 3 a\mathsf{T_2 = m_3g - m_3 a}-------3)T 2 =m 3 g−m 3 a−−−−−−−3)Put the value ofT_1, T_2T 1 ,T 2 in equation 2.→\mathsf{m_2g + m_3g - m_3a -(m_1g -m_1a) = m_2a}m 2 g+m 3 g−m 3 a−(m 1 g−m 1 a)=m 2 a→\mathsf{m_2g + m_3g - m_3a - m_1g - m_1a = m_2a }m 2 g+m 3 g−m 3 a−m 1 g−m 1 a=m 2 a→\mathsf{m_2g + m_3g - m_1 g = m_2a + m_3a + m_1a}m 2 g+m 3 g−m 1 g=m 2 a+m 3 a+m 1 aPut the masses of each one.→\mathsf{3g + g - 3g = 3a + a + 3a }3g+g−3g=3a+a+3a→\mathsf{ g = 7a }g=7a→\mathsf{a = \dfrac{g}{7}}a= 7g put the value of a in eq. 3→\mathsf{T_2 = m_3g - m_3a}T 2 =m 3 g−m 3 a→\mathsf{T_2 = g - a}T 2 =g−a→\mathsf{T_2 = g - \dfrac{g}{7}}T 2 =g− 7g →\mathsf{T_2 = \dfrac{7g-g}{7}}T 2 = 77g−g →\mathsf{T_2 = \dfrac{6g}{7}}T 2 = 76g hence,The acceleration is \dfrac{g}{7} 7g and Tension force between A and B is \dfrac{6g}{7} 76g