Answer: 2m1m2m1+m2 \large\frac{2m_1m_2}{m_1+m_2}gg

When the bodies are released, the heavier one moves downward and the lighter one moves upward.

Net force in the direction of motion of m1m_1 is F1=m1g−TF_1 = m_1g - T which is equal to m1am_1a as per Newton's second law.

⇒m1g−T=m1a(i)\Rightarrow m_1g - T = m_1 a \quad (i)

Net force in the direction of motion of m2m_2 is F2=T−m2gF_2 = T-m_2g which is equal to m2am_2a as per Newton's second law.

⇒T−m2g=m2a(ii)\Rightarrow T-m_2g = m_2a \quad (ii)

Solving (i)(i) and (ii)(ii) for acceleration, a→a=m1−m2m1+m2a \rightarrow a = \large\frac{m_1-m_2}{m_1+m_2}gg

Substituting in (ii)(ii), we get T=m2(a+g)=m2(gm1−m2m1+m2T = m_2 (a+g) = m_2 (g \large\frac{m_1-m_2}{m_1+m_2}+g) + g)

⇒T=gm2m1−m2+m1+m2m1+m2\Rightarrow T = gm_2 \large\frac{m_1 - m_2 + m_1 + m_2}{m_1+m_2}=2m1m2m1+m2 = \large\frac{2m_1m_2}{m_1+m_2}gg