A) \[\frac{7}{4}m\]
B) \[2m\]
C) \[\frac{9}{2}m\]
D) \[\frac{9}{4}m\]
Correct Answer: D
Solution :
[d] Acceleration when block B is in the liquid \[{{a}_{1}}=\frac{{{m}_{A}}g-({{m}_{B}}g-upthrust)}{({{m}_{A}}+3m)}\] \[=\frac{{{m}_{A}}g-\left( 3mg-\frac{3m}{2\rho }\rho g \right)}{({{m}_{A}}+3m)}\uparrow \] Acceleration when block B is outside if the liquid, \[{{a}_{2}}=\frac{3mg-{{m}_{A}}g}{({{m}_{A}}+3m)}(\downarrow )\] \[Given\,{{a}_{1}}={{a}_{2}}\], we get \[{{m}_{A}}g-\frac{3}{2}mg=3mg-{{m}_{A}}g\] \[\Rightarrow \,\,\,2{{m}_{A}}g-\frac{9}{2}mg\Rightarrow {{m}_{A}}=\left( \frac{9}{4} \right)m\]
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