In figure, the coefficient of static friction between the floor and block $B$ is $0.30$. The coefficient of static friction between the block $B$ and $A$ is $0.15$. The mass of block $A$ is $\frac{m}{2}\text{kg}$ and $B$ is $m\text{kg}$. What is the maximum horizontal force $F$ in $\text{Newtons}$ that can be applied to block $B$, so that the two blocks move together?

The correct option is C $0.675mg$
$m_A=m/2\text{kg},m_B=m\text{kg}$
${\mu }_A=0.15,{\mu }_B=0.3$
$\Rightarrow$Let both the blocks are moving together with common acceleration $a$. The maximum force will correspond to the maximum value of $f_1$, which can sustain the required acceleration for block $A$


From FBD of block,
maximum value of $f_1={\mu }_A{N}_A={\mu }_A={\mu }_a{m}_{A}g$

$\therefore a=\frac{{\mu }_A{m}_{A}g}{m_A}={\mu }_{A}g=0.15g{\text{m/s}}^2...\left(i\right)$
Now, on the system of blocks $A+B$, applying equation of dynamics from the FBD:


Maximum value of $f_2={\mu }_{B}N={\mu }_B(m_A+m_B)g$
Hence applying newton's $2^{\text{nd}}$ law,
$\Rightarrow F-{\mu }_B(m_B+m_A)g=(m_B+m_A)a$
putting $a=0.15g$ from Eq. $\left(i\right)$
$F=0.3(m_B+m_A)g+(m_B+m_A)0.15g$
$\Rightarrow F=0.45(m_B+m_A)g$
$F=(m+\frac{m}{2})0.45g$
$\therefore F=\frac{3}{2}m\times 0.45g=0.675mg\text{N}$