In the figure shown $m_A=10 kg, m_B=15 kg$. The maximum value of F is $\frac{125}{\mathrm{n}}$  below which the blocks move together. The value of n is ___

Assuming that both blocks move together, their common acceleration is $a_C=\frac{F}{10+15}=\frac{F}{25}$$FBD$ of block $B$ (on which no externally applied force acts).

The required friction force $f$ is equal to $f=m_B{a}_C=\frac{15F}{25}=\frac{3F}{5}$

Now, maximum static friction available is $f_L=\mu N_1=0.25\left(100\right)=25 N$ (here $N_1=m_{A}g=100 N$)

$\therefore f\le f_L\Rightarrow \frac{3F}{5}\le 25\Rightarrow F<\frac{125}{3}N\Rightarrow F_{max}=\frac{125}{3}N$