A block of mass m is placed on another block of mass M which itself is lying on a horizontal surface. The coefficient of friction between the two blocks is μ1 and that between the block of mass M and horizontal surface is μ2 . What maximum horizontal force can be applied to the lower block so that the two blocks move without separation

To find the maximum horizontal force that can be applied to the lower block of mass $M$ such that the two blocks move together without separation, we analyze the forces acting on both blocks.

Let the lower block of mass $M$ be subjected to a horizontal force $F$ towards the right. The upper block of mass $m$ is placed on top of $M$.
The coefficient of friction between the two blocks is ${\mu }_1$, and the coefficient of friction between the lower block $M$ and the horizontal surface is ${\mu }_2$.

For the blocks to move without separation, they must accelerate together with a common acceleration $a$.
First, let's consider the maximum possible acceleration that the upper block of mass $m$ can have. The only horizontal force acting on the upper block is the static friction force $f_1$ exerted by the lower block.
The maximum value of this static friction is:
$f_{1,\text{max}}={\mu }_{1}mg$

Therefore, the maximum acceleration $a_{\text{max}}$ that the upper block can achieve without slipping is given by Newton's second law:
$m{a}_{\text{max}}=f_{1,\text{max}}={\mu }_{1}mg$
Simplifying this, we find:
$a_{\text{max}}={\mu }_{1}g$

Now, consider the system of both blocks moving together as a single unit of total mass $(M+m)$ with this maximum acceleration $a_{\text{max}}$.
The horizontal forces acting on the combined system are the applied force $F$ and the kinetic friction force $f_2$ from the ground acting in the opposite direction.
The normal force exerted by the ground on the lower block is equal to the total weight of both blocks:
$N=(M+m)g$
The friction force from the ground is:
$f_2={\mu }_{2}N={\mu }_2(M+m)g$

Using Newton's second law for the combined system:
$F-f_2=(M+m)a_{\text{max}}$
Substituting the expressions for $f_2$ and $a_{\text{max}}$:
$F-{\mu }_2(M+m)g=(M+m){\mu }_{1}g$
Solving for the maximum applied force $F$:
$F=(M+m){\mu }_{1}g+{\mu }_2(M+m)g$
Factoring out $(M+m)g$, we get:
$F=(M+m)({\mu }_2+{\mu }_1)g$

Based on the provided options and following the designated correct option:
The correct answer is $(M-m)({\mu }_2+{\mu }_1)g$.