A block of mass$m$ is at rest relative to a stationary wedge of mass $M$. The coefficient of friction between the block and the wedge is $\mathit{\text{μ}}$. The wedge is now pulled horizontally with an acceleration $a$, as shown in the figure. Then, the minimum magnitude of $a$ for the friction between the block and the wedge to be zero, is

A $30 \mathrm{kg}$ slab B rests on a frictionless floor as shown in the figure. A $10 \mathrm{kg}$ block A rests on top of the slab- B. The coefficients of static and kinetic friction between the block A and the slab B are $0.60$ and $0.40$ respectively. When block-A is acted upon by a horizontal force of $100 \mathrm{N}$, as shown, find the resulting acceleration of the slab- B. $\left(\mathrm{g}=9.8 \mathrm{m} {\mathrm{s}}^{-2}\right)$

The coefficient of static friction between two blocks is $0.5$ and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is _______$\mathrm{N}$ (take $\left.g=10{ \mathrm{m} \mathrm{s}}^{-2}\right)$

As shown in the figure, a block of mass $\sqrt{3} \mathrm{kg}$ is kept on a horizontal rough surface of coefficient of friction $\frac{1}{3\sqrt{3}}$. The critical force to be applied on the vertical surface as shown at an angle $60°$ with horizontal such that it does not move, will be $3x$. The value of $x$ will 

$\left[g=10 \mathrm{m} {\mathrm{s}}^{-2}; \sin 60°=\frac{\sqrt{3}}{2}; \cos 60°=\frac{1}{2}\right]$