Maximum value of static friction is called

What is the maximum value of the force $F$ such that the block shown in the arrangement, does not move ........ $N$

In the figure, a ladder of mass $m$ is shown leaning against a wall. It is in static equilibrium making an angle $\theta$ with the horizontal floor. The coefficient of friction between the wall and the ladder is $\mu_1$ and that between the floor and the ladder is $\mu_2$. The normal reaction of the wall on the ladder is $N_1$ and that of the floor is $N_2$. If the ladder is about to slip, then

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$(A)$ $\mu_1=0 \mu_2 \neq 0$ and $N _2 \tan \theta=\frac{ mg }{2}$

$(B)$ $\mu_1 \neq 0 \mu_2=0$ and $N_1 \tan \theta=\frac{m g}{2}$

$(C)$ $\mu_1 \neq 0 \mu_2 \neq 0$ and $N _2 \tan \theta=\frac{ mg }{1+\mu_1 \mu_2}$

$(D)$ $\mu_1=0 \mu_2 \neq 0$ and $N _1 \tan \theta=\frac{ mg }{2}$

A block of mass $2 \,kg$ rests on a rough inclined plane making an angle of $30°$ with the horizontal. The coefficient of static friction between the block and the plane is $ 0.7$. The frictional force on the block is ....... $N$.

If a ladder weighing $250\,N$ is placed against a smooth vertical wall having coefficient of friction between it and floor is $0.3$, then what is the maximum force of friction available at the point of contact between the ladder and the floor ........ $N$

A block of mass $m$ (initially at rest) is sliding up (in vertical direction) against a rough vertical wall with the help of a force $F$ whose magnitude is constant but direction is changing. $\theta  = {\theta _0}t$  where $t$ is time in sec. At $t$ = $0$ , the force is in vertical upward direction and then as time passes its direction is getting along normal, i.e., $\theta  = \frac{\pi }{2}$ .The value of $F$ so that the block comes to rest when $\theta  = \frac{\pi }{2}$ , is