Minimum Force To Prevent Sliding

Here the applied horizontal force presses the block onto the wall, and the wall's normal reaction equals that applied force, $N=F$. The upward friction from the wall must support the block's full weight to prevent downward sliding, so at the verge of slipping friction reaches its maximum $\mu N$. Setting the limiting friction equal to the weight gives $\mu F=mg$, hence $F=\frac{mg}{\mu }=\frac{10\times 10}{0.5}=\frac{100}{0.5}=200\text{ N}$. The 100 N value equals just the weight and forgets to divide by the friction coefficient. The 50 N value multiplies by $\mu$ instead of dividing, which weakens rather than strengthens the needed push. The 150 N value does not satisfy the equilibrium condition. A subtle conceptual feature here is that the normal force is horizontal rather than vertical, unlike the usual block-on-a-table case, so it is the applied push, not gravity, that sets how much friction the wall can offer. This coupling between the pressing force and the available friction is what makes the problem self-limiting and gives a clean threshold condition. This is the NCERT block-against-a-wall friction problem. As a check, a smaller coefficient of friction would demand an even larger horizontal force, consistent with $F$ being inversely proportional to $\mu$.