An inclined plane is bent in such a way that the vertical cross-section is given by $y = \frac{x^2}{4}$,where $y$ is in the vertical direction and $x$ is in the horizontal direction. If the upper surface of this curved plane is rough with a coefficient of friction $\mu = 0.5$,the maximum height in $cm$ at which a stationary block will not slip downward is............$cm$.

Two vertical walls are separated by a distance of $2 \ m$. Wall $A$ is smooth while wall $B$ is rough with a coefficient of friction $\mu = 0.5$. $A$ uniform rod is placed between them as shown. The length of the longest rod that can be placed between the walls in equilibrium is equal to

$A$ block rests on a rough inclined plane making an angle of $30^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8$. If the frictional force on the block is $10 \text{ N}$, the mass of the block is $\left(g=10 \text{ ms}^{-2}\right)$ (in $\text{ kg}$)

$A$ block of mass $M$ is sliding down an inclined plane. An external force $F$ is applied vertically downwards on the block. The coefficient of static friction is $\mu_s$ and the coefficient of kinetic friction is $\mu_k$. The friction force acting on the block is:

$A$ block of mass $m_1=1 \ kg$ and another mass $m_2=2 \ kg$ are placed together (see figure) on an inclined plane with an angle of inclination $\theta$. Various values of $\theta$ are given in List $I$. The coefficient of friction between the block $m_1$ and the plane is always zero. The coefficient of static and dynamic friction between the block $m_2$ and the plane are equal to $\mu=0.3$. In List $II$,expressions for the friction on block $m_2$ are given. Match the correct expression of the friction in List $II$ with the angles given in List $I$,and choose the correct option. The acceleration due to gravity is denoted by $g$. [Useful information: $\tan(5.5^{\circ}) \approx 0.1; \tan(11.5^{\circ}) \approx 0.2; \tan(16.5^{\circ}) \approx 0.3$]

List $I$	List $II$
$P. \theta=5^{\circ}$	$1. m_2 g \sin \theta$
$Q. \theta=10^{\circ}$	$2. (m_1+m_2) g \sin \theta$
$R. \theta=15^{\circ}$	$3. \mu m_2 g \cos \theta$
$S. \theta=20^{\circ}$	$4. \mu(m_1+m_2) g \cos \theta$

$A$ wooden box lying at rest on an inclined surface of a wet wood is held at static equilibrium by a constant force $F$ applied perpendicular to the incline. If the mass of the box is $1 \ kg$,the angle of inclination is $30^{\circ}$ and the coefficient of static friction between the box and the inclined plane is $0.2$,the minimum magnitude of $F$ is (Use $g=10 \ m/s^2$)