Consider a uniform cubical box of side $a$ on a rough floor that is to be moved by applying a minimum possible force $F$ at a point $b$ above its centre of mass (see figure). If the coefficient of friction is $\mu = 0.4$,the maximum possible value of $100 \times \frac{b}{a}$ for a box not to topple before moving is

$A$ block of mass $5 kg$ is at rest on a rough inclined surface. If the angle of inclination of the plane is $60^{\circ}$,then the total force applied by the surface on the block is .......... $N$. (Take $g = 10 m/s^2$)

$A$ body is placed on a rough inclined plane of inclination $\theta$. As the angle $\theta$ is increased from $0^o$ to $90^o$,the contact force between the block and the plane

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

$A$ body takes just twice the time to slide down a plane inclined at $30^\circ$ to the horizontal as it would if the plane were frictionless. The coefficient of friction between the body and the plane is:

The minimum force required to move a body up an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is $\frac{1}{2 \sqrt{3}}$, then the angle of the inclined plane is (in $^{\circ}$)