$A$ block takes time $t$ to slide down a plane inclined at $45^\circ$ to the horizontal. If the surface is made smooth (frictionless), the block takes time $t/2$ to slide down the plane. The coefficient of friction between the block and the inclined plane is $\frac{\alpha}{100}$. The value of $\alpha$ is . . . . . . .

$A$ block of mass $1 \ kg$ is pushed up a surface inclined to the horizontal at an angle of $60^{\circ}$ by a force of $10 \ N$ parallel to the inclined surface as shown in the figure. When the block is pushed up by $10 \ m$ along the inclined surface,the work done against the frictional force is: $\left[g=10 \ m/s^2, \mu_s = 0.1\right]$

Find the reading of the weighing machine $(WM)$ in the following case. The system is in equilibrium.

$A$ disc arranged in a vertical plane has two grooves of the same length directed along the vertical chord $AB$ and the chord $CD$ as shown in the figure. Particles slide down along $AB$ and $CD$ starting from rest. The ratio of the time $t_{AB}/t_{CD}$ is

Two blocks $A$ and $B$ of equal masses are sliding down along straight parallel lines on an inclined plane of $45^{\circ}$. Their coefficients of kinetic friction are $\mu_A = 0.2$ and $\mu_B = 0.3$ respectively. At $t = 0$,both the blocks are at rest and block $A$ is $\sqrt{2} \text{ m}$ behind block $B$. Calculate the time and distance from the initial position where the front faces of the blocks come in line on the inclined plane as shown in the figure. (Use $g = 10 \text{ m/s}^2$)

The force required to move a body up a rough inclined plane is double the force required to prevent the body from sliding down the plane. The coefficient of friction, when the angle of inclination of the plane is $60^{\circ}$ is