A particle moves a distance $\mathrm{x}$ in time $\mathrm{t}$ according to equation $\mathrm{x}={\left(\mathrm{t}+5\right)}^{-1}$. The acceleration of particle is proportional to

Figure represents the position-time graph of a body of mass $4 \mathrm{kg}$. Impulse $\left(\mathrm{kg} \mathrm{m} {\mathrm{s}}^{-1}\right)$ imparted to the body at $t=0$ is

A solid sphere of mass $1 \mathrm{kg}$ and radius $1 \mathrm{m}$ rolls without slipping on a fixed inclined plane with an angle of inclination $\theta =30°$ from the horizontal. Two forces of magnitude $1 \mathrm{N}$ each, parallel to the incline, act on the sphere, both at distance $r=0.5 \mathrm{m}$ from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is $\mathrm{m} {\mathrm{s}}^{-2}$. (Take $g=10 \mathrm{m} {\mathrm{s}}^{-2}$)

If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.

Two inclined planes are placed as shown in figure.
A block is projected from the Point $A$ of inclined plane $AB$ along its surface with a velocity just sufficient to carry it to the top Point $B$ at a height $10 \mathrm{m}$. After reaching the Point $B$ the block slides down on inclined plane $BC$. Time it takes to reach to the point $C$ from point $A$ is $t\left(\sqrt{2}+1\right)\mathrm{s}$. The value of t is _____ (use $g=10 \mathrm{m} {\mathrm{s}}^{-2}$)

A block of mass $m$ lies on the wedge of mass $M$, which lies on the fixed horizontal surface. The wedge is free to move on the horizontal surface. A horizontal force of magnitude $F$ is applied on the block as shown, neglecting friction at all surfaces, the value of force $F$ such that block has no relative motion with respect to the wedge will be (Where $g$ is the acceleration due to gravity)

A container of liquid is released from the rest, on a smooth inclined plane as shown in the figure. Length of at the inclined plane is sufficient, and assume liquid finally equilibrium. Finally, the liquid surface makes an angle horizontal.......

In the figure shown a ring of mass $M$ and a block of mass $m$ are in equilibrium. The string is light and pulley $P$ dose not offer any friction and coefficient of friction between pole and $M$ is $\mu$. The frictional force offered by the pole on $M$ is

A $40 \mathrm{kg}$ slab rests on a frictionless floor. A $10 \mathrm{kg}$ block rests on top of the slab. The static coefficient of friction between the block and slab is $0.60$ while the kinetic coefficient is $0.40$. The $10 \mathrm{kg}$ block is acted upon by a horizontal force of $100 \mathrm{N}$. If $g=9.8 \mathrm{m} {\mathrm{s}}^{-2}$, the resulting acceleration of the slab will be-