In the given diagram, reading of spring balance will be $\left(g=10 \mathrm{m} {\mathrm{s}}^{-2}\right)$

The position-time graph of a particle of mass $2 \mathrm{kg}$ moving along $x$-axis is as shown in the figure. The magnitude of impulse on the particle at $\mathrm{t}=2 \mathrm{s}$ is

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

A block of mass $m,$ is kept on a wedge of mass $M,$ as shown in figure such that mass $m$ remains stationary $w.r.t.$ wedge. The magnitude of force $P$ is

Force acting on a body varies with time as shown below. If initial momentum of the body is $\overrightarrow{p},$ then the time taken by the body to retain its momentum $\overrightarrow{p}$ again is