If linear momentum of particle changes with time $t$ as $p=\left(3t^2+2t+1\right)\mathrm{kgm} {\mathrm{s}}^{-1},$ then force acting on particle at $t=3 \mathrm{s}$ will be

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