A block of mass $m=2 \mathrm{kg}$ is suspended with a light spring of spring constant $k=1000 \mathrm{N} {\mathrm{m}}^{-1}.$ If the lift is moving vertically upward with acceleration of $2 \mathrm{m} {\mathrm{s}}^{-2}$ then equilibrium elongation in the spring is $\left(g=10 \mathrm{m} {\mathrm{s}}^{-2}\right)$

Two blocks $A$ and $B$ of mass $2m$ and $m$ respectively are connected to a spring of spring constant $k.$ A horizontal force $F=5mg$ is applied to block $A$ as indicated. If acceleration of block $B$ at the instant is $g,$ then acceleration of block $A$ at that instant is

A block of mass $4 \mathrm{kg}$ is placed on a rough fixed inclined plane as shown. The minimum force $F$ which is parallel to inclined to keep the block at rest is $\left(g=10 \mathrm{m} {\mathrm{s}}^{-2}\right)$

The magnitude of relative acceleration of $A$ w.r.t. $B$ is (Assuming pulley and string are light and massless)

Two blocks $A$ and $B$ of masses $5 \mathrm{kg}$ and $10 \mathrm{kg}$ respectively are pushed on rough surface. The contact force between the blocks is $\left(g=10 \mathrm{m} {\mathrm{s}}^{-2}\right)$

A block of mass $m$ is placed over a wedge of mass $4 \mathrm{m}$ and the system is acted upon by a horizontal force $F.$ Friction is absent everywhere. The value of $F$ so that the block does not slide over the wedge, is

A uniform rod of mass $M$ and length $L$ is rotated in horizontal plane about a vertical axis with constant angular velocity $\omega$ as shown in figure. The tension in the rod at point $P$ is

In the given arrangement strings and pulley are light and smooth, what should be mass of block $A$ so that it could remain at rest?

In the given arrangement, friction is absent and strings, pulleys are ideal. The difference of tension $\left(T_1-T_2\right)$ is