In the given arrangement, if the lift is moving upwards with acceleration $\frac{g}{2},$ then force exerted by the string on ceiling is

The acceleration of block $A$ at the given instant is

The acceleration of block $C$ is

What should be the value of $M^{\text{'}}$ so that the block does not slide over the triangular wedge? All surfaces are smooth.

Blocks $A$ and $B$are initially in equilibrium. The springs and strings are ideal. The accelerations of the blocks $A$ and $B,$ just after the spring-$2$is cut, are respectively

A thin, uniform rod of mass $M$ and length $L$ is rotated with constant angular speed $\omega$ as shown in figure. Tension in the rod at position $A$ is

Two blocks of masses $10 \mathrm{kg}$ and $20 \mathrm{kg}$ are connected by a massless spring of spring constant $2000 \mathrm{N} {\mathrm{m}}^{-1}$ as shown in figure. A constant force of $200 \mathrm{N}$ acts on $20 \mathrm{kg}$ block. When the $10 \mathrm{kg}$  block has acceleration of $12 \mathrm{m} {\mathrm{s}}^{-2},$ the acceleration of $20 \mathrm{kg}$ block in $\mathrm{m} {\mathrm{s}}^{-2}$ is equal to

If hanging block is in equilibrium, then elongation in the spring in c.m. is equal to