$A$ string of negligible mass going over a clamped pulley of mass $m$ supports a block of mass $M$ as shown in the figure. The force on the pulley by the clamp is given by

$A$ helicopter of mass $2000 \,kg$ rises with a vertical acceleration of $15 \,m s^{-2}$. The total mass of the crew and passengers is $500 \,kg$. Calculate the magnitude and direction of the following (take $g = 10 \,m s^{-2}$):
$(a)$ Force on the floor of the helicopter by the crew and passengers.
$(b)$ Action of the rotor of the helicopter on the surrounding air.
$(c)$ Force on the helicopter due to the surrounding air.

$A$ block $B$ is placed on block $A$. The mass of block $B$ is less than the mass of block $A$. Friction exists between the blocks,whereas the ground on which the block $A$ is placed is taken to be smooth. $A$ horizontal force $F$,increasing linearly with time,begins to act on $B$. The acceleration $a_A$ and $a_B$ of blocks $A$ and $B$ respectively are plotted against $t$. The correctly plotted graph is

$A$ cold soft drink is kept on a balance. When the cap is opened,the weight reading on the balance:

On a pulley of mass $M$ hangs a rope with two masses $m_{1}$ and $m_{2}$ $(m_{1} > m_{2})$ tied at the ends as shown in the figure. The pulley rotates without any friction,whereas the friction between the rope and the pulley is large enough to prevent any slipping. Which of the following plots best represents the difference between the tensions in the rope on the two sides of the pulley as a function of the mass of the pulley?

In the figure shown,all surfaces are frictionless,and the mass of the block is $m = 1\, kg$. The block and wedge are held initially at rest. Now,the wedge is given a horizontal acceleration of $a = 10\, m/s^2$ by applying a force on the wedge,such that the block does not slip on the wedge. Calculate the work done by the normal force in the ground frame on the block in $\sqrt{3}$ seconds. (Assume $\tan \theta = a/g = 1$,so $\theta = 45^\circ$)