$A$ smooth sphere of radius $R$ moves along a straight line with a constant acceleration $a = g$. $A$ particle is placed at the top of the sphere. It is released from there with zero velocity relative to the sphere. What will be its speed relative to the sphere as a function of the angle $\theta$ as the particle slides?

$A$ man weighing $60 \text{ kg}$ is in a lift moving down with an acceleration of $1.8 \text{ ms}^{-2}$. The force exerted by the floor on him is:

$A$ wire is bent into the shape of a parabola given by $Y = Kx^2$. $A$ bead of mass $m$ is placed on the wire,which can slide on it without friction. Initially,when the wire is at rest,the bead is at the lowest point of the parabola. If the wire is now moved with a constant acceleration $a$ parallel to the $X$-axis,what will be the distance of the new equilibrium position of the bead from the $Y$-axis,such that it remains stationary relative to the wire?

$A$ wedge $Y$ with mass of $10 \text{ kg}$ has all frictionless surfaces, and the inclined surface makes an angle of $37^{\circ}$ with the horizontal. $A$ block $X$ with mass $2 \text{ kg}$ is placed at the highest point of the wedge as shown in the figure and is at rest. At $t=0$, the wedge $Y$ is pulled toward the right with a constant force $f$ of $24 \text{ N}$. Taking the block $X$ at rest at $t=0$, the time taken by it to slide down $8.8 \text{ m}$ on the slope, while $Y$ is in motion, is . . . . . . s. (Take $\tan(37^{\circ}) = 3/4$ and $g = 10 \text{ m/s}^2$)

$A$ man has a mass of $80 \text{ kg}$. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of $6 \text{ m/s}^2$. What would be his apparent weight in $\text{kg}$? (Take $g = 10 \text{ m/s}^2$)

$A$ lift is moving downwards with an acceleration equal to the acceleration due to gravity. $A$ body of mass $m$ kept on the floor of the lift is pulled horizontally. If the coefficient of friction is $\mu$,then the frictional resistance offered by the body is: