$A$ piece of wire is bent in the shape of a parabola $y=kx^2$ ($y$-axis vertical) with a bead of mass $m$ on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the $x$-axis with a constant acceleration $a$. The distance of the new equilibrium position of the bead,where the bead can stay at rest with respect to the wire,from the $y$-axis is

  • A
    $\frac{a}{gk}$
  • B
    $\frac{a}{2gk}$
  • C
    $\frac{2a}{gk}$
  • D
    $\frac{a}{4gk}$

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$A$ 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 $g = 10 \text{ m/s}^2$)

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