One-fourth length of a spring of force constant $K$ is cut away. The force constant of the remaining spring will be

$A$ bead of mass $m = 100 \,g$ is attached to one end of a spring of natural length $L$ and spring constant $k = \frac{(\sqrt{3}+1) mg}{L}$. The other end of the spring is fixed at point $A$ on a smooth vertical ring of radius $R$. The bead is at point $B$ such that the spring makes an angle of $30^{\circ}$ with the horizontal diameter. The normal reaction at $B$ just after it is released to move is (take $g = 9.8 \,ms^{-2}$): (in $\,N$)

The figure shows two cases. In the first case,a spring (spring constant $K$) is pulled by two equal and opposite forces $F$ at both ends. In the second case,it is pulled by a force $F$ at one end while the other end is fixed. The extensions $(x)$ in the springs will be:

$A$ spring of force constant $k$ is cut into two pieces such that one piece is three times the length of the other. The longer piece will have a force constant of

Find the acceleration of the $3 \text{ kg}$ mass when the acceleration of the $2 \text{ kg}$ mass is $2 \text{ ms}^{-2}$ as shown in the figure. (in $\text{ ms}^{-2}$)

If a spring of stiffness $k$ is cut into two parts $A$ and $B$ of length $l_{A}: l_{B}=2: 3$,then the stiffness of spring $A$ is given by