Assertion $(A)$: When a spring is cut into two equal parts,the spring constant of each piece is twice that of the original spring. Reason $(R)$: Spring constant is inversely proportional to the length of the spring.

Two blocks each of mass $m$ are connected to a spring of spring constant $k.$ If both are given velocity $v$ in opposite directions,then the maximum elongation of the spring is

Initially,a spring is at its natural length. $A$ block of mass $2 \, kg$ is attached to the lower end,and a block of mass $0.25 \, kg$ is at the top. If the system is released from the position where the spring is at its natural length,find the maximum force exerted by the system on the floor? (in $N$,take $g = 10 \, m/s^2$)

$A$ long spring is stretched by $2 \ cm$ and its potential energy is $U$. If the spring is stretched by $10 \ cm$,its potential energy will be

Two light vertical springs with equal natural lengths and spring constants $k_1$ and $k_2$ are separated by a distance $l$. Their upper ends are fixed to the ceiling and their lower ends to the ends $A$ and $B$ of a light horizontal rod $AB$. $A$ vertical downwards force $F$ is applied at point $C$ on the rod. $AB$ will remain horizontal in equilibrium if the distance $AC$ is

$A$ 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 $\,N$)