$A$ spring with $10$ coils has spring constant $k$. It is exactly cut into two halves,then each of these new springs will have a spring constant

$A$ body of mass $m$ is suspended from an ideal spring of force constant $k$. The expected change in the position of the body due to an additional force $F$ acting vertically downwards is

Two springs of spring constants $1500 \ N/m$ and $3000 \ N/m$ respectively are stretched with the same force. They will have potential energy in the ratio

$A$ force of $20\,dyne$ applied to the end of a spring increases its length by $1\,mm$. What will be the force constant of the spring?

$A$ spring of force constant $k$ is cut into two equal halves. The force constant of each half 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 $\,N$)