$A$ 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?

$A$ 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

The length of a spring is $\alpha$ when a force of $4\,N$ is applied on it and the length is $\beta$ when $5\,N$ force is applied. Then the length of the spring when $9\,N$ force is applied is

$A$ mass of $10 \ kg$ is suspended from a spring balance. It is pulled aside by a horizontal string so that it makes an angle of $60^{\circ}$ with the vertical. The new reading of the balance is

$A$ spring has length $L$ and force constant $K$. It is cut into two springs of length $L_1$ and $L_2$ such that $L_1 = N L_2$ ($N$ is an integer). The force constant of the spring of length $L_1$ is:

When a weight is applied to a spring,it stretches by an amount $x$. What is the energy stored in it? ($T$ is the tension force developed in the spring and $k$ is the spring constant.)