$A$ spring of force constant $k$ is cut into three equal pieces. If these three pieces are connected in parallel,the force constant of the combination will be

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 springs of force constants $K_1$ and $K_2$ are loaded with weights $W_1$ and $W_2$ respectively. Assume that the length of each spring is increased by the same amount. If $K_1 = 2 K_2$, then the ratio $\frac{W_2}{W_1}$ is

$A$ spring of force constant $k$ is cut into lengths of ratio $1:2:3$. They are connected in series and the new force constant is $k'$. Then they are connected in parallel and the force constant is $k''$. Then $k':k''$ is

When a $1.0\,kg$ mass hangs attached to a spring of length $50\,cm$,the spring stretches by $2\,cm$. The mass is pulled down until the length of the spring becomes $60\,cm$. What is the amount of elastic energy stored in the spring in this condition,if $g = 10\,m/s^2$? (in $Joule$)

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