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

$A$ body of mass $m$ is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is $1\, cm$. If the angular velocity is doubled,the elongation in the spring is $5\, cm$. The original length of the spring is ......... $cm$

Define spring constant and write its unit.

$A$ 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 $Joule$)

When the acceleration of the $5\ kg$ block is $2\ m/s^2$ as shown in the figure,what will be the acceleration of the $15\ kg$ block? (in $m/s^2$)