$A$ 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

$A$ 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

$A$ massless spring gets elongated by an amount $x_1$ under a tension of $5 \ N$. Its elongation is $x_2$ under a tension of $7 \ N$. For an elongation of $(5x_1 - 2x_2)$,the tension in the spring will be: (in $N$)

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 a body is suspended from a spring and released slowly,the extension in the spring is $x$. If the body is released suddenly,the extension in the spring is:

Two masses of $1 \, kg$ and $2 \, kg$ respectively are connected by a massless spring as shown in the figure. $A$ force of $20 \, N$ acts on the $2 \, kg$ mass. At the instant when the $1 \, kg$ mass has an acceleration of $10 \, m/s^2$ towards the right,the acceleration of the $2 \, kg$ mass is ......... $m/s^2$.