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

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

$A$ 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 whose unstretched length is $\ell$ has a force constant $k$. The spring is cut into two pieces of unstretched lengths $\ell_1$ and $\ell_2$ where $\ell_1 = n\ell_2$ and $n$ is an integer. The ratio $k_1/k_2$ of the corresponding force constants $k_1$ and $k_2$ will be:

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 ......... $m/s^2$.

Two springs $A$ and $B$ are fixed at the top and are stretched by $8 \,cm$ and $16 \,cm$ respectively, when loads of $20 \,N$ and $10 \,N$ are suspended at the lower ends. The ratio of the spring constants of the springs $A$ and $B$ is (in $: 1$)