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:
1. $1:9$
2. $1:11$
3. $1:14$
4. $1:6$

When a mass $m$ is connected individually to two springs $S_1$ and $S_2,$ the oscillation frequencies are $\nu_1$ and $\nu_2.$ If the same mass is attached to the two springs as shown in the figure, the oscillation frequency would be: 

A mass of $2.0$ kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released, the mass executes a simple harmonic motion. The spring constant is $200$ N/m. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan? 
(Take $g=10$ m/s²) 
                

When a mass is suspended separately by two different springs, in successive order, then the time period of oscillations is $t _1$ and $t_2$ respectively. If it is connected by both springs as shown in the figure below, then the time period of oscillation becomes $t_0.$ The correct relation between $t_0,$ $t_1$ & $t_2$ is:

1. ${{\mathrm{t}}_0}^2={{\mathrm{t}}_1}^2+{{\mathrm{t}}_2}^2$

2. ${{\mathrm{t}}_0}^{-2}={{\mathrm{t}}_1}^{-2}+{{\mathrm{t}}_2}^{-2}$

3. ${{\mathrm{t}}_0}^{-1}={{\mathrm{t}}_1}^{-1}+{{\mathrm{t}}_2}^{-1}$

4. ${\mathrm{t}}_0={\mathrm{t}}_1+{\mathrm{t}}_2$