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 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: ${}_{}$

         
1. \(v_2+v_2\)

2. \(\sqrt{v_1^2+v_2^2}\)

3. \(\frac{1}{v_1}+\frac{1}{v_1}^{-1}\)

4. \(\sqrt{v_1^2-v_2^2}\)

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