Two spherical bobs of masses \(M_A\) and \(M_B\) are hung vertically from two strings of length \(l_A\) and \(l_B\) respectively. If they are executing SHM with frequency as per the relation \(f_A=2f_B,\) Then: 
1. $l_A=\frac{l_B}{4}$

2. $l_A=4l_B$

3. $l_A=2l_B$ $\&$ $M_A=2M_B$

4. $l_A=\frac{l_B}{2}$ $\&$ $M_A=\frac{M_B}{2}$

The frequency of a simple pendulum in a free-falling lift will be:
1. zero
2. infinite
3. can't say
4. finite

The period of oscillation of a simple pendulum of length \(\mathrm{L}\) suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $\theta$, is given by:

1.   $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{gcos\theta }}}$               

2.  $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{gsin}\theta }}$

3.  $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g} }}$                     

4. $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{gtan\theta }}}$

A simple pendulum hanging from the ceiling of a stationary lift has a time period T₁. When the lift moves downward with constant velocity, then the time period becomes T₂. It can be concluded that: