If the length of a pendulum is made 9 times and mass of the bob is made 4 times, then the value of time period will become:

1. 3T

2. 3/2T

3. 4T

4. 2T

In a simple pendulum, the period of oscillation T is related to length of the pendulum l as:
1. $\frac{\mathrm{l}}{T}$= constant
2. $\frac{{\mathrm{l}}^2}{T}$= constant
3. $\frac{\mathrm{l}}{T^2}$= constant
4. $\frac{{\mathrm{l}}^2}{T^2}$= constant

A pendulum has time period T. If it is taken on to another planet having acceleration due to gravity half and mass 9 times that of the earth, then its time period on the other planet will be:

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

A simple pendulum attached to the ceiling of a stationary lift has a time period of 1 s. The distance y covered by the lift moving downward varies with time as y = 3.75 ${\mathrm{t}}^2$, where y is in meters and t is in seconds. If g = 10 $\mathrm{m}/{\mathrm{s}}^2$, then the time period of the pendulum will be:

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: 

Two simple pendulums of length 1 m and 16 m are in the same phase at the mean position at any instant. If T is the time period of the smaller pendulum, then the minimum time after which they will again be in the same phase will be:

1.  $\frac{3\mathrm{T}}{2}$

2.  $\frac{3\mathrm{T}}{4}$

3.  $\frac{2\mathrm{T}}{3}$

4.  $\frac{4\mathrm{T}}{3}$

A simple pendulum of mass m swings about point B between extreme positions A and C. Net force acting on the bob at these three points is correctly shown by:

1.		2.
3.		4.