There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is T. If the resultant acceleration becomes g/4, then the new time period of the pendulum is 

1. 0.8 T

2. 0.25 T

3. 2 T

4. 4 T

­­A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. If the lift accelerates upwards with an acceleration $\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ , then the period of the pendulum will be

1. T

2. $\frac{\mathrm{T}}{4}$

3. $\frac{2\mathrm{T}}{\sqrt{5}}$

4. $2\mathrm{T}\sqrt{5}$

The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle $\theta$ and then released. The bob will then pass through its equilibrium position with a speed v, where v equals

1. $\sqrt{2\mathrm{gl}(1-\mathrm{sin\theta })}$

2. $\sqrt{2\mathrm{gl}(1+\mathrm{cos\theta })}$

3. $\sqrt{2\mathrm{gl}(1-\mathrm{cos\theta })}$

4. $\sqrt{2\mathrm{gl}(1+\mathrm{sin\theta })}$