If the mass of a bob of a pendulum is increased by $9$ times, the period of the pendulum will?

The path length of oscillation of a simple pendulum of length $1 \,m$ is $16 \,cm$. Its maximum velocity is (Take $g = \pi^2 \,m/s^2$).

$A$ simple pendulum has a time period $T_1$. The point of suspension is now moved upward according to the equation $y = kt^2$,where $k = 1 \ m/s^2$. If the new time period is $T_2$,then the ratio $\frac{T_1^2}{T_2^2}$ will be (take $g = 10 \ m/s^2$):

In the following table, the relation of the graph in column-$I$ and the shape of the graph in column-$II$ is shown. Match them appropriately.

Column-$I$	Column-$II$
$(a)$ $T^2 \to l$	$(i)$ Linear
$(b)$ $T^2 \to g$	$(ii)$ Parabolic
$(c)$ $T \to l$	$(iii)$ Hyperbolic

$A$ pendulum bob has a speed of $3\, m/s$ at its lowest position. The pendulum is $0.5\, m$ long. The speed of the bob,when the length makes an angle of $60^o$ with the vertical,will be ..... $m/s$ (If $g = 10\, m/s^2$)

$A$ simple pendulum of length $l$ is made to oscillate with an amplitude of $45^{\circ}$. The acceleration due to gravity is $g$. Let $T_0 = 2 \pi \sqrt{l / g}$. The time period of oscillation of this pendulum will be