If a simple pendulum is taken to a place where $g$ decreases by $4\%$,then the time period

$A$ 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 $g = 10\, m/s^2$)

At a place,the length of the oscillating simple pendulum is made $\frac{1}{4}$ times keeping the amplitude the same. Then,the total energy will be:

$A$ pendulum clock is set to give correct time at sea level. This clock is moved to a hill station at an altitude of $2500 \ m$ above sea level. In order to keep correct time at the hill station,the length of the pendulum:

The length of a seconds pendulum is $1 \,m$ on the earth. If the mass and diameter of a planet are double that of the earth, then the length of the seconds pendulum on the planet will be: (in $\,m$)

$A$ simple pendulum consisting of a mass $M$ attached to a string of length $L$ is released from rest at an angle $\alpha$. $A$ pin is located at a distance $l$ below the pivot point. When the pendulum swings down,the string hits the pin as shown in the figure. The maximum angle $\theta$ which the string makes with the vertical after hitting the pin is: