The length of a simple pendulum is increased by $2\%$. Its time period will

$A$ seconds pendulum is placed in a space laboratory orbiting round the Earth at a height $3R$ from the Earth's surface. The time period of the pendulum will be ($R =$ radius of Earth).

What is the length of a simple pendulum,which ticks seconds (in $m$)?

The periodic time of a simple pendulum of length $1\, m$ and amplitude $2\, cm$ is $5\, s$. If the amplitude is made $4\, cm$,its periodic time in seconds will be

The sum of kinetic energy and potential energy of a simple pendulum bob is $0.02 \text{ J}$. The speed of the simple pendulum bob at the equilibrium position is approximately: (Consider mass of the bob = $20 \text{ g}$) (in $\text{ m/s}$)

The bob of a simple pendulum executes simple harmonic motion in water with a period $t$,while the period of oscillation of the bob is ${t_0}$ in air. Neglecting the frictional force of water and given that the density of the bob is $(4/3) \times 1000 \ kg/m^3$. What relationship between $t$ and ${t_0}$ is true?