A simple pendulum doing small oscillations at a place $$R$$ height above earth surface has time period of $$T_1=4 \mathrm{~s}$$. $$\mathrm{T}_2$$ would be it's time period if it is brought to a point which is at a height $$2 \mathrm{R}$$ from earth surface. Choose the correct relation [$$\mathrm{R}=$$ radius of earth] :

In simple harmonic motion, the total mechanical energy of given system is $$E$$. If mass of oscillating particle $$P$$ is doubled then the new energy of the system for same amplitude is:

The bob of a pendulum was released from a horizontal position. The length of the pendulum is $$10 \mathrm{~m}$$. If it dissipates $$10 \%$$ of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is:

[Use, $$\mathrm{g}: 10 \mathrm{~ms}^{-2}$$]