If the amplitude of linear S.H.M. is decreased then

In an n-p-n transistor 200 electrons enter the emitter in 10$$^{-8}$$ second. If 1% electrons are lost in the base, then the current that enters the emitter and the current amplification factor are respectively [e = 1.6 $$\times$$ 10$$^{-19}$$ C]

In the given figure potential at point 'A' is 900 volt and point 'B' is earthed. What will be the potential at point 'P' ?

Kinetic energy of a proton is equal to energy '$$E$$' of a photon. Let '$$\lambda_1$$' be the de-Broglie wavelength of proton and '$$\lambda_2$$' is the wavelength of photon. If $$\frac{\lambda_1}{\lambda_2} \alpha E^n$$, then the value of '$$n$$' is