Physics

              $\Delta L$ (change in length) $=\frac{F\mathcal{l}}{AE}=\frac{mg\mathcal{l}}{AE}$  

              $\Delta \mathcal{l}\propto g$  

              $\frac{\Delta {\mathcal{l}}_1}{g_1}=\frac{\Delta {\mathcal{l}}_2}{g_2}\Rightarrow \frac{10^{-4}}{10}=\frac{6*10^{-5}}{g_1}$  

              $g_1=\frac{6*10^{-4}}{10^{-4}}=$ 6 m/sec²

$Y=\frac{\sigma }{\in }=\frac{20}{10^{-10}},$ Y → young's modules of elasticity.

              Stress $\left(\sigma \right)$ at strain $\left(\in \right)$  5 * 10^-4

              $\sigma =\in Y$  

              $=5*10^{-4}*20*10^{10}$  

              = $10^2*10^6$  

              = 10⁸

              $U_d\left(Energydensity\right)=\frac{1}{2}\sigma \in$  

              $=\frac{1}{2}*10^8*5*10^{-4}$  

              $\begin{array}{l}=2.5*10^4\\ =25*10^3=25kJ/m^3\end{array}$  

Let 'h' be the height at which velocity becomes equal to magnitude of Acceleration

v = g = 10

v = u + at

10 = 0 + 10t

t = 1 sec

$h=ut+\frac{1}{2}a{t}^2$

$=0*1+\frac{1}{2}*10*1*1$

= 5m

In amplitude modulation, the amplitude of the career signal is varied in according with the modulating signal.

Dynamic Resistance at 2v

 = $\frac{\Delta v}{\Delta I}=\frac{2.1-2}{\left(10-5\right)*10^{-3}}=\frac{0.1}{5}*10^3$  

Dynamic Resistance at 4 v

= $\frac{\Delta v}{\Delta I}=\frac{4.2-4}{\left(250-200\right)}=\frac{0.2}{50*10^{-3}}=\frac{0.2}{50}*10^3$  

$\frac{R_{2v}}{R_{4v}}=\frac{0.1*10^3*50}{5*0.2*10^3}=5:1$  

${\lambda }_e=\frac{h}{mv}=\frac{h}{p_e}$    -(1)

${\lambda }_{Ph}=\frac{h}{P_{Ph}}$              -(2)

A/C to question

${\lambda }_e={\lambda }_{Ph}$  

$P_e=P_{Ph}\Rightarrow \frac{P_P}{P_{Ph}}=1$              -(3)

$k.E_e=E_e=\frac{1}{2}m{v}^2$  

$=\frac{1}{2}Pev$              -(4)

$k.E_{Ph}=E_{Ph}=m{c}^2$  

   = mc c

q = P_Ph c               -(5)

$\left(4\right)÷\left(5\right)$  

$\frac{E_e}{E_{Ph}}=\frac{v}{2c}$  

If size of object is very small as compare to wave length of EM wave in free space then, scattering will happen.