Physics

$\begin{array}{l}{f}_l=\mu N=0.6*mgcos\theta \\ f_l=0.6*2*10*\frac{4}{5}=9.6N\end{array}$

Component of weight along incline.

$=mgsin\theta =2*10*\frac{3}{5}=12N$

$\Rightarrow mgsin\theta >f_{l}soT=2.4N$

Elastic potential energy stored in the spring

$U=\frac{1}{2}k{x}^2$

$=\frac{1}{2}Fx=\frac{F^2}{2k}$

$Hence\frac{U_1}{U_2}=\frac{k_2}{k_1}\left. [Fissameforbothsprings\right]$

$=\frac{500}{1200}=\frac{5}{12}$

$K.E.\propto t^2$

$\begin{array}{l}\Rightarrow \frac{1}{2}m{v}^2\propto t^2\\ \Rightarrow v\propto t\\ \Rightarrow \frac{dv}{dt}=constant\\ \Rightarrow F=ma=\frac{mdv}{dt}=constant\end{array}$

From work energy theorem

$W_{sp}+W_E=?K$

$-\frac{1}{2}k{x}_0^2+QE.x_0=0$

$?x_0=\frac{2QE}{k}$

$\Delta V_L=V_{0\gamma \Delta T}$

$\Rightarrow \Delta V_s=V_0*3*\frac{\gamma }{3}*\Delta T$

$\Rightarrow \Delta V_L-\Delta V_S=0$

Applying Bernoulli's theorem

$\Rightarrow P_0+\rho g2h+2\rho gh+0=P_0+\frac{1}{2}2\rho v^2$

$\Rightarrow V^2=4gh$

$\Rightarrow v=2\sqrt[]{gh}$

Flux through a surface $\mathrm{?}=\stackrel{?}{E}.\stackrel{?}{A}$  

$\mathrm{?}=\left(6\stackrel{?}{i}-4\stackrel{?}{j}+7\stackrel{?}{k}\right).200\stackrel{?}{i}$

$=1200unit$

_θR = $\frac{1}{60°}$  for human eye

$\Rightarrow \frac{{\theta }_R}{2}=\frac{6.28}{2*x}$

$\Rightarrow x=21.6km$

For no emission of electron

$\lambda >{\lambda }_0$

$?{\lambda }_0=\frac{hc}{w_0},so\lambda >\frac{hc}{w_0}$

Resistance of whole wire is

$R=r^{\text{'}}*l$

$20=1*l$

$l=20m$

Let length of the shorter section = a

$\frac{1}{R}=\frac{1}{a}+\frac{1}{\left(20-a\right)}$

$\frac{1}{1.8}=\frac{20-a+a}{a\left(20-a\right)}?20a-a^2=36$

$a=2and18$

$?a=2m$