Physics

Using magnetic field due to straight wire :

$B=\frac{{\mu }_{0}i}{4\pi r}\left(sin\alpha +sin\beta \right)$

$=\frac{10^{-7}*1.5}{\left(\frac{0.09}{2\sqrt[]{3}}\right)}*\left(sin60°+sin60°\right)=10^{-5}T$

So, magnetic field due to three wires

= 3 * 10^-5 T

inside the plane

Comparing E = 20 cos (2 * 10¹⁰ t – 200x) V/m to

$E=E_{0}cos\left(\omega t-kx\right)v/m$              

$\omega =2*10^{10},K=200$               

Speed = $\frac{2*10^{10}}{200}=10^{8}m/s$  

R.I. $=\frac{C}{speed}=\frac{3*10^8}{10^8}=3$  

$NowR.I.=\sqrt[]{{\epsilon }_r{\mu }_r}$

$3=\sqrt[]{{\epsilon }_r*1}$

${\epsilon }_r=9$              

Capacitor makes potential difference constant.

Using faraday's law magnetic field should be outward and decreasing with time

Work done is equal to change in K.E.

$So{W}_1+W_2=\frac{1}{2}M{\left(0.8\sqrt[]{gh}\right)}^2-0$ W₁ ^-> work done by mg

$mgh+W_2=\frac{1}{2}m*0.64gh$  W₂ -> work done by air friction

$W_2=0.32mgh-mgh=-0.68mgh$              

W₂ = -0.68 mgh

Time period T = 2π $\sqrt[]{\frac{l}{g_{eff}}}$

$\Rightarrow T_i=T=2\pi \sqrt[]{\frac{l}{g}}$    

$\Rightarrow T_f=T\text{'}=2\pi \sqrt[]{\frac{4l/3}{g\left(1-\frac{\rho }{\sigma }\right)}}=2\pi \sqrt[]{\frac{16l}{9g}}$

$\Rightarrow T\text{'}=\frac{4}{3}T$

Energy required to melt

Q = $MS\Delta T+ML$  

$\Rightarrow 10^{-1}*2*10^3*10+10^{-1}*3.33*10^5$      

->3.53 * 10⁴ J

Heat produce in wire

H = l²RT

  $Q=3.53*10^4={\left(\frac{1}{2}\right)}^2*\left(4*10^3\right)*t$

$t=\frac{3.53*10^4*4}{4*10^3}=35.3sec$             

$\overrightarrow{x}=\left|\overrightarrow{P}+\overrightarrow{Q}\right|=\sqrt[]{P^2+Q^2}$          

=> $\overrightarrow{y}=\left|\overrightarrow{P}-\overrightarrow{Q}\right|=\sqrt[]{P^2+Q^2}$

$\left|\overrightarrow{x}+\overrightarrow{y}\right|=\sqrt[]{x^2+y^2+2xy}$

$\Rightarrow \sqrt[]{3\left(P^2+Q^2\right)}=\sqrt[]{{\left(\sqrt[]{P^2+Q^2}\right)}^2+{\left(\sqrt[]{P^2+Q^2}\right)}^2+2\left(P^2+Q^2\right)cos{\theta }_1}$

$\Rightarrow {\theta }_1=60°$

=>Using same formula : θ₂ = 90°

Suppose acceleration of wedge is a and acceleration of block w. r.t. wedge is a₁ then N cos60° = Ma = 16 a -> N = 32 a

For block w.r.t. wedge

N + 8a sin 30° = 8g cos 30°

N = 8g cos 30° - 8a sin 30°

->32a = 8g cos 30° - 8a sin 30°

->a = $\frac{\sqrt[]{3}}{9}g$  

Now for 8 kg,

$8gsin30°+8acos30°=m{a}_1$  

$a_1=g*\frac{1}{2}+\frac{\sqrt[]{3}}{9}g*\frac{\sqrt[]{3}}{2}$  

$=\frac{2}{3}g$  

Using conservation of linear momentum:

=> 40 * 3m = 60 * m + v * 2m

=> v = 30 m/s

$K{E}_i=\frac{1}{2}*3m*{\left(40\right)}^2$      

$K{E}_f=\frac{1}{2}*m*{\left(60\right)}^2+\frac{1}{2}*2m*{\left(30\right)}^2$     

$\Rightarrow \frac{K{E}_f}{K{E}_i}=\frac{54}{48}$

Fractional change in kinetic energy = 1 $-\frac{K{E}_f}{K{E}_i}=\frac{1}{8}$