Physics

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{4}+\frac{1}{4}$  

 R_eq = 2

$\frac{\Delta R_{eq}}{R_{eq}^2}=\frac{\Delta R_1}{R_1^2}+\frac{\Delta R_2}{R_2^2}$

$\begin{array}{l}\frac{\Delta R_{eq}}{4}=\frac{0.8}{16}+\frac{0.4}{16}+\frac{1.2}{16}\\ R_{eq}=0.3\end{array}$                

$Y=\frac{Mg{L}^3}{4b{d}^3\delta }$

For significant error in Y

$=\frac{\Delta M}{M}+\frac{3\Delta L}{L}+\frac{\Delta b}{b}+3\frac{\Delta d}{d}+\frac{\Delta \delta }{\delta }$              

$=\frac{1*10^{-3}}{2}+\frac{3*10^{-3}}{1}+\frac{10^{-2}}{4}+3*\frac{0.01*10^{-1}}{0.4}+\frac{10^{-2}}{5}$             

= 0.0155

$Y=\frac{Mg{L}^3}{4b{d}^3?}$              

For significant error in Y

$=\frac{?M}{M}+\frac{3?L}{L}+\frac{?b}{b}+3\frac{?d}{d}+\frac{??}{?}$           

$=\frac{1*10^{?3}}{2}+\frac{3*10^{?3}}{1}+\frac{10^{?2}}{4}+3*\frac{0.01*10^{?1}}{0.4}+\frac{10^{?2}}{5}$              

= 0.0155

Direction of E in the direction of y axes so flux is only due to top and bottom surface for bottom surface y = 0 E = 0

and for top surface y = 0.5 m      So

$E=150*{\left(0.5\right)}^2=\frac{150}{4}$

$fluxflowing?=EA=\frac{150}{4}*{\left(0.5\right)}^2$              

$=\frac{150}{16}$             

Gausses law $?=\frac{q}{{\epsilon }_0}$  

  $\frac{150}{16}=\frac{q}{{\epsilon }_0}$            

$=8.3*10^{-11}C$              

In given circuit inductor behave as a simple wire so resultant circuit will be

R_ef = 2 + 1 = 3 $\Omega$  

V = IR

$l=\frac{30}{3}=10A$  

$l=\frac{V_{0}B\mathcal{l}}{R}=\frac{V_0*5*20*10^{-2}}{4+1}$         

  $2*10^{-3}=V_0*20*10^{-2}$

$V_0=\frac{2*10^{-3}}{20*10^{-2}}=\frac{2}{2}*10^{-2}=1*10^{-2}m/s$

V₀ = 1 cm/s

Object is moving in upward direction with constant velocity so in upward motion (+2N) and for downward motion (-2N) So option (1) is correct representation.

Adiabatic equation : $P_1{V}^{\gamma }=P_2{V}^{\gamma }\Rightarrow \left(200\right){\left(1200\right)}^{\gamma }=P{\left(300\right)}^{\gamma }$

$\Rightarrow P=\left(200\right){\left(4\right)}^{\frac{3}{2}}=1600KPa$

$W=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=\frac{240-480}{1.5-1}=-480J$  

Battery is connected while dielectric is inserted so potential difference will be  remains same.

$U_i=\frac{1}{2}c{V}^2$

$U_f=\frac{1}{2}Kc{V}^2$

$\Rightarrow \Delta U=\frac{1}{2}\left(K-1\right)c{V}^2=\frac{1}{2}*1*200*10^{-9}*200^2=4$

from given information : I = A = 60°, r = $\frac{A}{2}=30°$ =>  $\mu =\sqrt[]{3}=\frac{c}{v}$

$\Rightarrow v=\frac{c}{\sqrt[]{3}}\Rightarrow time=\frac{\left(\frac{5\sqrt[]{3}}{100}\right)}{\left(\frac{3*10^8}{\sqrt[]{3}}\right)}=5*10^{-10}s$