Electric Current

Divided into 10 parts

$R=\frac{\rho l}{A}$

$R^{\prime }=\frac{\rho l}{10A}=\frac{R}{10}$

$R_S=5*\frac{R}{10} [series]$

$R_S=50$

$R_P=\frac{R}{50}\left[ parallel \right]$

$R_{eq }=R_S+R_P$

$=52\Omega$

For uniformly charged spherical shell,

$V=\frac{kq}{R}\left( For r\le R\right)$

$\therefore V_C=V_P$

$V_C-V_P= Zero$

In first condition R₁ = 36 $\Omega$  

In second condition R₂ = 18  $\Omega$

$P_1=\frac{V^2}{R_1}=\frac{{\left(240\right)}^2}{36}$               

$P_2=\frac{V^2}{R_2}+\frac{V^2}{R_2}=\frac{{\left(240\right)}^2}{18}+\frac{{\left(240\right)}^2}{18}$               

$P_2=\frac{{\left(240\right)}^2}{9}$               

So   $\frac{P_1}{P_2}=\frac{{\left(240\right)}^2/36}{{\left(240\right)}^2/9}=\frac{1}{4}$

x = 4

$Current in circuit i=\frac{10}{4+1}=2 A$

$Terminal voltage =E-iR$

$=10-2*1=8 V$

$l=JAcos\theta \Rightarrow 5=J*0.04*cos60°\Rightarrow J=\frac{5}{0.02}=250A/m^2$

$J=\frac{E}{\rho }\Rightarrow E=\rho J=44*10^{-8}*250=11*10^{-5}V/m$              

$l_1=\frac{50}{2000}=25mA,andl=\frac{V_i-V_Z}{1000}=\frac{100-50}{1000}=50mA$

$\Rightarrow l_z=l-l_1=25mA$

Method-I : Using Kirchhoff's Law for loop $\overrightarrow{BDABB},$  we can write

$-140+20l+6l_1=0\Rightarrow 10l+3l_1=70.......\left(i\right)$              

Using Kirchhoff's Law for loop $\overrightarrow{ACBBA},$  

we can write

$5\left(l-l_1\right)+90-6l_1=0$              

->-5l + 11l₁ = 90 .(ii)

Adding equation (1) with twice the equation (2), we have

$25l_1=250\Rightarrow l_1=\frac{250}{25}=10A$              

Method-II : Using concept of equivalent cell, we can write

$V_{AB}=\frac{\frac{140}{20}+\frac{0}{6}+\frac{90}{5}}{\frac{1}{20}+\frac{1}{6}+\frac{1}{5}}=\frac{2100+0+5400}{15+50+60}$

$=\frac{7500}{125}=60Volt$

$\Rightarrow l_1=\frac{60}{6}=10A$              

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

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

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

  $2*10^{-3}=V_0*20*10^{-2}$

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

V₀ = 1 cm/s

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,

$8gsin30°+8acos30°=m{a}_1$  

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

$=\frac{2}{3}g$