Class 12th

i_battery= $\frac{10?3}{1000}$  = 7mA

i_2kW = $\frac{3}{2000}$  =1.5 mA

i_z = (7 – 1.5) mA

= 5.5mA

For maximum value of s, initially, electron must move away from plate.

ut + $\frac{1}{2}$ at² = s

t = 1u = 1m/s               s = –1 m

1 * 1 – $\frac{1}{2}$ a * 1² = – 1

-> a = 4m/s²

$\frac{qE}{m}=4$      

  $\frac{q\sigma }{2{\epsilon }_{0}m}=4$     

$\sigma =\frac{4*2*9*10^{-12}*9*10^{-31}}{1.6*10^{-19}}$

$=\frac{8*81}{1.6}*10^{-24}$      

= 4.05 * 10^–22C/m²

${\displaystyle \int I.dt}={\displaystyle \underset{0}{\overset{10}{\int }}\left(20+3t\right)dt}$

$={\left(20t\right)}_0^{10}+3{\left(\frac{t^2}{2}\right)}_0^{10}=350C$

$\frac{1}{20}=\left(1.5-1\right)\left(\frac{2}{R}\right)$

R = 20cm

$\frac{1}{f\text{'}}=\left(\frac{1.5}{1.6}-1\right)\left(\frac{2}{R}\right)$            

$=\frac{-1}{16}*\frac{2}{20}$                  

f' = – 160cm

Equivalent circuit

$i=\frac{9}{4+1}=1.8\text{?}A$

$\frac{i}{2}=0.9\text{?}A$

(V_P– V_Q) = 0.9 * 6 – 0.9 * 2

V_C = 3.6V

U =  $\frac{1}{2}$ CV² =   $\frac{1}{2}$ * 4 * 3.6 * 3.6mJ

= 25.92mJ

$I=Q_{0.\omega }$            

$=\frac{CV}{\sqrt[]{LC}}=V\sqrt[]{\frac{C}{L}}$

$=12\sqrt[]{\frac{100*10^{-6}}{10*10^{-3}}}$

= 1.2 A

$\frac{h}{p_1}=\frac{h}{p_2}$

$\sqrt[]{2m_1{k}_1}=\sqrt[]{2m_2{k}_2}$

$\frac{k_2}{k_1}=\frac{m_1}{m_2}=1835$

$I=9{\displaystyle \underset{0}{\overset{9}{\int }}\left[\sqrt[]{\frac{10x}{x+1}}\right]dx}$

$=9\left[{\displaystyle \underset{0}{\overset{1/9}{\int }}0dx}+{\displaystyle \underset{1/9}{\overset{2/3}{\int }}dx}+{\displaystyle \underset{2/3}{\overset{9}{\int }}2dx}\right]$

$=9\left[\frac{2}{3}-\frac{1}{9}+2\left[9-\frac{2}{3}\right]\right]$

$=9\left[\frac{5}{9}+2*\frac{25}{3}\right]$

= 5 + 6 * 25

= 5 + 150

= 155

If x = 0, y = 6, 7, 8, 9, 10

If x = 1, y = 7, 8, 9, 10

If x = 2, y = 8, 9, 10

If x = 3, y = 9, 10

If x = 4, y = 10

If x = 5, y = no possible value

Total possible ways = (5 + 4 + 3 + 2 + 1) * 2

= 30

Required probability  $=\frac{30}{11*11}=\frac{30}{121}$

Given $\left|\overrightarrow{a}\right|=1$ , $\left|\overrightarrow{b}\right|=4$ , $\overrightarrow{a}\cdot \overrightarrow{b}=2$

$\overrightarrow{c}=2\left(\overrightarrow{a}*\overrightarrow{b}\right)-3\overrightarrow{b}$  

Dot product with  $\overrightarrow{a}$ on both sides

$\overrightarrow{c}\cdot \overrightarrow{a}=-6$ . (1)

Dot product with  $\overrightarrow{b}$  on both sides

$\overrightarrow{b}\cdot \overrightarrow{c}=-48$ . (2)

$\overrightarrow{c}\cdot \overrightarrow{c}=4{\left|\overrightarrow{a}*\overrightarrow{b}\right|}^2+9{\left|\overrightarrow{b}\right|}^2$

${\left|\overrightarrow{c}\right|}^2=4\left[{\left|\overrightarrow{a}\right|}^2\cdot {\left|\overrightarrow{b}\right|}^2-{\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)}^2\right]+9{\left|\overrightarrow{b}\right|}^2$

${\left|\overrightarrow{c}\right|}^2=4\left[\left(1\right){\left(4\right)}^2-\left(4\right)\right]+9\left(16\right)$

${\left|\overrightarrow{c}\right|}^2=4\left[12\right]+144$

${\left|\overrightarrow{c}\right|}^2=48+144$

${\left|\overrightarrow{c}\right|}^2=192$

$cos\theta =\frac{\overrightarrow{b}\cdot \overrightarrow{c}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|}$

$cos\theta =\frac{-48}{\sqrt[]{192}\cdot 4}$

$cos\theta =\frac{-48}{8\sqrt[]{3}\cdot 4}$

$cos\theta =\frac{-3}{2\sqrt[]{3}}$

$cos\theta =\frac{-\sqrt[]{3}}{2}$ $\theta =co{s}^{-1}\left(\frac{-\sqrt[]{3}}{2}\right)$