Class 12th

$\underset{h\to 0}{lim}\frac{{\displaystyle {\int }_{\left(\frac{\pi }{2}-h\right)}^{{\left(\frac{\pi }{2}\right)}^3}cos\left(t^{1/3}\right)dt}}{h^2}$

$=\underset{h\to 0}{lim}\frac{0+3{\left(\frac{\pi }{2}-h\right)}^{2}cos\left(\frac{\pi }{2}-h\right)}{2h}$

$=\underset{h\to 0}{lim}\frac{3{\left(\frac{\pi }{2}-h\right)}^{2}sinh}{2h}$

$=\frac{3{\pi }^2}{8}$

P (2 obtained on even numbered toss) = k (let)

P (2) = $\frac{1}{6}$  

P (  $\overline{2}$ )= $\frac{5}{6}$  

$k=\frac{5}{6}*\frac{1}{6}+{\left(\frac{5}{6}\right)}^3*\frac{1}{6}+{\left(\frac{5}{6}\right)}^5*\frac{1}{6}+...$

$=\frac{\frac{5}{6}*\frac{1}{6}}{1-{\left(\frac{5}{6}\right)}^2}$

$=\frac{5}{11}$

f = 15

$m=-\frac{v}{u}=+\frac{1}{2}$            

  $v=-\frac{u}{2}$           

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$            

$\frac{2}{-u}+\frac{1}{u}=\frac{1}{f}$            

u = – f = – 15cm

$\left|\Delta E_0\right|=\left(-136\left\{1-\frac{1}{4}\right\}\right)eV$

|DE₀| = –10.2

$\lambda =\frac{12400}{10.2}*10^{-10}m$

$\rho =\frac{h}{\lambda }=\frac{6.63*10^{-34}*10.2}{12400*10^{-10}}$                  

$?$ $mv=\frac{h}{\lambda }$            

  $1.8*10^{-27}$           

$v=\frac{6.63*10.2*10^{-34}}{12400*10^{-10}}$

$v=\frac{6.63*10.2}{12400*1.8}*10^3$            

$=\frac{6.63*102}{124*1.8}=3.02$ ]

= 3 m/s

From Gauss law

$?=\frac{q_{enclosed}}{{\epsilon }_0}=\frac{5q}{{\epsilon }_0}$

Since Gand S are in parallel

-> V_G = V_S

->3mA * 10 = 8A * R_S

->R_S = 3.75mW

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