Class 12th

$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}=r$

For 'B', x = 2r + 1, y = 3r – 1, z = -2r + 1

As AB is perpendicular to the line,

$\left\{\left(2r+1\right)-0\right\}2+\left\{\left(3r-1\right)-1\right\}3+\left\{\left(-2r+1\right)-2\right\}$

$r=\frac{2}{17}\Rightarrow B\left(\frac{2}{17},\frac{-11}{7},\frac{13}{17}\right)$     

direction ratios of AB

(2r + 1, 3r – 2, -2r – 1)

Equation of AB

$\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$

Kindly go through the solution

 $y=\overline{\overline{A}+B}=A.\overline{B}$

ax² + bx + c = 0

D = b² – 4ac

D = 0

b² – 4ac = 0

b² = 4ac

(i) AC = 1, b = 2 (1, 2, 1) is one way

(ii) AC = 4, b = 4

$\begin{array}{l}a=4c=1\\ a=2c=2\\ a=1c=4\end{array}\}3ways$

(iii) AC = 9, b = 6, a = 3, c = 3 is one way

1 + 3 + 1 = 5 way

Required probability = $\frac{5}{216}$   

All the charge given to a conducting sphere resides on outer surface.

Let sin = t

${\displaystyle \frac{}{}}$ ${\displaystyle \int \frac{sin\theta \left(2sin\theta .cos\theta \right)\left(si{n}^6\theta +si{n}^4\theta +si{n}^2\theta \right)\sqrt[]{2si{n}^4\theta +3si{n}^2\theta +6}}{2si{n}^2\theta }}$  d

sin = t

cos . d = dt

${\displaystyle \int \frac{u^{1/2}}{12}}du=\frac{u^{3/2}}{18}+C=\frac{{\left(2t^6+3t^4+6t^2\right)}^{3/2}}{18}+C=\frac{{\left(2si{n}^6\theta +3si{n}^4\theta +6si{n}^2\theta \right)}^{3/2}}{18}+C$

l + m – n = 0

l + m = n . (i)

l² + m² = n²

Now from (i)

l² + m² = (l + m)²

=> 2lm = 0

=>lm = 0

l = 0 or m = 0

=> m = n Þ l = n

if we take direction consine of line

$0,\frac{1}{\sqrt[]{2}},\frac{1}{\sqrt[]{2}}and\frac{1}{\sqrt[]{2}},0,\frac{1}{\sqrt[]{2}}$

cos a = $\frac{1}{2}$              

$si{n}^4\alpha +co{s}^4\alpha ={\left(\frac{\sqrt[]{3}}{2}\right)}^4={\left(\frac{1}{2}\right)}^4=\frac{9}{16}+\frac{1}{16}=\frac{10}{16}=\frac{5}{8}$  

  $\Delta E=13.6\left(\frac{1}{1^2}-\frac{1}{5^2}\right)=13.6*\frac{24}{25}eV$

$\Rightarrow \frac{hc}{\lambda }=13.6*\frac{24}{25}eV..........\left(1\right)$          

With the help of conservation of linear momentum, we can write

$\frac{h}{\lambda }=m_H{v}_H\Rightarrow \frac{hc}{\lambda }=c{m}_H{v}_H\Rightarrow v_H=\frac{\frac{hc}{\lambda }}{c{m}_H}=\frac{13.6*\frac{24}{25}*1.6*10^{-19}}{3*10^8*1.67*10^{-27}}=4.17m/s$           

xyz = 24

24 = 2³ * 3

Let's distribute 2, 3 among 3 variables. No. of positive integral solution =

No. of ways to distribute =

$\therefore In\Delta AQP$

$tan30°=\frac{PQ}{AQ}$

$\frac{1}{\sqrt[]{3}}=\frac{h}{x+y}$

x + y = $\sqrt[]{3}h.....\left(i\right)$ $\text{exception 25:}$

$\therefore ln\Delta BQP$

$tan45°=\frac{h}{y}$

h = y . (ii)

(i) & (ii) x + y = $\sqrt[]{3}y$ $\text{exception 25:}$

$\Rightarrow x=\left(\sqrt[]{3}-1\right)y.......\left(iii\right)$

Let the speed be S

$\frac{x}{S}=20$

x = 20.S

from (iii)

$\frac{y}{S}=10\left(\sqrt[]{3}+1\right)$

  $R_i=\frac{\rho \mathcal{l}}{A}$
$R_f=\frac{\rho \left(1.25\mathcal{l}\right)}{\left(A/1.25\right)}={\left(1.25\right)}^2*\frac{\rho \mathcal{l}}{A}$

$\Rightarrow R_f=1.5625*R_i$

$\Rightarrow \frac{R_f-R_l}{R_l}*100=56.25\mathrm{\%}$