Class 12th

In air $F=\frac{1}{4\pi {\in }_0}\frac{q_1{q}_2}{r^2}$  

In medium $F\text{'}=\frac{1}{4\pi \left(k{\in }_0\right)}\frac{q_1{q}_2}{r^2}$  

$F\text{'}=\frac{F_0}{K}$

$\frac{d?}{dt}=10t-3$            

at t = 2, V = 17

$i=\frac{V}{R}=\frac{17}{8}A$              

  $X=\overline{\left(A+\overline{B}\right)+\left(\overline{A}+B\right)}$

$\left(\overline{A+\overline{B}}\right)\cdot \left(\overline{\overline{A}+B}\right)$

$\left(\overline{A}\cdot \overline{B}\right)\cdot \left(\overline{A}\cdot \overline{B}\right)$

$\left(\overline{A}\cdot B\right)\cdot \left(A\cdot \overline{B}\right)=\overline{A}\cdot B\cdot A\cdot \overline{B}=0$  

Order of reactivity towards electrophile

Strength of +M/+R : –OH > –CH₃ > –Cl

In case of halogens, their –I effect dominates over

+M hence –Cl is deactivating and  is lesser reactive than  for incoming electrophile.

  $\frac{1}{f}=\left(\mu -1\right)\left(\frac{2}{R}\right)$                    f = 20 cm

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

$\frac{1}{v}+\frac{1}{10}=\frac{1}{20}$

Kindly go through the solution

Since $\frac{f}{2}<f_0$  

->current = 0

= IV cosf

$=\frac{20}{\sqrt[]{2}}*\frac{2}{\sqrt[]{2}}*$  cos 60°

= 10 W

Kindly go through the solution

p = i²R

P' = 0.64 i²R

By Brewsters law

m = tani

m =   $\sqrt[]{3}$

$1*\frac{\sqrt[]{3}}{2}=\sqrt[]{3}*sinr$

$sinr=\frac{1}{2}$    

r = 30°

|A| = 2

$\underset{2024times}{\underset{?}{adj\left(adj\left(adj....\left(a\right)\right)\right)}}={\left|A\right|}^{{\left(n-1\right)}^{2024}}$            

$={\left|A\right|}^{{\left(n-1\right)}^{2024}}$                                             

$=2^{2^{2024}}$                                          

$2^{2024}=\left(2^2\right)2^{2022}=4{\left(8\right)}^{674}=4{\left(9-1\right)}^{674}$             

-> $2^{2024}\equiv 4\left(mod9\right)$

->,  $2^{2024}\equiv 9m+4$  m ¬ even

$2^{9m+4}\equiv 16\cdot {\left(2^3\right)}^{3m}\equiv 16\left(mod9\right)$

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