Class 12th

Kindly go through the solution

R_eq = r

$P=\frac{V^2}{r}=\frac{4}{2}=2W$

The balanced reaction is.

            Cr₂  + 14H⁺ + 6e^– ® 2Cr³⁺ + 7H₂O

            x = 14

            y = 6

            A = 7

$\frac{1}{2}{\epsilon }_0{E}^2=\frac{B^2}{2{\mu }_0}$

$?E=CBandC=\frac{1}{{\mu }_0{\epsilon }_0}$          

$?=\frac{h}{p}\left[??\frac{1}{p}\right]$

Þ        lp = h (constant)

            So, the plot is a rectangular hyperbola.

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