Class 12th

Z = $\sqrt[]{R^2+{\left(x_C-x_L\right)}^2},$ if only L & C are present then R = 0 then p = 0

$r=\frac{mv}{qB}=\frac{\sqrt[]{2mk.E}}{qB}$  

$r\propto \frac{\sqrt[]{m}}{q}$                

m_p = m

q_p = q

m_d = 2m

q_d = q

m_a = 4m

q_a = 2q

$\frac{r_p}{r_d}=\frac{\sqrt[]{m_p}*q_d}{q_P*\sqrt[]{m_d}}$

$\frac{r_d}{r_{\alpha }}=\sqrt[]{2}$

$r_p:r_d:r_{\alpha }=1:\sqrt[]{2}:1$         

High permeability and low retentivity

Given

$\begin{array}{l}A=2\Omega \\ B=4\Omega \\ C=6\Omega \\ R_{eq}=\frac{2*4}{2+4}+6\end{array}$

$R_{eq}=\frac{22}{3}\Omega$

$f\left(-2\right)+f\left(3\right)=0.$ One root of f (x) = 0 is (-1)

Let's assume other root to be a

$\therefore f\left(x\right)=a\left(x+1\right)\left(x-\alpha \right)$

Given that f (-2) + f (3) = 0

a (-2 + 1) (-2 -a) + a (3 + 1) (3 - a) = 0

Þ 14 -3a = 0 Þ a =    $\frac{14}{3}$

$\therefore$ sum of roots =   $\frac{14}{3}-1=\frac{11}{3}$

$E\alpha \mathcal{l}$

$\Rightarrow \frac{E_1}{E_2}=\frac{{\mathcal{l}}_1}{{\mathcal{l}}_2}$

$\Rightarrow \frac{3}{2}=\frac{75}{{\mathcal{l}}_2}$

${\mathcal{l}}_2=\frac{75*2}{3}$

${\mathcal{l}}_2=50cm$

$\&{\mathcal{l}}_1=75cm$

$\Delta \mathcal{l}={\mathcal{l}}_1-{\mathcal{l}}_2=25cm$

R₁ =  $\left\{\left(a,1\right)\in N*N:\left|a-b\right|\le 13\right\}$

 $\left(a,a\right)\in R_{1}as\left|a-a\right|\le 13\left(Reflexive\right)$

$\left(a,b\right)\&\left(b,a\right)\in R_{1}as\left|a-b\right|=\left|b-a\right|\to \left(symmetric\right).$

But it is not necessary that if (a,b) & (b, c)  $\in R,then\left(a,c\right)\in R$

Eg   $-\left(21,10\right)\in R\&\left(10,1\right)\in Rbut\left(21,1\right)\notin R_1$

R₂ =   $\left\{\left(a,b\right)\in N*N:\left|a-b\right|\ne 13\therefore R_2\to Notequivalencesolutoin.\right\}$

$\left(a,b\right)\&\left(b,a\right)\in R_{2}as\left|a-b\right|=\left|b-a\right|$

But it is not necessary that if (a, b) & (b, c)  $\in R_2$  then (a, c) also $\in R_2$ .

Eg – (21, 1)   $\in R_2\&\left(1,8\right)\in R_{2}but\left(21,8\right)\notin R_2$

$\mu mg=\frac{k{q}_1{q}_2}{L^2}$

$\Rightarrow 0.25*0.01*10=\frac{9*10^9*2*10^{-7}*2*10^{-7}}{L^2}$

$=\frac{9*2^2*10^{-5+4}}{25*10}$

L = 12 cm

hu = hu₀ + K.E

Cases u = 2u₀

h2u₀ = hu₀ + K.E₁

K.E₁ = hu₀

$\frac{1}{2}m{v}_1^2=h{\upsilon }_0$

$v_1^2=\frac{2h{\upsilon }_0}{m}$  

$v_1=\sqrt[]{\frac{2h{\upsilon }_0}{m}}$ - (1)      

Now, cases 2

h 5u₀ = hu₀ + k.E₂

k.E₂ = 4hu₀

$\frac{1}{2}m{v}_2^2=4h{\upsilon }_0$

v₂ =     $\sqrt[]{\frac{8h{\upsilon }_0}{m}}$   - (2)

$\frac{v_2}{v_1}=\sqrt[]{\frac{8}{2}=2}$          

v₂ = 2v₁

n-factor of KMnO₄ in acidic medium = 5

n-factor of Mohr's salt = 1

meq of KMnO₄ = meq of Mohr's salt

0.01 * 5 * V = 0.05 * 1 * 20

Volume of KMnO₄ used, V = 20 mL

So; Volume of KMnO₄ left in burette = 50 -20mL

= 30 mL