Class 12th

For A : 100 gm solution 2 gm solute A

$\therefore Molality=\frac{2/M_A}{0.098}$

For B : 100 gm solution 8 gm solute B

$\therefore Molality=\frac{8/M_B}{0.092}$

$\frac{1}{4.261}=\frac{M_A}{M_B}$

$\therefore M_B=4.261*M_A$

SO₂ is absorbed to a greater extent then CH₄ on activated charcoal under same Gases with higher critical temperature are readily absorbed by activated charcoal.

$O_2,C{u}^{2+}andF{e}^{3+}$

are paramagnetic

$$ $\therefore$  Weakly attracted by magnetic field

NaCl and H2O are diamagnetic

$$ $\therefore$  Weakly repelled by magnetic field.

$-1\le \frac{x^2-3x+2}{x^2+2x+7}\le 1$                

$\Rightarrow 0\le \frac{2x^2-x+9}{x^2+2x+7}\&\frac{-5x-5}{x^2+2x+7}\le 0$                       

$x\in R\&-1\le x<\infty$

Electric Displacement

$\begin{array}{l}\overrightarrow{D}=\in o\overrightarrow{E}\\ \left[D\right]=\left[{\in }_0\overrightarrow{E}\right]=\left[{\in }_0\frac{\sigma }{{\in }_0}\right]\\ \left[D\right]=\left[\sigma \right]\end{array}$

Electric displacement $\left(\overrightarrow{D}\right)$ $\stackrel{}{}$ and surface change density

$C_6{H}_{12}{O}_6\to Glucose$

$\frac{MassofC}{massofglucose}=\frac{72}{180}$

% of C = 10.8 = $\frac{MassofC}{massofsolution}*100$

$\frac{10.8*250}{100}massofC$

$\therefore Molarity=\frac{0.375}{0.1825}=2.055=2.06$

Given : $\widehat{a}\cdot \widehat{b}=\widehat{b}\cdot \widehat{c}=\widehat{c}\cdot \widehat{a}=cos\theta \left(say\right)$  

$\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|=14$

$\left(\overrightarrow{a}*\overrightarrow{b}\right)\cdot \left(\overrightarrow{b}*\overrightarrow{c}\right)$            

  $=\overrightarrow{a}\cdot \left[\left(\overrightarrow{b}\cdot \overrightarrow{c}\right)\overrightarrow{b}-\left(\overrightarrow{b}\cdot \overrightarrow{b}\right)\overrightarrow{c}\right]$             

$=\left(\overrightarrow{a}\cdot \overrightarrow{b}\right)\left(\overrightarrow{b}\cdot \overrightarrow{c}\right)-{\left|\overrightarrow{b}\right|}^2\overrightarrow{a}\cdot \overrightarrow{c}$          

  $=\left|\overrightarrow{a}\right|{\left|\overrightarrow{b}\right|}^2\left|\overrightarrow{c}\right|\left(co{s}^2\theta -cos\theta \right)=14\left|\overrightarrow{b}\right|\left(co{s}^2\theta -cos\theta \right)$            

Similarly,  $\left(\overrightarrow{b}*\overrightarrow{c}\right)\cdot \left(\overrightarrow{c}*\overrightarrow{a}\right)$

=  $\left|\overrightarrow{b}\right|{\left|\overrightarrow{c}\right|}^2\left|\overrightarrow{a}\right|\left(co{s}^2\theta -sin\theta \right)=14\left|\overrightarrow{c}\right|\left(cos2\theta -sin\theta \right)\&\left(\overrightarrow{c}*\overrightarrow{a}\right)\cdot \left(\overrightarrow{a}*\overrightarrow{b}\right)$  

$=\left|\overrightarrow{c}\right|{\left|\overrightarrow{a}\right|}^2\left|\overrightarrow{b}\right|\left(co{s}^2\theta -sin\theta \right)=14\left|\overrightarrow{a}\right|\left(co{s}^2\theta -sin\theta \right)$

Given : 14  $\left(co{s}^2\theta -cos\theta \right)\left(\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|\right)=168$  

$\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|=\frac{12}{co{s}^2\theta -cos\theta }=\frac{12}{\frac{1}{4}-\left(-\frac{1}{2}\right)}=\frac{12}{\frac{3}{4}}=16$

Given :  $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$  are coplanar & pair wise equal angle.

$\left|z-1+i\right|\ge \left|z\right|$  

$\left|z+i\right|=\left|z-1\right|$

W = (2x, y) = (a, y)

Let S represent the line segment AB

For 'B'

x² + y² = 4

x = -y

-> x² = 2

$x=\pm \sqrt[]{2}$

-> $B\left(-\sqrt[]{2},\sqrt[]{2}\right)$  

$A\left(\frac{1}{2},-\frac{1}{2}\right)$

W (2x, y) lies on AB

-> $-\sqrt[]{2}<2x\le \frac{1}{2}$

$\left(\left|z\right|<2\right)$

$-\frac{1}{\sqrt[]{2}}<x\le \frac{1}{4}$

$\begin{array}{ccc}\hfill 3R\hfill & \hfill 2R\hfill & \hfill 4B\hfill \\ \hfill 5B\hfill & \hfill 3W\hfill & \hfill 2W\hfill \end{array}$  

I             II

Let

E₁ : a red ball is transferred from I to II

E₂ : a black is transferred from I to II

E₃ :a white transferred from I to II

E : a black ball is drawn from 2^nd bag after a ball from I to II was transferred.

$P\left(\frac{E_1}{E}\right)=\frac{P\left(E_1\cap E\right)}{P\left(E\right)}$  

$P\left(E\right)=P\left(E_1\cap E\right)+P\left(E_2\cap E\right)+P\left(E_3\cap E\right)$  

=  $P\left(E_1\right)\cdot P\left(\frac{E}{E_1}\right)+....+.....$  

=   $\frac{3}{10}\cdot \frac{5}{10}+\frac{4}{10}\cdot \frac{6}{10}+\frac{3}{10}\cdot \frac{5}{10}=\frac{54}{100}$

$P\left(E_1/E\right)=\frac{15/100}{54/100}=\frac{5}{18}$

Final range = 3 Initial Range

$\Rightarrow d_f=3d_i$ $\begin{array}{l}\Rightarrow \sqrt[]{2h\text{'}R}=3\sqrt[]{2hR}\\ \Rightarrow h\text{'}=9h=900m\end{array}$