Class 12th

For sphere 'C' after contacting with 'A'. q_A = q_C = $\frac{q}{2}.$

offer contacting with 'B'. q_B = q_C = $\frac{3q}{4}$

$F_{Net}=\left|F_1-F_2\right|$

F₂ = $\frac{K\cdot 3q^2*4}{r^{2}8}=\frac{3}{2}\frac{k{q}^2}{r^2}$

$F_1=\frac{9k{q}^2*4}{16r^2}=\frac{9k{q}^2}{4r^2}=\frac{9}{4}F$

$\therefore F_{Net}=\left|\frac{9}{4}F-\frac{3}{2}F\right|$

= $\frac{9-6}{4}F=\frac{3}{4}F$

  $\frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-2\left(3a-4b+12c+19\right)}{3^2+{\left(-4\right)}^2+12^2}$

$\frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-6a+8b-24c-38}{169}$               

$\left(x,y,z\right)\equiv \left(a-6,\beta ,\gamma \right)$               

  $\frac{\left(a-b\right)-a}{3}=\frac{\beta -b}{-4}=\frac{\gamma -c}{12}=\frac{-6a+8b-24c-38}{169}$              

  $\frac{\beta -b}{-4}=-2$              

$\Rightarrow \beta =8+b$               

$\Rightarrow 3a-4b+12c=150$                   ….(i)

a + b + c = 5

$\Rightarrow 3a+3b+3c=15$ ….(ii)

Applying (i) – (ii), we get :

= 56 + 216 + 7b – 9c = 56 + 216 – 135 = 137

Based upon the properties and uses of chemical substances.

First we arrange 5 red cubes in a row and assume  number of blue cubes between them

$Here,x_1+x_2+x_3+x_4+x_5+x_6=11$            

and    $x_2,x_3,x_4,x_5\ge 2$

so $x_1+x_2+x_3+x_4+x_5+x_6=3$  

No. of solutions = ⁸C₅ = 56

$\left|adj\left(adj\left(A\right)\right)\right|={\left|A\right|}^{2^2}={\left|A\right|}^4$

$\therefore {\left|A\right|}^4=\left|\begin{array}{ccc}\hfill 14\hfill & \hfill 28\hfill & \hfill -14\hfill \\ \hfill -14\hfill & \hfill 14\hfill & \hfill 28\hfill \\ \hfill 25\hfill & \hfill -14\hfill & \hfill 14\hfill \end{array}\right|$

$={\left(14\right)}^3\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right|$

$={\left(14\right)}^3\left(3-2\left(-5\right)-1\left(-1\right)\right)$

${\left|A\right|}^4={\left(14\right)}^4\Rightarrow \left|A\right|=14$

In sucrose, glycosidic linkage is between C₁ of α-glucose and C₂ of β-fructose.

Based upon the properties and uses of chemical substances.

$f\left(x\right)=\frac{2e^{2x}}{e^{2x}+e^x}andf\left(1-x\right)=\frac{2e^{2-2x}}{e^{2-2x}+e^{1-x}}$

$\therefore \frac{f\left(x\right)+f\left(1-x\right)}{2}=1$

i.e. f (x) + f (1 – x) = 2

$\therefore f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+....+f\left(\frac{99}{100}\right)$

${\displaystyle \sum _{x=1}^{49}f\left(\frac{x}{100}\right)+f\left(1-\frac{x}{100}\right)+f\left(\frac{1}{2}\right)}$

= 49 * 2 + 1 = 99

Consider the below image