Class 12th

On decreasing pressure of NO by a factor of '2' the rate of reaction decreases by a factor of '4'

$\therefore$ Order of reaction w.r.t 'NO' = 2

Meq of K₂Cr₂O₇ = Meq of Fe²⁺

-> (Molarity * Volume * nf) of K₂Cr₂O₇ = (molarity * volume * nf) of Fe²⁺

->0.02 * 20 * 6 = M * 10 * 1

M = 0.24 M

Molarity = 24 * 10^-2 M

The current lags the voltage by an angle of 90 Degree (? /2 radians). This is due to the inductor opposing the change in current.

Radius, R = R₀A^1/3

take z = x + iy

$z^2+\overline{z}=0$               

$\Rightarrow x^2-y^2+x+i2xy-yi=0$               

$\Rightarrow x^2-y^2+x=0andy\left(2x-1\right)=0$             

if y = 0 Þ x = 0, -1

$ifx=\frac{1}{2}\Rightarrow y=\pm \frac{\sqrt[]{3}}{2}$               

$\Sigma \left(Re\left(z\right)+lm\left(z\right)\right)=\left(0-1+\frac{1}{2}+\frac{1}{2}\right)+\left(0+0+\frac{\sqrt[]{3}}{2}-\frac{\sqrt[]{3}}{2}\right)=0$               

Let A = $\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill d\hfill & \hfill e\hfill & \hfill f\hfill \\ \hfill 9\hfill & \hfill h\hfill & \hfill i\hfill \end{array}\right]$  

Now A^TA

trace will be    $a^2+b^2+c^2+d^2+e^2+f^2+9^2+h^2=6$

total ways = 

$\left|f\left(x\right)\right|\le 800\Rightarrow 2n^2-n-1\le 800$  

$\Rightarrow 2n^2-n-801\le 0$                

${\displaystyle \sum _{x\in S}^{}f\left(x\right)={\displaystyle \sum _{}^{}\left(2x^2-x-1\right)}}$                

$=2\left(19^2+18^2+........1^2+0^2+1^2+.....+20^2\right)$                

= 10620

y⁵ – 9xy + 2x = 0

differentiate 5y⁴ – 9x $\frac{dy}{dx}$ - 9y + 2 = 0

$\frac{dy}{dx}=\frac{9y-2}{5y^4-9x}$        

 For horizontal tangent $\frac{dy}{dx}=0\Rightarrow y=\frac{2}{9}$  which does not satisfy the equation so no horizontal

For vertical tangent -> $5y^4-9x=0$  

->m = 0, N = 2

$sin\left(2x^2\right).109_e\left(tan{x}^2\right)dy+4xydx=4\sqrt[]{2}.x.\left(sin{x}^{2}cos\frac{\pi }{4}-cos{x}^{2}sin\frac{\pi }{4}\right)dx$  

$\Rightarrow \mathcal{l}n\left(tan{x}^2\right)dy+\frac{4x}{sin\left(2x^2\right)}ydx=\frac{4x\left(sin{x}^2-cos{x}^2\right)}{sin\left(2x^2\right)}dx$

Integrate

$\Rightarrow y.\mathcal{l}n\left(tan{x}^2\right)=2.\mathcal{l}n\left(\frac{sin{x}^2+cos{x}^2-1}{sin{x}^2+cos{x}^2+1}\right)+C$

$x=\sqrt[]{\frac{\pi }{6}},y=1$  calculate C.