Class 11th

Potential energy is best defined as a stored energy in a system. This energy is stored due to the configuration or position of the objects within a conservative force field. Potential energy in physics tells us that it's the capacity of the work done when the system is allowed to move from that position to the reference point.  

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{{\left(-4\sqrt[]{\frac{2}{5}}\right)}^2}{a^2}+\frac{32}{b^2}=1$                

$\Rightarrow \frac{32}{5a^2}+\frac{9}{b^2}=1$ ……. (i)

From (i)

$\frac{6}{b^2}+\frac{9}{b^2}=1\Rightarrow b^2=15\&a^2=16$

$a^2+b^2=15+16=31$

  $\begin{array}{l}2x+y=4\\ 2x+6y=14\end{array}\}y=2,x=3$              

B (1, 2)

Let C (k, 4 – 2k)

Now AB² = AC²

->5k² – 24k + 19 = 0

$\alpha =\frac{6+1+\frac{10}{5}}{3}=\frac{18}{5}$    

Now 15 (a + b)

$15\left(\frac{17}{5}\right)=51$                

  $f\left(x\right)=x^4-4x+1=0$              

$f\text{'}\left(x\right)=4x^3-4$

$=4\left(x-1\right)\left(x^2+1+x\right)$              

$\Rightarrow$ Two solution

  $\frac{dy}{dx}=\frac{ax-by+a}{bx+cy+a}$

$=bxdy+cydy+ady=axdx-bydx+adx$                

$=\frac{c{y}^2}{2}+ay-\frac{a{x}^2}{2}-ax+bxy=k$              

$a{x}^2+a{y}^2+2ax-2ay=k$            

$\Rightarrow x^2+y^2+2x-2y=\lambda$              

Short distance of (11,6)

= $\sqrt[]{12^2+5^2}-5$

= 13 – 5

= 8

  $x={\displaystyle \sum _{n=0}^{\infty }{a}^n}=\frac{1}{1-a}y={\displaystyle \sum _{n=0}^{\infty }{b}^n=\frac{1}{1-b}{\displaystyle \sum _{n=0}^{\infty }{c}^n=\frac{1}{1-c}}}$               

Now,

a, b, c -> AP

1 – a, 1 – b, 1 – c -> AP

$\frac{1}{1-a},\frac{1}{1-b},\frac{1}{1-c}\to HP$

x, y, z -> HP

  $\overline{z}=i{z}^2$

Let z = x + iy

x – iy = I (x² – y² + 2xiy)

Case-I

x = 0

-y² = -y

y = 0, 1

Case – II

$y=-\frac{1}{2}$

$\Rightarrow x^2-\frac{1}{4}=\frac{1}{2}\Rightarrow x=\pm \frac{\sqrt[]{3}}{2}$

Area of polygon

$=\frac{1}{2}\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill \frac{\sqrt[]{3}}{2}\hfill & \hfill \frac{-1}{2}\hfill & \hfill 1\hfill \\ \hfill \frac{-\sqrt[]{3}}{2}\hfill & \hfill \frac{-1}{2}\hfill & \hfill 1\hfill \end{array}\right|=\frac{1}{2}\left|-\sqrt[]{3}-\frac{\sqrt[]{3}}{2}\right|=\frac{3\sqrt[]{3}}{4}$  

Mol wt A₂B = Mol wt of AB₃

 Hence (2A + B) = (A + 3B)

  $\therefore A=2B$

Atomic wt of A is two times of atomic wt of B.

$2NOC{l}_{\left(g\right)}?2NO\left(g\right)+C{l}_2\left(g\right)$

t = 0       2 mol/Lit             0            0

t_eq          2 – x                    x            x/2

$K_C=\frac{{\left[NO\right]}^2*\left[C{l}_2\right]}{{\left[NOCl\right]}^2}=\frac{x^2*x}{{\left(2-x\right)}^2*2}=\frac{x^3}{{\left(2-x\right)}^2*2}$      

$=\frac{0.4*0.4*0.4}{1.6*1.6*2}$ = 0.0125 = 125 * 10^-4

$\Delta V=2.4*10^{-26}m/sec$

And $\Delta x=10^{-7}m$  

According to Heisenberg's uncertainty

$\Delta v*\Delta x=\frac{h}{4\pi *m}$  

$\therefore m=\frac{h}{4\pi *\Delta v*\Delta x}=\frac{6.626*10^{-34}}{4*3.14*2.4*10^{-26}*10^{-7}}$

$\therefore$  m = 0.02198 Kg

Or 21.98 gm

Or 22 gm in the nearest integer.