Class 12th

Total possibilities = 2⁵ * 2⁵

Farounable case = ⁵C₂ * 3³ = 10 * 3³

$\therefore requiredprobability=\frac{10*3^3}{2^5*2^5}=\frac{5*27}{2^9}=\frac{135}{2^9}$

$2x{y}^2-y)dx+xdy=0$

$\Rightarrow \frac{dy}{dx}+2y^2-\frac{y}{x}=0$

$Put\frac{1}{y}=z$

Then $-\frac{1}{y^2}\frac{dy}{dx}=\frac{dz}{dx}$

$\frac{dz}{dx}+\frac{1}{x}z=2$

Point (2, 1) c = 2 – 4 = 2 y = $\frac{x}{x^2-2}$

$\left|y\left(1\right)\right|=1$

Aluminium - Kaolinite

Iron - Siderite

Copper - Malachite

Zinc  - Calamine

Blood – Negatively charged colloid.

According to Hardy – Schulze Rule

FeCl₃ is more efficient for blood clotting

${\mathcal{l}}_1:\frac{x-3}{1}=\frac{y+1}{2}=\frac{z-4}{2}=t$

${\mathcal{l}}_2:\frac{x-3}{2}=\frac{y-3}{2}=\frac{z-2}{1}=s$

$\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill \widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 2\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right|=\left(-2\widehat{i}+3\widehat{j}-2\widehat{k}\right)$

$\mathcal{l}:\overrightarrow{r}=\overrightarrow{0}+\lambda \left(-2\widehat{i}+3\widehat{j}-2\widehat{k}\right)$

$\mathcal{l}\&{\mathcal{l}}_1\to$

Point of intersection $\mathcal{l}\&{\mathcal{l}}_1$ is P (2, 3, 2)

Point Q on ${\mathcal{l}}_2$ is (3 + 2s, 3 + 2s, 2 + s).

Valium – Tranquilizer

Morphine – Analgesic

Norethindrone – Antifertility drug

Vitamin B₁₂ – Pernicious anaemia

Density = $\frac{}{}$ $\frac{mass}{volume}$

Cu > Co > Fe > Cr > Zn

  $\frac{a+20}{2}=3+7r$

a + 20 = 6 + 14r     . (i)

b = -2 + 10r           . (ii)

a = 18r – 2              . (iii)

Solving (i) and (iii) we get

20 + 18r – 2 = 6 + 14r

r = -3

$\therefore$   a = 14 + 14 (-3) = -56 and b = -2 -30 = -32

$\left|a+b\right|\left|-56-32\right|=88$

x – 2y = 1, x – y + kz = -2, ky + 4z = 6

 x – 2y + 0. z – 1 = 0

x – y + kz + 2 = 0

0x + ky + 4z – 6 = 0

0x + ky + 4z – 6 = 0

${\Delta }_1=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 0\hfill \\ \hfill -2\hfill & \hfill -1\hfill & \hfill k\hfill \\ \hfill 6\hfill & \hfill k\hfill & \hfill 4\hfill \end{array}\right|=-\left(k+10\right)\left(k+2\right)$   

For no solution

$\Delta =0,{\Delta }_1\ne 0$       

k = 2

radius r = 0.5 mm

$\sigma =5*10^{7}s/m$

$\overrightarrow{E}=10*10^{-3}V/m=10^{-2}V/m$

As we know

$J=\sigma E$

$J=\frac{i}{A}=\sigma E$

$i=\sigma EA=5*10^7*10^{-2}*\pi *{\left(0.5*10^{-3}\right)}^2=1.25\pi *10^{-1}A=125\pi mA=x^3\pi mA$

$$ $\therefore$  x = 5