Class 12th

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

f = 3GHz = 3 * 10⁹ Hz

${\lambda }_{vacuum}=\frac{v}{f}=\frac{3*10^{8}m/s}{3*10^9/s}=10^{-1}m=0.1m$

${\lambda }_{vacuum}=\frac{{\lambda }_{vacuum}}{{\mu }_{medium}}=\frac{0.1}{{\mu }_{medium}}$

${\mu }_{medium}=\sqrt[]{{\mu }_r{\in }_r}=\sqrt[]{1*2.25}=1.5$

$\therefore {\mu }_{medium}=\frac{0.1m}{1.5}=0.0667m=6.67cm=667*10^{-2}cm$

$l={\displaystyle \underset{1}{\overset{3}{\int }}\left[x^2-2x+1-3\right]dx=}{\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2-3\right]dx={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-3{\displaystyle \underset{1}{\overset{3}{\int }}dx}}}$  .(A)

$l_1={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dxPut{\left(x-1\right)}^2=t}$

$l_1=\frac{1}{2}\left[0{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\sqrt[]{t}}}{\displaystyle \underset{1}{\overset{2}{\int }}\frac{dt}{\sqrt[]{t}}}+2{\displaystyle \underset{2}{\overset{3}{\int }}\frac{dt}{\sqrt[]{t}}}+3{\displaystyle \underset{3}{\overset{4}{\int }}\frac{dt}{\sqrt[]{t}}}\right]=\frac{1}{2}\left\{{\left|\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_1^2{\left|+2\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_2^3{+3\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}]}_3^4\right\}$      = $5-\sqrt[]{2}-\sqrt[]{3}$  

Hence from (A)

= $5-\sqrt[]{2}-\sqrt[]{3}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$

2^nd method       

${\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-6..............\left(A\right)}$

From (A), $l=5-\sqrt[]{2}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$  

Sound level is given in dB = 10 log $\frac{{}_{}}{{}_{}}\frac{}{{}_{}}$ $\left(\frac{P_0}{P_i}\right)or10log\left(\frac{l}{l_0}\right)$

As sound level decreases 5 dB every km,

So, in 20 km sound level will decrease by 20 * 5 = 100 dB.

$\Delta \beta ={\beta }_2-{\beta }_1=10log\left(\frac{l_2}{l_1}\right)$

$-100=10log\left(\frac{l_2}{l_1}\right)$

$10^{-10}=\frac{l_2}{l_1}\Rightarrow l_2=10^{-10}{l}_1$

$P_2=10^{-10}{P}_1$

$P_2=10^{-10}*\left(0.1*10^3\right)\Rightarrow P_2=10^{-8}W=10^{-x}W$

$$ $\therefore$  x = 8