Class 12th

n-factor of KMnO₄ in acidic medium = 5

n-factor of Mohr's salt = 1

meq of KMnO₄ = meq of Mohr's salt

0.01 * 5 * V = 0.05 * 1 * 20

Volume of KMnO₄ used, V = 20 mL

So; Volume of KMnO₄ left in burette = 50 -20mL

= 30 mL

.For maxima = y = (2n + 1) $\frac{\lambda }{2}\frac{D}{a}$ $\frac{}{}\frac{}{}$

For 1st maxima for l1 wavelength (n = 1)

${}_{}\frac{{}_{}}{}\frac{}{}$  $y_1=\frac{3{\lambda }_1}{2}\frac{D}{a}$ - (1)

First maxima for l2 wavelength

${}_{}$ $y_2=\frac{3}{2}{\lambda }_2\frac{D}{a}$  - (2)

$\Rightarrow y_2-y_1=\frac{3}{2}\frac{D}{a}\left[{\lambda }_2-{\lambda }_1\right]$

$=\frac{3}{2}*\frac{2*5*10^{-9}}{0.5*10^{-3}}$

$=\frac{3}{10^{-3}}*10^{-8}$

$\Delta y=3*10^{-5}m$

Moles of C = moles of CO₂

=  $\frac{0.793}{44}mol$

Mass of C =  $\frac{0.793}{44}*12g$

= 0.261g

Moles of H = 2 *  $\frac{0.442}{18}mol$

Mass of H =  $2*\frac{0.442}{18}*1g$

Total mass of compound = 0.492g (given)

So; mass of O = (0.492 – 0.216 – 0.049) g

= 0.227g

% of O = $\frac{0.227}{0.492}*100$  

= 46.14%

the nearest integer = 46

Let l = l₀ cos wt

Then v = v₀ sinwt

at t = 0, v = 0

but l = l₀

$l_{rms}=\frac{l_0}{\sqrt[]{2}}$

$V_{rms}=l_{rms}z$      

$\Rightarrow 220=\frac{l_0}{\sqrt[]{2}}\left(X_L\right)$        

$\Rightarrow 220=\frac{l_0}{\sqrt[]{2}}\left(2\pi *50*200*10^{-3}\right)$        

$\Rightarrow 220=\frac{l_0}{\sqrt[]{2}}\left(20\pi \right)$

$l_0=\frac{220\sqrt[]{2}}{20\pi }$  

$l_0=\frac{11\sqrt[]{2}}{\pi }=\frac{\sqrt[]{a}}{\pi }$

a = 121 * 2

a = 242

Bromination through free radical mechanism occurs at allylic carbon.

I_b = 10 µA

I_C = 1.5 mA

R_L = 50 kW or (Rc)

Base – emitter voltage = 10 mv

$R_B=\frac{V_B}{I_B}=\frac{10*10^{-3}}{10*10^{-6}}=10^3\Omega$

$A_v=\left(\frac{\Delta l_C}{\Delta l_B}\right)*\left(\frac{R_C}{R_B}\right)$

= $\frac{1.5*10^{-3}}{10*10^{-6}}*\frac{5*10^3}{10^3}$  

A_v = 750

f₁ = 15 cm          

$P_1=\frac{1}{15}*100=\frac{100}{15}D$

$P_3=\frac{100}{15}D$

$\frac{1}{f^2}=\left({\mu }_2-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$=\left(1.25-1\right)*\frac{-2}{R}$

$\frac{1}{f_2}=\frac{0.25*-2}{15}$

$\frac{1}{f_2}=\frac{-1}{30}c{m}^{-1}$

$\begin{array}{l}{p}_2=\frac{1}{f_2}=-\frac{100}{30}\\ P_R=P_1+P_2+P_3\\ P_R=\frac{100}{15}-\frac{100}{30}+\frac{100}{15}\\ P_R=10D\end{array}$

$P_R=\frac{1}{f_R}$

$f_R=\frac{1}{10}*100cm$

$f_R=10cm$        

$f\left(x\right)=a{x}^2+bx+c$  

g (x) = px + q

$f\left(g\left(x\right)\right)=a{\left(px+q\right)}^2+b\left(p\right)\left(+q\right)+c$                

$\Rightarrow 8x^2-2x=a{\left(px+q\right)}^2+b\left(px+q\right)+c$                

Compare 8 = ap² …………… (i)

-2 = a (2pq) + bp

0 = aq² + bq + c

$9\left(f\left(x\right)\right)=p\left(a{x}^2+bx+c\right)+q$                

? 4x² + 6x + 1 = apx² + bpx + cp + q

? Andhra Pradesh = 4 ……………. (ii)

6 = bp

1 = cp + q

From (i) & (ii), p = 2, q = -1

? b = 3, c = 1, a = 2

f (x) = 2x² + 3x + 1

f (2) = 8 + 6 + 1 = 15

g (x) = 2x – 1

g (2) = 3

(a) $CoC{\mathcal{l}}_3.4N{H}_{3}is\left[CoC{l}_2{\left(N{H}_3\right)}_4\right]Cl$ will show cis- trans isomerism as;

(b) $CoC{\mathcal{l}}_3.5N{H}_{3}is\left[Co{\left(N{H}_3\right)}_{5}C\mathcal{l}\right]C{\mathcal{l}}_2$ will not show cis- trans isomerism.

(c) $CoC{\mathcal{l}}_3.6N{H}_{3}is\left[Co{\left(N{H}_3\right)}_6\right]C{\mathcal{l}}_3$ will not show cis- trans isomerism.

(d) $CoC\mathcal{l}{\left(N{O}_3\right)}_2.5N{H}_{3}is\left[Co{\left(N{H}_3\right)}_{6}C\mathcal{l}\right]{\left(N{O}_3\right)}_2$  will not show cis-trans isomerism.

  $B_C=\frac{{\mu }_{0}l}{2r}=B$ ${}_{}\frac{{}_{}}{}$ -(1)

$\begin{array}{l}{B}_P=\frac{{\mu }_{0}l{r}^2}{2{\left(r^2+\frac{r^2}{4}\right)}^{3/2}}\\ B_P=\frac{{\mu }_{0}l{r}^2}{2*r^3{\left(\frac{5}{4}\right)}^{3/2}}\end{array}$

$B_P=\frac{{\mu }_{0}l}{2r{\left(\frac{5}{4}\right)}^{3/2}}$

$\Rightarrow B_P=\frac{B{4}^{3/2}}{5^{3/2}}$

$B_P=B\frac{2^3}{{\left(\sqrt[]{5}\right)}^3}$

$B_P={\left(\frac{2}{\sqrt[]{5}}\right)}^{3}B$