Class 12th

Melting point of monocarboxylic acids with even number of carbon is more than that with odd number of carbon due to lattice energy.

With increase in molar mass of monocarboxylic acids size of alkyl group (Hydrophobic portion) increases and therefore solubility in water decreases.

 $h{\upsilon }_1=h{\upsilon }_0+K.E_1$

$\Rightarrow 3.8eV=0.6eV+K.E_1$

K. E₁ = 3.2 eV (i)

$\Rightarrow h{\upsilon }_2=h{\upsilon }_{01}+K.E_2$

1.4 = 0.6 + K.E₂

KE₂ = 0.8 eV (ii)

$\frac{K.E_1}{K.E_2}=\frac{3.2}{0.8}=4$

$\frac{v_1}{v_2}=2:1$

Kindly consider the following figure

  $f\left(x\right)+f\left(x+k\right)=n\forall x\in R,\&k>0............\left(i\right)$

Replace x by x + k.          

$f\left(x+k\right)+f\left(x+2k\right)=n...............\left(ii\right)$

From (i) & (ii), f(x + 2k) = f(x).

$\therefore f\left(x\right)$  is periodic with period = 2k.

$I_1={\displaystyle \underset{0}{\overset{4nk}{\int }}f\left(x\right)}dx=2x{\displaystyle \underset{0}{\overset{2k}{\int }}f\left(x\right)dx}...................\left(ii\right)$

$I_2={\displaystyle \underset{-k}{\overset{3k}{\int }}f\left(x\right)}dx$ put x = t + k

$={\displaystyle \underset{-2k}{\overset{2k}{\int }}t\left(t+k\right)dt}=2{\displaystyle \underset{0}{\overset{2k}{\int }}f\left(t+k\right)dt}$

$=2n{\displaystyle \underset{0}{\overset{2k}{\int }}ndx}=4n^{2}k.$

I_R = I_B (due to magnetic field)

(due to Radiation)

$\frac{P}{A}*Efficiency=\frac{B_0^2}{2{\mu }_0}c$

$\Rightarrow \frac{200}{4\pi *4^2}*\frac{3.5}{100}=\frac{B_0^2}{2*4\pi *10^{-7}}*3*10^8$

$\Rightarrow B_0=1.71*10^{-8}T$

$F_C=F_q$

$\Rightarrow \frac{m{v}^2}{R}=\frac{\rho R}{2{\in }_0}q$

$m{v}^2=\frac{\rho R^{2}q}{2{\epsilon }_0}\Rightarrow \frac{1}{2}m{v}^2=k.E.=\frac{1}{4}\frac{\rho R^{2}q}{{\in }_0}$

$U_i=\frac{Q^2}{2C}$

Uf = 1.44 Ui

Qf = Q + 2

$1.44U_i=\frac{{\left(Q+2\right)}^2}{2C}$

$\Rightarrow 1.44\frac{Q^2}{2C}=\frac{{\left(Q+2\right)}^2}{2C}$

$\Rightarrow 1.2Q=Q+2$

0.2 Q = 2

Q = 10 C

  $f\left(\frac{r}{4}\right)=\sqrt[]{2},f\left(\frac{r}{2}\right)=0\&f\text{'}\left(\frac{r}{2}\right)=1$

g(x)   ${\displaystyle \underset{}{\overset{}{\int }}\left(f\text{'}\left(t\right)sect+tantsectf\left(t\right)\right)dt}$

$={\left[f\left(t\right)sect\right]}_x^{\pi /4}$

=   $\sqrt[]{2}*\sqrt[]{2}-f\left(x\right)secx$

$=\frac{2cosx-f\left(x\right)}{cosx}$

=   $\frac{2sinx+1}{sinx}=3$

Sphalarite           ZnS

Calamine            ZnCO₃

Galena                PbS

Siderite               FeCO₃

$f\left(x\right)=\{\begin{array}{cc}\hfill \left[x\right],x<0\hfill & \hfill \left|1-x\right|,x\ge 0\hfill \end{array}$ $g\left(x\right)=\{\begin{array}{cc}\hfill c^x-x,x<0\hfill & \hfill {\left(x-1\right)}^2-1,x\ge 0\hfill \end{array}$

fog

$fog(\begin{array}{}\end{array}\left[e^x-x\right],\left(e^x-x<0\right)\left(x<0\right)\left[{\left(x-1\right)}^2-1\right]{\left(x-1\right)}^2-1<0x\ge 0\left|1-e^x+x\right|,e^x-x\ge 0x<0\left|1-{\left(x+1\right)}^2+1\right|,{\left(x-1\right)}^2-1\ge 0x\ge 0,x\in \left(2,\infty \right)($

 Not possible as of inequalities give $?.$  

e^x – x < 0, x < 0                 (x 1)² – 1 < 0

Not possible                     (x – 1 + 1)(x – 1 – 1) < 0

(x)(x – 2) < 0

x   $\in \left(0,2\right)$

continuous  $\forall$ x < 0

$fog\{\begin{array}{ccc}\hfill \left|1-e^x+x\right|,x<0\hfill & \hfill \left|1-{\left(x-1\right)}^2+1\right|,x=0\hfill & \hfill \left|1-{\left(x-1\right)}^2+1\right|,x\ge 2\hfill \end{array}$ Discontinuous at 0

continuous   $\forall x$

  $\therefore$  fog is discontinuous at 0