Class 12th

Molality itself a strength representing terms for solution which does not depend upon the temperature.

Given, hT = 25m, R = 6400 km

hR = 49m

dm (Maximum distance for satisfactory communication)

dm = $\sqrt[]{2R{h}_T}+\sqrt[]{2R{h}_R}$ $\text{exception 25:}\text{exception 25:}$

dm = $\sqrt[]{2*6400*10^3*25}+\sqrt[]{2*6400*10^3*49}$ $\text{exception 25:}\text{exception 25:}$

= $\sqrt[]{5*10^4*6400}+\sqrt[]{6400*10^3*49*2}$ $\text{exception 25:}\text{exception 25:}$

= $192\sqrt[]{5}*10^2$ $\text{exception 25:}{}^{}$

= $k\sqrt[]{5}*10^2$ $\text{exception 25:}{}^{}$

k= 192

J (current density) = 4 * 10⁶ Am^-2

Area between radial distance  $\frac{R}{2}toR$

A =  $\pi \left[R^2-\frac{R^2}{4}\right]=\frac{3R^2}{4}\pi$

I = AJ

=  $\frac{3R^2}{4}\pi *4*10^6$

= 3R² * 10⁶ πAmp

= 3 (4 * 10^-3)² * 10⁶ πAmp

= 3 * 16 * 10^-6 * 10⁶π Amp

= 48πA

R shunt Resistance

Given

CD = 3m

CF = 2m

Current through potentiometer wire

I = $\frac{v}{R_0}=\frac{VA}{\rho L}$ $\frac{}{{}_{}}\frac{}{}$ (1)

A Area of potentiometer wire

$$  $\rho \to$ Resistivity of the wire

L Length of the potentiometer wire

Case 1 When 8 $\Omega$ $$ shunt is bal aced at 3m length

$V_{CD}=I{R}_{CD}=\frac{V}{\rho L}A*\frac{\rho CF}{A}$

$V_{CD}=\frac{V\mathcal{l}}{L}=\frac{V*3}{L}$

VCD = VAB = E – I1r

= $-\frac{E}{R+r}r=\frac{V}{L}*3$ $\frac{}{}\frac{}{}$

$$ $\Rightarrow E\left[1-\frac{r}{8+r}\right]=\frac{V}{L}*3$  (2)

Case 2 When 4 $$ $\Omega$ shunt is balanced at 2m length

$V_{CF}=I{R}_{CF}=\frac{V}{\rho L}A*\frac{\rho CF}{A}=\frac{V*2}{L}$

$V_{CF}=V_{AB}=E-I_{2}r$

= $E-\frac{E}{R+r}r$ $\frac{}{}$

= $E\left[1-\frac{r}{4+r}\right]$ $\frac{}{}$

$\frac{}{}\frac{}{}$  $\Rightarrow E\left(1-\frac{r}{4+r}\right)=\frac{V*2}{L}$ (3)

$\left(2\right)÷\left(3\right)$

$\frac{1-\frac{r}{8+r}}{1-\frac{r}{4+r}}=\frac{3}{2}$

$\Rightarrow \left(\frac{8}{8+r}\right)*\left(\frac{4+8}{4}\right)=\frac{3}{2}$

$\Rightarrow 4\left(4+r\right)=3\left(8+r\right)$

$\Rightarrow 16+4r=24+3r$

r = 24 – 16

r = 8 $\Omega$

Aniline has higher viscosity due to intermolecular H -bonding.

We know

Energy on any orbit is given by,

$E_n=\frac{-13.6}{n^2}{z}^2$

${}_{}\frac{}{}$  $\Rightarrow E_1=\frac{-13.6}{12}*9$ [n1 = 1] (1)

For 3rd orbit

$$ $E_3=\frac{-13.6}{9}*$   [n2 = 3]

E3 = 13.6ev

$\Delta E=E_3-E_1$

= 13.6 – (13.6 * 9)

$\Delta E=8*13.6ev=photonenergy$

$8*13.6ev=\frac{hc}{\lambda }$

$\Rightarrow 8*13.6ev=\frac{1242ev}{\lambda }nm$

$\lambda =11.415*10^9$

= 114.15 * 1010m

3d = 0.6mm

D = 80cm

= 800mm

Path difference is given by

BP – Andhra Pradesh = Dx

$\frac{}{}$ $\Rightarrow \frac{dy}{D}=\left(2n+1\right)\frac{\lambda }{2}$   [for Dark fringe at P]

n = 0, for first dark fringe

$\frac{dy}{D}=\frac{\lambda }{2}$

$\lambda =\frac{2d}{D}y$

$$ $=\frac{2d}{D}*\frac{d}{2}[y=\frac{d}{2},Given$  first dark fringe is observed on the screen directly opposite to one of the slits]

$\lambda =\frac{2*0.6*0.6}{800*2}$

$\lambda =450mm$

In strong ligand field  ${\mathrm{C}\mathrm{o}}^{3+}$ will have ${\mathrm{t}}_{2\mathrm{g}}^6{\mathrm{e}}_{\mathrm{g}}^0$ of configuration and $\mathrm{\Delta }\mathrm{t}=\frac{4}{9}{\mathrm{\Delta }}_0$

for lamp,   $P=\frac{v^2}{R}$

$\Rightarrow R=\frac{25*25}{5}=125\Omega$              

$l_{max}=\frac{v}{R}=\frac{25}{125}=\frac{1}{5}A$    

$\Rightarrow I_{max}=\frac{220}{125+R}=\frac{1}{5}$

$\Rightarrow 1100=125+R$        

R = 975  $\Omega$