24 Jan 2023 Shift 2

Potential at centre$\;V=\;{ (\lambda \cdot \pi {\text{R}}_2) }/{4 \pi \epsilon_0 {\text{R}}_2}+\;{ (\lambda \cdot \pi {\text{R}}_1) }/{4 \pi \epsilon_0 {\text{R}}_1}$ $\;=\;\frac{\lambda }{2 \epsilon_0}$

For monoatomic gas $\gamma _1=\;\frac{5}{3}$For diatomic gas at low temperatures$\gamma _2=\;\frac{7}{5}$ $\therefore \;\frac{\gamma _1}{\gamma _2}=\;\frac{\;\frac{5}{3}}{\frac{7}{5}}=\;\frac{25}{21}$

$\;\lambda _{ {\text{D}}}=\;{ {\text{h}}}/{ {\text{p}}}=\;{ {\text{h}}}/{\sqrt{2 {\text{mK}}}}$ $\; \therefore \lambda \propto \;{1}/{\sqrt{ {\text{m}}}}$ $\; \because {\text{m}}_alpha > {\text{m}}_{ {\text{p}}} > {\text{m}}_{ {\text{e}}}$ $\;\lambda _{ {\text{e}}} > \lambda _{ {\text{p}}} > \lambda _alpha$

$\; {\text{T}}^2 \propto {\text{R}}^3 \Rightarrow (\;{ {\text{T}}_1}/{ {\text{T}}_2}) ^2= (\;{ {\text{R}}_1}/{ {\text{R}}_2}) ^3$ $\;\Rightarrow (\;\frac{1}{2.83}) ^2= (\;{1.5 \times 10^6}/{ {\text{R}}_2}) ^3$ \;\Rightarrow {\text{R}}_2= \[(2.83)^2 × (1.5 × 10^6) ^3]$ ^{1 / 3}$$;=8^{1 / 3} \times 1.5 \times 10^6=3 \times 10^6 {\text{km}}$

AM Broadast $\to 540-1600 {\text{KHz}}$FM Broadcast $\to 88-108 {\text{MHz}}$Television $\to 54-890 {\text{MHz}}$Salellite communication $\to 3.7-4.2 {\text{GHz}}$ $\therefore$ A-II, B-I, C-IV, D-III

${\text{Y}}=\overline{ {\text{A}}_1 ⊙ {\text{B}}_1}$ NAND

$\; {\text{B}}=\mu _0 {\text{nI}}$ $\;=4 \pi \times 10^{-7} \times 70 \times 10^2 \times 2$ $\;=56 \pi \times 10^{-4} {\text{T}}$ $\;=176 \times 10^{-4} {\text{T}}$

$\; {\text{g}}^{'}=\;{ {\text{g}}}/{ (1+\;{ {\text{h}}}/{ {\text{R}}}) ^2}$ $\; {\text{g}}^{'}= {\text{g}} \{1-\;{ {\text{d}}}/{ {\text{R}}}\}$Statement I is correct & Statement II is incorrect

$\;\int {\text{d}} \epsilon=\int {\text{B}}(\omega {\text{x}}) {\text{dx}}$ $\;\epsilon= {\text{B}} \omega int_{0}^{ {\text{L}}} {\text{xd}} {\text{x}}=\;{ {\text{B}} \omega {\text{L}}^2}/{2}$

Based on fact.

\;{[ {\text{T}}^{-1}]\ = $[ {\text{L}}^1]$ ^{ {\text{a}}} $[ {\text{M}}^1 {\text{L}}^{-3}]$ ^{ {\text{b}}} $[;{ {\text{MLT}}^{-2}}/{ {\text{L}}}]$ ^{ {\text{c}}}}$$;\Rightarrow {\text{T}}^{-1}= {\text{M}}^{ {\text{b}}+ {\text{c}}} ⋅ {\text{L}}^{ {\text{a}}-3 {\text{b}}} ⋅ {\text{T}}^{-2 {\text{c}}}$$; {\text{c}}=;\frac{1}{2}, {\text{b}}=-;\frac{1}{2}, ;; {\text{a}}-3 {\text{b}}=0$$; {\text{a}}+;\frac{3}{2}=0 \Rightarrow {\text{a}}=-;\frac{3}{2}$

Natural length $= {\text{L}}_0$Extension $=\text{bo} {x}$ $\; {\text{kx}}= {\text{m}} ( {\text{L}}_0+ {\text{x}}) \omega ^2$ $\;\Rightarrow 12.5 {\text{x}}=\;\frac{1}{5} ( {\text{L}}_0+ {\text{x}}) 25 \Rightarrow 1.5 {\text{x}}= {\text{L}}_0$ $\;\Rightarrow \;{ {\text{x}}}/{ {\text{L}}_0}=\;\frac{2}{3}$

$\; {\text{R}}_{ {\text{eq}}}=\;\frac{400 \times 100}{500}+100$ $\;=180 Ω$ $\; {\text{i}}=\;\frac{90}{180}=\;\frac{1}{2} {\text{A}}$ $\;\;\text{ Reading }\;=\;\frac{1}{2} \times \;\frac{400}{500} \times 100$ $\;=40 \;\text{ volt }\;$

${\text{C}}=\;{\omega }/{ {\text{k}}}=\;{ {\text{E}}_0}/{ {\text{B}}_0}$

Displacement $=$ area $=16-8+16-8=16 {\text{m}}$Distance $=Σ \mid$ area $\mid =48 {\text{m}}$ $\;\frac{\;\text{ displacement }\;}{\;\text{ Distance }\;}=\;\frac{1}{3}$

Concept based

${\text{T}} \propto \;{1}/{\sqrt{ {\text{g}}}}$

$\;{ {\text{hc}}}/{\lambda }= [1-\;\frac{1}{16}] (13.6 {\text{eV}})$So, $\lambda =94.1 {\text{nm}}$

For isothermal process ${\text{P}}- {\text{V}}$ graph is rectangular hyperbolaAs dotted line is isobaric line which implies $T_3 > T_2 > T_1$ as volume is increasing.

\;{ \vec{\text{P}} \cdot { \vec{\text{Q}}=0 \;(\;{ {\text{i}}}^{\^}+2 {\text{m}} \;{ {\text{j}}}^{\^}+ {\text{m}} \;{ {\text{k}}}^{\^}) \cdot (4 \;{ {\text{i}}}^{\^}-2 \;{ {\text{j}}}^{\^}+ {\text{m}} \;{ {\text{k}}}^{\^})=0 $\;\Rightarrow 4-4 {\text{m}}+ {\text{m}}^2=0$ $\;\Rightarrow ( {\text{m}}-2)^2=0 \Rightarrow {\text{m}}=2$

$\;I_1=\;\frac{m_1 R^2}{2} \;\; I_2=\;{m_2( {\text{R}} / 2)^2}/{2}$ $\;\;\frac{I_1}{I_2}=\;\frac{4 m_1}{m_2}=\;\frac{4 \cdot \rho \pi R^2 ℓ}{\rho \cdot \;\frac{\pi R^2}{4} \times \;\frac{ℓ}{2}} \Rightarrow \;\frac{I_1}{I_2}=32$

$\; {\text{T}}=2 \pi \sqrt{\;{ {\text{m}}}/{ {\text{k}}}}=1$ $\; {\text{T}}^{'}=2 \pi \sqrt{\;{ {\text{m}}+3}/{ {\text{k}}}}=2$ $\;\;{ {\text{T}}}/{ {\text{T}}^{'}}=\sqrt{\;{ {\text{m}}}/{ {\text{m}}+3}}=\;\frac{1}{2}$ $\;\Rightarrow \;{ {\text{m}}}/{ {\text{m}}+3}=\;\frac{1}{4}$ $\; {\text{m}}=1$

No. of mole $=\;\frac{120}{240}=\;\frac{1}{2}$No. of molecules $=\;\frac{1}{2} \times {\text{N}}_{ {\text{A}}}$Energy released $=\;\frac{1}{2} \times 6 \times 10^{23} \times 200$ $=6 \times 10^{25} {\text{MeV}}$

$\; {\text{C}}_0=\;{\in _0 {\text{A}}}/{ {\text{d}}}=15 {\text{pF}}$ $\; {\text{C}}=\;{ {\text{K}} \in _0 {\text{A}}}/{2 {\text{d}}}=\;\frac{3.5}{2} \times 15 {\text{pF}}=\;\frac{105}{4} {\text{pF}}$

{ {\text{F}}}^{\to }= {\text{ti}}+3 {\text{t}}^2 \;{ {\text{j}}}^{\^} \;{ {\text{md}} { {\text{v}}}^{\to }}/{ {\text{dt}}}= {\text{t}} \;{ {\text{i}}}^{\^}+3 {\text{t}}^2 \;{ {\text{j}}}^{\^} {\text{m}}=1 {\text{kg}}, int_{0}^{{ {\text{v}}}^{\to }} {\text{dv}}=int_{0}^{ {\text{t}}} {\text{tdt}} \;{˙{i}}^{\^}+int_{0}^{ {\text{t}}} 3 {\text{t}}^2 {dt \;{j }^{\^} { {\text{v}}}^{\to }=\;{ {\text{t}}^2}/{2} \;{ {\text{i}}}^{\^}+ {\text{t}}^3 \;{ {\text{j}}}^{\^} $\;\text{ Power }\;={ {\text{F}}}^{\to } \cdot { {\text{V}}}^{\to }=\;{ {\text{t}}^3}/{2}+3 {\text{t}}^5$ $\;\text{ At }\; {\text{t}}=2, \;\text{ power }\;=\;\frac{8}{2}+3 \times 32$ $=100$

As volume is constant,So resistance $\propto$ (length $)^2$ $\Rightarrow \% \;\text{ change \in resistance }\;=20+20+\;\frac{400}{100}=44 \%$

Force on $5 {\text{cm}}$ side is\;|{ \vec{\text{F}}|= {\text{ILB}} sin \;\theta $\;=(2) (5 \times 10^{-2}) \times \;\frac{3}{4} \times \;\frac{12}{13}=\;\frac{9}{130} {\text{N}}$So, $x=9$

After long time an inductor behaves as a resistance-less path.So current through cell${\text{I}}=\;{12}/{ {\text{R}} / 3}=3 {\text{A}}\{ \because {\text{R}}=12 Ω\}$

$\;\;{ {\text{I}}}/{ {\text{f}}_{ {\text{H}}_2 {\text{O}}}}= (\;{\mu _{ {\text{g}}}}/{\mu _{ {\text{H}}_2 {\text{O}}}}-1) (\;{2}/{ {\text{R}}})$ $\;=\;\frac{1}{8} (\;{2}/{ {\text{R}}})$ $\;=\;{1}/{ (4 {\text{f}}_{ {\text{air}}}) }$So, ${\text{f}}_{ {\text{H}}_2 {\text{O}}}=4 {\text{f}}_{\;\text{air }\;}=72 {\text{cm}}$So change in focal length $=72-18=54 {\text{cm}}$

When the ball attain terminal velocity$\; {\text{F}}_{ {\text{v}}}= ( {\text{mg}}- {\text{F}}_{ {\text{B}}}) ( \because {\text{a}}=0)$ $\;= {\text{V}} \sigma_{ {\text{b}}} {\text{g}}- {\text{V}} \rho _{ℓ} {\text{g}}$ $\;={\text{Vg}} (\sigma_{ {\text{b}}}-\rho _{ℓ})$ $\;=\;\frac{4}{3} \pi (10^{-3}) ^3 \times 9.8(10.5-1.5) \times 10^3$ $\;=3696 \times 10^{-7} {\text{N}}$ $\;\;\text{ So, }\; {\text{x}}=7$

${\text{Ce}}^{+4}$ and ${\text{Tb}}^{+4}$ act as oxidising agent.

${\text{N}}_2$ $\;\sigma 1 {\ s}^2 \sigma^* 1 {\ s}^2 \sigma 2 {\ s}^2 \sigma^* 2 {\ s}^2 \pi 2 {\text{p}}_{ {\text{x}}}^2=\pi 2 {\text{p}}_{ {\text{y}}}^2 \;{\sigma 2 {\text{p}}_{ {\text{z}}}^2}/{\;\text{ HOMO }\;}$ $\; {\text{N}}_2^{+}-\sigma 1 {\ s}^2 \sigma^* 1 {\ s}^2 \sigma^* 2 {\ s}^2 \sigma^* 2 {\ s}^2 \pi 2 {\text{p}}_{ {\text{x}}}^2=\pi 2 {\text{p}}_{ {\text{y}}}^2 \;{\sigma 2 {\text{p}}_{ {\text{z}}}^1}/{\;\text{ HOM }\;}$ $\; {\text{O}}_2-\sigma 1 {\ s}^2 \sigma^* 1 {\ s}^2 \sigma 2 {\ s}^2 \sigma^* 2 {\ s}^2 \sigma 2 {\text{p}}_{ {\text{z}}}^2$ $\; \;\; \pi 2 {\text{p}}_{ {\text{x}}}^2=\pi 2 {\text{p}}_{ {\text{y}}}^2$ $\;\pi^* 2 {\text{p}}_{ {\text{x}}}^1=\pi^* 2 {\text{p}}_{ {\text{y}}}^1( {\text{HOMO}})$ $\; {\text{O}}_2^{+}-\sigma {\ s}^2 \sigma^* 1 {\ s}^2 \sigma 2 {\ s}^2 \sigma^* 2 {\ s}^2 \sigma 2 {\text{p}}_{ {\text{z}}}^2 \pi 2 {\text{p}}_{ {\text{x}}}^2=\pi 2 {\text{p}}_{ {\text{y}}}^2$ $\;\pi^* 2 {\text{p}}_{ {\text{x}}}^1=\pi^* 2 {\text{p}}_{ {\text{y}}}^0( {\text{HOMO}})$ $\; {\text{N}}_2 \Rightarrow 0 \;\text{ unpaired }\; {\text{e}}^{-} \;\text{\in HOMO }\;$ $\; {\text{N}}_2^{+} \Rightarrow 1 \;\text{ unpaired }\; {\text{e}}^{-} \;\text{\in HOMO }\;$ $\; {\text{O}}_2 \Rightarrow 2 \;\text{ unpaired }\; {\text{e}}^{-} \;\text{\in HOMO }\;$ $\; {\text{O}}_2^{+} \Rightarrow 1 \;\text{ unpaired }\; {\text{e}}^{-} \;\text{\in HOMO }\;$

Crystal field theory introduce spectrochemical series based upon the experimental values of $\Delta$ but can't explain it's order. While other three points are explained by CFT. Specially when the CFSE increases thermodynamic stability of the complex increases.

$\; {\text{t}}_{1 / 2} \propto ( {\text{P}}_{ {\text{o}}}) ^{1- {\text{n}}}$ $\;\;{ ( {\text{t}}_{1 / 2}) _1}/{ ( {\text{t}}_{1 / 2}) _2}=\;{ ( {\text{P}}_0) _1^{1- {\text{n}}}}/{ ( {\text{P}}_{0_2}) _2^{1- {\text{n}}}}$ $\;\Rightarrow (\;\frac{4}{2}) = (\;\frac{50}{100}) ^{1- {\text{n}}}$ $\;\Rightarrow 2= (\;\frac{1}{2}) ^{1- {\text{n}}}$ $\;\Rightarrow 2=(2)^{ {\text{n}}-1}$ $\;\Rightarrow {\text{n}}-1=1$ $\;\Rightarrow {\text{n}}=2$

${\text{Z}}=55[ {\text{Cs}}] \Rightarrow [ {\text{Xe}}] 6 {\ s}^1$ $[ {\text{Cs}}^{+}] \Rightarrow [ {\text{Xe}}]$ i.e. upto $5 {\ s}$ count ${\text{e}}^{-}$of s-subshell i.e. $1 {\ s}, 2 {\ s}, 3 {\ s}, 4 {\ s}, 5 {\ s} \Rightarrow 10$ electrons

Ag.$\;4 {\text{Ag}}+8 {\text{CN}}^{-}+ {\text{O}}_2+2 {\text{H}}_2 {\text{O}} \to 4 [ {\text{Ag}}( {\text{CN}})_2] ^{-}+4 {\text{OH}}^{-}$ $\;2 [ {\text{Ag}}( {\text{CN}})_2] ^{-1}+ {\text{Zn}} \to 2 {\text{Ag}} ↓+ [ {\text{Zn}}( {\text{CN}})_4] ^{-2}$

${\text{Be}}$ has less negative value compared to other AEM. However it's reducing nature is due to large hydration energy associated with the small size of ${\text{Be}}^{2+}$ ion and relatively large value of the atomization enthalpy of metal.

Theoretical, NCERT based.

Assertion - A : Benzene is more stable than cyclohexatriene (True)Reason $-\text{bo} {R}$ : Delocalised $\pi- {\text{e}}$ cloud lies B.M.O so more attracted by nuclei of carbon atom.(True & Correct Explanation)

${{\text{C}}_6 {\text{H}}_5 {\text{COOH}}}_{(\text{WA})}+ {\text{NaOH}} _{(\text{SB})}\to {{\text{C}}_6 {\text{H}}_5 {\text{COONa}}}_{(\text{Salt})}+ {\text{H}}_2 {\text{O}}$(A) $\to$ (B) Free ${\text{H}}^{+}$ions are replaced by ${\text{Na}}^{⊕}$ which decreases conductance.(B) $\to$ (C) Un-dissociated benzoic acid reacts with ${\text{NaOH}}$ and forms salt which increases ions & conductance increases.(C) $\to$ (D) After equivalence point at (3), ${\text{NaOH}}$ added further increases {\text{Na}}^{⊕} & {\text{OH}}^{⊙} ions which further increases the conductance.

Theoretical.

Statement $-1$ is (True)Pure aniline is colourless liquidStatement - 2 is (False)Aniline becomes dark brown due to action of air and light [oxidation]

Most easily deprotonation

(1) ${\text{Na}} > {\text{Cs}} > {\text{Li}}-$ true $\{$ If considered with sign $\}$ The low solubility of CsI is due to smaller hydration enthalpy of it's two ions${\text{Li}}_2 {\text{CO}}_3$ is highly stable to heat - falseIn Conc. ${\text{NH}}_3, {\text{K}}$ formed blue solution - trueAll the alkali metal hydrides are ionic solid (True).

$[ {\text{Co}} ( {\text{NH}}_3) _6] ^{+3}- {\text{d}}^2 {\ sp}^3$, diamagnetic

$3 {\text{SO}}_2+ {\text{Cr}}_2 {\text{O}}_7^{2-}+2 {\text{H}}^{+} \to 3 {\text{SO}}_4^{2-}+2 {\text{Cr}}^{+3}+ {\text{H}}_2 {\text{O}}$

For physisorptions(a)Decreases with increase in temperature(b)No appreciable activation energy is required

Peroxodisulphuric acid - {\text{H}}- {\text{O}}- {\text{S}}_{||}_{\text{O}}^{||}^{\text{O}}- {\text{O}}- {\text{O}}- {\text{S}}_{||}_{\text{O}}^{||}^{\text{O}}- {\text{OH}} No. of $\pi$ - bonds $=4$ Pyro sulphuric acid - {\text{H}}- {\text{O}}- {\text{S}}_{||}_{\text{O}}^{||}^{\text{O}}- {\text{O}}- {\text{S}}_{||}_{\text{O}}^{||}^{\text{O}}- {\text{OH}} No. of $\pi-$ bonds $=4$Total $\pi-$ bonds $=8$

Total numbers of isomer $=03$

No. of possible tripeptide :Val & Pro is $2^3$(1) val - val - val(2)pro - pro - pro(3) val - pro - pro(4)pro - val - pro(5) val - val - pro(6)val - pro - val(7)pro - pro - val(8)pro - val - val

Mass percent, mole fraction, molarity, ppm, molality are used for measuring concentration terms.

At(a) $\to {\text{CO}}_2$ exist as gas(b) $\to$ liquefaction of ${\text{CO}}_2$ starts(c) $\to$ liquefaction ends(d) $\to {\text{CO}}_2$ exist as liquid.Between (b) & (c) $\to$ liquid and gaseous ${\text{CO}}_2$ co-exist.As volume changes from (b) to (c) gas decreases and liquid increases.(A), (C) $\to$ correct

Let V.P. of pure ${\text{A}}$ be ${\text{P}}_{ {\text{A}}}^0$ Let V.P of pure $B$ be ${\text{P}}_{ {\text{B}}}^0$When {\text{X}}_{ {\text{A}}}=0.7 & {\text{X}}_{ {\text{B}}}=0.3 $\; {\text{P}}_{ {\ s}}=350$ $\;\Rightarrow {\text{P}}_{ {\text{A}}}^0 \times 0.7+ {\text{P}}_{ {\text{B}}}^0 \times 0.3=350$ . . . (i)When {\text{X}}_{ {\text{A}}}=0.2 & {\text{X}}_{ {\text{B}}}=0.8 $\; {\text{P}}_{ {\ s}}=410$ $\;\Rightarrow {\text{P}}_{ {\text{A}}}^0 \times 0.2+ {\text{P}}_{ {\text{B}}}^0 \times 0.8=410$ . . . (ii)Solving (i) and (ii)$\; {\text{P}}_{ {\text{A}}}^0=314 {\text{mm}} {\text{Hg}}$ $\; {\text{P}}_{ {\text{B}}}^0=434 {\text{mm}} {\text{Hg}}$ $\;=(314)$

$\;1 \to 2 \Rightarrow \;\text{ Isobaric process }\;$ $\;2 \to 3 \Rightarrow \;\text{ Isochoric process }\;$ $\;3 \to 1 \Rightarrow \;\text{ Isothermal process }\;$ $\; {\text{W}}= {\text{W}}_{1 \to 2}+ {\text{W}}_{2 \to 3}+ {\text{W}}_{3 \to 1}$ $\;= (- {\text{P}} ( {\text{V}}_2- {\text{V}}_1) +0 [- {\text{P}}_1 {\text{V}}_1 \ln (\;{ {\text{V}}_2}/{ {\text{V}}_1}) ] )$ $\;= [-1 \times (40-20)+0+ [-1 \times 20 \ln (\;\frac{20}{40}) ] ]$ $\;=-20+20 \ln 2$ $\;=-20+20 \times 2.3 \times 0.3$ $\;=-6.2 \;\text{ bar }\; {\text{L}}$ $\;| {\text{W}}|=6.2 \;\text{ bar }\; 1=620 {\text{J}}$

Concentration of calcium lactate $=0.005 {\text{M}}$,: concentration of lactate ion $=(2 \times 0.005) {\text{M}}$. Calcium lactate is a salt of weak acid $+$ strong base $\therefore$ Salt hydrolysis will take place.$\; {\text{pH}}=7+\;\frac{1}{2}( {\text{pKa}}+\log {\text{C}})$ $\;=7+\;\frac{1}{2}(5+\log (2 \times 0.005))$ $\;=7+\;\frac{1}{2}[5-2 \log 10]=7+\;\frac{1}{2} \times 3=8.5=85 \times 10^{-1}$