15 Apr 2023 Shift 1

The electromagnetic spectrum is ordered by wavelength. Microwaves have longer wavelengths than infrared, which is longer than ultraviolet, which is longer than X-rays. Based on the provided ranges:
(A) Microwave corresponds to the longest wavelength range, (IV) $0.1 {\text{ m}}$ to $1 {\text{ mm}}$.
(D) Infra-red follows, corresponding to (III) $1 {\text{ mm}}$ to $700 {\text{ nm}}$.
(B) Ultraviolet has shorter wavelengths, corresponding to (I) $400 {\text{ nm}}$ to $1 {\text{ nm}}$.
(C) X-Ray has the shortest wavelengths, corresponding to (II) $1 {\text{ nm}}$ to $10^{-3} {\text{ nm}}$.
The correct matching is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).

$E=\;\frac{K P}{r^3} \propto \;\frac{1}{r^3}$

Work done $=$ Area of graph${\text{W}}\;=\;\frac{1}{2}(400-100)(4-2)$ ${\text{W}}\;=300 {\text{J}}$

$\; {\text{N}}= {\text{N}}_0 (\;\frac{1}{2}) ^{ {\text{t}} / {\text{T}}_{/ 2}}$ $\; {\text{N}}= {\text{N}}_0 (\;\frac{1}{2}) ^{15 / 5}$ $\; {\text{N}}=\;{ {\text{N}}_0}/{8}$ $\;\;\text{ Decayed nuclei }\;= {\text{N}}_0-\;{ {\text{N}}_0}/{8}$ $\;=\;{7 {\text{N}}_0}/{8}$

Given, $\vec{r}=10 t \hat{i}+15 t^2 \hat{j}+7 \hat{k}$ \;{ {\text{v}}}^{\to }=\;\frac{\vec{d r}}{d t}=10 \hat{j}+30 { {\text{j}}^{\^}} \;{ {\text{a}}}^{\to }=\;{ {\text{d}} { {\text{v}}}^{\to }}/{ {\text{dt}}}=30 { {\text{j}}^{\^}} $\;{ {\text{F}}}^{\to }= {\text{m}} { {\text{a}}}^{\to } \to \;\text{ along }\;(+) {\text{y}} \;\text{-axis }\;$

$\; {\text{d}}=\sqrt{2 R {\text{h}}_{ {\text{T}}}}+\sqrt{2 {\text{Rh}}_{ {\text{R}}}}$ $\; {\text{d}}=\sqrt{2 {\text{R}}}({\sqrt{180}}+{\sqrt{245}})$ $\; {\text{d}}={\sqrt{2 \times 64 \times 10^5}}({\sqrt{180}}+{\sqrt{245}})$ $\; {\text{d}}=3577.7(13.416+15.652) {\text{m}}$ $\; {\text{d}}=104 {\text{km}}$

For minima, $a sin \;\theta=\lambda$ $\; {\text{a}}=\;\frac{\lambda }{ sin \;30^{\circ}}=2 \lambda$ $\; {\text{a}}=1200 {\text{nm}}$ $\; {\text{a}}=1.2 \mu {\text{m}}$

$\; {\text{e}}=- {\text{L}} \;{ {\text{di}}}/{ {\text{dt}}}$ $\; {\text{i}}_1=\;{ {\text{E}}}/{ {\text{R}}}=\;\frac{12}{6}=2 {\text{A}}$ $\;20=- {\text{L}} (\;\frac{0-2}{10^{-3}})$ $\; {\text{L}}=10 {\text{mH}}$

$\;F_C=m \omega ^2 a$ $\;\;\frac{G m^2}{(2 a)^2}=m \omega ^2 a$ $\;\omega =\sqrt{\;\frac{G m}{4 a^3}}$

By energy conservation,$\; {\text{K}}_1+ {\text{U}}_1= {\text{K}}_2+ {\text{U}}_2$ $\; {\text{O}}-\;{ {\text{GMm}}}/{2 {\text{R}}}=\;\frac{1}{2} {\ mv}^{2-}\;{ {\text{GMm}}}/{ {\text{R}}}$ $\; {\text{V}}=\sqrt{\;{ {\text{GM}}}/{ {\text{R}}} \times \;{ {\text{R}}}/{ {\text{R}}}}$ $\; {\text{V}}=\sqrt{ {\text{gR}}}$

For voltmeter,$\; {\text{I}}=\;{50}/{ {\text{R}}+54}=0.001 {\text{A}}$ $\; {\text{R}}=50 {\text{k}} Ω$For Ammeter,$\; {\text{I}}_{ {\text{r}}}=10-1=9 {\text{mA}}$ $\; {\text{V}}_{ {\text{G}}}= {\text{V}}_{ {\text{r}}}$ $\;1 {\text{mA}} \times 54=9 {\text{mA}} \times {\text{r}}$ $\; {\text{r}}=6 Ω$

In series,$\; {\text{R}}_{\;\text{eq }\;}= {\text{R}}_1+ {\text{R}}_2+ {\text{R}}_3+. . . .$ $\; {\text{R}}_{\;\text{eq }\;}> {\text{R}}_{\;\text{Greatest }\;}$Hence, statement-I is false.Resistivity of conductor increases with temperature.Hence, statement-II is also false

$\;\lambda =\;{ {\text{h}}}/{ {\text{p}}}=\;{ {\text{h}}}/{\sqrt{2 {\text{mE}}}}$ $\;\;\frac{\lambda _1}{\lambda _2}=\sqrt{\;{ {\text{E}}_2}/{ {\text{E}}_1}}=\sqrt{\;\frac{1}{4}}=\;\frac{1}{2}$ $\;\lambda _2=2 \lambda$

$\; {\text{A}}_{ {\text{y}}}= {\text{A}} \cos 30^{\circ}=2 {\sqrt{3}}$ $\; {\text{A}}=4$ $\; {\text{A}}_{ {\text{x}}}= {\text{A}} sin \;30^{\circ}=2$

By dimensional analysis,$\; {\text{V}}=\lambda ^{ {\text{a}}} {\text{g}}^{ {\text{b}}} \rho ^{ {\text{c}}}$ \;{[ {\text{L}}^1 {\text{T}}^{-1}]\ = $[ {\text{L}}^1]$ ^{ {\text{a}}} $[ {\text{L}}^1 {\text{T}}^{-2}]$ ^{ {\text{b}}} $[ {\text{M}}^1 {\text{L}}^{-3}]$ ^{ {\text{c}}}}$$;{[ {\text{L}}^{ {\text{l}}} {\text{T}}^{-1}]$ = $[ {\text{M}}^{ {\text{c}}} {\text{L}}^{ {\text{a}}+ {\text{b}}-3 {\text{c}}} {\text{T}}^{-2 {\text{b}}}]$ }$On comparing respective powers,$; {\text{c}}=0,-2 {\text{b}}=-1, {\text{a}}+ {\text{b}}-3 {\text{c}}=1,$$; {\text{b}}=;\frac{1}{2}, {\text{a}}=;\frac{1}{2}$

In given figure, ${\text{D}}_1, {\text{D}}_3 \to$ Forward Biased ${\text{D}}_2 \to$ Reversed Biased$\; {\text{R}}_{ {\text{eq}}}=\;\frac{(20)(10)}{30}=\;\frac{20}{3} Ω$ $\; {\text{I}}=\;{ {\text{V}}}/{ {\text{R}}}=\;\frac{10}{ (\;\frac{20}{3}) }=1.5 {\text{A}}$

In SHM,$\; {\text{F}} \propto - {\text{x}} \to {\text{A}}$ $\; {\text{a}} \propto - {\text{x}} \to {\text{B}}$ $\; {\text{V}}_{\;\text{mean }\;} \to \;\text{ max imum }\; \to {\text{c}}$ $\; {\text{a}}_{\;\text{extreme }\;} \to \;\text{ maximum }\;$Hence, (A), (B) and (C) are true.

By hooke's law,$\;\gamma =\;{ {\text{Fl}}}/{ {\text{A}} \Delta {\text{l}}}$ $\;\Delta {\text{l}} \propto \;{ {\text{Fl}}}/{ {\text{A}}}$ $\;\;{\Delta {\text{l}}_2}/{\Delta {\text{l}}_1}=\;{ {\text{F}}_2 {\text{l}}_2}/{ {\text{F}}_1 {\text{l}}_1} \times \;{ {\text{A}}_1}/{ {\text{A}}_2}$ $\;=\;{2 {\text{f}} \times 2 {\text{L}}}/{ {\text{f}} \times {\text{L}}} \times \;{\pi( {\text{r}})^2}/{\pi(2 {\text{r}})^2}$ $\;\Delta {\text{l}}_1=\Delta {\text{l}}_2$ $\;\Delta {\text{l}}_2=1$

$\;\;\text{ Kinetic energy per molecule }\;=\;\frac{3}{2} {\text{k}}_{ {\text{B}}} {\text{T}}$ $\;\propto \;{ {\text{k}}_{ {\text{Ar}}}}/{ {\text{k}}_{ {\text{H}}_2}}=\;{ {\text{T}}}/{ {\text{T}}}=1$

$\;\;\text{ Given, }\; \;\; x= (5 t^{2-}4 t+5) m$ $\;V=\;\frac{d x}{d t}=10 t-4$ $\;t=2, V=20-4$ $\;V=16 m / s$

Biot-Savart law,$\; {\text{B}}=\;\frac{\mu _0}{4 \pi} \times \;{ {\text{qV}}}/{ {\text{r}}^2}$ $\; {\text{B}}=10^{-7} \times \;\frac{1.6 \times 10^{-19} \times 6.76 \times 10^6}{ (0.52 \times 10^{-10}) ^2}$ $\; {\text{B}}=40 {\text{T}}$

A $20 {\text{cm}}$ long metallic rod is rotated with $210 {\text{rpm}}$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field $0.2 {\text{T}}$ parallel to the axis exists everywhere. The emf developed between the centre and the ring is ${\text{mV}}$. (Take $\pi=22 / 7$ )

For ${\text{A}}, {\text{n}}=2$ $\; {\text{B}}, {\text{n}}=3$ $\; {\text{C}}, {\text{n}}=4$ $\;\;\frac{1}{\lambda }= {\text{R}} (\;{1}/{ {\text{n}}_1^2}-\;{1}/{ {\text{n}}_2^2})$ $\;\;\frac{1}{\lambda _2}= {\text{R}} (\;\frac{1}{3^2}-\;\frac{1}{4^2})$ $\;\;\frac{1}{\lambda _2}=\;{7 {\text{R}}}/{144}$ . . . (1)$\;\;\frac{1}{\lambda _1}= {\text{R}} (\;\frac{1}{2^2}-\;\frac{1}{3^2})$ $\;\;\frac{1}{\lambda _1}=\;{5 {\text{R}}}/{36}$ . . . (2)(1) and (2)$\;\;\frac{\lambda _1}{\lambda _2}=\;\frac{7}{20}=\;\frac{7}{4 \times 5}$ $\; {\text{n}}=5$

For minimum deviation${\text{r}}=\;{ {\text{A}}}/{2}=\;\frac{60}{2}=30^{\circ}$ $\;1 sin \; {\text{i}}={\sqrt{2}} sin \; {\text{r}}$ $\; sin \; {\text{i}}={\sqrt{2}} \times sin \;30^{\circ}$ $\; sin \; {\text{i}}=\;{1}/{{\sqrt{2}}}$ $\; {\text{i}}=45^{\circ}$ $\;\delta _{\min }=2 {\text{i}}- {\text{A}}=90-60=30^{\circ}$ $\; \;\; \delta _{\min }=30^{\circ}$

Work done, ${\text{w}}=\int {\text{Fdx}}$ ${\text{w}}\;=int_{2}^{4} 5 {\text{xdx}}$ {\text{w}}\;=\;\frac{5}{2} \[ {\text{x}}^2]$ _2^4$${\text{w}};=;\frac{5}{2}(16-4)$${\text{w}};=30 {\text{J}}$

$\;\mu =\;{ {\text{m}}}/{ℓ}=20 {\text{gm}} / {\text{m}}$ $\;\;{18 {\text{gm}}}/{ℓ}=20 {\text{gm}} / {\text{m}}$ $\;ℓ=\;\frac{9}{10} {\text{m}}$For fundamental mode,$\; {\text{f}}=\;{ {\text{V}}}/{2 ℓ}$ $\; {\text{V}}=50 \times \;\frac{18}{10}=90 {\text{m}} / {\ s}$

$\; {\text{I}}_{\;\text{sphere }\;}=\;\frac{2}{5} {\text{MR}}^2= {\text{Mk}}^2$ $\; {\text{k}}_{ {\ sph}}=\sqrt{\;\frac{2}{5}} {\text{R}}$ $\; {\text{I}}_{\;\text{cylinder }\;}=\;{ {\text{MR}}^2}/{2}= {\text{Mk}}^2$ $\; {\text{k}}_{\;\text{cyl }\;}=\;{ {\text{R}}}/{{\sqrt{2}}}$ $\;\;{ {\text{k}}_{ {\ sph}}}/{ {\text{k}}_{ {\text{cyl}}}}=\;{2}/{{\sqrt{5}}}$ $\; {\text{x}}=5$

$\; {\text{R}}_{ {\text{eq}}}=\;\frac{6}{2}=3 Ω$ $\; {\text{I}}=\;{ {\text{V}}}/{ {\text{R}}}=\;\frac{9}{3}=3 {\text{A}}$ $\; {\text{V}}_{ {\text{A}}}+(1.5)(2)-(1.5) 4= {\text{V}}_{ {\text{B}}}$ $\; {\text{V}}_{ {\text{A}}}- {\text{V}}_{ {\text{B}}}=3 {\text{V}}$

All the capacitors are in parallel$\; {\text{C}}_{ {\text{eq}}}=(5+ {\text{x}}) \mu {\text{F}}$ $\; {\text{Q}}= {\text{C}}_{ {\text{aq}}} {\text{V}}$ $\;100=(5+ {\text{x}})(10)$ $\; {\text{x}}=5$

$P_{i n}=P_0+\rho g h+\;\frac{2 T}{r}$ $\;P_{\;\text{\in }\;}-P_0=1000 \times 10 \times 0.1+\;\frac{2 \times 0.075}{0.001}$ $\;=1000+150$ $\;P_{\;\text{\in }\;}-P_0=1150 {\text{Pa}}$

Calcination is the process in which the ore of the metal is heated to high temperature in the absence or limited supple of air or oxygen. Anhydrous iron is formed during iron calcination.${\text{Fe}}_2 {\text{O}}_3 \cdot {\text{xH}}_2 {\text{O}} {\to }^{\Delta } {\text{Fe}}_2 {\text{O}}_3+ {\text{xH}}_2 {\text{O}}$

Water gas shift reaction${\text{CO}}+ {\text{H}}_2 {\text{O}} \to {\text{CO}}_2+ {\text{H}}_2$

The reaction sequence involves two main steps.

1. Diazotization: Aniline reacts with ${\text{NaNO}}_2$ and ${\text{HCl}}$ (implied by the low temperature) at $0-5^\circ{\text{C}}$ to form benzenediazonium chloride. This is intermediate 'A'.
2. Azo Coupling: The benzenediazonium salt is an electrophile that reacts with the electron-rich N,N-dimethylaniline. The coupling occurs at the para position of N,N-dimethylaniline, which is highly activated by the $-{\text{N}}({\text{CH}}_3)_2$ group. The final product 'B' is an azo dye, p-(dimethylamino)azobenzene.

Be and mg have abnormally small size when compared to other alkali metals. Due to this, electrons in the atom are tightly hold and thus, they have high Ionisation energy. Hence, Be and mg donot undergo transition and don't impart characteristic colour to the flame.

The reaction shown is the hydrolysis of a cyclic anhydride. The water molecule ({\text{H}}_2{\text{O}}$) acts as a nucleophile and attacks one of the carbonyl carbons of the anhydride. This nucleophilic acyl substitution leads to the cleavage of one of the C-O bonds in the ring. The anhydride ring opens up to form a dicarboxylic acid. The solution also indicates the formation of an intramolecular hydrogen bond between the two adjacent carboxylic acid groups, which stabilizes the product.

The ratio should be between 2.5 to 4 : 1 so ans $\to 3$

For ${\text{CaCl}}, {\text{Cs}}^{⊕}$ is present at Body center and ${\text{Cl}}^{Θ}$ at all corner. $\;{\sqrt{3 {\text{a}}}}/{2}= {\text{r}}_{ {\text{cs}}^{⊕}}+ {\text{r}}_{ {\text{Cl}}^{Θ}}$

According to Bohr's model the angular momentum is quantised and equal to $\;{ {\text{nh}}}/{2 \pi}$. Heisenberg uncertainty principle explains orbital concept, which is based on probability of finding electron.

The correct sequence of reagents is given in option (B).

1. (i) ${\text{Br}}_2/{\text{Fe}}$: Electrophilic bromination of 4-nitrotoluene occurs ortho to the activating ${\text{CH}}_3$ group, yielding 2-bromo-4-nitrotoluene.
2. (ii) $\frac{\text{Fe}}{\text{H}}^{+}$: The nitro group ({\text{NO}}_2$) is reduced to an amino group ({\text{NH}}_2$).
3. (iii) ${\text{HONO}}$: The amino group is converted to a diazonium salt ({\text{N}}_2^+$).
4. (iv) ${\text{CuCl}}$: A Sandmeyer reaction replaces the diazonium group with a chlorine atom.
5. (v) ${\text{KMnO}}_4$: The methyl group ({\text{CH}}_3$) is oxidized to a carboxylic acid (-COOH), yielding the final product.

The reaction proceeds in four steps:

1. Hydroboration-Oxidation: Propene ($CH_3-CH=CH_2$) reacts with $B_2H_6$ followed by $H_2O_2, NaOH$ to give propan-1-ol ($CH_3CH_2CH_2OH$) via anti-Markovnikov addition.

2. Oxidation: The primary alcohol, propan-1-ol, is oxidized by the mild oxidizing agent PCC to form the aldehyde, propanal ($CH_3CH_2CHO$).

3. Grignard Reaction: The Grignard reagent, $CH_3MgBr$, attacks the carbonyl carbon of propanal to form an intermediate magnesium alkoxide.

4. Hydrolysis: Subsequent hydrolysis (work-up) of the intermediate yields the final product, a secondary alcohol, butan-2-ol ($CH_3CH_2CH(OH)CH_3$).

Srinage, Jammu & Kashmir in January photo chemical smog is produced when sunlight reacts with nitrogen oxides.$\; {\text{NO}}_2( {\text{g}}) {\to }^{ {\text{hu}}} {\text{NO}}( {\text{g}})+ {\text{O}}( {\text{g}})$ $\; {\text{O}}( {\text{g}})+ {\text{O}}_2 \to {\text{O}}_3( {\text{g}})$ $\; {\text{NO}}+ {\text{O}}_3 \to {\text{NO}}_2+ {\text{O}}_2$

$\; {\text{N}}_2+ {\text{O}}_2$ $\; {\text{N}} \equiv {\text{N}} {\text{O}}= {\text{O}}$ $\;\;\text{ B.O. }\;-3 > 2$ $\;\;\text{ B.L. }\; 3 < 2$ Isoelectronic haveIdentical bond order

The solution provides the structures for the three phosphorus oxyacids.

1. H₄P₂O₇ (Pyrophosphoric acid): The structure shows two PO₄ tetrahedra linked by sharing one oxygen atom. This results in a single P-O-P bond.
2. (HPO₃)₃ (Trimetaphosphoric acid): This is a cyclic compound where three PO₄ tetrahedra are linked in a ring. Each phosphorus atom is connected to two other phosphorus atoms via an oxygen bridge, forming three P-O-P bonds in total.
3. P₄O₁₀ (Phosphorus pentoxide dimer): This has a cage-like structure. There are six oxygen atoms that bridge the four phosphorus atoms, resulting in six P-O-P bonds.
   Thus, the respective numbers of P-O-P bonds are 1, 3, and 6.

Arginine exist is zwitterion, so solid nature and soluble in polar solvent.(i) Polar, so not high soluble in benzene(ii) It has $3 {\text{pKa}}$ values(iii) True(iv) High molecular mass to high M.P.

The crystal field splitting energy (Δ₀) depends on the central metal ion, as the ligand (H₂O) is the same in all given complexes. The central metal ions and their d-electron configurations are:

• [Mn(OH₂)₆]³⁺: Mn³⁺ is a d⁴ ion.
• [Fe(OH₂)₆]³⁺: Fe³⁺ is a d⁵ ion.
• [Cr(OH₂)₆]³⁺: Cr³⁺ is a d³ ion.
• [Ti(OH₂)₆]³⁺: Ti³⁺ is a d¹ ion.
  The Crystal Field Stabilization Energy (CFSE) is calculated for each. The complex with Cr³⁺ (d³) has the most negative CFSE value of -1.2 Δ₀, as the three electrons occupy the lower energy t₂g orbitals without any electron in the higher energy eg orbitals. This indicates the highest magnitude of Δ₀.

Fact

(A) ${\text{pH}}$ of $10^{-8} {\text{M}} {\text{HCl}}$ in acidic range (6.98).(B) Conjugate Base of ${\text{H}}_2 {\text{PO}}_4^{-}$is ${\text{HPO}}_4^{2-}$(C) ${\text{K}}_{ {\text{w}}}$ increases with increasing Temperature, as the temperature increases, the dissociation of water increases.(D) At half neutralization point, half of the acid is present in the from salt.${\text{pH}}= {\text{Pk}}_{ {\text{a}}}+\log \;\frac{1}{1}= {\text{Pk}}_{ {\text{a}}}$

Chlorine is ortho/para directing, para is major.

Higher the electron density on Benzene Ring, Higher its Reactivity towards electrophilic substitution Reaction Rate $\propto {\text{EDG}} \propto \;{1}/{ {\text{EWG}}}$

The matching of monomers to their respective polymers is based on standard polymerization reactions.

• (A) Tetrafluoroethene (F₂C=CF₂) undergoes addition polymerization to form (III) Teflon, a polymer with the repeating unit -[CF₂-CF₂]-.
• (B) Acrylonitrile (CH₂=CHCN) polymerizes to form (I) Orlon (polyacrylonitrile).
• (C) Caprolactam, a cyclic amide, undergoes ring-opening polymerization to form (IV) Nylon-6.
• (D) Isoprene (2-methyl-1,3-butadiene) is the monomer unit for (II) Natural rubber.
  Therefore, the correct matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).

$( {\text{Co}} ( {\text{H}}_2 {\text{O}}) _6) ^{+2} \to {\text{CO}}^{+2} \to 3 {\text{d}}^7$u.e. in $t_2 g$ is $=(1)$

$\Delta {\text{S}}\;=\;{\Delta {\text{H}}}/{ {\text{T}}_{ {\mp}}}$ $28.4\;=\;{30.4 \times 1000}/{ {\text{T}}_{ {\mp}}}$ ${\text{T}}_{ {\mp}}\;=1070.422 {\text{K}} .$

Isoelectronic species are atoms or ions that have the same number of electrons. To find the total number of isoelectronic species in the given set, we calculate the number of electrons for each.

• ${\text{O}}^{2-}$ has $8 + 2 = 10$ electrons.
• ${\text{F}}^{-}$ has $9 + 1 = 10$ electrons.
• ${\text{Al}}$ has $13$ electrons.
• ${\text{Mg}}^{2+}$ has $12 - 2 = 10$ electrons.
• ${\text{Na}}^{+}$ has $11 - 1 = 10$ electrons.
• ${\text{O}}^{+}$ has $8 - 1 = 7$ electrons.
• ${\text{Mg}}$ has $12$ electrons.
• ${\text{Al}}^{3+}$ has $13 - 3 = 10$ electrons.
• ${\text{F}}$ has $9$ electrons.
  The species with 10 electrons are ${\text{O}}^{2-}, {\text{F}}^{-}, {\text{Mg}}^{2+}, {\text{Na}}^{+},$ and ${\text{Al}}^{3+}$. There are 5 such species.

$\; {\text{E}}_{ {\text{a}}}=300 {\text{kJ}} {\text{mol}}^{-1}$ $\;\;{ {\text{E}}_{ {\text{a}}}}/{ {\text{T}}}=\;{ {\text{E}}_{ {\text{a}}}^{'}}/{ {\text{T}}^{'}}$(Since rate of catalyzed and uncatalysed reaction is same)$\;\;\frac{300}{600}=\;\frac{E_{a, f}^{'}}{300}$ $\;E_{a, f}^{'}=150$ $\;20=150$ $\;20=150-E_{a, b}^{'}$ $\;E_{a, b}^{'}=130$