24 Jun 2022 Shift 2

Velocity gradient $=\;\frac{d v}{d x}$ $\Rightarrow$ Dimensions are \;\frac{[L T^{-1}]\}{[L]}=[T^{-1}]$Decay constant$λ$has dimensions of$[T^{-1}]$because of the relation$;\frac{d N}{d t}=-λ N$$\Rightarrow $ Velocity gradient and decay constant have same dimensions.

We know that${\text{T}}^2 \propto {\text{R}}^3$ $\Rightarrow (\;\frac{T^{'}}{T})^2=(\;\frac{3 R}{R})^3$ $\Rightarrow \;\frac{T^{'}}{T}=3 \sqrt{3}$ $\Rightarrow T^{'}=3 \sqrt{3} \;\text{ years }\;$

At any $\theta: T-m g \cos \theta=\;\frac{m v^2}{R}$ $\Rightarrow T=m g \cos \theta+\;\frac{m v^2}{R}$Since $v$ is constant,$\Rightarrow {\text{T}}$ will be minimum when $\cos \theta$ is minimum.$\Rightarrow \theta=180^{\circ}$ corresponds to ${\text{T}}_{\;\text{minimum }\;}$

According to given information: $\;{ {\text{kq}}^2}/{ {\text{L}}^2}=\mu {\text{mg}} \Rightarrow {\text{L}}=\sqrt{\;{ {\text{k}}}/{\mu {\text{mg}}} {\text{q}}}$Putting the values, we get ${\text{L}}=12 {\text{cm}}$

Efficiency $η=1-\;\frac{T_L}{T_H}$ $=1-\;\frac{400}{1000}=0.6$ $\Rightarrow 0.6=\;\frac{W}{Q}$ $\Rightarrow W=0.6 Q=3000 {\text{kcal}}=12.6 \times 10^6 {\text{J}}$

$\omega _1 A_1=\omega _2 A_2$ $\Rightarrow \;\frac{A_1}{A_2}=\;\frac{\omega _2}{\omega _1}$ $=\sqrt{\;\frac{k_2}{m_2}} \times \sqrt{\;\frac{m_1}{k_1}}=\sqrt{\;\frac{9 k}{100} \times \;\frac{50}{2 k}}=\;\frac{3}{2}$

$R_{e q}=\;\frac{2 \times 4}{2+6}+6=\;\frac{22}{3}$ $\Rightarrow A$ and $B$ are in parallel and $C$ is in series.

Electromagnet requires high permeability and Iow retentivity.

$\therefore r=\;\frac{m v}{q B}=\;\frac{\sqrt{2 m(K E)}}{q B}$ $\Rightarrow r_1: r_2: r_3=\;\frac{\sqrt{m_1}}{q_1}: \;\frac{\sqrt{m_2}}{q_2}: \;\frac{\sqrt{m_3}}{q_3}$ $=\;\frac{\sqrt{1}}{1}: \;\frac{\sqrt{2}}{1}: \;\frac{\sqrt{4}}{2}$ $=1: \sqrt{2}: 1$

$X=|X_C-X_L|$So, it can be zero if $X_C=X_L$And, average power in ac circuit can be zero.

For equilibrium$-\;\frac{d U}{d r}=0=\;\frac{10 A}{r^{11}}-\;\frac{5 B}{r^6}$ $\Rightarrow r^5=\;\frac{2 A}{B}$And $r=(\;\frac{2 A}{B})^{1 / 5}$

Let time taken to ascent is $t_1$ and that to descent is $t_2$. Height will be same so$H=\;\frac{1}{2} \times 12 t_1^2=\;\frac{1}{2} \times 8 t_2^2$ $\Rightarrow \;\frac{t_1}{t_1}=\;\frac{\sqrt{2}}{\sqrt{3}}$

$\theta_1=\;\frac{1}{2} \alpha(2 \times 1-1)=5 {\text{rad}}$ $\Rightarrow \alpha=10 {\text{rad}} / {\sec}^2$So $\theta_2=\;\frac{1}{2} \times \alpha(2 \times 2-1)=15 {\text{rad}}$

$\;\frac{1}{2} \times 1.5 \times 60^2 \times \;\frac{1}{4}=100 \times 0.42 \times \Delta T$ $\Delta T=\;\frac{1.5 \times 60^2}{8 \times 100 \times 0.42}=16.07^{\circ} {\text{C}}$

Let initially the charge is q so$\;\frac{1}{2} \;\frac{q^2}{C}=U_i$And $\;\frac{1}{2} \;\frac{(q+2)^2}{C}=U_f$Given $\;\frac{U_f-U_i}{U_i} \times 100=44$ $\;\frac{(q+2)^{2-}q^2}{q}=.44$ $\Rightarrow q=10 C$

$\;\frac{m v^2}{r}=\;\frac{2 k \rho \times \pi R^2 q}{r}$ $\Rightarrow \;\frac{1}{2} m v^2=\;\frac{\rho R^2 q}{4 \epsilon_0}$

$200 \times \;\frac{1}{4 \pi \times 16} \times \;\frac{3.5}{100}=\;\frac{B_0^2}{2 \mu _0} C$ $\mu _0=4 \pi \times 10^{-7} {\text{Tm}} / {\text{A}}$ $C=3 \times 10^8 {\text{m}} / {\sec}$ $\Rightarrow B_0=1.71 \times 10^{-8} {\text{T}}$

$3.8=0.6+\;\frac{1}{2} m v_1^2$ $1.4=0.6+\;\frac{1}{2} m v_2^2$ $\Rightarrow \;\frac{v_1^2}{v_2^2}=\;\frac{3.2}{0.8}=\;\frac{4}{1}$ $\Rightarrow \;\frac{v_1}{v_2}=\;\frac{2}{1}$

$\;\frac{I_{\max }}{I_{\min }}=(\;\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}})^2=(\;\frac{5}{1})^2$ $=\;\frac{25}{1}$

$T . E .=\;\frac{-Z^2 m e^4}{8(n h \epsilon_0)^2}$ $P . E .=\;\frac{-Z^2 m e^4}{4(n h \epsilon_0)^2}$ $K . E .=\;\frac{Z^2 m e^4}{8(n h \epsilon_0)^2}$As electron makes transition to higher level, total energy and potential energy increases (due to negative sign) while the kinetic energy reduces.

$\Rightarrow v \cos \alpha=u \cos 45^{\circ} . . . . . \;\text{ (i) }\;$ $\& v \sin \alpha=u \sin 45^{\circ}-g t . . . . \;\text{ (ii) }\;$Solve for u we get$u=20 \sqrt{2} {\text{m}} / {\text{s}}$ $\Rightarrow H=\;\frac{u^2 \sin ^2 45^{\circ}}{20}=20 {\text{m}}$

We know that $v=f \lambda$Putting the values,$\;\frac{3 \times 10^8}{\sqrt{6.25}}=f \times 20 \times 10^{-3}$ $\Rightarrow f=6 \times 10^9 {\text{Hz}}$

$\therefore E=I \times (\;\frac{20}{4})=\;\frac{25}{(30+20)} \times (\;\frac{20}{4})$ $=\;\frac{1}{2} \times 5=2.5$ volts $=\;\frac{25}{10}$ volts

$A=|10 \cos (\pi x)|$At $x=\;\frac{4}{3}$ $A=|10 \cos (\pi \times \;\frac{4}{3})|$ $=|-5 {\text{cm}}|$ $\therefore$ Amp $=5 {\text{cm}}$

${\text{V}}_{ {\text{L}}}=5 {\text{V}}$ as ${\text{V}}_{ {\text{Z}}}=5 {\text{V}}$ $\therefore I_L=\;\frac{V_L}{R_L}=\;\frac{5}{10^3}=5 {\text{mA}}$

$N_1=\;{(\frac{10^{-2}}{1})}/{2^4}$ $N_2=\;{(\frac{10^{-2}}{2})}/{2^2}$ $\Rightarrow \;\frac{N_1}{N_2}=\;\frac{1}{2}$ $\therefore$ Mass ratio of A and B,$\;\frac{m_1}{m_2}=\;\frac{N_1}{N_2} \times (\;\frac{M_1}{M_2})$ $=\;\frac{1}{2} \times (\;\frac{1}{2})$ $=\;\frac{1}{4}$ $=\;\frac{25}{100}$ $\therefore x=25$

$1 \times \sin 60^{\circ}=\sqrt{3} \times \sin r$ $\Rightarrow r=30^{\circ}$ $\therefore l_1=4 \sqrt{3} \times 2$ $=8 \sqrt{3} {\text{cm}}$ $\therefore$ Thickness, $t=l_1 \cos 30^{\circ}$ $=8 \sqrt{3} \times \;\frac{\sqrt{3}}{2}$ $=4 \times 3$ $=12 {\text{cm}}$

$V_{\max }=N A B \omega$ $=1000 \times 1 \times 0.07 \times (2 \pi \times 1)$ $≃ 440$ volts

By $1^{\;\text{st }\;}$ law,$\Delta U=\Delta Q-\;\frac{\Delta Q}{4}=\;\frac{3}{4} \Delta Q$ $\Rightarrow n C_v \Delta T=\;\frac{3}{4} n C \Delta T$ $\Rightarrow C=\;\frac{4 C_v}{3}=2 R$

Temperature of surrounding $=20^{\circ} {\text{C}}$For $0 \to 6$ minutes, average temp. $=70^{\circ} {\text{C}}$ $\to$ Rate of cooling $\propto 70^{\circ} {\text{C}}-20^{\circ} {\text{C}}=50^{\circ} {\text{C}}$For $6 \to t_2$ minutes, average temp. $=50^{\circ} {\text{C}}$ $\to$ Rate of cooling $\propto 30^{\circ} {\text{C}}$ $\Rightarrow t_2-6=\;\frac{5}{3}(6$ minutes $)$ $\Rightarrow {\text{t}}_2=16$ minutes

$C_x H_y+(x+\;\frac{y}{4}) O_2 \to x C O_2+\;\frac{y}{2} {\text{H}}_2 {\text{O}}$From the reaction,Produced ${\text{CO}}_2= {\text{x}} {\text{mol}}$and produced ${\text{H}}_2 {\text{O}}=\;\frac{y}{2} {\text{mol}}$Given produced ${\text{CO}}_2=330 {\text{g}}$ $\therefore$ moles of ${\text{CO}}_2=\;\frac{330}{44}=\;\frac{30}{4}=x$Also given produced ${\text{H}}_2 {\text{O}}=270 {\text{gm}}$ $\therefore$ Moles of ${\text{H}}_2 {\text{O}}=\;\frac{270}{18}=15=\;\frac{y}{2}$.$\Rightarrow {\text{y}}=30$ $\therefore x: y=\;\frac{30}{4}: 30=1: 4$ Formula of the compound $=( {\text{CH}}_4)_n$ $\therefore$ Weight of ${\text{C}}$ in $( {\text{CH}}_4)_n=12 {\text{n}}$Weight of ${\text{H}}$ in $( {\text{CH}}_4)_n=4 {\text{n}}$ $\therefore$ Weight ratio of ${\text{C}}$ and ${\text{H}}$ $=12 {\text{n}}: 4 {\text{n}}$ $=3: 1$ $\therefore \%$ of $C=\;\frac{3}{4} \times 100=75$and $\%$ of ${\text{H}}=\;\frac{1}{4} \times 100=25$

Energy of one photon$E=\;\frac{1240}{\lambda (n m)} {\text{eV}}$ $=\;\frac{1240}{300}$ $=4.1333 {\text{eV}}$ $\therefore \;\text{ Energy of one mole of photon }\;$ $=4.1333 \times 6.02 \times 10^{23} {\text{eV}}$ $=4.1333 \times 6.02 \times 10^{23} \times 1.6 \times 10^{-19} {\text{J}}$ $=\;\frac{4.1333 \times 6.02 \times 10^{23} \times 1.6 \times 10^{-19}}{1000} {\text{kJ}}$ $=399 {\text{kJ}} / {\text{mol}}$

(1) Bond order $=\;\frac{1}{2}[ {\text{N}}_{ {\text{b}}}- {\text{N}}_{ {\text{a}}}]$ ${\text{N}}_{ {\text{b}}}=$ No of electrons in bonding molecular orbital${\text{N}}_{ {\text{a}}}=$ No of electrons in anti bonding molecular orbital(2) upto 14 electrons, molecular orbital configuration isHere ${\text{N}}_{ {\text{a}}}=$ Anti bonding electron $=4$ and ${\text{N}}_{ {\text{b}}}=10$(3) After 14 electrons to 20 electrons molecular orbital configuration is -Here $N_a=10$and ${\text{N}}_{ {\text{b}}}=10$In ${\text{O}}$ atom 8 electrons present, so in ${\text{O}}_2, 8 \times 2=16$ electrons present.in $O_2^{2-}$ no of electrons $=18$(A) Molecular orbital configuration of ${\text{O}}_2^{2-}(18$ electrons) is$\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \sigma_{2 p_z^2} \pi_{2 p_x^2}=\pi_{2 p_y^2} \pi_{2 p_x^2}^*=\pi_{2 p_y^2}^*$ $\therefore \;\; N_b=10$ $N_a=8$ $\therefore \;\; B O=\;\frac{1}{2}[10-8]=1$ (B) $C_2^{2-}$ has 14 electrons.Moleculer orbital configuration of $C_2^{2-}$ is$\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \pi_{2 p_x^2}=\pi_{2 p_y^2} \sigma_{2 p_z^2}$ $\therefore {\text{N}}_{ {\text{a}}}=4$ ${\text{N}}_{ {\text{b}}}=10$ $\therefore \;\; {\text{BO}}=\;\frac{1}{2}[10-4]=3$(C) $N_2^{2-}$ has 16 electrons.Moleculer orbital configuration of $N_2^{2-}$ is$\sigma_{1 s^2} \sigma_{1 s^2}^* \sigma_{2 s^2} \sigma_{2 s^2}^* \pi_{2 p_x^2}=\pi_{2 p_y^2} \sigma_{2 p_z^2} \pi_{2 p_x^1}^*=\pi_{2 p_y^1}^*$ $\therefore N_a=6$ ${\text{N}}_{ {\text{b}}}=10$ $\therefore \;\; {\text{BO}}=\;\frac{1}{2}[10-6]=2$The correct order of bond orders of $C_2^{2-}, N_2^{2-}$ and $O_2{ }^{2-}$ ${\text{O}}_2{ }^{2-}< {\text{N}}_2{ }^{2-}< {\text{C}}_2{ }^{2-}$

$2 {\text{C}}$ (graphite) $+3 {\text{H}}_2( {\text{g}}) \to {\text{C}}_2 {\text{H}}_6( {\text{g}})$ $\Delta {\text{H}}_{ {\text{r}}}=+1560+2(-394)+3(-286)$ $=-86.0 {\text{kJ}} {\text{mol}}^{-1}$Enthalpy of formation of ${\text{C}}_2 {\text{H}}_6( {\text{g}})=-86.0 {\text{kJ}}$ mol $-1$

${\text{A}} \to$ ProductsFor a first order reaction,${\text{t}}_{1 / 2}=\;{\ln 2}/{ {\text{k}}}=\;{0.693}/{ {\text{k}}}$Time for $90 \%$ conversion,$t_{90 \%}=\;\frac{1}{k} \ln \;\frac{100}{10}=\;\frac{\ln 10}{k}=\;\frac{2.303}{k}$ $t_{90 \%}=\;\frac{2.303}{0.693} t_{1 / 2}=3.32 t_{1 / 2}$

Melting points of the given metals${\text{Hg}}:-38.83^{\circ} {\text{C}}$ ${\text{Ag}}: 961.8^{\circ} {\text{C}}$ ${\text{Ga}}: 29.76^{\circ} {\text{C}}$ ${\text{Cs}}: 28.44^{\circ} {\text{C}}$ $\therefore$ Metal having highest melting point is Ag.

Hall-Herault process is used for the extraction of aluminium by electrolysis molten ${\text{Al}}_2 {\text{O}}_3$ $2 {\text{Al}}_2 {\text{O}}_3+3 {\text{C}} \to 4 {\text{Al}}+3 {\text{CO}}_2$

Molecular hydrogen is produced as a byproduct in the industrial production of ${\text{NaOH}}$ by electrolysis of aq ${\text{NaCl}}$ solution${\text{NaCl}} \to {\text{Na}}^{+}+ {\text{Cl}}^{-}$ ${\text{H}}_2 {\text{O}} ⇌ {\text{H}}^{+}+ {\text{OH}}^{-}$Cathode : $2 {\text{H}}_2 {\text{O}}+2 {\text{e}} \to {\text{H}}_2+2 {\text{OH}}^{-}$Anode $: 2 {\text{Cl}}^{-} \to {\text{Cl}}_2+2 {\text{e}}$ ${\text{NaOH}}$ is crystallised from the remaining part of electrolyte.

Baking soda $( {\text{NaHCO}}_3)$ is used in certain type of fire extinguishers because it decomposes at high temperature to produce ${\text{CO}}_2$ which extinguishes fire$2 {\text{NaHCO}}_3( {\text{s}}) {\longrightarrow }^{\Delta } {\text{Na}}_2 {\text{CO}}_3( {\text{s}})+ {\text{H}}_2 {\text{O}} ↑+ {\text{CO}}_2 ↑$

${\text{PCl}}_5$ is well known but ${\text{NCl}}_5$ is not because nitrogen does not have vacant d-orbitals in its valence shell.So, nitrogen cannot expand its octet. On the other hand, phosphorus has vacant d-orbitals in its valence shell which enables it to expand its octet.

Crystal field splitting $(\Delta _0)$ for octahedral complexes depends on oxidation state of the metal as well as to which transition series the metal belongs. For the same oxidation state, the crystal field splitting $(\Delta _0)$ increases as we move from $3 d \to 4 d \to 5 d$ ${\text{Cr}}^{3+}$ and ${\text{Fe}}^{3+}$ belong to $3 d$ series, ${\text{Mo}}^{3+}$ belongs to $4 d$ series and ${\text{Os}}^{3+}$ belongs to $5 d$ series.Therefore crystal field splitting $(\Delta _0)$ is highest for $[ {\text{Os}}( {\text{H}}_2 {\text{O}})_6]^{3+}$.

Among the given gases, the green house gases which are responsible for heating the atmosphere are ${\text{CH}}_4$, water vapour and ozone. Nitrogen is not a green house gas. ${\text{CH}}_4, {\text{O}}_3$ and ${\text{H}}_2 {\text{O}}$ causes global warming in Tropospheric level.${\text{N}}_2$ does not cause global warming.

Carbocation (A) is stabilised by hyperconjugation due to $4 \alpha$ hydrogen atoms. Carbocation (C) is also stabilised by hyperconjugation due to 4 a hydrogen atoms but destabilised by $-1$ effect of $O$-atom. Carbocation (B) is most stable as it is stabilised by resonance.$\therefore$ Correct decreasing order of stability is ${\text{B}}> {\text{A}}> {\text{C}}$

The $\pi$-bond present is alkenes is weaker than $\sigma$ bond present in alkanes. That makes alkenes less stable than alkanes. Therefore, statement-I is correct.Carbon-carbon double bond is stronger than Carbon-carbon single bond because more energy is required to break 1 sigma and 1 pi bond than to break 1 sigma bond only. Therefore, statement-II is also correct.