29 Jan 2023 Shift 2

$\; ( {\text{N}}_0) {\text{A}}=\;\frac{320}{16}=20 \;\text{ moles }\;$ $\;( {\text{N}}_0) {\text{B}}=\;\frac{320}{32}=10 \;\text{ moles }\;$ $\; {\text{N}}_{ {\text{A}}}=\;{ ( {\text{N}}_0) _{ {\text{A}}}}/{(2)^{2 / 1}}=\;\frac{20}{4}=5$ $\; {\text{N}}_{ {\text{B}}}=\;{ ( {\text{N}}_0) _{ {\text{B}}}}/{(2)^{2 / .5}}=\;\frac{10}{2^4}=0.625$ $\;\;\text{ Total }\; {\text{N}}=5.625$ $\;\;\text{ No. of atoms }\;=5.625 \times 6.023 \times 10^{23}$ $\; \;\; =3.38 \times 10^{24}$

$\;\sqrt{\;\frac{3 R T}{M}}=\sqrt{\;\frac{\alpha+5}{\alpha}} \sqrt{\;\frac{8}{\pi} \;\frac{R T}{M}}$ $\;3=\;\frac{\alpha+5}{\alpha} \;\frac{8}{\pi}$ $\;\alpha=28$

$\;\;{\lambda _alpha}/{\lambda _{ {\text{p}}}}=\;{\;{ {\text{h}}}/{\sqrt{2 {\text{m}}_alpha {\text{q}}_alpha {\text{V}}}}}/{{ {\text{h}}}/{\sqrt{2 {\text{m}}_{ {\text{p}}} {\text{q}}_{ {\text{p}}} {\text{V}}}}}$ $\;\;{\lambda _alpha}/{\lambda _{ {\text{p}}}}=\sqrt{\;\frac{1}{8} \;\; {\text{m}}=8}$

$\; {\text{a}}_1= {\text{g}} sin \;\theta= {\text{g}} / {\sqrt{2}}$ $\; {\text{a}}_2= {\text{g}} sin \;\theta- {\text{Kg}} \cos \theta=\;{ {\text{g}}}/{{\sqrt{2}}}-\;{ {\text{Kg}}}/{{\sqrt{2}}}$ \; {\text{t}}_2= {\text{nt}}_1 & \;\; {\text{a}}_1 {\text{t}}_1^2= {\text{a}}_2 {\text{t}}_2^2 $\;\;{ {\text{g}}}/{{\sqrt{2}}} {\text{t}}_1^2= (\;{ {\text{g}}}/{{\sqrt{2}}}-\;{ {\text{kg}}}/{{\sqrt{2}}}) {\text{n}}^2 {\text{t}}_1^2$ $\; {\text{K}}=1-\;{1}/{ {\text{n}}^2} \;\; \;\text{ Ans. }\; 4$

$\;50=V \times 10$ $\;V=5 {\text{m}} / {\ s}$ $\; {\text{V}}=0+ {\text{a}} \times 20$ $\;5= {\text{a}} \times 20$ $\; {\text{a}}=\;\frac{1}{4} {\text{m}} / {\ s}^2$ $\; {\text{F}}= {\text{ma}}=20 \times \;\frac{1}{4}=5 {\text{N}}$

$\; {\text{P}}_2 {\text{A}}- {\text{P}}_1 {\text{A}}=5.4 \times 10^5 \times {\text{g}}$ ${\text{P}}_2- {\text{P}}_1=\;\frac{5.4 \times 10^6}{500}=5.4 \times 2 \times 10^2 \times 10$ ${\text{P}}_2+0+\;\frac{1}{2} \rho {\text{V}}_2^2= {\text{P}}_1+0+\;\frac{1}{2} \rho {\text{V}}_1^2$ ${\text{P}}_2- {\text{P}}_1=\;\frac{1}{2} \rho ( {\text{V}}_1^{2-} {\text{V}}_2^2) =\;\frac{1}{2} \rho ( {\text{V}}_1- {\text{V}}_2) ( {\text{V}}_1+ {\text{V}}_2)$ $10.8 \times 10^3=\;\frac{1}{2} \times 1.2 ( {\text{V}}_1- {\text{V}}_2) \times 2 \times 3 \times 10^2$ $10.8 \times 10=3.6 ( {\text{V}}_1- {\text{V}}_2)$ ${\text{V}}_1- {\text{V}}_2=30$ $(\;{ {\text{V}}_1- {\text{V}}_2}/{ {\text{V}}}) \times 100=\;\frac{30}{300} \times 100=10 \%$

Work done by a force is given by $W = \vec{F} \cdot \vec{d} = Fd \cos \theta$, where $\theta$ is the angle between the force vector $\vec{F}$ and the displacement vector $\vec{d}$. Work is positive if the force has a component in the direction of displacement and negative if it has a component opposite to the displacement.

(A) The man applies an upward force, and the bucket's displacement is upward. The angle $\theta = 0^\circ$, so the work done is positive. This statement is incorrect.

(B) Gravitational force acts downward, while the bucket's displacement is upward. The angle $\theta = 180^\circ$, so the work done by gravity is negative ($W_g < 0$). This statement is correct.

(C) Frictional force always opposes motion. As the body slides down, friction acts up the incline, opposite to the displacement. The angle $\theta = 180^\circ$, so the work done by friction is negative. This statement is incorrect.

(D) For an object moving with uniform velocity, the net force is zero, meaning the applied force equals the friction force in magnitude. The work done by the applied force is $W_{app} = F_{app}d \cos(0^\circ) > 0$, as the force is in the direction of motion. This statement is incorrect.

(E) Air resistance is a dissipative force that always opposes the velocity of the pendulum bob. Therefore, the angle between the force of air resistance and the displacement is always $180^\circ$, making the work done negative. This statement is correct.

Thus, the correct statements are (B) and (E).

$\; {\text{a}}_{ {\text{c}}}=\;{ {\text{V}}^2}/{ {\text{r}}}=\;\frac{4^2}{2}=\;\frac{16}{2}=8 {\text{m}} / {\ s}^2$ \;{ \vec{\text{V}}=4 \;{ {\text{j}}}^{\^} \;{ {\text{a}}_{ \vec{\text{c}}}=8 \;{ {\text{i}}}^{\^}

\;\omega _{ {\text{E}}}={q \vec{\text{E}} \cdot { \vec{\text{S}} \;=2 \times 10^{-2}[30 \;{ {\text{i}}}^{\^} \cdot (-\;{ {\text{i}}}^{\^})] $\;=2 \times 10^{-2}(-30)$ $\;=-60 \times 10^{-2}$ $\;=-\;\frac{60}{100}=-0.6 {\text{J}}$ $\;=-600 {\text{mJ}}$

$\;\mu =\;\text{ modulation index }\;=\;\frac{A_{\max }-A_{\min }}{A_{\max }+A_{\min }}$ $\;=\;\frac{14-6}{14+6}=0.4$

$\; {\text{B}}=\;{\mu _0 {\text{i}}}/{2 {\text{R}}} \times 4$ $\; {\text{B}}^{'}=\;{\mu _0 {\text{i}}}/{2 {\text{R}}^{'}}$ $\; {\text{R}}^{'}=4 {\text{R}}$ $\; {\text{B}}^{'}=\;{\mu _0 {\text{i}}}/{8 {\text{R}}}$ $\;\;{ {\text{B}}^{'}}/{ {\text{B}}}=\;\frac{1}{16}$ $\; {\text{B}}^{'}=2 {\text{T}}$

Sensitivity of potentiometer wire is inversely proportional to potential gradient.

${\text{P}}=\;\frac{2 \mu sin \;\theta}{1.22 \lambda }$

Statement-I is correct as EMW are neutral.Statement-II is wrong.${\text{E}}_0=\sqrt{\;\frac{1}{\mu _0 \epsilon_0}} {\text{B}}_0$

$\;\Delta {\text{Q}}=184 \times 10^3$ $\; {\text{m}}=0.600 {\text{kg}} \;\text{ at }\;-12^{\circ} {\text{C}}$ $\; {\text{S}}=222.3 {\text{J}} / {\text{kg}} /{ }^{\circ} {\text{C}}$ $\; {\text{L}}=336 \times 10^3 {\text{J}} / {\text{kg}}$ $\; {\text{Q}}_1=0.600 \times 2222.3 \times 12=16000.56 {\text{J}}$ $\;\;\text{ Remaining heat }\; \Delta {\text{Q}}_1=184000-16000.56$ $\;=167999.44 {\text{J}}$ $\;\;\text{ For meeting at }\; 0{ }^{\circ} {\text{C}}$ $\;\Delta {\text{Q}}_2=0.600 \times 336000=201600 {\text{J}} \;\text{ needed }\;$ $\; \therefore 100 \% \;\text{ ice is not melted }\;$ $\;\;\text{ Amount of ice melted }\;$ $\;167999.44= {\text{m}} \times 336000=0.4999 {\text{kg}}$ $\; \therefore \;\text{ mass of water }\;=0.4999 {\text{kg}}$ $\;\;\text{ Mass of ice }\;=0.1001$ $\; \therefore \;\text{ Ratio }\;=\;\frac{0.1001}{0.4999} \approx 1: 5$

${\text{X}}_{ {\text{L}}}$ is not equal to ${\text{X}}_{ {\text{C}}}$. So rms current in (b) can never be larger than (a).

Sol. ${\text{T}}^2 \propto {\text{R}}^3$ $\;\;{ {\text{T}}_1^2}/{ {\text{T}}_2^2}=\;{ {\text{R}}_1^3}/{ {\text{R}}_2^3} \Rightarrow (\;{ {\text{T}}_1}/{ {\text{T}}_2}) ^2= (\;{ {\text{R}}}/{{ {\text{R}}}/{4}}) ^3$ $\; \therefore \;\; \;{ {\text{T}}_1^2}/{ {\text{T}}_2^2}=64$ $\; \therefore \;\; {\text{T}}_2^2=\;{ {\text{T}}_1^2}/{64}$ $\; \therefore \;\; T_2=\;\frac{24}{8}=3$ $\;$

$\;( {\text{x}}- {\text{At}})^{2+} ( {\text{y}}-\;{ {\text{t}}}/{ {\text{B}}}) ^2= {\text{a}}^2$ $\;{[ {\text{At}}]= {\text{A}} \times \;{1}/{ {\text{T}}}= {\text{L}}}$ $\; \therefore \;\; [ {\text{A}}]= {\text{T}}^1 {\text{L}}^1$ $\; \;\; \;{ {\text{t}}}/{ {\text{B}}} \;\text{ is \in meters }\;$ $\; \therefore \;\; \;{1}/{ {\text{T}}[ {\text{B}}]}= {\text{L}}$ $\; \therefore \;\; [ {\text{B}}]= {\text{T}}^{-1} {\text{L}}^{-1}$ $\; \therefore \;\; \;\text{ Correct Ans. (2) }\;$

Sol. $ℓ=50 {\text{cm}}$ ${\text{t}}=1 {\ sec}$ $\therefore {\text{V}}=\;\frac{0.05}{1}=0.05 {\text{m}} / {\ s} \;\text{. }\;$ $\;i=\;\frac{40 \times 0.05 \times 0.05}{10}=0.01 {\text{A}}$ $\therefore \;\; {\text{F}}= {\text{B}}_1 ℓ=40 \times 0.01 \times 0.05$ $\; {\text{F}}=0.02 {\text{N}}$ $\; \therefore \;\; {\text{W}}=0.02 \times ℓ=0.02 \times .05$ $\; \therefore \;\; {\text{W}}=1 \times 10^{-3} {\text{J}}$

$\;{ {\text{R}}_1+ {\text{R}}_2}/{10}=\;\frac{60}{40}=\;\frac{3}{2} \Rightarrow {\text{R}}_1+ {\text{R}}_2=15$Now $\;\frac{R_1 R_2}{ (R_1+R_2) \times 3}=\;\frac{40}{60}=\;\frac{2}{3} \Rightarrow R_1 R_2=30$

$\;\;\text{ Use }\; \Delta {\text{L}}=int_{0}^{ {\text{t}}} \tau {\text{dt}}$ $\; {\text{L}}_0=int_{0^2 {\text{mg}} ( {\text{v}}}^{ {\text{x}}} {\text{t}}) {\text{dt}}$ $\;=m g v_x \;\frac{t^2}{2}=(0.1)(10)(10 {\sqrt{2}}) \;\frac{2^2}{2}$ $\;=20 {\sqrt{2}}$ $\;={\sqrt{800}} {\text{kgm}}^2 / {\ s}$ $\;$

$\; {\text{u}}=\;{ {\text{h}}}/{ {\text{h}}^{'}}=\;\frac{5.25}{5.00}$ $\;\;\text{ Least count }\;=\;\frac{1}{20} {\text{cm}}-\;\frac{49}{50} \cdot \;\frac{1}{20} {\text{cm}}$ $\;=\;\frac{1}{50} \times \;\frac{1}{20} {\text{cm}}=0.01 {\text{mm}}$ $\;\ln {\text{u}}=\ln {\text{h}}-\ln {\text{h}}^{'}$ $\;\;{ {\text{du}}}/{ {\text{u}}}=\;{ {\text{dh}}}/{ {\text{h}}}-\;{ {\text{dh}}^{'}}/{ {\text{h}}^{'}}$ $\; {\text{du}}= [\;\frac{0.01}{5.25}+\;\frac{0.01}{5.00}] \;\frac{5.25}{5.00}$ $\;=\;\frac{41}{10} \times 10^{-3}$ $\;\;\text{ Ans. }\;=41$

${\text{E}}=-\;{ {\text{dV}}}/{ {\text{dr}}}=-4 {\text{ar}} \equiv \;{\rho {\text{r}}}/{3 \epsilon_0} \;\text{ (compare) }\;$Result inside uniformly charged solid sphere.$\rho =-12 a \epsilon_0$

\; {\text{v}} \;{ {\text{dv}}}/{ {\text{dx}}}=\;{ {\text{v}}^2}/{ {\text{R}}} \Rightarrow int_{15}^{ {\text{v}}} \;{ {\text{dv}}}/{ {\text{v}}}=\;{1}/{ {\text{R}}} int_{0^{ {\text{x}}} {\text{dx}} $\;v=15 e^{x / R}$ $\;\;\frac{d x}{d t}=15 e^{x / R}$ $int_{0^{\;\frac{\pi R}{2}} e}^{-x / R} \text{dx}=15 int_{0}^{t_0} \text{dt}$ $t_0=40 (1-e^{-\pi / 2})$

$\;\;{1}/{2 \pi {\text{fC}}}=2 \pi {\text{fL}}$ $\; {\text{C}}=\;{1}/{4 \pi^2 {\text{f}}^2 {\text{L}}}=\;\frac{1}{4 \times \pi^2 \times 49 \times 10^6 \times 2 \times 10^{-6}}$ $\; {\text{C}}=\;\frac{1}{3872} {\text{F}}$ $\; {\text{x}}=3872 \;\text{ Ans. }\;$

$\;| {\text{F}}|=η {\text{A}} \;{\Delta {\text{v}}}/{\Delta {\text{h}}}: 0.1=5 \times 10^{-3} \times 0.2 \times \;{ {\text{v}}}/{.25 \times 10^{-3}}$ $\; {\text{v}}=0.025 {\text{m}} / {\ s} \;\text{ or }\; {\text{v}}=25 \times 10^{-3} {\text{m}} / {\ s}$

$\;\;\frac{1}{4} {\text{a}}=-25 {\text{x}} ; {\text{a}}=-100 {\text{x}}$ $\;\omega ^2=100 \;\; \omega =10,$ $\omega {\text{A}}=4 \;\; {\text{A}}=\;\frac{4}{10}=0.4 {\text{m}}$ ${\text{A}}=40 {\text{cm}}$

$\;\mu _1={\sqrt{2.8 \times 1}}={\sqrt{2.8}}$ $\;\mu _2={\sqrt{6.8 \times 1}}={\sqrt{6.8}}$ $\;\mu _1 sin \;i=\mu _2 \cos i \;\; \tan i=\;\frac{\mu _2}{\mu _1}=\sqrt{\;\frac{6.8}{2.8}}$ $\;\tan i= (\;\frac{2.8+4}{2.8}) ^{1 / 2} \;\; i=\tan ^{-1} (1+\;\frac{10}{7}) ^{1 / 2}$ $\;\theta=7 \;\text{ Ans. }\;$

$\;\;\frac{\epsilon}{r+5} \times 5=200 x$ . . . (1)$\;\;\frac{\epsilon \times 15}{r+15}=300 x$ . . . (2)$\;\Rightarrow r=5 Ω$Ans. 5

The first ionization energies (as in NCERT) are as follows:${\text{B}}: 801 {\text{kJ}} / {\text{mol}}$ ${\text{Al}}: 577 {\text{kJ}} / {\text{mol}}$ ${\text{Ga}}: 579 {\text{kJ}} / {\text{mol}}$ ${\text{Ga}}:[ {\text{Ar}}] 3 {\text{d}}^{10} 4 {\ s}^2 4 {\text{p}}^1$

$[ {\text{FeF}}_6] ^{3-}: {\text{Fe}}^{3+}=3 {\text{d}}^5 \Delta _0 < {\text{P}}$Number of unpaired ${\text{e}}^{-}=5 \therefore \mu ={\sqrt{35}} {\text{BM}}$ $[ {\text{CoF}}_6] ^{3-}: {\text{Co}}^{3+}=3 {\text{d}}^6 (\Delta _0 < {\text{P}})$Number of unpaired ${\text{e}}^{-}=4 \therefore \mu ={\sqrt{24}} {\text{BM}}$ $[ {\text{Co}} ( {\text{C}}_2 {\text{O}}_4) _3] ^{3-}: {\text{Co}}^{3+}=3 {\text{d}}^6 (\Delta _{ {\text{O}}} > {\text{P}})$ Number of unpaired ${\text{e}}^{-}=0 \therefore \mu =0 {\text{BM}}$

A. Osmosis III B. Reverse osmosis I C. Electro osmosis IV D. Electrophoresis II

(i), (ii) and (iv) correct.Manganese exhibits $+7$ oxidation state in its oxide. $( {\text{Mn}}_2 {\text{O}}_7)$ {\text{Ru}} & {\text{Os}} from ${\text{RuO}}_4$ & ${\text{OsO}}_4$ oxide in $+8$ oxidation stateCr in $+6$ oxidation act is oxidizing.Sc does not show $+4$ oxidation state.

Neoprene: Elastomer Polyester: Fibre Polystyrene : Thermoplastic Urea-Formaldhyde Resin: Thermosetting polymer

${\text{I}}^{-}+ {\text{H}}_2 {\text{O}}_2 \to { {\text{I}}_2}_{(\text{A})}+ {\text{H}}_2 {\text{O}}$ ${\text{I}}_2+\text{Starch}_{(\text{Indicator})} \to \text{Blue}$

Theory based.

{\text{CH}}_3- {\text{CH}}_2- {\text{CH}}^{|}^{\text{CH}_3}- {\text{CH}}_2- {\text{Br}} \to C-C-C^{|}^{\text{C}}=C (1) {\text{CH}}_3- {\text{CH}}^{|}^{\text{Br}}- {\text{CH}}_3 \to {\text{CH}}_3 {\text{CH}}= {\text{CH}}_2(1) {\text{CH}}_3- {\text{CH}}_2- {\text{CH}}_2- {\text{CH}}^{|}^{\text{Br}}- {\text{CH}}_3 \to ${\text{C}}- {\text{C}}- {\text{C}}- {\text{C}}= {\text{C}}+ {\text{C}}- {\text{C}}- {\text{C}}= {\text{C}}- {\text{C}}$, cis & trans (3) {\text{C}}- {\text{C}}- {\text{C}}^{|}^{\text{CH}_3}_{|}_{\text{CH}_3}- {\text{C}}_{|}_{\text{Br}}- {\text{C}} \to {\text{C}}- {\text{C}}- {\text{C}}_{|}_{\text{C}}^{|}^{\text{C}}- {\text{C}}= {\text{C}}(1)

$\; {\text{C}}_{ {\text{x}}} {\text{H}}_{ {\text{y}}}+ ( {\text{x}}+\;{ {\text{y}}}/{4}) {\text{O}}_2 \to {\text{xCO}}_2+\;{ {\text{y}}}/{2} {\text{H}}_2 {\text{O}}$ $\; {\text{x}}+\;{ {\text{y}}}/{4}=9.5$ $\;\;{ {\text{y}}}/{2}=3$ $\;\Rightarrow {\text{x}}=8, {\text{y}}=6$

Theory based.

A solution of ${\text{CrO}}_5$ in amyl alcohol has a blue colour. So, option (4) is correct.

The growth of fish gets inhibited if the concentration of dissolved Oxygen in water is less than $6 {\text{ppm}}$ and Biochemical Oxygen demand in clean water should be less than $5 {\text{ppm}}$.

Hydrolysis of the given tetrapeptide will give the following:

Only (B) and (C) are correct.(B) ${\text{G}}= {\text{H}}- {\text{TS}}$At constant $T$ $\Delta {\text{G}}=\Delta {\text{H}}- {\text{T}} \Delta {\text{S}}$(A) First law is given by$\Delta {\text{U}}= {\text{Q}}+ {\text{W}}$If we apply constant ${\text{P}}$ and reversible work.$\Delta {\text{U}}= {\text{Q}}- {\text{P}} \Delta {\text{V}}$(C)By definition of entropy change${\text{dS}}=\;{ {\text{dq}} q_{\;\text{rev }\;}}/{ {\text{T}}}$At constant $T$ $\Delta {\text{S}}=\;{ {\text{q}}_{\;\text{rev }\;}}/{ {\text{T}}}$(D) ${\text{H}}= {\text{U}}+ {\text{PV}}$For ideal gas${\text{H}}= {\text{U}}+ {\text{nRT}}$At constant $T$ $\Delta {\text{H}}=\Delta {\text{U}}+\Delta {\text{nRT}}$

Calamine: ${\text{ZnCO}}_3$Siderite: ${\text{FeCO}}_3$Sphalerite: ${\text{ZnS}}$Malachite: ${\text{CuCO}}_3 \cdot {\text{Cu}}( {\text{OH}})_2$

Statement-I is correct.Ni is used in Hydrogenation of unsaturated fat to make edible fats.Statements-II is false as hydride of silicon is electron precise & neither electron deficient nor electron rich.

(A) van't Hoff factor, i$i=\;\frac{\;\text{ Normal molar mass }\;}{\;\text{ Abnormal molar mass }\;}$(B) ${\text{k}}_{ {\text{f}}}=$ Cryoscopic constant(C) Solutions with same osmotic pressure are known as isotonic solutions.(D) Solutions with same composition of vapour over them are called Azeotrope.

$2 {\text{Li}} {\text{NO}}_3 {\to }^{\Delta } {\text{Li}}_2 {\text{O}}+2 {\text{NO}}_2+\;\frac{1}{2} {\text{O}}_2$Hence three products ${\text{Li}}_2 {\text{O}}, {\text{NO}}_2$ and ${\text{O}}_2$