28 Jun 2022 Shift 1

$[$ Pressure $][$ Time $]= [\;\frac{\;\text{ Force }\;}{\;\text{ Area }\;}] [\;\frac{\;\text{ distance }\;}{\;\text{ Area }\;}]$[Coefficient of viscosity $]= [\;\frac{\;\text{ Force }\;}{\;\text{ Area }\;}] [\;\frac{\;\text{ distance }\;}{\;\text{ Area }\;}]$Statement ' $A$ ' is trueBut Statement ' $R$ ' is false are coefficient of viscosity$=\;\frac{\;\text{ Force }\;}{\;\text{ Area }\; \times \;\text{ Velocity gradient }\;}$

$a_r=k^2 r t^2=\;\frac{v^2}{r}$ $\Rightarrow v^2=k^2 r^2 t^2$ or $v=k r t$and $\;\frac{d|v|}{d t}=k r$ $\Rightarrow a_t=k r$ $\Rightarrow |\overline{F} \cdot \overline{v}|=(m k r)(k r t)$ $=m k^2 r^2 t=$ power delivered

$x=4 sin \; (\;\frac{\pi}{2}-\omega t)$ $=4 \cos (\omega t)$ $y=4 sin \;(\omega t)$ $\Rightarrow x^{2+}y^2=4^2$ $\Rightarrow$ The particle is moving in a circular motion with radius of $4 {\text{m}}$.

(A) Moment of inertia of solid sphere of radius ${\text{R}}$ about a tangent $=\;\frac{2}{5} M R^{2+}M R^2=\;\frac{7}{5} M R^2$ $\Rightarrow A-(I I)$(B) Moment of inertia of hollow sphere of radius ${\text{R}}$ about a tangent $=\;\frac{2}{3} M R^{2+}M R^2=\;\frac{5}{3} M R^2$ $\Rightarrow B-(I)$(C) Moment of inertia of circular ring of radius $( {\text{R}})$ about its diameter $=\;\frac{ (M R^2) }{2}$ $\Rightarrow {\text{C}}-( {\text{IV}})$(D) Moment of inertia of circular disc of radius ( $R$ ) about any diameter$=\;\frac{M R^2 / 2}{2}=\;\frac{M R^2}{4}$ $\Rightarrow {\text{D}}-( {\text{III}})$

$T_A=2 T_B$Now $T_A^2 \propto r_A^3$ $\Rightarrow (\;\frac{r_A}{r_B}) ^3= (\;\frac{T_A}{T_B}) ^2$ $\Rightarrow r_A^3=4 r_B^3$

$r^{'}=\;\frac{r}{4}$ $\Rightarrow \Delta E=T(\Delta S)$ $=T \times 4 \pi (n r^{' 2}-r^2) , n=64$ $=T \times 4 \pi \times (4-1) r^2$ $\Rightarrow \Delta E=0.075 \times 4 \times 3.142(3) \times 10^{-4} {\text{J}}$ $=2.8 \times 10^{-4} {\text{J}}$

$W=\;\frac{\mu R (T_2-T_1) }{1-r}$ for a polytropic process for adiabatic process ${\text{r}}=\gamma$ $\Rightarrow$ Statement I is true.In an adiabatic process$\Delta {\text{U}}=-\Delta {\text{W}}$ $\Rightarrow$ If work is done on the gas$\Rightarrow \Delta {\text{W}}$ is negative$\Rightarrow \Delta U$ is positive or temperature increases$\Rightarrow$ Statement II is true

As one moves closer to a positive charge (isolated) the density of electric field line increases and so does the electric field intensity$\Rightarrow$ Statement I is trueAs opposite poles of an electric dipole would experience equal and opposite forces so net force on a dipole in a uniform electric field will be zero$\Rightarrow$ Statement II is true

$|E_0{{}/{}|} =\;\frac{k q / 2}{a^2} {\sqrt{2}}+\;{k q}/{(a {\sqrt{2}})^2}$ $=\;{k q}/{{\sqrt{2}} a^2}+\;\frac{k q}{2 a^2}$ $=\;\frac{k q}{a^2} (\;{1}/{{\sqrt{2}}}+\;\frac{1}{2}) , k=\;\frac{1}{4 \pi \epsilon_0}$

Inside a hollow cylindrical conductor with uniform current distribution net magnetic field is zero in hollow space.But outside the cylindrical conductor $B \propto \;\frac{1}{r}$ $\Rightarrow$ Graph in option D would be a correct one

$E_0=2.25 {\text{V}} / {\text{m}}$ $B_0=1.5 \times 10^{-8} {\text{T}}$ $\Rightarrow \;\frac{E_0}{B_0}=1.5 \times 10^8 {\text{m}} / {\ s}$ $\Rightarrow$ Refractive index $=2$Distance to be travelled $=6 {\text{km}}$ $\;\text{ Time taken }\;=\;\frac{6 \times 10^3}{1.5 \times 10^8}=4 \times 10^{-5} {\ s}$

$\mu =\;\frac{ sin \; (\frac{\delta _m+A}{2}) }{ sin \;(A / 2)}=\cot A / 2$ $\Rightarrow \cos A / 2= sin \; (\;\frac{\delta _m+A}{2})$ $\Rightarrow \;\frac{\pi}{2}-\;\frac{A}{2}=\;\frac{\delta _m+A}{2}$ $\Rightarrow \pi-2 A=\delta _m$

R.P. $=\;\frac{1}{1.22 \lambda / a}$ $=\;\frac{24.4 \times 10^{-2}}{1.22 \times 2440 \times 10^{-10}}$ $=8.2 \times 10^5$

$\lambda _p=\;\frac{h}{p}=\;\frac{h c}{E} . . . . . . \;\text{ (i) }\;$ $\lambda _e=\;{h}/{{\sqrt{2 m E}}} . . . . . . \;\text{ (ii) }\;$From (i) and (ii)$\lambda _p \propto \lambda _e^2$

$K_p>0$If $Q$ is released $\Rightarrow Q>0$ $\Rightarrow {\text{K}}_{ {\text{p}}}+ {\text{Q}}>0$Even the particle has to be given kinetic energy greater than magnitude of $Q$ to maintain momentum conservation.$\Rightarrow {\text{K}}+ {\text{Q}}>0$

This is NAND gate ABY001101011110

In forward bias diode conductsIn revers bias it does not conducts

(A) is true as $n-p-n$ transistor permits more current than $p-n-p$ transistor as electrons which are majority charge carriers in n-p-n have higher mobility than holes which are majority carriers in p-n-p transistor.$\Rightarrow$ Statement ${\text{R}}$ is correct explanation of statement ${\text{A}}$.

Television signal $\Rightarrow 6 {\text{MHz}}$Radio signal $\Rightarrow 2 {\text{MHz}}$High Quality music $\Rightarrow 20 {\text{kHz}}$Human speech $\Rightarrow 3 {\text{kHz}}$

$\;\frac{v}{4.08}-\;\frac{v}{4.16}=\;\frac{40}{12}$ $v=\;\frac{40}{12} \times \;\frac{4.08 \times 4.16}{0.08}$ $=707.2 {\text{m}} / {\ s}$

$\;\frac{1}{2} m v^2=m g l(1-\cos \theta)$ $\Rightarrow v={\sqrt{2 g l(1-\cos \theta)}}$ $=\sqrt{2 \times 10 \times 2.5 \times \;\frac{1}{2}}$ $=5 {\text{m}} / {\ s}$

Using the conditions of a balanced wheat stone bridge and adding the end correction.$\;\frac{15}{(43+2)}=\;\frac{R}{(102-45)} \Rightarrow R=\;\frac{57}{45} \times 15$ $R=19 Ω$

Using the condition of a balanced wheat stone bridge,$\Rightarrow \;{ {\text{R}}}/{3}=\;\frac{4}{6} \Rightarrow {\text{R}}=2 Ω$So the effective resistance of the circuit is${\text{R}}_{ {\text{eq}}}=\;\frac{6 \times 9}{6+9}=\;\frac{18}{5} Ω$ ${\text{i}}=\;{36}/{ {\text{R}}_{ {\text{eq}}}}=10 {\text{A}}$

${\text{L}}=100 \times 10^{-3} {\text{H}}$ ${\text{C}}=100 \times 10^{-6} {\text{F}}$ ${\text{R}}=120 Ω$ $\omega {\text{L}}=10 Ω$ $\;\frac{1}{\omega C}=\;\frac{1}{10^4 \times 10^{-6}}=100 Ω$ $\Rightarrow {\text{X}}_{ {\text{C}}}- {\text{X}}_{ {\text{L}}}=90 Ω$ $\Rightarrow Z={\sqrt{90^{2+}120^2}}=150 Ω$ $\Rightarrow I_{r m s}=\;\frac{20}{150}=\;\frac{2}{15} {\text{A}}$For heat resistance by $16^{\circ} {\text{C}}$ heat required $=32 {\text{J}}$ $\Rightarrow (\;\frac{2}{15}) ^2 \times (120) \times t=32$ $t=\;\frac{32 \times 15 \times 15}{4 \times 120}=15$

$|\vec{i}|=|\vec{r} \times (m \vec{v})|$ $=|(3 \hat{i}-\hat{j}) \times (3 \hat{j}+\hat{k})|$ $=|-\hat{i}-3 \hat{j}+9 \hat{k}|$ $={\sqrt{91}}$

Taking the system as man and trolley and using conservation of linear momentum.$60 \times {\text{v}}=(60+120) \times 2$ $\Rightarrow {\text{v}}=6 {\text{m}} / {\ s}$

$a=\;\frac{M g}{4 M+M}=\;\frac{g}{5}$ (in upward direction)$T=M (g+\;\frac{g}{5}) =\;\frac{6 M g}{5}$ $\Rightarrow x=6$

$U=2 (\;\frac{3}{2} R) 300$ $=3 \times 8.31 \times 300$ $=7479 {\text{J}}$

$R=\;\frac{m v}{q B}$ $R=\;{{\sqrt{2 m K E}}}/{q B}$ $=\;{\sqrt{2 \times 24 \times 1.67 \times 10^{-27} \times 5 \times 1.6 \times 10^{-16}}}/{1.6 \times 10^{-19} \times 0.5}$ $=10.009 {\text{cm}}=10 {\text{cm}}$

Total capacitance $=0.01 \times 100=1 \mu {\text{F}}$ $\omega =500 \times 2 \pi=1000 \pi {\text{rad}} / {\ s}$ $\omega L=\;\frac{1}{\omega C}$ $\Rightarrow L=\;\frac{1}{\omega ^2 C}=\;\frac{1}{1^6 \pi^2 \times 10^{-6}}=\;\frac{1}{10} H=100 {\text{mH}}$

The vacancy defect increases the density of substance.It does not change the density of the crystal. It only creates cationic vacancies. Frenkel Defect causes vacancy defect at its original site and an interstitial defect at its new location. Therefore, it does not change the density of the solid.

The potential difference between the fixed layer and the diffused layer of opposite charges is called zeta potential.It is related to the charge on the surface of colloidal particles.

Element ${\text{E}}$ is Selenium (Se)Electronic configuration of $E$ is $[A r] 3 d^{10} 4 s^2 4 p^4$The element which is just above ' ${\text{E}}$ ' in periodic table is sulphur, its electronic configuration is [Ne]$3 s^2 3 p^4$

Magnesium can reduce ${\text{Al}}_2 {\text{O}}_3$ at a temperature below $1350^{\circ} {\text{C}}$ while above $1350^{\circ} {\text{C}}$ aluminium can reduce ${\text{MgO}}$ because below $1350^{\circ} {\text{C}} \Delta {\text{G}}$ of ${\text{MgO}}$ (formation) is more negative and above $1350^{\circ} {\text{C}} \Delta$ ${\text{G}}$ of ${\text{Al}}_2 {\text{O}}_3$ (formation) is more negative.The melting and boiling point of magnesium are lower than those of aluminium but it is not the correct reason.

${\text{CuO}}+ {\text{H}}_2 \to {\text{Cu}}+ {\text{H}}_2 {\text{O}}$ (under hot conditions)

${\text{Ba}} ( {\text{N}}_3) _2 \;{\longrightarrow }^{\Delta } {\text{Ba}}+3 {\text{N}}_2$

As +ve oxidation state increases, EN of element increases hence acidic character increases. Down the group, non-metallic character decreases, acidic character decreases.Acidic character : ${\text{E}}_2 {\text{O}}_5> {\text{E}}_2 {\text{O}}_3$Down the group, acidic character of ${\text{E}}_2 {\text{O}}_3$ decreases

Electronic configuration of Europium (Eu) is $[ {\text{Kr}}] 4 {\text{f}}^7 6 {\ s}^2$. It can loose two electron to gain the half filled stable state. Hence, $+2$ oxidation state is most stable.

$[ {\text{Ni}}( {\text{CN}})_4] ^{2-}: {\text{d}}^8$ configuration, SFL, sq. planar splitting (dsp ${ }^2$ ), diamagnetic.$[ {\text{Ni}}( {\text{CO}})_4] : {\text{d}}^{10}$ config (after excitation), SFL, tetrahedral splitting $( {\ sp}^3)$, diamagnetic.$[ {\text{NiCl}}_4] ^{2-}: {\text{d}}^8$ config, WFL, tetrahedral splitting $( {\ sp}^3)$, paramagnetic $(2.$ unpaired ${\text{e}}^{-})$.

${\text{NaAsO}}_2$ (Sodium arsenite) is an insecticide, antibacterial agent herbicide.

TLC is a technique used to separate mixture of compounds based on differences in polarity. In TLC a glass plate coated with a stationary phase is spotted with the mixture to be separated.

$A$ and $B$ are aromatic as they are cyclic, planar and has $4 n+2 \pi e^{-}(n=1)$

is inactive towards ${\text{S}}_{ {\text{N}}} 1$ as halogen is attached to bridge head carbon atom, where formation of carbocation is not possible.

${\text{R}}- {\text{NH}}_2 {\longrightarrow } ^{{\text{NaNO}}_2}_{+ \text{HCl}} {\text{R}}- {\text{N}}_2^{+} \to {\text{R}}^{+} {\longrightarrow } _{ {\text{H}}_2 {\text{O}}} {\text{R}}- {\text{OH}}$

Buna-S, PHBr and Butadiene-styrene are copolymer. Only neoprene is namopolymer.

Mostly ${\text{H}}$-bonding is responsible for the stability of $\alpha$-helix form.

${\text{S}}^{2-}+ [ {\text{Fe}}( {\text{CN}})_5 {\text{NO}}) ] ^{2-} \to [ {\text{Fe}}( {\text{CN}})_5( {\text{NOS}})] ^{4-}$

First Step :${\text{MnO}}_2+4 {\text{HCl}} \to {\text{MnCl}}_2+ {\text{Cl}}_2+2 {\text{H}}_2 {\text{O}}$Here $1 {\text{mol}}$ of ${\text{MnO}}_2$ produce $1 {\text{mol}}$ of ${\text{Cl}}_2$ $\therefore$ Mole ratio of $n_{M_{n O_2}}: n_{C l_2}=1: 1$Second Step :${\text{Cl}}_2+2 {\text{Kl}} \to 2 {\text{KCl}}+ {\text{I}}_2$Here, $1 {\text{mol}}$ of ${\text{Cl}}_2$ produce $1 {\text{mol}}$ of ${\text{I}}_2$Mole ratio of $n_{C l_2}: n_{I_2}=1: 1$Third Step :${\text{I}}_2+2 {\text{Na}}_2 {\text{S}}_2 {\text{O}}_3 \to 2 {\text{Nal}}+ {\text{Na}}_2 {\text{S}}_2 {\text{O}}_3$ $1 {\text{mol}}$ of ${\text{I}}_2$ react with $2 {\text{mol}}$ of ${\text{Na}}_2 {\text{S}}_2 {\text{O}}_3$ Mole ratio of $n_{I_2}: n_{N_2 S_2 O_3}=1: 2$Given ${\text{Na}}_2 {\text{S}}_2 {\text{O}}_3$ is $60 {\text{mL}}$ of $0.1 {\text{M}}$ $\therefore$ Number of moles of ${\text{Na}}_2 {\text{S}}_2 {\text{O}}_3$ $= {\text{V}}($ in ${\text{L}}) \times {\text{M}}$ (Molarity)$=\;\frac{60}{1000} \times 0.1$ $=0.006 {\text{mol}}$ $\therefore$ Number of moles of ${\text{I}}_2$ $=\;\frac{1}{2}(0.006)$ $=0.003$ $\therefore$ Moles of ${\text{MnO}}_2=0.003$ (as mole ratio of ${\text{MnO}}_2$ and ${\text{Cl}}_2=1: 1$ )Molar mass of ${\text{MnO}}_2=55+32=87$ $\therefore \;\text{ Mass of }\; {\text{MnO}}_2=0.003 \times 87=0.261 {\text{gm}}$Given ${\text{MnO}}_2=2 {\text{g}}$ $\therefore \% \;\text{ of }\; {\text{MnO}}_2=\;\frac{0.261}{2} \times 100=13 \%$

Given,$\;\text{ Work function }\;=6.63 \times 10^{-19} {\text{J}}$ $=\;\frac{6.63 \times 10^{-19}}{1.6 \times 10^{-19}}$ $=4.14 {\text{eV}}$We know,$E=\;\frac{1240}{\lambda (n m)}$ $\Rightarrow 4.14=\;\frac{1240}{\lambda }$ $\lambda =300 {\text{nm}}$

${\text{PF}}_5 \Rightarrow {\ sp}^3 {\text{d}}$ hybridisation(5 sigma bonds, zero lone pair on central atom)Value of $y=1$