10 Apr 2023 Shift 2

Due to rotation of earth, $g_{\;\text{eff }\;}=g-\omega ^2 R \cos ^2 \theta$ Where ' $\theta$ ' is angle made with equator Also, At poles, $\theta=90^{\circ}$ $\;\Delta {\text{g}}=\omega ^2 {\text{R}} \cos ^2 \theta$ $\; \;\; =\omega ^2 {\text{R}} \cos ^2 90=0$[no effect on poles]${\text{g}}_{\;\text{eff }\;}= {\text{g}}-\omega ^2 {\text{R}} \cos ^2 \theta$for equator $\theta=0^{\circ}$So, $g_{\;\text{eff }\;}=g-w^2 R$ $\ \Delta {\text{g}}=\omega ^2 {\text{R}}$ (Which is maximum change)

$\; {\text{I}}_{ {\text{res}}}=4 {\text{I}}_{ {\text{o}}} \cos ^2 (\;\frac{\theta}{2})$ $\;\;\text{ If }\; \theta=\;\frac{\pi}{3}, {\text{I}}_{ {\text{res}}}=4 {\text{I}}_{ {\text{o}}} \cdot \cos ^2 [\;\frac{\pi}{6}]$ $\;=4 {\text{I}}_0 \cdot (\;{{\sqrt{3}}}/{2}) ^2$ $\;I_1= (4 I_0) (\;\frac{3}{4}) =3 I_0$ $\;\;\text{ If }\; \theta=\;\frac{\pi}{2}, I_{\;\text{res }\;}=4 I_0 \cdot \cos ^2 (\;\frac{\pi}{2})$ $\;=4 {\text{I}}_0 (\;{1}/{{\sqrt{2}}}) ^2$ $\;= (4 {\text{I}}_{ {\text{o}}}) (\;\frac{1}{2})$ $\; {\text{I}}_2=2 {\text{I}}_{ {\text{o}}}$ $\;\;\frac{I_1}{I_2}=\;\frac{3}{2}$ $\;$

$\;T^2=\;\frac{4 \pi^2 r^3}{G M}$ $\;T^2=\;{4 \pi^2(2 {\text{R}})^3}/{G M}$ $\;=4 \times 8 \times \;\frac{\pi 2 \times R^3}{G M}$ $\;=4 \times 8 \times \;{(g) (R^3) }/{ {\text{GR}}^2}$At surface of earth $\Rightarrow {\text{g}}=\;{ {\text{GM}}}/{ {\text{R}}^2}$So, ${\text{GM}}= {\text{g}} \cdot {\text{R}}^2$Also $\pi^2=g$ $\; {\text{T}}^2=32 {\text{R}}$ $\;\Rightarrow {\text{T}}=\sqrt{32 {\text{R}}}$

Electrons moves in curved path because there velocity ${ {\text{u}}}^{\to }$ may make any angle $\theta$ with acceleration $\vec{a}$ between time interval of two successive collisions.Also electron moves from lower potential to higher potential.

Bond width of Amplitude modulated signal (AM) $\;=2 \times {\text{f}}_{\;\text{(message signal) }\;}$ $\;=2 \times 3 {\text{KHz}}$ $\;=6 {\text{KHz}}$

The zero error in verniel scale is $=6 \times 0.1 {\text{mm}}=0.6 {\text{mm}}$ (+ve zero error)Note: +ve zero error will have to be subtractedFrom the reading of the object.Now, the diameter measured with the help of Vernier scale isGiven by $\to$ M.S.D + V.S.D $\times$ L.S$\;\Rightarrow 3.2 {\text{cm}}+0.1 {\text{mm}} \times 4$ $\;=3.24 {\text{cm}}$The actual diameter is $\Rightarrow 3.24 {\text{mm}}$ - (zero error) $=3.24-0.6=3.18 {\text{cm}}$

In projectile motion, ${\text{H}}_{\;\text{many }\;}=\;{ {\text{u}}^2 sin \;^2 \theta}/{2 {\text{g}}}$ at $\theta=30^{\circ}, \;\; {\text{H}}_1=\;{ {\text{u}}^2 sin \;^2 30^{\circ}}/{2 {\text{g}}}$ at $\theta=60^{\circ}, \;\; {\text{H}}_2=\;{ {\text{u}}^2 sin \;^2 60^{\circ}}/{2 {\text{g}}}$ $\;{H_1}/{ {\text{H}}_2}=\;\frac{ sin \;^2 30^{\circ}}{ sin \;^2 60^{\circ}}=\;{ (\frac{1}{2}) }/{ (\;{{\sqrt{3}}}/{2}) ^2}=\;\frac{1}{3}$

Inertia is the property of mass due to which the object continues to move until any external force do not stops it. In the case of rotation of fan, if we switch off then also it moves for some time as air resistance takes time to stop it and due to inertia of fan it moves for some time.

${\text{t}}=\;{2 {\text{d}}}/{3}$ ${\text{K}}=\infty$ for metals${\text{C}}_1=\;{\epsilon_0 {\text{A}}}/{ {\text{d}}}\;\;\;$ $\;\;\; {\text{C}}_2=\;{\epsilon_0 {\text{A}}}/{ {\text{d}}- {\text{t}}+\;{ {\text{t}}}/{ {\text{k}}}}=\;{\epsilon_0 {\text{A}}}/{ {\text{d}}-\;{2 {\text{d}}}/{3}+0}=\;{3 \epsilon_0 {\text{A}}}/{ {\text{d}}}$ $=\;{ {\text{C}}_2}/{ {\text{C}}_1}=\;{\;{3 \epsilon_0 {\text{A}}}/{ {\text{d}}}}/{{\epsilon_0 {\text{A}}}/{ {\text{d}}}}=\;\frac{3}{1}$

$\;\text{ In electromagnetic wave, }\; E_0= {\text{B}}_0 {\text{C}}\;$ ${\text{E}}_0=6 \times 10^{-7} \times 3 \times 10^8\; {\text{E}}_0 \to \;\text{ Amplitude of electric field }\;$ $=18 \times 10^1\; {\text{B}}_0 \to \;\text{ Amplitude of magnetic field }\;$ $=180 {\text{v}} / {\text{m}}\; {\text{C}} \to \;\text{ Speed of light }\;$

Here we can see that ${\text{D}}_1$ and ${\text{D}}_3$ conducts but ${\text{D}}_2$ is reversed biased.Current $I_1$ will be equally distributed among $I_3$ and ${\text{I}}_4$ and ${\text{I}}_3=0$ $\; {\text{I}}_1= {\text{I}}_2+ {\text{I}}_4+ {\text{I}}_3$ $\; {\text{I}}_1=2 {\text{I}}_2$ $\;\;{ {\text{I}}_1}/{ {\text{I}}_2}=2$

Internal energy of ${\text{O}}_2=\;\frac{5}{2} {\text{nRT}}=\;\frac{5}{2} \times 2 {\text{RT}}=5 {\text{RT}}$Internal energy of ${\text{Ne}}= {\text{nRT}}=\;\frac{3}{2} \times {\text{nRT}}=\;\frac{3}{2} \times 4 {\text{RT}}=6 {\text{RT}}$ Total energy of mixture $($ system $)=5 R T+6 R T=11 R T$

If equation of SHM is in the form $y=a sin \;(\omega t)+B \cos (\omega t)$Then its amplitude is $=\sqrt{ {\text{A}}^{2+} {\text{B}}^2}$Here ${\text{A}}= {\text{B}}=1$ in equation ${\text{y}}= sin \;(\omega {\text{t}})+({\text{cost}})$Therefore, Amplitude $={\sqrt{(1)^{2+}(1)^2}}={\sqrt{2}}$

Time taken b / w A & B \Rightarrow t_1=\;\frac{x}{v_1}Time between b / w \;\; B & C \Rightarrow t_2=\;{ {\text{x}}}/{ {\text{v}}_2}Average velocity $(v)=\;\frac{\;\text{ Total displacement }\;}{\;\text{ Total time }\;}=\;\frac{x+x}{t_1+t_2}=\;\frac{2 x}{\frac{x}{v_1}+\;\frac{x}{v_2}}$(v) $=\;{2 {\text{v}}_1 {\text{v}}_2}/{ {\text{v}}_1+ {\text{v}}_2} \;\;$ or $\;\; \;{2}/{ {\text{v}}}=\;{1}/{ {\text{v}}_1}+\;{1}/{ {\text{v}}_2}$

If $\;\frac{7}{8}$ th is disintegrated it means only $\;\frac{1}{8}$ th part is radioactive active no. of nuclears after ' ${\text{n}}$ ' half lives$\;\Rightarrow \;{ {\text{N}}_{ {\text{o}}}}/{2^{ {\text{n}}}}=\;{ {\text{N}}_{ {\text{o}}}}/{8}$ $\;2^{ {\text{n}}}=8= {\text{n}}=3$So, the elapsed is 3 half lives $=3 {\text{T}}$

In Adiabatic process, $\Delta {\text{Q}}=0$If gas is compressed, then w (by gas) $\ne 0$Therefore by $1^{\;\text{st }\;}$ law$\;\Delta {\text{Q}}=\Delta {\text{u}}+ {\text{w}}$ $\;0=\Delta {\text{u}}+ {\text{w}}$ $\;\Delta {\text{u}}=- {\text{w}} \ne 0$It implies in adiabatic compression, internal energy of gas changes.

Diamagnetic substances have the property due to which they tends to move away from stronger magnetic field to weaker magnetic field, as their magnetic susceptibility is negative.Therefore both statements are correct.

$\;{ {\text{W}}}/{ {\text{A}}}= {\text{Y}} \cdot \;\frac{\Delta ℓ_1}{ℓ}\;\;{ {\text{W}}}/{3 {\text{A}}}=4 {\text{Y}} \cdot \;\frac{\Delta ℓ_2}{ℓ}$ $\Delta {\text{l}}_1=\;{ {\text{W}} ℓ}/{ {\text{AY}}}\;\Delta {\text{l}}_2=\;{ {\text{W}} ℓ}/{12 {\text{AY}}}$ $\;\frac{\Delta ℓ_1}{\Delta ℓ_2}=\;\frac{12}{1}\;$

$\;\;\text{ Work function }\;(\varphi )= {\text{hv}}^{\;\text{th }\;}$ $\;=6.6 \times 10^{-34} \times 5 \times 10^{14}$ $\;=33 \times 10^{-20}$ $\;\varphi =3.3 \times 10^{-19} {\text{J}}$ $\;=\;\frac{3.3 \times 10^{-19}}{1.6 \times 10^{-19}} {\text{ev}} \Rightarrow 2.07$

According to lenz's law, the rate of charge of flum produced by bar magnet will be approused by the conducting loops.

Lowest wavelength of by may sense will be obtained for trasition $n=\infty \to n=1$ and for balmer series, Lyman Series ${\text{n}}=\infty \to {\text{n}}=2$for Lyman, ${\text{E}}_0=\;{ {\text{hC}}}/{917 Å}$for balmer, $\;{ {\text{E}}_0}/{4}=\;{ {\text{hC}}}/{\lambda (Å)}$using this$\lambda =917 \times 4=3668$

$\;\;\text{ emf induced \in solenoid }\; \Rightarrow \;\text{ BAWN }\; sin \;(\omega {\text{t}}), {\text{w}}=700 {\text{rad}} / {\ s}$ $\;\;\text{ Amplitude }\; \Rightarrow \;\text{ BAWN }\;$ $\;\;\text{ Area }\;( {\text{A}})=2 {\text{cm}} \times 2 {\text{cm}}$ $\;=4 {\text{cm}}^2$ $\;=4 \times 10^{-4} {\text{m}}^2$ $\;( {\text{B}})_{\;\text{solenoid }\;}=\mu _0 {\ni}$ $\;=4 \pi \times 10^{-7} \times 5000 \times 2.5$ $\;=5 \pi \times 10^{-3}$ $\; {\text{n}}=\;{50 \;\text{ turns }\;}/{ {\text{cm}}}=\;{5000}/{ {\text{m}}}$ $\; {\text{i}}=2.5$Amplitude of emf $= (5 p \times 10^{-3}) (4 \times 10^{-4}) (700)(1)$ $=5 \times \;\frac{22}{7} \times 4 \times 700 \times 10^{-7}=44 \times 10^{-4}$

$\;\rho =3 \times 10^{-7} Ω- {\text{cm}}$ $\; {\text{R}}=\rho \cdot \;{1}/{ {\text{A}}}$ $\;=\;{3 \times 10^{-7} \times (10^{-2} {\text{m}}) }/{ (100 \times 10^{-4} {\text{m}}^2) }=3 \times 10^{-7}$

$\vec{\tau }={ {\text{r}}}^{\to } \times { {\text{F}}}^{\to }$ $\;\vec{\tau }=\;\text{ head }\;- \;\text{ tail }\;$ $\;=(0-2) \hat{i}+(0(-3)) \hat{j}$ $\;=-2 \hat{i}+3 \hat{j}$ $\;\tau =(-2 \hat{i}+3 \hat{j}) \times (-p \hat{k})$ $\;=-2 \hat{p}-3 p \hat{i}$ $\;=-p(3 p \hat{i}+2 \hat{j}) \;\frac{a}{b}=\;\frac{3}{2}=\;\frac{x}{2} \;\; x=3$

$\; {\text{B}}_{\;\text{total }\;}= {\text{B}}_{ {\text{I}}}+ {\text{B}}_{ {\text{II}}}+ {\text{B}}_{\;\text{III }\;}$ $\; {\text{B}}_{ {\text{I}}}=0$ $\; {\text{B}}_{\;\text{III }\;}=0$Because $\vec{d l} \times \vec{r}=0$Now, magnetic field due to semicirclaur ring at its center is given by$\; {\text{B}}_{\;\text{II }\;}=\;{\mu _0 {\text{i}}}/{4 {\text{R}}}$ $\;=\;\frac{4 \pi \times 10^{-7} \times 14}{4 \times 2.2 \times 10^{-2}}$ $\;=\;\frac{22}{7} \times \;\frac{10^{-7} \times 14}{22 \times 10^{-3}}$ $\;=2 \times 10^{-7}$ $\;=2$

By equation of continuity$\; {\text{A}}_1 {\text{V}}_1= {\text{A}}_2 {\text{V}}_2$ $\; (2 {\text{cm}}^2) (4 {\text{cm}} / {\ s})= (10 \times 10^{-2} {\text{cm}}^2) ( {\text{v}})$ $\;\;{8 {\text{cm}}^3}/{ {\ s}}=10^{-1} {\text{cm}}^2( {\text{v}})$ $\; {\text{V}}=80 {\text{cm}} / {\ s}$

When an object executes S.H.M, its morning acceleration is given by ${\text{a}}_{\max }=\omega ^2 {\text{A}}$Where $\omega =\;\frac{2 \pi}{T}$Therefore, ${\text{a}}_{\max }=\;{4 \pi^2 {\text{A}}}/{ {\text{T}}^2}$(Max force) ${\text{F}}_{\max }= {\text{ma}}_{\max }=5 \times \;\frac{4 \times 3.14 \times 3.14}{3.14 \times 3.14} \times (1)$ $=20 {\text{N}}$

In uniform circular motion$\;F_c=m a_c$ $\;\;\text{ (q) }\;(E)=\;{ {\ mv}^2}/{ {\text{r}}}$(e) $(\;{2 {\text{k}} \lambda }/{ {\text{r}}}) =\;{ {\ mv}^2}/{ {\text{r}}}$ $\; {\text{v}}^2=\;{( {\text{e}})(2 {\text{k}} \lambda )}/{ {\text{m}}}=\;\frac{ (1.6 \times 10^{-19}) \times 2 \times (9 \times 10^9) \times (2 \times 10^{-8}) }{9 \times 10^{-31}}$ $\;=1.6 \times 4 \times 10^{13}$ $\; {\text{V}}^2=16 \times 4 \times 10^{12} \Rightarrow {\text{v}}=8 \times 10^6 {\text{m}} / {\ s}$Ans. 8

Due to lens ${\text{L}}_1$ $\; {\text{u}}=-6 {\text{m}}$ $\; {\text{f}}=+24 {\text{m}}$ $\;\;{1}/{ {\text{v}}}-\;{1}/{ {\text{u}}}=\;\frac{1}{f}$ $\;\;\frac{1}{v}=\;\frac{1}{24}-\;\frac{1}{6}=\;\frac{1-4}{24} \Rightarrow v=-8 m$Due to lens ${\text{L}}_2$ $\; {\text{U}}=-18 {\text{m}}$ $\; {\text{F}}=+9 {\text{m}}$ $\;\;{1}/{ {\text{v}}}-\;{1}/{ {\text{u}}}=\;{1}/{ {\text{f}}} {\text{x}}$ $\;\;{1}/{ {\text{v}}}-\;\frac{1}{9}=\;\frac{1}{18}$ $\; {\text{V}}=\;{1}/{ {\text{v}}}=\;\frac{2-1}{18}$ $\; {\text{V}}=18 {\text{m}}$

Tension in the string $\Rightarrow 16000 {\text{N}}$ Maximum power $=( {\text{F}})( {\text{V}})$ $\;=16000 \times 3$ $\;=48000$ $\;=48 {\text{kW}}$Ans. 48

${\text{FCC}} \;\;\;$ $\;\;\;\;\;\;\;\ \; {\text{BCC}}$ ${\sqrt{2}} a=4 {\text{r}} \;$ $\;{\sqrt{3}} {\text{a}}=4 {\text{r}}$ ${\text{a}}=\;{4 {\text{r}}}/{{\sqrt{2}}} \times \;{{\sqrt{2}}}/{{\sqrt{2}}}\;$ $a=2 {\sqrt{2}} {\text{r}}\;$

The process of making is soap is saponification.Ester + Base $\to$ Alcohol + SoapIn saponification, triglycerides are combine with strong base and form fatty acid so this is alkaline Hydrolysis reaction.

$\;16 {\text{g}} {\text{CH}}_4=\;\text{ mole }\;=1$ $\; {\text{e}}-=60.0 \times 10^{23}$ $\;19 {\text{Hz}}=0.5 {\text{mole}}=11.4( {\text{L}}) {\text{STP}}$ $\;1 {\text{mole}} {\text{N}}_2=2 {\text{rg}}$ $\;0.5 {\text{mol}} {\text{SO}}_2=\;\text{ weights }\; 32 {\text{g}} .$

So ${\text{K}} > {\text{Ca}} > {\text{Be}}$

Acidity $\propto$ stability of conjugate base Stability orderActivity $\to {\text{E}} > {\text{C}} > {\text{D}} > {\text{A}} > {\text{B}}$

$\;\;\text{ CFSE }\;=-0.4 \times {\text{t}}_{2 {\text{g}}}+0.6 \times {\text{eg}}+ {\text{xp}}$ $\;=-0.4 \times 2+0.6 \times 0+ {\text{xp}}$(A) $=-0.8 \to {\text{Ti}}^{+2}$(B) $\; {\text{V}}^{+2} \to 3 {\text{d}}^3$ $\;\;\text{ CFSE }\;=-0.4 \times {\text{t}}_{2 {\text{g}}}+0.6 \times {\text{eg}}+ {\text{xp}}$ $\;=-0.4 \times 3+0.6 \times 0+ {\text{xp}}$ $\;=-1.2$ $\;$ $\;\text{ CFSE }\;=\;-0.4 \times {\text{t}}_{2 {\text{g}}}+0.6 \times {\text{eg}}+ {\text{xp}}$ $\;-0.4 \times 3+0.6 \times 1+ {\text{xp}}$ $=\;-1.2+0.6=0.6$ ${\text{CFSE}}\;=-0.4 \times {\text{t}}_{2 {\text{g}}}+0.6 \times {\text{eg}}+ {\text{xp}}$ $\;=-0.4 \times 3+0.6 \times 2$ $\;=-1.2+1.2$ $\;=0$

Carious method is used for quantitative analysis of sulfurSo Methionine is correct answer

Washing Soda $\to {\text{Na}}_2 {\text{CO}}_3 \cdot 10 {\text{H}}_2 {\text{O}}$0.2Soda Ash $\to {\text{Na}}_2 {\text{CO}}_3$No. of water $=10+0=(10)$

${\text{FeO}}+ {\text{C}} \to {\text{Fe}}+ {\text{CO}}_2$At $600^{\circ} {\text{C}} \Delta {\text{G}}$ of Reaction is $- {\text{Ve}}$

The Physical properties of isotope of Hydrogen are different due to Large mass difference

Stability of carbocation $\propto \;\frac{1}{\;\text{ Hydride affinity }\;}$