29 Jun 2022 Shift 1

Let us assume that they meet at ${\text{t}}= {\text{t}}_0$ $A: 80=\;\frac{1}{2} g t_0^2 . . . . . . \;\text{ (i) }\;$ $B: 80=u (t_0-2) +\;\frac{1}{2} g (t_0-2) ^2$From (i), $t_0=4$ $\Rightarrow 80=2 u+5(2)^2$ $\Rightarrow u=30 {\text{m}} / {\ s}$

Conserving momentum :$m(30 {i}^{∧})+m(40 {j}^{∧})+2 m(\vec{v})=\vec{0}$ $\Rightarrow \overline{v}=-15 {i}^{∧}-20 {j}^{∧}$ $\Rightarrow |v|=25 {\text{m}} / {\ s}$

$\;{ {\text{A}}}/{ {\text{A}}_0}=\;{ {\text{N}}}/{ {\text{N}}_0}$ $\;\frac{2 \times 10^{-5}}{2.56 \times 10^{-3}}=\;{ {\text{N}}}/{ {\text{N}}_0}$ $\;{ {\text{N}}}/{ {\text{N}}_0}=\;\frac{1}{128} \Rightarrow {\text{N}}=\;{ {\text{N}}_0}/{128}$After 7 half life activity comes down to given value ${\text{T}}=7 \times 5$ $=35 \;\text{ days }\;$

${\text{L}}_0=$ angular momentum of shell about ${\text{O}}$.As shell is rolling$\;\text{ so }\; V_{c m}=\omega R$ $L_0=m V_{c m} R+I \omega$ $=1 \times \omega {\text{R}} \times {\text{R}}+\;\frac{2}{3} {\text{R}}^2 \omega$ $=\;\frac{5}{3} {\text{R}}^2 \omega$so $a=5$

No of moles $=\;\frac{44.8}{22.4}=2$Gas is mono atomic so ${\text{C}}_{ {\text{v}}}=\;\frac{3}{2} {\text{R}}$ $\Delta {\text{Q}}= {\text{nC}}_{ {\text{v}}} \Delta {\text{T}}$ $=2 \times \;\frac{3}{2} {\text{R}}(20)$ $=60 {\text{R}}$ $=60 \times 8.3$ $=498 {\text{J}}$

By Hooke's Lawso ${\text{F}} \alpha \Delta {\text{L}}$ $\;{ {\text{F}}_1}/{ {\text{F}}_2}=\;{\Delta {\text{L}}_1}/{\Delta {\text{L}}_2}$ $\;\frac{10}{20}=\;{ ( {\text{L}}_1- {\text{L}}) }/{ ( {\text{L}}_2- {\text{L}}) }$ ${\text{L}}=2 {\text{L}}_1- {\text{L}}_2$

To free the electron from metal surface minimum energy required, is equal to the work function of that metal.So Assertion A, is correct.$h v=w_0+K . E \cdot _{\max }$ $\;\text{ if }\; h v=w_0$ $\Rightarrow K \cdot \cdot _{\max }=0$Hence reason R, is correct, But ${\text{R}}$ is not the correct explanation of A. When energy of incident radiation is equal to the work function of the metal, then the KE of photoelectrons would be zero. But this statement does not comment on the situation when energy is less than the work function.

$W_{\;\text{total }\;}=\Delta K$ $=\;\frac{1}{2} (\;\frac{1}{2}) [ \{b(4)^{5 / 2}\} ^{2-}0]$ $=\;\frac{b^2}{4} \times 4^5$ $\Rightarrow W_{\;\text{total }\;}=16 J$

$R=\;\frac{m v}{B q}=\;{{\sqrt{2 m K}}}/{B q}$ $\Rightarrow R \propto {\sqrt{K}}$ $\Rightarrow$ ratio $=2: 1$

We know that $X_C=\;\frac{1}{\omega C}$ and $X_L=\omega L$Also, at $\omega =\omega _r: X_L=X_C$ $\Rightarrow$ For $\omega < \omega _r$ : capacitiveand $\omega =\omega _r: z={\sqrt{R^{2+} (X_L-X_C) ^2}}=R$

${\text{F}}=\rho {\text{v}}^2 {\text{a}}$ $\Rightarrow 10 \times 1=2 \times$ acceleration$\Rightarrow$ Acc. $=5 {\text{m}} / {\ s}^2$

From the equation{[a] \equiv \[P V^2]$ }$${[b] \equiv [V]}$$\Rightarrow [;\frac{a}{b}] \equiv [P V]$

${\sqrt{A^{2+}A^{2+}2 A^2 \cos \theta}}=2 {\sqrt{A^{2+}A^{2+}2 A^2(-\cos \theta)}}$ $\Rightarrow 2 A^{2+}2 A^2 \cos \theta=8 A^{2+}8 A^{2-}(-\cos \theta)$ $\Rightarrow 5 \cos \theta=3$ $\Rightarrow \theta=\cos ^{-1} (\;\frac{3}{5})$

$v_{e s c}-\sqrt{\;\frac{2 G M}{R}}=\sqrt{\;\frac{2 G}{R} \times \rho \times \;\frac{4}{3} \pi R^3}$ $\Rightarrow v_{e s c} \propto R {\sqrt{\rho }}$ $\Rightarrow \;\frac{ (v_{e s c}) _B}{ (v_{e s c}) _A}=1$ $\Rightarrow (v_{e s c}) _B=12 {\text{km}} / {\ s}$

$B_H=B \cos \theta$ $\therefore B=\;\frac{B_H}{\cos \theta}=\;{0.5 {\text{G}}}/{\cos 30^{\circ}} \Rightarrow \;{G}/{{\sqrt{3}}}$

Particle velocity $=\;\frac{∂ x}{∂ t}$ $\Rightarrow$ Maximum particle velocity $=(2 \pi n)(10)$ $\Rightarrow (2 \pi n)(10)=(n \lambda )(4)$ $\Rightarrow \lambda =5 \pi$

$U=\;\frac{1}{2} (k C_0) V^2$ $\Rightarrow \;\frac{U^{'}}{U}=1.5$ $\Rightarrow$ Energy increases by $50 \%$

$v^{2-}u^2=2 a s$ $\Rightarrow 0^{2-}200^2=2 (\;\frac{-q E}{m}) (S)$ \Rightarrow -200^2=2 \[;\frac{-40 × 10^{-6} × 10^5}{100 × 10^{-6}}]$ [S]$$\Rightarrow S=;\frac{4}{2 \times 4} {\text{m}}=0.5 {\text{m}}$

Fringe width $=\;\frac{\lambda D}{d}$ $\Rightarrow$ Fringe width $\propto \;\frac{\lambda }{d}$ $\Rightarrow$ New fringe width $=0.5 {\text{mm}} \times \;\frac{1.2}{2}=0.3 {\text{mm}}$

$v=f \lambda$ $\Rightarrow f=\;\frac{v}{\lambda }=\;\frac{3 \times 10^8}{1000 \times 10^{-9}} {\text{Hz}}=3 \times 10^{14} {\text{Hz}}$ $\Rightarrow$ Channels $=\;\frac{\;\frac{2}{100} \times 3 \times 10^{14}}{8 \times 10^3}=75 \times 10^7$

$H=\;\frac{V^2}{R} \cdot \Delta t$ $\Rightarrow H=\;\frac{V^2}{R_1} \cdot 20=\;\frac{V^2}{R_2} \cdot 60 . . . .$. (i)Also, $H=\;\frac{V^2}{ [\;\frac{R_1 R_2}{R_1+R_2}] } \cdot \Delta t$ $=\;\frac{4}{3} \cdot \;\frac{V^2}{R_1} \cdot \Delta t [ \because R_2=3 R_1]$ $\Rightarrow \Delta t=15$

$I=\;\frac{1}{2} \epsilon_0 E_0^2 \cdot c=\;\frac{1}{2} \epsilon_0 (c B_0) ^2 c$ $\Rightarrow I=\;\frac{1}{2} \epsilon_0 c^3 B_0^2$ $\Rightarrow 0.22=\;\frac{1}{2} (8.85 \times 10^{-12}) (3 \times 10^8) ^3 B_0^2$ $\Rightarrow B_0 ≃ 43 \times 10^{-9} {\text{T}}$

$\;\frac{L_1}{K_1 A_1}+\;\frac{L_2}{K_2 A_2}=\;{L_1+L_2}/{K_{\;\text{eff }\;} A_{\;\text{eff }\;}}$ $\Rightarrow \;\frac{4}{K}+\;\frac{2.5}{2 K}=\;{6.5}/{K_{\;\text{eff }\;}}$ $\Rightarrow \;\frac{10.5}{2 K}=\;{6.5}/{K_{\;\text{eff }\;}}$ $\Rightarrow K_{e f f}=\;\frac{13 K}{10.5}= (1+\;\frac{5}{21}) K$ $\Rightarrow \alpha=21$

$v=\omega {\sqrt{A^{2-}y^2}}$ $\Rightarrow 3 \omega {\sqrt{10^{2-}5^2}}=\omega \sqrt{ (A^{'}) ^{2-}5^2}$ $\Rightarrow 9 \times 75= (A^{'}) ^{2-}25$ $\Rightarrow A^{'}={\sqrt{28 \times 25}} {\text{cm}}$ $\Rightarrow x=700$

Resistance $=\tan 45^{\circ}=1 Ω$ $\Rightarrow 1=\;\frac{p I}{A}$ $\Rightarrow p=\;{\pi(1.2 {\text{cm}})^2}/{31.4 {\text{cm}}}=1.44 \times 10^{-1} Ω {\text{cm}}$ $\Rightarrow x=144$

$η=\;\frac{W}{Q}=\;\frac{300-225}{300}$ $\Rightarrow \;\frac{75}{300}=1-\;\frac{T_L}{T_H}$ $\Rightarrow T_L=\;\frac{3}{4} T_H=\;\frac{3}{4}(500)=375 {\text{K}}$ $\Rightarrow T_L=102^{\circ} {\text{C}}$

Impact parameter $\propto \cot \;\frac{\theta}{2}$ $\Rightarrow \sqrt{\;\frac{d_1}{d_2}}=\;{{\sqrt{3}}}/{1}$ $\Rightarrow d_1=3 d_2$ $\Rightarrow x=3$

Power gain $= [\;\frac{\Delta i_C}{\Delta i_B}] ^2 \times \;\frac{R_o}{R_i}$ = \[;\frac{10^{-2}}{10^{-4}}]$ ^2 \times ;\frac{2}{1}$$=2 \times 10^4$$\Rightarrow x=2$

1st refraction: $\;\frac{1.5}{v_1}-0=\;\frac{0.5}{15}$ $\Rightarrow v_1=45 {\text{cm}}$2nd refraction : $\;\frac{1}{v_2}-\;\frac{1.5}{15}=\;\frac{-0.5}{-15}$ $\Rightarrow \;\frac{1}{v_2}=\;\frac{1}{30}+\;\frac{1}{10}$ $=\;\frac{4}{30}$ $\Rightarrow v_2=+7.5 {\text{cm}}$ $\Rightarrow$ Distance from centre $=22.5 {\text{cm}}=225 {\text{mm}}$

$V_B-V_A=i \times 2$ $=\;\frac{15}{1+2} \times 2$ $\Rightarrow V_B-V_A=10 \;\text{ volts }\;$

Given,Mass of ${\text{Fe}}_3 {\text{O}}_4=4.640 {\text{kg}}=4640 {\text{gm}}$Molar mass of ${\text{Fe}}_3 {\text{O}}_4=56 \times 3+16 \times 4=232 {\text{g}}$ $\therefore$ Moles of $F_3 {\text{O}}_4=\;\frac{4640}{232}=20$Also, givenMass of ${\text{CO}}=2.520 {\text{kg}}=2520 {\text{gm}}$Molar mass of ${\text{CO}}=12+16=28 {\text{gm}}$ $\therefore$ Molar of $C O=\;\frac{2520}{28}=90$Here ${\text{Fe}}_3 {\text{O}}_4$ is limiting reagent as to find limiting reagent, divide the given moles of reactants with their respective stoichiometric coefficient and reactant for which this ratio is minimum will be limiting reagentFor ${\text{Fe}}_3 {\text{O}}_4, \;\frac{\;\text{ moles }\;}{\;\text{ stoichiometric coefficient }\;}=\;\frac{20}{1}$For $C O, \;\frac{\;\text{ moles }\;}{\;\text{ stoichiometric coefficient }\;}=\;\frac{90}{4}=22.5$ $\therefore {\text{Fe}}_3 {\text{O}}_4$ is limiting reagent.Now produced ${\text{Fe}}=20 \times 3=60 {\text{mol}}$ $\therefore$ Weight of ${\text{Fe}}=60 \times 56=3360 {\text{g}}$

$\;\text{ (A) }\; {\text{Cr}}(24)=1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^5 4 s^1$ $=[A r] 3 d^5 4 s^1$(B) Magnetic quantum number $(m)$ values ranging from $-I$ to $+\mid$ including zero.$\therefore$ It can have negative value.(C) According to Aufbau rule, electrons are filled first in these orbitals which have low energy.$\therefore$ Statement $C$ is correct.(D) We know,Number of Radial nodes $= {\text{n}}- {\text{I}}-1$and number of Angular nodes $=1$ $\therefore$ Total nodes $= {\text{n}}- {\text{I}}-1+1= {\text{n}}-1$

Covalent character $\propto$ polarising power of cationCorrect decreasing order of covalent character${\text{LiCl}}> {\text{NaCl}}> {\text{KCl}}> {\text{CsCl}}$

In deionized water no common ion effect will take place so maximum solubility.

If there are similar charges on colloidal particles, neutralization does not take place and the colloid becomes stable. So, option $(B)$ is correct.Brownian movement helps to counter the force of gravity acting on colloidal particles and thus doesn't allow them to settle down.Therefore, option (D) is also correct.When two oppositely charged sols are mixed, coagulation takes place due to neutralization. So, option (A) is wrong.When dispersed phase is added in excess, it forms a separate layer and thus the emulsion gets destabilized. So, option (C) is also wrong.

Atomic number of ${\text{Pt}}$ is 78Electronic configuration is $-{ }_{78} {\text{Pt}} \to [ {\text{Xe}}] 4 {\text{f}}^{14} 5 {\text{d}}^9 6 {\ s}^1$ (Exceptional electronic configuration)

In the extraction of Silver and Gold, ${\text{NaCN}}$ is used to leach the metal.$4 {\text{Au}}( {\ s})+8 C \overline{N}(a q)+2 {\text{H}}_2 {\text{O}}+ {\text{O}}_2(g)$ $\;\; \to 4 [ {\text{Au}}( {\text{CN}})_2] ^{-}(a q)+4 O H^{-}(a q)$ $2 [ {\text{Au}}( {\text{CN}})_2] ^{-}( {\text{aq}})+ {\text{Zn}} \to 2 {\text{Au}}( {\ s})+ [ ( {\text{Zn}}( {\text{CN}})_4) ] ^{2-}( {\text{aq}})$In case of ore containing ${\text{ZnS}}$ and ${\text{PbS}}$, the depressant used is ${\text{NaCN}}$.

A reducing agent is a substance that undergoes oxidation itself, causing another substance to be reduced. For hydrogen peroxide ({\text{H}}_2{\text{O}}_2$) to act as a reducing agent, the oxidation state of its oxygen atoms must increase from -1 to 0 (\in${\text{O}}_2$). The reaction must also take place \in a basic medium, indicated by the presence or formation of${\text{OH}}^{-}$ ions.

Let's analyze the correct reaction:
$2 \overset{+7}{\text{Mn}}\text{O}_4^{-} + 3 {\text{H}}_2 \overset{-1}{\text{O}}_2 \to 2 \overset{+4}{\text{Mn}}\text{O}_2 + 3 \overset{0}{\text{O}}_2 + 2 {\text{H}}_2 {\text{O}} + 2 {\text{OH}}^{-}$

In this reaction, the oxidation state of oxygen in ${\text{H}}_2{\text{O}}_2$ increases from -1 to 0 in ${\text{O}}_2$, meaning ${\text{H}}_2{\text{O}}_2$ is oxidized and acts as a reducing agent. The oxidation state of manganese decreases from +7 to +4, meaning ${\text{MnO}}_4^{-}$ is reduced. The presence of ${\text{OH}}^{-}$ ions in the products confirms the basic medium. In contrast, option A is acidic, and in options B and D, ${\text{H}}_2{\text{O}}_2$ acts as an oxidizing agent.

MetalLiNaKRbCsColourCrimson redYellowVioletRed VioletBlue λ / nm670.8589.2766.5780.0455.5

Caesium is used in devising photoelectric cells.Boron fibres are used in making bullet-proof vest.Silicones being surrounded by non-polar alkyl groups are water repelling in nature.Gallium is less toxic and has a very high boiling point, so it is used in high temperature thermometers.

${\text{P}}_4+3 {\text{NaOH}}+3 {\text{H}}_2 {\text{O}} \to {\text{PH}}_3+3 {\text{NaH}}_2 {\text{PO}}_2$oxoacid $= {\text{H}}_3 {\text{PO}}_2$ (hypo phosphorus acid) or (phosphinic acid)

${\text{CaCO}}_3+ {\text{H}}_2 {\text{SO}}_4 \to {\text{CaSO}}_4+ {\text{H}}_2 {\text{O}}+ {\text{CO}}_2$Industries like oil refinery releases ${\text{SO}}_2$ which causes air pollution. It reacts with water to form acid rain when ${\text{SO}}_2 {\text{mix}}$ with water it forms ${\text{H}}_2 {\text{SO}}_4$ (Sulphuric acid).

{\text{R}}- {\text{OH}}+ {{\text{R}}- { {\text{C}}^{||}^{O}}- {\text{OH}} \longrightarrow {\text{R}}- {\text{O}}- {\text{C}}^{||}^{O}- {\text{R}}}_{\text{nucleophilic acyl substitution}} electron with drawing group on carboxylic acidwill increase the rate of esterification

Buna - N is synthetic rubber which can be stretched and retains its original status on releasing the force.

DNA contains $\Rightarrow \beta- {\text{D}}-2-$ deoxyriboseRNA contains $\Rightarrow \beta- {\text{D}}$ - ribose

Histamine is nitrogenous compound it does notcontain sulpher

Phenol are weakly acidic. Phenol is more acidic than alcohol & ${\text{H}}_2 {\text{O}}$ statement (I) is correct. (II) is incorrect.

From ideal gas equation we know${\text{PV}}= {\text{nRT}}$ $\Rightarrow P V=\;\frac{W}{M} R T$ $\Rightarrow P=\;\frac{W}{V} \cdot \;\frac{R T}{M}$ $\Rightarrow P=d \cdot \;\frac{R T}{M} [ \because d=\;\frac{W}{V}]$We know, $760 {\text{mm}}$ of ${\text{Hg}}=1 {\text{atm}}$ $\therefore 100 {\text{mm}}$ of ${\text{Hg}}=\;\frac{100}{760} {\text{atm}}$ $\therefore$ Pressure $(P)=\;\frac{100}{760} {\text{atm}}$Density $(d)=0.46$ ${\text{R}}=0.082 {\text{L}} {\text{atm}} {\text{K}}^{-1} {\text{mol}}^{-1}$ ${\text{T}}=(257+273) {\text{K}}=530 {\text{K}}$Putting the values in above equation, we get$\;\frac{100}{760}=\;\frac{0.46 \times 0.082 \times 530}{M}$ $\Rightarrow M=152$