29 Jul 2022 Shift 1

[\;\frac{\rho r^3}{T^{3 / 2}}] =\;\frac{[M L^{-3}]\ $[L^3]$ }{[M L^0 T^{-2}]$ ^{3 / 2}} \ne [T]$As the equation for first statement is wrong dimensionally.$\Rightarrow A$is false and$R$ is true.

Max. Height $=h=\;\frac{u^2}{2 g}$ $\Rightarrow {\text{u}}=\sqrt{2 {\text{gh}}}$ ${\text{S}}= {\text{ut}}+\;\frac{1}{2} a {\text{at}}^2$ $\;{ {\text{h}}}/{3}={\sqrt{2 g h}} {\text{t}}+\;\frac{1}{2}(- {\text{g}}) {\text{t}}^2$ $\;{ {\text{gt}}^2}/{2}-{\sqrt{2 g h t}}+\;\frac{h}{3}=0 \;\; (.$ Roots are $t_1 \& t_2)$ $\;{ {\text{t}}_2}/{ {\text{t}}_1}=\;{\sqrt{2 {\text{gh}}}+\sqrt{2 {\text{gh}}-4 \times { {\text{g}}}/{2} \times { {\text{h}}}/{3}}}/{\sqrt{2 {\text{gh}}}-\sqrt{2 {\text{gh}}-4 \times \;{ {\text{g}}}/{2} \times \;{ {\text{h}}}/{3}}}=\;{\sqrt{2 {\text{gh}}}+\sqrt{{4 {\text{gh}}}/{3}}}/{\sqrt{2 {\text{gh}}}-\sqrt{\;{4 {\text{gh}}}/{3}}}=\;{{\sqrt{3}}+{\sqrt{2}}}/{{\sqrt{3}}-{\sqrt{2}}}$

${\text{t}}=\sqrt{ {\text{x}}}+4$ $\Rightarrow {\text{x}}=( {\text{t}}-4)^2= {\text{t}}^{2-}8 {\text{t}}+16$ $\Rightarrow \;{ {\text{dx}}}/{ {\text{dt}}}=2 {\text{t}}-8$ $\Rightarrow \;{ {\text{dx}}}/{ {\text{dt}}}{{}/{}|} _{ {\text{t}}=4}=2 \times 4-8=0$

$N=\;\frac{m v^2}{r}$ $\Rightarrow$ The graph given in option A suits the best for the above relation.

${\text{E}}=\;\frac{1}{2} {\mu}^2$At Highest point, Velocity ${\text{V}}= {\text{u}} \cos 60^{\circ}=\;{ {\text{u}}}/{2}$ $\therefore$ K.E at topmost point $=\;\frac{1}{2} {\text{m}} (\;{ {\text{u}}}/{2}) ^2=\;{ {\text{E}}}/{4}$

$\overline{r}_{c o m}=\;\frac{m_1 \overline{r}_1+m_2 \overline{r}_2}{m_1+m_2}$ $=\;\frac{(1-9) \hat{i}+(2-6) \hat{j}+(1+3) \hat{k}}{4}$ $=\;\frac{-8 \hat{i}-4 \hat{j}+4 \hat{k}}{4}$ $\overline{r}_{c o m}=-2 \hat{i}-\hat{j}+\hat{k}$ $|\overline{r}|={\sqrt{4+1+1}}={\sqrt{6}}$ $|\hat{i}+2 \hat{j}+\hat{k}|={\sqrt{6}}$

Due to obtuse angle of contact the water doesn't wet the oiled surface properly and cannot wash it also.$\Rightarrow$ Assertion is correct and Reason given is a correct explanation.

Young's modulus of matter depends on material of wire and is independent of the dimensions of the wire. As the material remains same so Young's modulus also remain same.

${\text{g}}_{ {\text{cff}}}= {\text{g}} \cos \alpha$ ${\text{T}}=2 \pi \sqrt{\;{ {\text{L}}}/{ {\text{g}} \cos \alpha}}$

By Gauss law ∮ { \vec{\text{E}}{\text{d}} { \vec{\ s}=\;{ {\text{Q}}_{ {\in }}}/{\epsilon_{ {\text{o}}}} ${\text{E}} .4 \pi {\text{r}}^2=\;{int_{0}^{{} {\text{r}}} \rho _{ {\text{o}}} (\frac{3}{4}-{ {\text{r}}}/{ {\text{R}}}) 4 \pi {\text{r}}^2 {\text{dr}}}/{\epsilon_0}$ ${\text{E}} 4 \pi {\text{r}}^2=\;{\rho _{ {\text{o}}} 4 \pi}/{\epsilon_{ {\text{o}}}} (\;\frac{3}{4} \;{ {\text{r}}^3}/{3}-\;{ {\text{r}}^4}/{4 {\text{R}}})$ ${\text{E}} {\text{r}}^2=\;{\rho _{ {\text{o}}} {\text{r}}^3}/{4 \epsilon_{ {\text{o}}}} \{1-\;{ {\text{r}}}/{ {\text{R}}}\}$ ${\text{E}}=\;{\rho _{ {\text{o}}} {\text{r}}}/{4 \epsilon_{ {\text{o}}}} \{1-\;{ {\text{r}}}/{ {\text{R}}}\}$

(Properties of conductor)Statement - I, true as body of conductor acts as equipotential surface.Statement $-2$ True, as conductor is equipotential.Tangential component of electric field should be zero. Therefore electric field should be perpendicular to surface.

Let length of wire be ' $ℓ$ 'Area of wire as 'A'For equivalent wire length $=2 ℓ \&$ area will be ${\text{A}}$ Thermal resistance${\text{R}}_{ {\text{eq}}}= {\text{R}}_1+ {\text{R}}_2$ $\;{2 ℓ}/{\sigma_{ {\text{eq}}} {\text{A}}}=\;{ℓ}/{\sigma_1 {\text{A}}}+\;{ℓ}/{\sigma_1 {\text{A}}}$ $\;{2 ℓ}/{\sigma_{ {\text{eq}}}}=\;\frac{ℓ}{\sigma_1}+\;\frac{ℓ}{\sigma_2} \Rightarrow \sigma_{ {\text{eq}}}=\;\frac{2 \sigma_1 \sigma_2}{\sigma_1+\sigma_2}$

${\text{E}}=440 {\text{Sin}} 100 \pi {\text{t}}, {\text{L}}=\;{{\sqrt{2}}}/{\pi} {\text{H}}$ ${\text{X}}_{ {\text{L}}}=\omega {\text{L}}=100 \pi \;{{\sqrt{2}}}/{\pi}=100 {\sqrt{2}} Ω$Peak current ${\text{I}}_0=\;{ {\text{E}}_0}/{ {\text{X}}_{ {\text{L}}}}=\;{440}/{100 {\sqrt{2}}}=2.2 {\sqrt{2}} {\text{A}}$AC ammeter reads ${\text{RMS}}$ value therefore reading will be ${\text{I}}_{ {\text{rms}}}$ ${\text{I}}_{ {\text{rms}}}=\;{ {\text{I}}_0}/{{\sqrt{2}}}=2.2 {\text{A}}$

Given circuit is ${\text{R}}- {\text{L}}$ growth circuit ${\text{i}}=\;{ {\text{E}}}/{ {\text{R}}} (1- {\text{e}}^{- {\text{t}} / \tau })$ ${\text{i}}=\;{ {\text{E}}}/{2 {\text{R}}}=\;{ {\text{E}}}/{ {\text{R}}} (1- {\text{e}}^{- {\text{t}} / \tau )})$Solving ${\text{t}}=\tau \ln 2$ ${\text{t}}=\;{1}/{ {\text{R}}} \ln 2=\;\frac{1}{100} 0.693=0.00693$ $=7 {\text{ms}}$ ${\text{i}}(15 {\text{ms}})=\;{ {\text{E}}}/{ {\text{R}}} (1- {\text{e}}^{-\;\frac{15}{10}})$ ${\text{i}}=\;\frac{6}{100}(1-1 / 4)=\;\frac{3}{4} \times \;\frac{6}{100}$ ${\text{U}}=\;\frac{1}{2} {\text{LI}}^2$by solving we get $U=1 {\text{mJ}}$.

(a) uv rays - used for water purification(b) x-rays used for diagnosing fracture(c) Microwaves are used for mobile and radar communication(d) Infrared waves show less scattering therefore used in foggy days$(a-i i),(b-i),(c-i i i),(d-i v)$

${\text{E}}=\;{ {\text{hc}}}/{\lambda }-\varphi -\;\text{ (i) }\;$ $2 {\text{E}}=\;{ {\text{hc}}}/{\lambda ^{'}}-\varphi -\;\text{ (ii) }\;$(ii) $-$ (i)${\text{E}}= {\text{hc}} (\;\frac{1}{\lambda ^{'}}-\;\frac{1}{\lambda })$ $\Rightarrow \lambda ^{'}=\;{ {\text{hc}} \lambda }/{ {\text{E}} \lambda + {\text{hc}}}$

${\text{E}}_{ {\text{n}}}=\;{-13.6}/{ {\text{n}}^2} {\text{ev}}$ $\Rightarrow \;{ {\text{E}}_2- {\text{E}}_1}/{ {\text{E}}_{\infty }- {\text{E}}_1}=\;\frac{13.6 (1-\frac{1}{4}) }{13.6}=\;\frac{3}{4}$

Modulation index: $m=\;\frac{A_m}{A_c}$Given $2 {\text{A}}_{ {\text{m}}}=8$ ${\text{A}}_{ {\text{m}}}+ {\text{A}}_{ {\text{c}}}=9 \Rightarrow {\text{A}}_{ {\text{c}}}=5$ $\therefore {\text{m}}=\;\frac{4}{5}=0.8$

$1 {\text{MSD}}=\;\frac{1}{20} {\text{cm}}$ $1 {\text{VSD}}=\;\frac{24}{25} {\text{MSD}}=\;\frac{24}{25} \times \;\frac{1}{20} {\text{cm}}$ $\therefore$ Least count $=\;\frac{1}{20} (1-\;\frac{24}{25}) {\text{cm}}$ $=\;\frac{1}{20} \times \;\frac{1}{25}=\;\frac{1}{500} {\text{cm}}$ $=0.002 {\text{cm}}$

${\text{MSR}}=2.5 {\text{mm}}$ ${\text{CSR}} \;\; =45 \times \;\frac{0.5}{50} {\text{mm}}$ $=0.45 {\text{mm}}$ $\;\text{ Diameter reading }\; \;\; = {\text{MSR}}+ {\text{CSR}}-\;\text{ zero error }\;$ $\;\; =2.5+0.45-(-0.03)$ $\;\; =2.98 {\text{mm}}$

$\theta=45^{\circ}$ $R_1=\;\frac{R}{2}$ $\;\frac{u^2 \ sin 2 \theta_1}{g}=\;\frac{u^2 \ sin (90^{\circ}) }{2 g}$ $\Rightarrow 2 \theta_1=30^{\circ}$ $\theta_1=15^{\circ}$

$g (1-\;\frac{2 h}{R}) =g (1-\;\frac{d}{R})$ $\;\frac{2 h}{R}=\;\frac{d}{R}$ $\alpha h=d$ $\alpha=2$

$P_1 V_1^\gamma =P_2 V_2^2$ $\;\frac{P_1}{d_1^\gamma }=\;\frac{P_2}{d_2^\gamma }$ $\;\frac{d_1 T_1}{d_1^\gamma }=\;\frac{d_2 T_2}{d_2^\gamma }$ $T_2= (\;\frac{d_2}{d_1}) ^{\gamma -1} T_1$ $=(32)^{\;\frac{2}{5}} T_1$ $T_2=4 T_1$

$C_V=\;\frac{f}{2} R$total degree of freedoms$=1 \times 3+3 \times 5=18$ $\;\frac{\alpha^2}{4}=\;\frac{18}{2 n}=\;\frac{18}{2 \times 4}$ $\Rightarrow \alpha^2=9$ $\alpha=3$

All $9 Ω$ resistances are in parallel$R_{e q}=3 Ω$ $I=\;\frac{6}{3} A=2 A$

$B=\;\frac{\mu _0 n I}{2 R}$ $37.68 \times 10^{-4}=\;\frac{4 \pi \times 10^{-7} 100 I}{2 \times 5 \times 10^{-2}}$ $I=\;\frac{300 A}{100}$ $=3 {\text{A}}$

$I_A= ({\sqrt{I_1}}+{\sqrt{I_2}}) ^2=25 I$ $I_B= ({\sqrt{I_1}}-{\sqrt{I_2}}) ^2=I$So, $I_A-I_B=24 I$

$\;\frac{314}{100}=2 \pi {\text{R}} \;\; {\text{R}}=0.5 {\text{m}}$Magnetic Moment $=$ IA$=14 \times \pi {\text{R}}^2$ $=14 \times (3.14) \times \;\frac{1}{4}$ $=10.99 \approx 11.00$

{ \vec{\text{A}}=4 {\sqrt{3}} { {\text{i}}^{\^}-3 {\sqrt{3}} { {\text{j}}^{\^}-5 { {\text{k}}^{\^} $\mu _1 \ sin {\text{i}}=\mu _2 \ sin {\text{r}}$ As incident vector A makes i angle with normal ${\text{z}}$-axis & refracted vector ${\text{R}}$ makes ${\text{r}}$ angle with normal ${\text{z}}$ - axis with help of direction cosine$i=\cos ^{-1} (\;\frac{A_z}{A}) =\cos ^{-1} (\;{5}/{\sqrt{(4 {\sqrt{3}})^{2+}(3 {\sqrt{3}})^{2+}5^2}})$ $=\cos ^{-1} (\;\frac{5}{10}) \Rightarrow i=60^{\circ}$ ${\sqrt{2}} sin \;60={\sqrt{3}} \times sin \;r$ $r=45^{\circ}$Difference between $i \& {\text{r}}=60-45=15$

${\text{E}}=\;{ {\text{V}}}/{ {\text{d}}}=\;\frac{\;\text{ Potential barrier Across Junction }\;}{\;\text{ width of Depletion layer }\;}$ $=\;{0.6 {\text{V}}}/{6 \times 10^{-6} {\text{m}}}=1 \times 10^5 {\text{V}} / {\text{m}}$ $=1 \times 10^5 {\text{N}} / {\text{C}}$

(A) ${\text{BCl}}_3$ - Even electron molecules${\text{SF}}_6$ - Expended octet molecules(B) ${\text{NO}}$ - Odd electron molecules ${\text{H}}_2 {\text{SO}}_4$ - Expanded octet(C) ${\text{SF}}_6$ - Even electron molecules ${\text{H}}_2 {\text{SO}}_4$ - Expanded octet(D) ${\text{BCl}}_3$ - Even electron molecules ${\text{NO}}$ - odd electron molecules

\;{{ {\text{N}}_{2( {\text{g}})}_{20 {\text{g}}}}}+\;{{3 {\text{H}}_{2(g)}}}_{5 {\text{g}}} ⇌ 2 {\text{NH}}_{3( {\text{g}})}Ideally $28 {\text{g}} {\text{N}}_2$ reacts with $6 {\text{g}} {\text{H}}_2$ limiting reagent is ${\text{N}}_2$ $\therefore$ Amount of ${\text{NH}}_3$ formed on reacting $20 {\text{g}} {\text{N}}$ is, $=\;\frac{34 \times 20}{28}=24.28 {\text{g}}$ $=1.42$ moles

Albumin from the egg was poured into $100 {\text{mL}}$ of $5 \%( {\text{w}} / {\text{v}}) {\text{NaCl}}$ solution in water. This would result in the formation of Iyophilic sol. Albumin molecules get dispersed in water the colloidal particles of albumin are stabilised by hydrogen bond with water molecules.

I.E. ${\text{Na}}< {\text{Al}}< {\text{Mg}}< {\text{Si}}$ $\because 496<$ l.E. $( {\text{Al}})< 737$Option (C), matches the conditioni.e. I.E. $( {\text{Al}})=577 {\text{kJ}} {\text{mol}}^{-1}$

Earthy and undesired materials present in the ore other then the desired metal is known as gangue.

Zinc dissolves in excess of aqueous alkali.${\text{Zn}}+2 {\text{OH}}^{-}+2 {\text{H}}_2 {\text{O}} \longrightarrow [ {\text{Zn}}( {\text{OH}})_4] ^{2-}+ {\text{H}}_2 ↑($ Tetrahydroxozincate(II) ion)However, this reaction in NCERT is given as${\text{Zn}}+2 {\text{NaOH}} \longrightarrow {\text{Na}}_2 {\text{ZnO}}_2+ {\text{H}}_2 ↑$ ${\text{ZnO}}_2^{2-}$ is anhydrous form of $[ {\text{Zn}}( {\text{OH}})_4] ^{2-}$.${\text{ZnO}}_2^{2-}+2 {\text{H}}_2 {\text{O}} ⇌ [ {\text{Zn}}( {\text{OH}})_4] ^{2-}$ So in aqueous medium best answer of this question is $[ {\text{Zn}}( {\text{OH}})_4] ^{2-}$.

${\text{Li}}_2 {\text{O}}, {\text{NaNO}}_2$As per NCERT lithium nitrate when heated gives lithium oxide, ${\text{Li}}_2 {\text{O}}$. Whereas other alkali metal nitrates decompose to give the corresponding nitrite.$4 {\text{LiNO}}_3 \longrightarrow 2 {\text{Li}}_2 {\text{O}}+4 {\text{NO}}_2+ {\text{O}}_2$ $2 {\text{NaNO}}_3 \longrightarrow 2 {\text{NaNO}}_2+ {\text{O}}_2$However, the decomposition product of ${\text{NaNO}}_3$ is temperature dependent process as shown in the below reaction. ${\text{NaNO}}_3 \;{{\;{\longrightarrow }^{\Delta }}}_{{500^{\circ} {\text{C}}}} {\text{NaNO}}_2( {\ s})+\;\frac{1}{2} {\text{O}}_2( {\text{g}})$ $\Delta ↓, 800^{\circ} {\text{C}}$ $\; {\text{Na}}_2 {\text{O}}( {\ s})+ {\text{N}}_2( {\text{g}})+ {\text{O}}_2( {\text{g}})$As the temperature is not mentioned, we can go by the answer. (C)

The number of lone pair of electrons in the central atom of ${\text{SCl}}_2, {\text{O}}_3, {\text{ClF}}_3$ and ${\text{SF}}_6$ are $2,1,2$ and 0 respectivelyTheir structures are as,

${\text{Sc}}^{+3}$ and ${\text{Zn}}^{+2}$ are colourless as they contain no unpaired electron. Whereas the transition metal ions ${\text{Cu}}^{+2}, {\text{Ti}}^{+3}, {\text{V}}^{+2}$ and ${\text{Mn}}^{+2}$ are coloured as they contain unpaired electrons.The unpaired electron from lower energy $d$ orbital gets excited to a higher energy $d$ orbital on absorbing light of frequency which lies in visible region. The colour complementary to light absorbed is observed.

In neutral or Faintly alkaline medium, thiosulphate is oxidised almost quantitatively to sulphate ion according to reaction given below,$8 {\text{MnO}}_4^{-}+3 {\text{S}}_2 {\text{O}}_3^{2-}+ {\text{H}}_2 {\text{O}} \to 8 {\text{MnO}}_2+6 {\text{SO}}_4^{2-}+2 {\text{OH}}^{-}$Here the ${\text{Mn}}$ changes from ${\text{Mn}}^{+7}$ to ${\text{Mn}}^{+4}$Thus overall change in its oxidation number would be of 3 .

Both sodium chlorate and sodium arsenate behave as herbicide.

is the strongest base among the given compounds due to the maximum $+1$ effect and the lone pair of ${\text{N}}$ is not in dynamic state so it can be donated easily.

O-xylene has different resonating structures which will produce different ozonolysis products.m-xylene p-xylene and toluene have identical resonating structures which will give identical ozonolysis products.

In ${\text{NaCN}}$; carbon is more nucleophilic atom.Whereas in ${\text{AgCN}} ; {\text{Ag}}- {\text{C}}$ has covalent bond.

Amytal is hypnotic drug used to treat sleeping disorder.

$\;\frac{l}{A}=129 {\text{m}}^{-1}$ ${\text{KCl}}$ solution $1 \Rightarrow 74.5 {\text{ppm}}, {\text{R}}_1=100 Ω$ ${\text{KCl}}$ solution $2 \Rightarrow 149 {\text{ppm}}, {\text{R}}_2=50 Ω$Here, $\;\frac{p p m_1}{p p m_2}=\;\frac{M_1}{M_2}= (\;\frac{w_{1 / M_0}}{V} \times \;\frac{V}{w_{2 / M_0}})$ $\;\frac{Lambda_1}{Lambda_2}=\;\frac{k_1 \times \frac{1000}{M_1}}{k_2 \times \;\frac{1000}{M_2}}$ $=\;\frac{k_1}{k_2} \times \;\frac{M_1}{M_2}$ $=\;\frac{50}{100} \times 2$ $=\;\frac{Lambda_1}{Lambda_2}=1000 \times 10^{-3}$ =1000