6 Apr 2023 Shift 2

$\;\;\text{ using }\; v_{ {\text{rms}}}=\sqrt{\;{3 R T}/{ {\text{m}}}}$ $\; {\text{v}}_0=\sqrt{\;{3 R \times 200}/{ {\text{m}}}}$ . . . (1)$\; ( {\text{v}}^{'}) =\sqrt{\;{3 R \times 800}/{ {\text{m}}}}$ . . . (2)dividing (2) by (1)$\;\;{ {\text{v}}^{'}}/{ {\text{v}}_0}=\sqrt{\;\frac{800}{200}}={\sqrt{4}}=2$ $\;\;\text{ or }\; {\text{v}}^{'}=2 {\text{v}}_0$

Both the statements are trueAs we know speed of light in a medium${\text{v}}=\;{ {\text{c}}}/{\mu } \;\text{ or }\; {\text{f}} \lambda =\;{ {\text{c}}}/{\mu }$therefore $\lambda \propto \;\frac{1}{\mu }$when light will travel through two different mediums their phase difference will change $\Delta {\text{Q}}=\;\frac{2 \pi}{\lambda } \Delta {\text{x}}$and ${\text{R}}$ is correction explanation

Given, modulation index $=60 \%=0.6$ $\;{ {\text{A}}_{ {\text{m}}}}/{ {\text{A}}_{ {\text{c}}}}=\;\frac{0.6}{1}$Using componendo - dividendo, we can write$\;\;{ {\text{A}}_{ {\text{m}}}+ {\text{A}}_{ {\text{c}}}}/{ {\text{A}}_{ {\text{m}}}- {\text{A}}_{ {\text{c}}}}=\;\frac{0.6+1}{0.6-1}=\;\frac{1.6}{-0.4}$ $\; {\text{A}}_{ {\text{m}}}+ {\text{A}}_{ {\text{c}}}=\;\frac{1.6}{-0.4} \times ( {\text{A}}_{ {\text{m}}}- {\text{A}}_{ {\text{c}}})$ $\;=\;\frac{1.6}{-0.4} \times (-3)=12 {\text{V}}$

$\;\;\text{ Using }\; {\text{v}}=\sqrt{\;{\gamma {\text{RT}}}/{ {\text{m}}}}$ $\;\;{ {\text{U}}_{ {\text{H}}_2}}/{ {\text{v}}_{ {\text{O}}_2}}=\sqrt{\;{ {\text{m}}_{ {\text{O}}_2}}/{ {\text{m}}_{ {\text{H}}_2}}}=\sqrt{\;\frac{32}{2}}=\sqrt{\;\frac{16}{1}}=4: 1$(since both hydrogen and oxygen are di-atomic, $\gamma$ will be same)

In this case, since masses of both charges are not same, therefore, we need to find center of mass (COM), about which dipole will oscillate and then we will find moment of Inertia about this axis, to find torque & hence $\omega$. As we know, COM will divide length in the inverse ratio of the masses, therefore, COM will be at a distance of $\;{ {\text{L}}}/{3}$ from 2 {\text{m}} & \;{2 {\text{L}}}/{3} from ${\text{m}}$MI about this axis$I=2 m (\;\frac{L}{3}) ^{2+} (\;\frac{2 L}{3}) ^2$Or I $=\;{2 {\text{mL}}^2}/{ {\text{a}}}+\;{4 {\text{mL}}^2}/{ {\text{a}}}=\;{6 {\text{mL}}^2}/{ {\text{a}}}=\;{2 {\text{mL}}^2}/{3}$Using \omega =\;{2 {\text{mL}}^2}/{3} & {\text{p}}= {\text{qL}} $\omega =\sqrt{\;{ {\text{qLE}}}/{{2 {\text{L}}^2}/{3}}}=\sqrt{\;{3 {\text{qE}}}/{2 {\text{mL}}}}$None of these given option is correct. (BONUS)

As per pascal's law, when we apply pressure to an ideal liquid it is equally distributed in the entire liquid and to the walls as well.Since due to applied pressure, every morning, the tooth paste does not get compressed and we can safely consider it on incompressible liquid.Therefore both statements are true and ${\text{R}}$ is correct explanation of ${\text{A}}$.

This question is based on the conceptual clarity that we should connect ammeter in series and voltmeter in parallel to measure current and potential difference, respectivelyAlso, when we use a galvanometer to create an ammeter, shunt resistance should be very small and should be in parallel.When we create a voltemeter shunt should be large and in series with galvanometer.All these criteria are satisfied in option (2)

Method-1Using exact law of cooling$\; {\text{T}}- {\text{T}}_{ {\ s}}= ( {\text{T}}_0- {\text{T}}_{ {\ s}}) {\text{e}}^{- {\text{Kt}}}$ $\;\;\text{ Case-I: }\;(40-10)=(60-10) {\text{e}}^{-7 {\text{K}}}$ $\;30=50 {\text{e}}^{-7 {\text{K}}}$ $\;\;\text{ Case-II: }\;( {\text{T}}-10)=(40-10) {\text{e}}^{-7 {\text{K}}} \;\text{ or }\; {\text{T}}-10=30 {\text{e}}^{-7 {\text{K}}}$ $\;\;\text{ Dividing }\;(2) \;\text{ by }\;(1)$ $\;\;{ {\text{T}}-10}/{30}=\;\frac{30}{50}$ $\;\Rightarrow {\text{T}}-10=\;\frac{30 \times 30}{50}=18$ $\;\;\text{ or }\; {\text{T}}=28^{\circ} {\text{C}}$Methode-2Using newton's average law of cooling$\;\;{ {\text{T}}_{ {\text{i}}}- {\text{T}}_{ {\text{f}}}}/{ {\text{t}}}= {\text{k}} (\;{ {\text{T}}_{ {\text{i}}}+ {\text{T}}_{ {\text{f}}}}/{2}- {\text{T}}_{ {\ s}})$ $\;\;\text{ Case-I:- }\; \;\frac{60-40}{7}= {\text{R}} [\;\frac{60+40}{2}-10] \Rightarrow \;\frac{20}{7}= {\text{k}}[40]$ . . . (i)$\;\;\text{ Case-II:- }\; \;{40- {\text{T}}}/{7}= {\text{R}} [\;{20+ {\text{T}}}/{2}]$ . . . (2) $\;\;\text{ Dividing (2) by (1) }\;$ $\;\;{40- {\text{T}}}/{20}=\;{20+ {\text{T}}}/{80}$ $\;160-4 {\text{T}}=20+ {\text{T}}$ $\;5 {\text{T}}=140$ $\; {\text{T}}=28^{\circ} {\text{C}}$

By theory of electromagnetic waves$\; {\text{U}}_{ {\text{E}}}=\;\frac{1}{2} \epsilon_0 {\text{E}}^2 \;\text{ and }\;$ $\; {\text{U}}_{ {\text{B}}}=\;\frac{1}{2} \;{ {\text{B}}^2}/{\mu _0}$

using newton's formula $F=\;\frac{G M m}{r^2}$at surface of earth, $100=\;{ {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ {\text{Re}}^2}$ . . . (1)$\;\;\text{ at }\; {\text{r}}= {\text{R}}_{ {\text{e}}}+\;{ {\text{R}}_{ {\text{e}}}}/{4}=\;\frac{5}{4} {\text{R}}_{ {\text{e}}}$ $\; {\text{F}}^{'}=\;{ {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ (\;\frac{5}{4} {\text{R}}_{ {\text{e}}}) ^2}=\;\frac{16}{25} \times \;{ {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ {\text{R}}_{ {\text{e}}}{ }^2}$ $\; {\text{F}}^{'}=\;\frac{16}{25} \times 100=64 {\text{N}}$

Given alternating ${\text{AC}}$ source {\text{E}}=36 sin \;(120 \pi {\text{t}}) {\text{v}} & capacitor ${\text{C}}=150 \mu {\text{F}}$ using ${\text{Q}}= {\text{CV}}$we can write ${\text{Q}}= ( {\text{CE}}_0 sin \;\omega {\text{t}})$Current ${\text{i}}=\;{ {\text{dQ}}}/{ {\text{dt}}}= ( {\text{CE}}_0 \omega \cos \omega {\text{t}})$max. value of current ${\text{i}}_0= {\text{CE}}_0 \omega$or ${\text{i}}_0=150 \times 10^{-6} \times 36 \times 120 \pi$ $=2.03 {\text{A}}$

Based on the data provided${\text{U}}=100-80=20 {\text{cm}}$ ${\text{V}}=180-100=80 {\text{cm}}$Using $\;{1}/{ {\text{f}}}=\;{1}/{ {\text{v}}}+\;{1}/{ {\text{u}}} \;\;$ or ${\text{f}}=\;{ {\text{uv}}}/{ {\text{u}}+ {\text{v}}}=\;\frac{20 \times 80}{20+80} \;\;$ or ${\text{f}}=16 {\text{cm}}$For error analysis,$\;{1}/{ {\text{f}}}=\;{1}/{ {\text{v}}}+\;{1}/{ {\text{u}}}$Differentiating$-\;{ {\text{Df}}}/{ {\text{f}}^2}=-\;{ {\text{Dv}}}/{ {\text{v}}^2}-\;{\Delta {\text{u}}}/{ {\text{u}}^2}$To calculate \Delta {\text{u}} & \Delta {\text{v}} ${\text{U}}=(100 \pm 2)-(80 \pm 0.2)=(20 \pm 0.4) {\text{cm}}$Therefore $\Delta {\text{u}}=0.4 {\text{cm}}$,Similarly $\Delta {\text{v}}=0.4 {\text{cm}}$.Now $\;{\Delta {\text{f}}}/{ {\text{f}}}= {\text{f}} [\;{\Delta {\text{v}}}/{ {\text{v}}^2}+\;{\Delta {\text{u}}}/{ {\text{u}}^2}]$ $\;{\Delta {\text{f}}}/{ {\text{f}}}=16 [\;\frac{0.4}{(80)^2}+\;\frac{0.4}{(20)^2}]$(Note: every data is in ${\text{cm}}$ )$\;\;{\Delta {\text{f}}}/{ {\text{f}}}=\;\frac{16 \times 0.4}{(20)^2} [\;\frac{1}{4^2}+1]$ $\;=\;\frac{16 \times 0.4}{20^2} \times \;\frac{17}{16}=\;\frac{17 \times 0.4}{400}$ $\;\% \;\text{ Error : }\; \;{\Delta {\text{f}}}/{ {\text{f}}} \times 100=\;\frac{17 \times 0.4}{400} \times 1000$ $\;=1.7$

Let potential of the junction be ${\text{x}}$ volts using junction law ${\text{i}}_{ {\text{i}}}+ {\text{i}}_2+ {\text{i}}_3=0$ or $\;\frac{x-30}{10}+\;\frac{x-12}{20}+\;\frac{x-2}{30}=0$ $\;\;\text{ or }\; \;\frac{1}{60}[6 x-180+3 x-36+2 x-4]=0$ $\;\;\text{ or }\; \;\frac{1}{60}[11 x-220]=0$ $\;\;\text{ or }\; x=\;\frac{220}{11}=20 {\text{V}}$current through $20 Ω$ is $=\;\frac{x-12}{20}$ ${\text{i}}_2=\;\frac{20-12}{20}=0.4 {\text{A}}$

Given $U=\;\frac{1}{2} m \omega ^2 r^2$, to find radius $r$ as $f(n)$, where ${\text{n}}$ is orbit Using Bohr's postulate : angular momentum ${\text{L}}= {\ mvr}=\;{ {\text{nh}}}/{2 \pi}$ or ${\text{mr}} \omega ^2=\;{ {\text{nh}}}/{2 \pi}$ $\Rightarrow {\text{r}} \propto \sqrt{ {\text{n}}}$

Statement ${\text{A}}$ is correct and Statement ${\text{R}}$ is wrong as per the theory of ${\text{p}}- {\text{n}}$ junction.

Since planets revolve around the sun in an elliptical orbit its linear speed is not constant, hence option 1 not correct (and right choice).Other statement are correct as per theory.

Given, ${\text{m}}=5 {\text{kg}}, {\text{R}}=2 {\text{m}}$ time ${\text{t}}$ for $1 {\text{rev}}=3.14 {\ sec}$ or $\pi {\ sec}$ $\theta$ for $1 {\text{rev}}=2 \pi {\text{rad}}$ Therefore $\omega =\;{\theta}/{ {\text{t}}}=\;\frac{2 \pi}{\pi}=2 {\text{rad}} / {\ s}$ centrifugal force ${\text{F}}= {\text{mR}} \omega ^2$ or ${\text{F}}=5 \times 2 \times 2^2=40 {\text{N}}$

Given speed ${\text{v}}=\pi {\text{m}} / {\ s}$or $R \omega =\pi$or $\omega =\;\frac{\pi}{R} {\text{rad}} / {\ s}$angular displacement $\theta=120^{\circ}$ or $\;\frac{2 \pi}{3}$uising $\theta=\omega {\text{t}}$ ${\text{t}}=\;\frac{\theta}{\omega }=\;{2 \pi / 3}/{\pi / {\text{R}}}=\;{2 {\text{R}}}/{3}$linear displacement $d=2 {\text{R}} sin \;(\theta / 2)$ ${\text{d}}=2 {\text{R}} sin \; (\;\frac{120}{2})$ $=2 {\text{R}} \times sin \;60=2 {\text{R}} \times \;{{\sqrt{3}}}/{2}$ $= {\text{R}} {\sqrt{3}}$average velocity $=\;{ {\text{d}}}/{ {\text{t}}}=\;{ {\text{R}} {\sqrt{3}}}/{2 {\text{R}} / 3}=\;{3 {\sqrt{3}}}/{2}$

Using ${\text{KE}}_{\max }= {\text{eV}}_{ {\ s}}= {\text{hf}}-\varphi _0$where $\varphi _0$ is work function, ${\text{V}}_{ {\ s}}$ is stopping potential and ${\text{f}}$ is frequency or ${\text{V}}_{ {\ s}}=\;{ {\text{h}}}/{ {\text{e}}} {\text{f}}-\;{\varphi _0}/{ {\text{e}}}$ therefore the slope ${\text{m}}$ will be same for all graphs and will be independent of $\varphi _0$.

Using ${\text{I}}^{ {\ st}}$ equation of motion$\;t=\;\frac{v-u}{a}$ $\;t=\;\frac{60-10}{2}=\;\frac{50}{2}=25 {\ sec}$

For pendulum${\text{T}}_{\max }= {\text{mg}}+\;{ {\ mv}^2}/{ {\text{L}}}$ . . . (1)Given ${\text{m}}=\;\frac{1}{4} {\text{kg}}, {\text{L}}=1 {\text{m}}, {\text{g}}=9.8 {\text{m}} / {\ s}^2$and amplitude ${\text{A}}=\;\frac{1}{10} {\text{m}}$For SHM, ${\text{KE}}_{\max }=\;\frac{1}{2} {\ mv}^2=\;\frac{1}{2} {\text{m}} \omega ^2 {\text{A}}^2$using $\omega =\sqrt{\;{ {\text{g}}}/{ {\text{L}}}}$ $m v^2=m (\sqrt{\;{ {\text{g}}}/{ {\text{L}}}}) ^2 {\text{A}}^2=\;{ {\text{mgA}}}/{ {\text{L}}}$ . . . (2)using (2) in (1)${\text{T}}_{\max }=2 {\text{mg}}+\;{ {\text{mgA}}^2}/{ {\text{L}}^2}$ =m g \[1+;\frac{1}{10^2}]$ =;\frac{1}{4} \times 9.8 \times ;\frac{101}{100}$or${\text{T}}_{\max }=;\frac{98.98}{40}$Therefore${\text{x}}=99$

$\;\;\text{ Using }\; {\text{E}}=\;{ {\text{ke}}^2}/{2 {\text{r}}}$ $\; {\text{r}}=\;{ {\text{Re}}^2}/{2 {\text{E}}}$Given ${\text{E}}=12.8 {\text{eV}}=12.8 \times {\text{e}}$ Joule$\; {\text{r}}=\;{9 \times 10^9 {\text{e}}^2}/{2 \times 12.8 {\text{e}}}=\;\frac{9 \times 10^9 \times 1.6 \times 10^{-19}}{2 \times 12.8}$ $\; {\text{r}}=\;\frac{9 \times 10^{-10}}{(2 \times 12.8 / 1.6)}=\;\frac{9 \times 10^{-10}}{10} {\text{m}}$Therefore ${\text{x}}=16$

Let $n_1$ maxima of $7000 Å$ coincides with $n_2$ maxima of $5500 Å$therefore ${\text{n}}_1 \beta_1= {\text{n}}_2 \beta_2$or $\;{ {\text{n}}_1}/{ {\text{n}}_2}=\;\frac{\lambda _2}{\lambda _1}=\;\frac{5500}{7000}=\;\frac{11}{14}$therefore $11^{\;\text{th }\;}$ maxima of $7000 Å$ will coincide with $14^{\;\text{th }\;}$ maximum of $5500 Å$To find the least distance of this${\text{y}}= {\text{n}}_1 \beta_1$or ${\text{y}}=\;{ {\text{n}}_1 \lambda _1 {\text{D}}}/{ {\text{d}}}=\;\frac{11 \times 7000 \times 10^{-10} \times 150 \times 10^{-2}}{2.5 \times 10^{-3}}$ $=\;\frac{11 \times 7 \times 5}{2.5} \times 10^{-5} {\text{m}}$or ${\text{y}}=462 \times 10^{-5} {\text{m}}$therefore ${\text{n}}=462$

Given$\; {\text{a}}=1000 {\text{cm}}$ $\; {\text{b}}=1 {\text{cm}}$ $\;\;\text{ or }\; {\text{b}} < < {\text{a}}$we will take larger coil as primary ${\text{B}}=\;{\mu _0 {\text{i}}_{ {\text{p}}} {\text{N}}}/{2 {\text{a}}}$flux $\varphi _{ {\ s}}= {\text{BA}}=\;{\mu _0 {\text{i}}_{ {\text{p}}} {\text{N}}}/{2 {\text{a}}} \times \pi {\text{b}}^2 \times {\text{n}}$Mutual inductance ${\text{M}}=\;{\varphi _{ {\ s}}}/{ {\text{i}}_{ {\text{p}}}}$ $\;M=\;{\mu _0 N n \pi b^2}/{2 \times {\text{a}}}$ $\;\;\text{ or }\; M=\;\frac{4 \pi \times 10^{-7} \times 200 \times 10 \times \pi \times 1 \times 10^{-4}}{2 \times 1000 \times 10^{-2}}$ $\;=4 \pi^2 \times 10^{-9}$ $\;\;\text{ or }\; M=4 \times 10^{-8} (\;\text{ using }\; \pi^2=10)$

As per the arrangement given, distance between the capacitor plates are ${\text{a}}$ and ${\text{b}}$ and ${\text{a}} \ne {\text{b}}$ using the diagram we can write ${\text{b}}=5- {\text{a}}-1=(4-a)$ in ${\text{mm}}$ as we know capacitance of capacitor $C=\;\frac{\epsilon_0 A}{d}$ and in series arrangement$\;{1}/{ {\text{C}}_{\;\text{eq }\;}}=\;{1}/{ {\text{C}}_1}+\;{1}/{ {\text{C}}_2}$ $\;{1}/{ {\text{C}}_{\;\text{eq }\;}}=\;{ {\text{a}}}/{\epsilon_0 {\text{A}}}+\;{4- {\text{a}}}/{\epsilon_0 {\text{A}}}=\;{4(\;\text{ \in mm }\;)}/{\epsilon_0 {\text{A}}}$or ${\text{C}}_{ {\text{eq}}}=\;{\epsilon_0 {\text{A}}}/{4( {\text{mm}})}$Given ${\text{A}}=200 {\text{cm}}^2$ ${\text{C}}_{ {\text{eq}}}=\;\frac{\epsilon_0 \times 200 \times 10^{-4}}{4 \times 10^{-3}}$ $=\epsilon_0 50 \times 10^{-1}$or ${\text{C}}_{ {\text{eq}}}=5 \epsilon_0$ farad therefore ${\text{n}}=5$

$\; {\text{B}}=\;\frac{\pi}{2} \times 10^{-3}$ $\;\;\text{ K.E. }\;=\;\frac{1}{2} {\text{mV}}^2$ $\;\Rightarrow {\text{V}}=\sqrt{\;{2 {\text{KE}}}/{ {\text{m}}}}$ $\;\;\text{ PItch }\;= {\text{v}} \cos 60^{\circ} \times \;\text{ time period of one rotation }\;$ $\;= {\text{v}} \cos 60^{\circ} \times \;{2 \pi {\text{m}}}/{ {\text{eB}}}$ $\;=\sqrt{\;\frac{2 \times 2 \times 1.6 \times 10^{-9}}{1.6 \times 10^{-27}}} \times \cos 60^{\circ} \times \;\frac{2 \pi \times 1.6 \times 10^{-27}}{1.6 \times 10^{-19} \times \;\frac{\pi}{2} \times 10^{-3}}$ $\;=2 \times 10^4 \times \;\frac{1}{2} \times 4 \times 10^{-5}$ $\;=4 \times 10^{-1} {\text{m}}=40 {\text{cm}}$

Let $u$ and $v$ be speeds, just before and after body strikes the ground.Given $\;{ {\text{u}}}/{ {\text{v}}}=\;\frac{4}{1}$ $\;\;\text{ loss \in }\; {\text{KE}}: \Delta {\text{KE}}=\;\frac{\;\frac{1}{2} {\mu}^2=\frac{1}{2} {\ mv}^2}{\frac{1}{2} {\mu}^2}$ $\;\Delta {\text{KE}}=1- (\;{ {\text{v}}}/{ {\text{u}}}) ^2=1-\;\frac{1}{16}=\;\frac{15}{16}$ $\;\;\text{ Percentage loss }\;=\;\frac{15}{16} \times 100=375$

Given radius of gyration is same for ring and solid sphere$\; {\text{K}}_{ {\text{R}}}= {\text{K}}_{ {\ ss}}$ $\; {\text{R}}_{ {\text{R}}}=\sqrt{\;\frac{2}{5}} {\text{R}}_{ {\ ss}}$ $\;\;\text{ or }\; \;{ {\text{R}}_{ {\text{R}}}}/{ {\text{R}}_{ {\ ss}}}=\sqrt{\;\frac{2}{5}}$ $\;\;\text{ therefore }\; {\text{x}}=5$

Method-I:$\; {\text{R}}_{ {\text{eq}}}=2+\;{5 {\text{R}}}/{5+ {\text{R}}}=\;{10+7 {\text{R}}}/{5+ {\text{R}}}$ $\; {\text{i}}=\;{3}/{ {\text{R}}_{ {\text{eq}}}}=\;{3(5+ {\text{R}})}/{10+7 {\text{R}}}$ $\; {\text{i}}_1=\;\frac{2}{5}, {\text{i}}_2=\;{2}/{ {\text{R}}}$ $\; {\text{i}}= {\text{i}}_1+ {\text{i}}^2$ $\;\;{3(5+ {\text{R}})}/{10+7 {\text{R}}}=\;\frac{2}{5}+\;{2}/{ {\text{R}}}=\;{2(5+ {\text{R}})}/{5 R}$ $\;15 {\text{R}}(5+ {\text{R}})=2(5+ {\text{R}})(10+7 {\text{R}})$ $\;75 {\text{R}}+15 {\text{R}}^2=2 (50+35 {\text{R}}+10 {\text{R}}+2 {\text{R}}^2)$ $\;15 {\text{R}}^{2+}75 {\text{R}}=14 {\text{R}}^{2+}90 {\text{R}}+100$ $\; {\text{R}}^{2-}15 {\text{R}}-100=0$ $\; {\text{R}}=\;{15 {\sqrt{225 \times 1 \times 100}}}/{2}$ $\;=\;{15 \pm {\sqrt{625}}}/{2}=\;\frac{15 \pm 25}{2}$ $\;R=20 Ω$Method-II: Given potential across $5 Ω$ and voltmeter is $2 {\text{V}}$. To find resistance ${\text{R}}$ of voltmeter.Let current in $5 Ω$ be ${\text{i}}_1$, and in ${\text{R}} {\text{i}}_2$.${\text{i}}_1=\;\frac{2}{5} \;\text{ and }\; {\text{i}}_2=\;{2}/{ {\text{R}}}$ ${\text{V}}$ across $2 Ω$ will be 1 volt and ${\text{i}}=\;\frac{1}{2} {\text{A}}$.Using junction law: ${\text{i}}= {\text{i}}_1+ {\text{i}}_2$ $\;\;\frac{1}{2}=\;\frac{2}{5}+\;{2}/{ {\text{R}}}$ $\;\;{2}/{ {\text{R}}}=\;\frac{1}{2}-\;\frac{2}{5}=\;\frac{1}{10}$ $\; {\text{R}}=20 Ω$

$\;\;\text{ Using stress }\;=\;\frac{\;\text{ force }\;}{\;\text{ area }\;}=\;{ {\text{mg}}}/{ {\text{A}}}$ $\;\Rightarrow {\text{m}}=\;{ {\text{S}} \times {\text{A}}}/{ {\text{g}}}=\;{7 \times 10^5 \times \pi {\text{R}}^2}/{ {\text{g}}}$ $\;=\;\frac{7 \times 10^5 \times \frac{22}{7} \times (7 \times 10^{-3}) ^2}{9.8} \;\text{ (Note: }\; 14 {\text{mm}} \;\text{ is diameter) }\;$ $\;=11 {\text{kg}}$

Hydration enthalpy $\propto \;\frac{1}{\;\text{ size }\;}$Down the group as size increases hydration enthalpy decreasesOrder : ${\text{Be}}^{2+}> {\text{Mg}}^{+2}> {\text{Ca}}^{+2}> {\text{Sr}}^{+2}> {\text{Ba}}^{+2}$

IUPAC name of ${\text{K}}_3 [ {\text{Co}} ( {\text{C}}_2 {\text{O}}_4) _3]$ is Potassium trioxalatocobaltate(III)

The one-letter codes for amino acids are standardized abbreviations. For some, the code is the first letter of their name. For others, a different letter is used to avoid duplication.

• (D) Alanine is assigned (III) A, its initial letter.
• Since A is taken, (A) Arginine is assigned (IV) R, a phonetically similar letter ("aRginine").
• For the "Asp" family, (B) Aspartic acid is assigned (I) D ("asparDic acid").
• (C) Asparagine is assigned (II) N to distinguish it from Aspartic acid, using the prominent 'N' sound in its name.
  Matching these pairs gives the sequence: (A)-IV, (B)-I, (C)-II, (D)-III.

Nessler reagent is $- {\text{K}}_2 [ {\text{HgI}}_4]$

In solid state ${\text{BeCl}}_2$ as polymer, in vapour state it form chloro-bridged dimer while above $1200 {\text{K}}$ it is monomer.

Strongest acid from the following is $- {\text{NO}}_2$ group has more EWG nature so more acidic,

As electropositive character increases basic character of oxide increases. ${{\text{B}}_2 {\text{O}}_3} _{\text{acidic}} < {{\text{Al}}_2 {\text{O}}_3< {\text{Ga}}_2 {\text{O}}_3} _{\text{amphoteric}}< {{\text{In}}_2 {\text{O}}_3< {\text{Tl}}_2 {\text{O}}_3}_{\text{basic}}$

Step with highest activation energy is RDS, so step II is RDSNo. of activated complex $=3$ ${\text{P}}$ and ${\text{Q}}$ are intermediates(Number of intermediates $=2$ )