1 Feb 2023 Shift 2

${\text{T}}_{ {\text{H}}}=99^{\circ} {\text{C}}=99+273$ $\;\;\;\; =372 {\text{K}}$ $1-\;{ {\text{T}}_{ {\text{C}}}}/{ {\text{T}}_{ {\text{H}}}}=\;\frac{1}{3}$ ${l}\;{ {\text{T}}_{ {\text{C}}}}/{ {\text{T}}_{ {\text{H}}}}=\;\frac{2}{3} \;\; (1) \Rightarrow {\text{T}}_{ {\text{C}}}=\;\frac{2}{3} \times 372$ $=2 \times 124=248 {\text{K}}$ $1-\;{ {\text{T}}_{ {\text{C}}}+ {\text{X}}}/{ {\text{T}}_{ {\text{H}}}}=\;\frac{1}{6}$ $\;\frac{5}{6}=\;{ {\text{T}}_{ {\text{C}}}+ {\text{X}}}/{ {\text{T}}_{ {\text{H}}}}$ $\;\frac{5}{6}=\;{248+ {\text{X}}}/{372}$ $248+ {\text{X}}=5 \times 62$ ${\text{X}}=310-248=62 {\text{K}}$

Potential of a conducting sphere is$\; {\text{V}}=\;{ {\text{KQ}}}/{ {\text{R}}} \;\text{ (Solid as well as hollow) }\;$ $\; {\text{V}}_1= {\text{V}}_2 \;\text{ and }\; {\text{R}}_1= {\text{R}}_2$ $\; \therefore {\text{Q}}_1= {\text{Q}}_2$

${\text{B}}_{ {\text{C}}}\;=\;{\mu _0 {\text{I}}}/{4 \pi {\text{R}}}(\pi)( {\text{B}} \;\text{ at centre of circular arc) }\;$ $\;=\;{\mu _0 {\text{I}}}/{4 {\text{R}}}=\;\frac{4 \pi \times 10^{-7} \times 3}{4 \times \;\frac{\pi}{10}}$ $\;=3 \times 10^{-6} {\text{T}}=3 \mu {\text{T}}$

\;\varphi ={ \vec{\text{B}} \cdot { \vec{\text{A}} $\;= {\text{BA}} \cdot \cos \theta$ Most suitable ans is 2 [Otherwise ${\text{ABCD}}$ ]

Modulating Index$\mu =\;\;\frac{A_m}{A_c}$ $\mu _1\;=\;\frac{X}{Y}$ $\mu _2\;=\;\frac{X}{2 Y}$ $\;\frac{\mu _1}{\mu _2}\;=\;\frac{2}{1}$

At highest point$\; {\text{V}}_{ {\text{y}}}=0$ $\; {\text{V}}_{ {\text{x}}}= {\text{u}}_{ {\text{x}}}= {\text{u}} \cos \theta$ $\; {\text{U}}_{ {\text{g}}}= {\text{mgh}}, \;\text{ it is maximum at }\; {\text{H}}_{\max } .$

by mirror formula$\;\;{1}/{ {\text{v}}}+\;{1}/{ {\text{u}}}=\;{1}/{ {\text{f}}}$ $\;\;{1}/{ {\text{v}}_1}+\;\frac{1}{-15}=\;\frac{1}{(-20)}$ $\;\;{1}/{ {\text{v}}_1}=-\;\frac{1}{20}+\;\frac{1}{15}$ $\;=\;\frac{-3+4}{60}$ $\; {\text{v}}_1=60 {\text{cm}}$ $\;\;{1}/{ {\text{v}}_2}+\;\frac{1}{(-25)}=\;\frac{1}{(-20)}$ $\;\;{1}/{ {\text{v}}_2}=\;\frac{-1}{20}+\;\frac{1}{25}$ $\;=\;\frac{-5+4}{100}=\;\frac{-1}{100}$ $\; {\text{v}}_2=-100 {\text{cm}}$ $\; {\text{d}}=60+100=160 {\text{cm}}$

Tension $(F)=m g$ $\; \;\; =4 \times \;\frac{10}{4}=10 {\text{N}}$ $\;\Delta {\text{L}}=\;{ {\text{FL}}}/{ {\text{AY}}}$ $\;=\;\frac{10 \times 6}{3 \times 10^{-6} \times 2 \times 10^{11}}$ $\;=10^{-4} {\text{m}}=0.1 {\text{mm}}$

Suppose resistance of each arm is $r$, then $r=R / n$ $\;R_{e q(A B)}=\;\frac{R_1 R_2}{R_1+R_2}$ $\;\;\frac{r(n-1) r}{r+(n-1) r}$ $\;=\;\frac{r(n-1) r}{n r}$ $\;=\;\frac{n-1}{n} r$ $\;=\;\frac{(n-1) R}{n^2}$

$\; {\text{N}}= {\text{Mg}}- {\text{F}} {\text{Sin}} 30^{\circ}$ $\;= {\text{mg}}-\;{ {\text{F}}}/{2}=100-\;{ {\text{F}}}/{2}=\;{200- {\text{F}}}/{2}$ $\; {\text{F}} {\text{Cos}} 30^{\circ}=\mu {\text{N}}$ $\;{\sqrt{3}} \;{ {\text{F}}}/{2}=0.25 \times (\;{200- {\text{F}}}/{2})$ $\;4 {\sqrt{3}} {\text{F}}=200- {\text{F}}$ $\; {\text{F}}=\;{200}/{4 {\sqrt{3}}+1}=25.22$

Error of voltmeter decreases with increase in its resistance.

Woks as voltage regulator in reverse bias and as simple P-n junction in forward bias.

$\; {\text{T}}=2 \pi \sqrt{\;{ℓ}/{ {\text{g}}}}$ $\; {\text{T}}^2=\;{4 \pi^2}/{ {\text{g}}} \times ℓ$ $\; {\text{T}}^2 \alpha ℓ$

$\; {\text{V}}_{ {\text{e}}}=\sqrt{\;{2 {\text{GM}}}/{ {\text{R}}}}=\sqrt{\;{2 {\text{G} \rho } \frac{4}{3} \pi {\text{R}}^3}/{ {\text{R}}}}= {\text{C}} {\sqrt{\rho }} \cdot {\text{R}}$ $\;\;{ {\text{V}}_{ {\text{e}}_1}}/{ {\text{V}}_{ {\text{e}}_2}}=\;{ {\text{R}}_1}/{ {\text{R}}_2} \sqrt{\;\frac{\rho _1}{\rho _2}}=\;\frac{1}{2}$ $\;\;{ {\text{R}}_1^2}/{ {\text{R}}_2^2} \times \;\frac{\rho _1}{\rho _2}=\;\frac{1}{4}$ $\;\;{ {\text{R}}_1}/{ {\text{R}}_2}=\;\frac{1}{3}$ $\; {\text{g}}=\;{ {\text{GM}}}/{ {\text{R}}^2}=\;{ {\text{G}} \frac{4}{3} \pi {\text{R}}^3 \times \rho }/{ {\text{R}}^2} {\text{C}} \cdot \rho {\text{R}}$ $\;\;{ {\text{g}}_1}/{ {\text{g}}_2}=\;{\rho _1 {\text{R}}_1}/{\rho _2 {\text{R}}_2}=\;\frac{1}{4} \;{ {\text{R}}_2^2}/{ {\text{R}}_1^2} \times \;{ {\text{R}}_1}/{ {\text{R}}_2}$ $\;=\;\frac{1}{4} \times \;{ {\text{R}}_2}/{ {\text{R}}_1}=\;\frac{3}{4}$

\;\Delta {\text{E}}=13.6 {\text{Z}}^2 \[;\frac{1}{2^2}-;\frac{1}{4^2}]$ {\text{eV}}$$;=13.6 \times (4)^2 (;\frac{1}{4}-;\frac{1}{16}) {\text{eV}}$$;=13.6[4-1] {\text{eV}}$$;=13.6 \times 3=40.8 {\text{eV}}$

Impulse $=$ Area under $F=t$ curve(a) $\;\frac{1}{2} \times 1 \times 0.5=\;\frac{1}{4} {\text{N}} . {\ s}$(b) $0.5 \times 2=1$ N.s (maximum)(c) $\;\frac{1}{2} \times 1 \times 0.75=\;\frac{3}{8} {\text{N}} . {\ s}$(d) $\;\frac{1}{2} \times 2 \times 0.5=\;\frac{1}{2}$ N.s

Say dimensional formale of mass is ${\text{H}}^x C^y G^z$ $\; {\text{M}}^1= ( {\text{ML}}^2 {\text{T}}^{-1}) ^{ {\text{x}}} ( {\text{LT}}^{-1}) ( {\text{M}}^{-1} {\text{L}}^3 {\text{T}}^{-2}) ^z$ $\; {\text{M}}^1 {\text{L}}^0 {\text{T}}^0= {\text{M}}^{ {\text{x}}- {\text{z}}} {\text{L}}^{2 {\text{x}}+ {\text{y}}+3 z} {\text{T}}^{- {\text{x}}- {\text{y}}-2 z}$on comparing both side$\;x-z=1$ $\;2 x+y+3 z=0$ $\;-x-y-2 z=0$On solving above equations we get$x=\;\frac{1}{2} \;\; {\text{y}}=\;\frac{1}{2} \;\; {\text{z}}=\;\frac{-1}{2}$

For isochoric process$\;{ {\text{P}}}/{ {\text{T}}}= {\text{n}} \;{ {\text{R}}}/{ {\text{V}}}=\;\text{ constan }\; {\text{t}}$ ${\text{P}}=\;{ {\text{nR}}}/{ {\text{V}}}( {\text{t}}+273)$If ${\text{P}}=0 \Rightarrow {\text{t}}=-273^{\circ} {\text{C}}$

${\text{u}}_{ {\text{E}}} > = {\text{u}}_{ {\text{B}}} > =\;\frac{1}{2}< {\text{u}}_{\;\text{total }\;} >..$So $\;{ u_E⟩}/{ u_{\;\text{total }\;}⟩}=\;\frac{1}{2}$

${\text{K}}_{\max }= {\text{hf}}- {\text{hf}}_0$For ${\text{f}}=2 {\text{f}}_0$ $\;\frac{1}{2} {\text{m}} {\text{V}}_1^2=2 {\text{h}} {\text{f}}_0- {\text{hf}} {\text{f}}_0= {\text{hf}}_0$For ${\text{f}}=5 {\text{f}}_0$ $\;\frac{1}{2} {\text{m}} {\text{V}}_2^2=5 {h{f } _ { 0 }}-{h {f } _ { 0 }}=4 {h {f } _ { 0 }}$ $\;{ {\text{V}}_1}/{ {\text{V}}_2}=\;\frac{1}{2}$

${\text{u}}=20 {\text{m}} / {\ s}, {\text{S}}_1=500 {\text{m}}, {\text{v}}=0$By third equation of mation$\;0=(20)^{2-}2 {\text{a}} .500 \Rightarrow {\text{a}}=\;\frac{4}{10} {\text{m}} / {\ s}^2$ $\; {\text{u}}=20 {\text{m}} / {\ s}, {\text{S}}_2=250 {\text{m}}, {\text{v}}=?$ $\; {\text{v}}^2=(20)^{2-}2 {\text{a}} .250$ $\;= {\text{v}}={\sqrt{200}} {\text{m}} / {\ s}$ $\; {\text{x}}=200$

$\;F=5+3 y^2$ $\;W=int_{2}^{5} (5+3 y^2) d y$ \;= \[5 y+;\frac{3 y^3}{3}]$ _2^5$$;=132 {\text{J}}$

$\; {\text{I}}= {\text{I}}_{ {\text{cm}}}+ {\text{Md}}^2$ $\;=\;{ {\text{MR}}^2}/{2}+ {\text{MR}}^2$ $\;=\;\frac{3}{2} {\text{MR}}^2$ $\; {\text{x}}=3$

For A mass number $=34$Total binding energy $=1.2 \times 34=40.8 {\text{MeV}}$For ${\text{B}}$ mass number $=26$total binding energy $=1.8 \times 26 {\text{MeV}}$ $=46.8 {\text{MeV}}$Difference of ${\text{BE}}=6 {\text{MeV}}$

$\; {\text{N}}=600, {\text{A}}=70 \times 10^{-4} {\text{m}}^2, {\text{B}}=0.4 {\text{T}}$ $\;\omega =\;\frac{500 \times 2 \pi}{60}=\;\frac{100 \pi}{6} {\text{rad}} / {\ s}$ \; {\text{E}}= {\text{NAB}} \omega sin \;\omega t \;\; \omega t \;\text{ is angle }\; {\text{b}} / {\text{w}} { \vec{\text{A}} & { \vec{\text{B}} $\;=600 \times 70 \times 10^{-4} \times 0.4 \times \;\frac{100 \pi}{6} \times \;\frac{1}{2}$ $\;=44 {\text{V}}$

$\; {\text{U}}_{ {\text{i}}}=\;\frac{1}{2} {\text{kx}}_0^2$ $\; {\text{K}}_{ {\text{i}}}=0$ $\; {\text{U}}_{ {\text{f}}}=\;\frac{1}{2} {\text{k}} (\;{ {\text{x}}_0}/{2}) ^2$ $\; {\text{K}}_{ {\text{f}}}=0.25 {\text{J}}$ $\;\;\frac{1}{2} {\text{kx}} {\text{x}}_0^{2+}0=\;\frac{1}{2} {\text{k}} \;{ {\text{x}}_0^2}/{4}+0.25$ $\;\;\frac{1}{2} {\text{kx}} {\text{x}}_0^2 \;\frac{3}{4}=\;\frac{1}{4}$ $\;\;\frac{1}{2} {\text{k}} \;\frac{3}{100}=1 \Rightarrow {\text{k}}=\;\frac{200}{3} {\text{N}} / {\text{m}}$ $\;=67 {\text{N}} / {\text{m}}$

$; {\text{I}}_1= {\text{I}}_2=\;\frac{20-10}{10}=1 {\text{A}}$ ${\text{I}}_3=1 {\text{A}}$ $|\;{ {\text{I}}_1+ {\text{I}}_3}/{ {\text{I}}_2}|=2$

$\;\;\text{ Fringe shift }\;=\;\frac{t(\mu -1)}{\lambda } B$ $\;=\;\frac{10 \times 10^{-6}(1.2-1)}{5 \times 10^{-7}} {\text{B}}$ $\;=\;\frac{10^{-5} \times 0.2}{5 \times 10^{-7}}=4$

\;{ \vec{\text{E}}= {\text{E}}_0 \times \;{ {\text{i}}}^{\^} $\;\varphi _{\;\text{net }\;}=\varphi _{ {\text{ABCD}}}= {\text{E}}_0 {\text{a}} \cdot {\text{a}}^2$ $\;\;{ {\text{q}}_{\;\text{en }\;}}/{E_0}= {\text{E}}_0 {\text{a}}^3$ $\; {\text{q}}_{ {\text{en}}}= {\text{E}}_0 E_0 {\text{a}}^3$ $\;=4 \times 10^4 \times 9 \times 10^{-12} \times 8 \times 10^{-6}$ $\;=288 \times 10^{-14} {\text{C}}$ $\; {\text{Q}}=288$Ans. 288

$\; {\text{A}}_1 {\text{V}}_1= {\text{A}}_2 {\text{V}}_2$ $\;750 \times 10^{-4} {\text{V}}_1=500 \times 10^{-6} \times 0.3$ $\; {\text{V}}_1=\;\frac{500 \times 3 \times 10^{-3}}{750} {\text{m}} / {\ s}$ $\;=2 \times 10^{-3} {\text{m}} / {\ s}$ $\;\;{ {\text{dh}}}/{ {\text{dt}}}=-2 \times 10^{-3} {\text{m}} / {\ s}$

(Bond enthalpy order${\text{C}}- {\text{C}} > {\text{Si}}- {\text{Si}} > {\text{Ge}}- {\text{Ge}} > {\text{Sn}}- {\text{Sn}})$

${\text{H}}$-bonding stabilised vitamin ${\text{C}}$

(It is SN1 reaction so rate of reaction depends on the concentration of alkyl halide only.

The magnetic field at point O is the vector sum of the fields due to the two straight segments and the semicircular arc.
The straight segments are semi-infinite wires whose extensions pass through point O. Thus, the magnetic field produced by them at O is zero.
The magnetic field at the center of a semicircular arc of radius $R$ carrying current $I$ is given by:
$B_{arc} = \frac{1}{2} \left( \frac{\mu_0 I}{2R} \right) = \frac{\mu_0 I}{4R}$
Given $I = 3 \text{ A}$ and assuming a standard radius $R = 1 \text{ cm} = 0.01 \text{ m}$ (often provided in problem context):
$B_{arc} = \frac{4\pi \times 10^{-7} \text{ T m/A} \times 3 \text{ A}}{4 \times 0.01 \text{ m}} = 3\pi \times 10^{-5} \text{ T}$
Using the right-hand rule, the current in the semicircle flows counter-clockwise relative to O, so the magnetic field direction at O is out of the plane of the paper.

The total magnetic field at O is $3\pi \times 10^{-5} \text{ T}$ (out of the page).

$( [ {\text{Co}} ( {\text{NH}}_3) _5 {\text{NO}}_2] ^{2+}.$Two linkage isomers possible ${\text{NO}}_2 \to$ Ambidentate ligand

Does not show esterification test.Presence of both sulphur and nitrogen give red colour in Lassigne's test.

(Gypsum is used for making fireproof wall boards.

(Nessler's Reagent $\to {\text{K}}_2 [ {\text{HgI}}_4]$

Assertion (A) : $\alpha$-halocarboxylic acid on reaction with dil. $\text{NH}_3$ gives good yield of $\alpha$-amino carboxylic acid whereas the yield of amines is very low when prepared from alkyl halides.
This assertion is correct. When an alkyl halide reacts with ammonia, the primary amine product formed (\text{R-NH}_2$) is generally more nucleophilic than ammonia itself. This leads to successive alkylations, forming secondary, tertiary amines, and even quaternary ammonium salts (overalkylation), resulting \in a low yield of the primary amine. In contrast, \in the reaction of an$\alpha$-halocarboxylic acid with ammonia, the product is an$\alpha$-amino acid. The amino group (\text{-NH}_2$) in the newly formed amino acid is immediately protonated by the carboxylic acid group (\text{-COOH}$) present \in the same molecule or by other acid molecules \in solution. This converts the amino group to a non-nucleophilic ammonium group (\text{-NH}_3^+$), thereby preventing further reaction with other $\alpha$-halocarboxylic acid molecules and stopping overalkylation, leading to a good yield.

Reason (R) : Amino acids exist in zwitter ion form in aqueous medium.
This reason is correct. Amino acids contain both an acidic carboxyl group (\text{-COOH}$) and a basic amino group (\text{-NH}_2$). In an aqueous medium, an intramolecular proton transfer occurs from the carboxyl group to the amino group, forming a dipolar ion called a zwitterion (\text{R-CH(NH}_3^+\text{)-COO}^-$).

Relationship between (A) and (R):
Both statements are individually correct. The existence of amino acids as zwitterions (where the amino group is protonated as $\text{-NH}_3^+$) means that the amino group is non-nucleophilic. This non-nucleophilicity is precisely why the amino acid product does not undergo further reaction with the $\alpha$-halocarboxylic acid, thereby preventing overalkylation and leading to a good yield as described in (A). Therefore, (R) provides the underlying chemical characteristic that ultimately explains the observation in (A). However, the statement in (R) describes a general property of amino acids in aqueous solution, rather than explicitly detailing the dynamic process of how overalkylation is prevented during the synthesis. While the zwitterionic form is the result of the protonation that stops overalkylation, (R) does not directly describe the mechanistic cause. Thus, (R) is a correct statement related to (A), but it is not considered the direct and complete explanation for the difference in yields.

The final answer is $\boxed{B}$

In urea production ${\text{NH}}_3$ and ${\text{CO}}_2$ consumed so least responsible for global warming.

$2 {\text{Cu}}^{+} \to {\text{Cu}}^{2+}+ {\text{Cu}}$The stability of ${\text{Cu}}^{2+}$ (aq) rather than ${\text{Cu}}^{+}( {\text{aq}})$, is due to the much more negative $\Delta _{ {\text{hyd}}} {\text{H}}$ of ${\text{Cu}}^{2+}$ (aq) than ${\text{Cu}}^{+}$(aq), which more than compensates for the second ionisation enthalpy of ${\text{Cu}}$.

$( {\text{K}}_2 {\text{S}}_2 {\text{O}}_8( {\ s})+2 {\text{D}}_2 {\text{O}}( {\text{l}}) \to 2 {\text{KDSO}}_4.$ (aq. $)+ {\text{D}}_2 {\text{O}}_2$