In the reaction $_7{{\rm{N}}^{14}} + \alpha { \to _8}{X^{17}}{ + _1}{p^1}$ identify $X$.

In the reaction $\;\;\;\;\;{\;_7}{N^{14}} + \,\,\alpha { \to _8}{X^{17}}{ + _1}{p^1}$ ${\rm{8}}$ is the atomic number of oxygen molecule. So, here $X$ is oxygen$\left( {{{\rm{O}}_2}} \right)$ molecule.

The correct Option is "${{\rm{O}}_2}$"

Partial pressure of ${{\rm{O}}_2}$ in the reaction $2{\rm{A}}{{\rm{g}}_2}{\rm{O}}\left( s \right)\;\leftrightharpoons\;4{\rm{Ag}}\left( s \right) + {{\rm{O}}_2}\left( {\rm{g}} \right)$

$2{\rm{A}}{{\rm{g}}_2}{\rm{O}}\left( s \right)\;\leftrightharpoons\;4{\rm{Ag}}\left( s \right) + {{\rm{O}}_2}\left( {\rm{g}} \right)$ ${K_p} = {p_{{{\rm{O}}_2}}}$ $({ \because }\;\;\;{\rm{Ag}}\;{\rm{ and }}\;{\rm{A}}{{\rm{g}}_2}{\rm{O }}\;{\rm{are }}\;{\rm{solids}})$

The correct Option is "${K_p}$"

In the reaction ${\rm{BrO}}_3^ - \left( {aq} \right) + 5{\rm{B}}{{\rm{r}}^ - }\left( {aq} \right) + 6{{\rm{H}}^ + } \to 3{\rm{B}}{{\rm{r}}_2}\left( l \right) + 3{{\rm{H}}_2}{\rm{O}}\left( l \right)$ The rate of appearance of bromine $({\rm{B}}{{\rm{r}}_2})$ is related to rate of disappearance of bromide ions as following:

$\frac{{d\left( {{\rm{B}}{{\rm{r}}_2}} \right)}}{{dt}} = - \frac{3}{5}\frac{{d\left( {{\rm{B}}{{\rm{r}}^ - }} \right)}}{{dt}}$

$\frac{{d\left( {{\rm{B}}{{\rm{r}}_2}} \right)}}{{dt}} = \frac{3}{5}\frac{{d\left( {{\rm{B}}{{\rm{r}}^ - }} \right)}}{{dt}}$

The correct Option is "$\frac{{d\left( {{\rm{B}}{{\rm{r}}_2}} \right)}}{{dt}} = - \frac{3}{5}\frac{{d\left( {{\rm{B}}{{\rm{r}}^ - }} \right)}}{{dt}}$"

$\frac{{d\left( {{\rm{B}}{{\rm{r}}_2}} \right)}}{{dt}} = - \frac{5}{3}\frac{{d\left( {{\rm{B}}{{\rm{r}}^ - }} \right)}}{{dt}}$

$\frac{{d\left( {{\rm{B}}{{\rm{r}}_2}} \right)}}{{dt}} = \frac{5}{3}\frac{{d\left( {{\rm{B}}{{\rm{r}}^ - }} \right)}}{{dt}}$

In the reaction, $2{N_2}{O_5} \to 4N{O_2} + {O_2}$ initial pressure is $500\;atm$ and rate constant $k$ is $3.38 \times {10^{ - 5}}\;{s^{ - 1}}$ after 10 min the final pressure of ${N_2}{O_5}$ is

From the unit of rate constant $(i.e.,\;{s^{ - 1}}),$ it is clear that the reaction is of first order. $2{N_2}{O_5} \to 4N{O_2} + {O_2}$ Hence, for first order reaction , $k = \frac{{2.303}}{t}\log \frac{{{p_0}}}{{{p_t}}}$ $\therefore 3.38 \times {10^{ - 5}} = \frac{{2.303}}{{10 \times 60}}\log \frac{{500}}{{{p_t}}}$ $or\;\log \frac{{500}}{{{p_t}}} = 0.00880$ $\therefore \frac{{500}}{{{p_t}}} = \;anti\;\log \;0.00880$ $ = 1.02$ ${p_t} = \frac{{500}}{{1.02}} = 490\;atm$

The correct Option is "490 atm"

Identify ‘A’ in the reaction:

${\rm{C}}{{\rm{H}}_3}{\rm{C}} \equiv {\rm{CH}} = {\rm{C}}{{\rm{H}}_2}$

Product [A] in the reaction :

In such cases where migrating group is cycloalkyl group, ring expansion may occur.

In the reaction, water is formed by the combination of :

Follow the mechanism of esterification

The correct Option is "Hydroxyl of acid with alcoholic hydroxyl hydrogen"

Hydroxyl of alcohol with carboxylic hydrogen

In the reaction, $A\mathop {\mathop \to \limits^{{{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_{7{\rm{\;\;}}}}} }\limits_{{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}} {\rm{acetone}}\mathop \to \limits^{{\rm{Oxidation\;\;\;}}} {\rm{acetic\;acid}},\;A$ is

The correct Option is "2-propanol"

In the reaction, ${\rm{C}}{{\rm{H}}_3}{\rm{OH}}\xrightarrow{{{\rm{Oxidation\;}}}}{\rm{A}}\xrightarrow{{{\rm{N}}{{\rm{H}}_3}}}{\rm{B}}$; $A$ and $B$ are

${\rm{HCHO}},{\rm{\;HCOON}}{{\rm{H}}_4}$

The correct Option is "${\rm{HCOOH}},{\rm{\;HCOON}}{{\rm{H}}_4}$"

${\rm{HCOOH}},{\rm{\;HCON}}{{\rm{H}}_2}$

${\rm{HCHO}},{\rm{\;HCON}}{{\rm{H}}_2}$

Product formed in the reaction;

Methyl thiomethyl group is inserted at $ortho$ position by heating phenol with dimethyl sulphoxide and pyridine $ - {\rm{S}}{{\rm{O}}_3}/{\left( {{\rm{C}}{{\rm{H}}_3}{\rm{CO}}} \right)_2}{\rm{O}}$.

The product obtained in the reaction

$Iso$-propyl magnesium bromide reduces di-$iso$-propyl ketone to secondary alcohol. However, only $-{{\rm{H}}^ + }$ ion adds to ketone in spite of bulky alkyl group due to steric hinderance

What is the product in the reaction ${\rm{C}}{{\rm{H}}_3}{\rm{CON}}{{\rm{H}}_2}\xrightarrow{{{\rm{NaO}}{{\rm{H}}_2}/{\rm{HCl}}}}X?$

Amides, on treating with ${\rm{HN}}{{\rm{O}}_2}$, give acids. $\begin{array}{l} {\rm{C}}{{\rm{H}}_3}{\rm{CON}}{{\rm{H}}_2}\xrightarrow[{\left( {{\rm{HN}}{{\rm{O}}_2}} \right)}]{{{\rm{NaN}}{{\rm{O}}_2}/{\rm{HCl}}}}{\rm{C}}{{\rm{H}}_3}{\rm{COOH}} + {{\rm{N}}_2} + {{\rm{H}}_2}{\rm{O}}\\ acetic\;acid \end{array}$

The correct Option is "${\rm{C}}{{\rm{H}}_3}{\rm{COOH}}$"

${\rm{C}}{{\rm{H}}_3}{\rm{N}}{{\rm{H}}_2}$

${\rm{C}}{{\rm{H}}_3}{\rm{CHO}}$

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{C}}{{\rm{H}}_3}\mathop \to \limits^{\left[ {\rm{O}} \right]} {{\rm{C}}_6}{{\rm{H}}_5}{\rm{COOH}}\mathop \to \limits^{{\rm{NaOH}}} {{\rm{C}}_6}{{\rm{H}}_5}{\rm{COONa}}\mathop \to \limits^{{\rm{CaO}} + {\rm{NaOH}}} {{\rm{C}}_6}{{\rm{H}}_6}$

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{OH}}$

The correct Option is "${{\rm{C}}_6}{{\rm{H}}_6}$"

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{COONa}}$

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{ONa}}$

The organic product formed in the reaction; ${{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{5}}}{\rm{COOC}}{{\rm{H}}_{\rm{3}}}\xrightarrow[{{\rm{(II)}}\,{{\rm{H}}_{\rm{2}}}{\rm{O}}}]{{{\rm{(I)}}\,\,{\rm{LiAl}}{{\rm{H}}_{\rm{4}}}}}:$

The correct Option is "${{\rm{C}}_6}{{\rm{H}}_5}{\rm{C}}{{\rm{H}}_2}{\rm{OH}}$ and ${\rm{C}}{{\rm{H}}_3}{\rm{OH}}$"

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{COOH}}$ and ${\rm{C}}{{\rm{H}}_4}$

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{C}}{{\rm{H}}_3}$ and ${\rm{C}}{{\rm{H}}_3}{\rm{OH}}\,$

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{C}}{{\rm{H}}_3}$ and ${\rm{C}}{{\rm{H}}_4}$

The products obtained in the reaction

p-keto acids are the only carboxylic acids that decarboxylate under mild heat

The major product obtained in the reaction,

When addition of HCN takes place at $\alpha ,\beta - $ unsaturated carbonyl compounds, it gives $\beta - $ cyano compounds

Reduction of ${\rm{N}}{{\rm{O}}_2}$ group to ${\rm{N}}{{\rm{H}}_2}$ group is taking place by ${\rm{Fe}}/{\rm{HCI}}.$

${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

The correct Option is "${\rm{Fe}}/{\rm{HCl}}$"

Amine is not formed in the reaction (A) Hydrolysis of $RCN$ (B) Reduce of $R{\rm{CH}} = {\rm{NOH}}$ (C) Hydrolysis of $R{\rm{NC}}$ (D) Hydrolysis of $R{\rm{CON}}{{\rm{H}}_2}$ The correct answer is

$R{\rm{CN\;}}\mathop \to \limits^{{\rm{Hydrolysis\;}}} R{\rm{COOH}} + {\rm{N}}{{\rm{H}}_3}$ $R{\rm{CH}} = {\rm{NOH}}\mathop \to \limits^{{\rm{Reduction\;\;}}} R{\rm{C}}{{\rm{H}}_2}{\rm{N}}{{\rm{H}}_2} + {{\rm{H}}_2}{\rm{O}}$ $R{\rm{CN}} + 2{{\rm{H}}_2}{\rm{O}}\mathop \to \limits^{{\rm{\;\;}}\Delta {\rm{\;\;\;}}} R{\rm{N}}{{\rm{H}}_2} + {\rm{HCOOH}}$ $R{\rm{CON}}{{\rm{H}}_2}\mathop \to \limits^{{\rm{Hydrolysis\;}}} R{\rm{COOH}} + {\rm{N}}{{\rm{H}}_3}$

The product $\left[ A \right]$ formed in the reaction; $2{{\rm{C}}_{\rm{5}}}{{\rm{H}}_5} - {\rm{CN}}\xrightarrow[{{\rm{E}}{{\rm{t}}_{\rm{2}}}{\rm{O}}}]{{{\rm{Na}}}}\left[ A \right]$ is:

This is thorpe nitrile condensation involving only $\alpha - {\rm{H}} - $atom of nitrile due to strong attraction nature of ${\rm{CN}}$ gp.

In the reaction, $2{\rm{KI}} + {{\rm{H}}_2}{{\rm{O}}_2} + {{\rm{O}}_3} \to 2{\rm{KOH}} + {{\rm{O}}_2} + \,A,$ the compund $A$ is:

$2{\rm{KI}} + {{\rm{H}}_2}{\rm{O}} + {{\rm{O}}_3} \to 2{\rm{KOH}} + {{\rm{O}}_2} + {{\rm{I}}_2}$

The correct Option is "${{\rm{I}}_2}$"

In the reaction, $HN{O_3} + {P_4}{O_{10}} \to 4HP{O_3} + X$ the product X is

$4HN{O_3} + {P_4}{O_{10}} \to \mathop {4HP{O_3} + 2{N_2}{O_5}}\limits_{{\rm{Dinitrogen pentoxide}}} $ The product is dinitrogen pentoxide $\left( {{N_2}{O_5}} \right)$

The correct Option is "${N_2}{O_5}$"

Which of the following applies in the reaction, ${\rm{C}}{{\rm{H}}_3}{\rm{CHBrC}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_3}{\rm{\;}}\xrightarrow{{{\rm{Alc}}.{\rm{KOH\;}}}}{\rm{\;}}$ (i)${\rm{C}}{{\rm{H}}_3}{\rm{CH}} = {\rm{CHC}}{{\rm{H}}_3}$ (major product) (ii) ${\rm{C}}{{\rm{H}}_2} = {\rm{CHC}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_3}$ (minor product)

Alkyl halides on heating with alcoholic KOH give dehydrogenation reaction to yield alkene. If in reaction, more than one alkenes are formed, then according to Saytzeff, the most highly substituted alkene is the major product.

The correct Option is "Saytzeff’s rule"

In the reaction, $2{\rm{Al}}\left( s \right) + 6{\rm{HCl}}\left( {aq} \right) \to 2{\rm{A}}{{\rm{l}}^{3 + }}\left( {aq} \right) + 6{\rm{C}}{{\rm{l}}^ - }\left( {aq} \right) + 3{{\rm{H}}_2}\left( {\rm{g}} \right)$

$2{\rm{Al}}\left( s \right) + 6{\rm{HCl}}\left( {aq} \right) \to 2{\rm{A}}{{\rm{l}}^{3 + }}\left( {aq} \right) + 6{\rm{C}}{{\rm{l}}^ - }\left( {aq} \right) + 3{{\rm{H}}_2}\left( g \right)$ $3 \times 22.4{\rm{L}}{{\rm{H}}_2}\left( {\rm{g}} \right)$ at STP is produced by 6 moles of $HCl{\rm{ }}(aq)$ Hence, 11.2 L ${{\rm{H}}_2}\left( {\rm{g}} \right)$ at STP is produced by 1 mole HCl (aq).

6 L HCl (aq) is consumed for every $3{\rm{L}}{{\rm{H}}_2}\left( {\rm{g}} \right)$ produced

$33.6{\rm{L}}{{\rm{H}}_2}\left( {\rm{g}} \right)$ is produced regardless of temperature and pressure for every mole Al that reacts

$67.2{\rm{L}}{{\rm{H}}_2}\left( {\rm{g}} \right)$ at STP is produced for every mole Al that reacts

The correct Option is "$11.2{\rm{L}}{{\rm{H}}_2}\left( {\rm{g}} \right)$ at STP is produced for every mole HCl(aq) consumed"

In the reaction, The major product A is

This reaction follows benzyne mechanism.

An insect of mass m = 2g is inside a vertical drum of radius 1m that is rotating with an angular velocity of 10 rad/s. If the insect does not fall off then the minimum coefficient of friction is

$f \ge mg \Rightarrow \mu mr{\omega ^2} \ge mg \Rightarrow \mu = \frac{g}{{r{\omega ^2}}} = \frac{{10}}{{1 \times 100}} = 0.1$

Two light rays having the same wavelength $\lambda $ in vacuum are in phase initially. Then the first ray travels a path ${{L_1}}$ through a medium of refractive index ${{n_1}}$ while the second ray travels a path of length ${{L_2}}$ through a medium of refractive index ${{n_2}}$. The two waves are then combined to produce interference. The two waves are then combined to produce interference. The phase difference between the two waves is

The optical path between any two points is proportional to the time of travel. The distance traversed by light in a medium of refractive index $\mu $ in time $t$ is given by $d = vt \ldots \ldots \ldots .\left( i \right)$ Where v is velocity of light in the medium. The distance traversed by light in a vacuum in this time. ${\rm{\Delta }} = ct$ $ = c{\rm{ }} \cdot \frac{d}{v}\left[ {From\;{\rm{ }}Eq.\left( i \right)} \right]$ $ = d\frac{c}{v} = \mu d.......\left( {ii} \right)$ $\left( {{\rm{Since}},\;\mu = \frac{c}{v}} \right)$ This distance is the equivalent distance in vacuum and is called optical path. Here, optical path for first ray $ = {n_1}{L_1}$ Optical path for second ray $ = {n_2}{L_2}$ Path difference $ = {n_1}{L_1} - {n_2}{L_2}$ Now, phase difference $ = \frac{{2\pi }}{\lambda } \times $ path difference $ = \frac{{2\pi }}{\lambda } \times \left( {{n_1}{L_1} - {n_2}{L_2}} \right)$

$\frac{{2\pi }}{\lambda }\left( {{L_2} - {L_1}} \right)$

The correct Option is "$\frac{{2\pi }}{\lambda }\left( {{n_1}{L_1} - {n_2}{L_2}} \right)$"

$\frac{{2\pi }}{\lambda }\left( {{n_2}{L_1} - {n_1}{L_2}} \right)$

$\frac{{2\pi }}{\lambda }\left( {\frac{{{L_1} - {L_2}}}{{{n_1} - {n_2}}}} \right)$

A point charge q is placed at a distance r from the centre of an uncharged conducting sphere of Radius R. The potential at any point on the sphere is

Potential V= same $\therefore \,V = \frac{1}{{4\pi { \in _{\rm{o}}}}}\frac{q}{r} + {V^1}$ ${V^1} = $ potential at centre due to induced charge = O $\therefore V = \frac{1}{{9\pi { \in _{\rm{o}}}}}\frac{q}{r}$

The correct Option is "$\frac{1}{{4\pi {\varepsilon _ \circ }}}\frac{q}{r}$"

$\frac{1}{{4\pi {\varepsilon _ \circ }}}\frac{{qR}}{{{r^2}}}$

$\frac{1}{{4\pi {\varepsilon _ \circ }}}\frac{{q{r^2}}}{R}$

Primary constant of a transmission line are

A transmission line has four primary constants, resistance, inductance, capacitance and conductance

The correct Option is "Both (a) and (b)"

An electric heater rated 220 V and 550 W is connected to A.C mains. The current drown by it is

$P = VI \Rightarrow I = \frac{P}{V} = \frac{{550}}{{220}} = 2.5A$

An electron of ${K_\alpha }$ spectral line of an atom is $59\;keV,$ then the wavelength of ${K_\alpha }$ line will be

$\lambda = \frac{{hc}}{{eV}} = \frac{{12375}}{{59000}} = 0.20\mathop A\limits^ \circ \;\;\left[ {{ \because }\frac{{hc}}{e} = 12375} \right]$

The correct Option is "$0.20\;\;\mathop A\limits^ \circ $"

$0.42\;\mathop A\limits^ \circ $

$0.31\;\mathop A\limits^ \circ $

$0.62\;\mathop A\limits^ \circ $

A medium shows relation between $i$ and $r$ as shown. If speed of light in the medium is $nc$ then value of $n$ is .

From graph it is clear that $\tan 30^\circ = \frac{{\sin r}}{{\sin i}}$ $ \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{\sin \,\,r}}{{\sin \,\,i}} = \frac{1}{\mu } \Rightarrow \mu = \sqrt 3 $ Also $v = \frac{c}{\mu } = nc \Rightarrow n = \frac{1}{\mu } = \frac{1}{{\sqrt 3 }} = {\left( 3 \right)^{ - 1/2}}$

The correct Option is "${3^{ - 1/2}}$"

The potential energy of a certain spring when stretched through a distance s is 10 J. The amount of work (in joule) that must be done on this spring to stretch it through additional distance s will be

$U = \frac{1}{2}k{s^2} = 10\;{\rm{J}}\,,\,U' = \frac{1}{2}k{\left( {s + s} \right)^2} = 4\left( {\frac{1}{2}k{s^2}} \right) = 40 J $ , $W = U' - U = 40 - 10 = 30{\rm{\;J}}$