datavorous/entrance-exam-dataset huggingface dataset:
An organic compound A upon reacting with NH3 gives B. On heating, B gives C. C in presence of KOH reacts with Br2 to give CH3CH2NH2 . A is

Options:
A CH3-C|H-CH3 COOH
B CH3CH2CH2COOH
C CH3COOH
D CH3CH2COOH

The correct answer is:
CH3CH2COOH

Solution:
CH3CH2COOHA+NH3→CH3CH2CO2NH4B→ Δ CH3CH2CONH2C
→ KOH+Br2Hoffmann Bromamide reaction CH3CH2NH2+H2O

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
An organic compound A upon reacting with NH3 gives B. On heating, B gives C. C in presence of KOH reacts with Br2 to give CH3CH2NH2. A is :

Options:
A
B CH 3 CH 2 COOH
C CH 3 COOH
D CH 3 CH 2 CH 2 COOH

The correct answer is:
CH 3 CH 2 COOH

Solution:
Ammonia reacts with carboxylic acid to form ammonium carboxylate (B). Which on heating form amide (C). Amide on reaction with bromine and KOH undergo Hoffmann Bromamide reaction to form amine with one carbon less.

CH3−CH2−CO||−OH+NH3(A)→CH3CH2−CO||−O⊖(B)N⊕H4CH3CH2−CO||−O⊖(B)N⊕H4→ΔCH3CH2−CO||−NH2(C)CH3CH2−CO||−NH2(C)→(Hoffmann bromamide degradation)Br2/KOHCH3CH2−NH2

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
An organic compound A upon reacting with NH3 gives B. On heating, B gives C. C in presence of KOH reacts with Br2 to give CH3CH2NH2. A is :

Options:
A
B CH 3 CH 2 COOH
C CH 3 COOH
D CH 3 CH 2 CH 2 COOH

The correct answer is:
CH 3 CH 2 COOH

Solution:
Ammonia reacts with carboxylic acid to form ammonium carboxylate (B). Which on heating form amide (C). Amide on reaction with bromine and KOH undergo Hoffmann Bromamide reaction to form amine with one carbon less.

CH3−CH2−CO||−OH+NH3(A)→CH3CH2−CO||−O⊖(B)N⊕H4CH3CH2−CO||−O⊖(B)N⊕H4→ΔCH3CH2−CO||−NH2(C)CH3CH2−CO||−NH2(C)→(Hoffmann bromamide degradation)Br2/KOHCH3CH2−NH2

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2013 (07 Apr)
An organic compound, ' $A$ ' with molecular formula $\mathrm{C}_8 \mathrm{H}_8 \mathrm{O}$ on reaction with $\mathrm{I}_2 / \mathrm{KOH}$ gives salt of carboxylic acid ' $B$ ' and a halo-compound ' $C$ '. Compound ' $C$ ' on the reaction with silver powder gives ' $D$ '. The structures of ' $A$ ' and ' $D$ ' are

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
An organic compound answers Molisch's test as well as Benedict's test. But it does not answer Scliwanoff's test. Most probably, it is

Options:
A Sucrose
B Protein
C Fructose
D Maltose

The correct answer is:
Maltose

Solution:
An organic compound answers Molisch's test as well as Benedict's test. But it does not answer Scliwanoff's test. Most probably, it is Maltose. It is a reducing sugar. Molisch's test is used to detect all kinds of carbohydrates. The Benedict's test identifies reducing sugars (monosaccharides and some disaccharides), which have free ketone or aldehyde functional groups. Scliwanoff's test is simple test that is used for the detection/presence of fructose or keto group in a sugar solution.

Topic: Chemistry, Biomolecules, JEE Main
An organic compound $\left(\mathrm{C}_3 \mathrm{H}_9 \mathrm{~N}\right)(A)$, when treated with nitrous acid, gave an alcohol and $\mathrm{N}_2$ gas was evolved. $(A)$ on warming with $\mathrm{CHCl}_3$ and caustic potash gave (C) which on reduction gave iso-propylmethylamine. Predict the structure of $(A)$.

Options:
A

B

C

D

The correct answer is:

Solution:
On treating with nitrous acid primary a mine gives an alcohol and $\mathrm{N}_2$ gas. So, ' $A$ ' should be a primary amine.

Topic: Chemistry, Amines, NEET, NEET 2012 (Mains)
An organic compound $\left(\mathrm{C}_3 \mathrm{H}_9 \mathrm{~N}\right)(A)$, when treated with nitrous acid, gave an alcohol and $\mathrm{N}_2$ gas was evolved. $(A)$ on warming with $\mathrm{CHCl}_3$ and caustic potash gave (C) which on reduction gave iso-propylmethylamine. Predict the structure of $(A)$.

Options:
A

B

C

D

The correct answer is:

Solution:
On treating with nitrous acid primary a mine gives an alcohol and $\mathrm{N}_2$ gas. So, ' $A$ ' should be a primary amine.

Topic: Chemistry, Amines, JEE Main
An organic compound contains \(69 \%\) carbon and 4. \(8 \%\) hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when \(0.20 \mathrm{~g}\) of this substance is subjected to complete combustion.

Options:

None

Solution:
We know that \(\% \mathrm{C}=12 / 44 \times\left(\right.\) Mass of \(\mathrm{CO}_2\) formed \() /(\) Mass or substance taken \() \times 100\)
Substituting the values of \(\%\) of \(\mathrm{C}\) and mass of the substance taken, we have,
\(69=\frac{12}{44} \times \frac{\text { mass of } \mathrm{CO}_2 \text { formed } \times 100}{\text { mass of substance taken }}\)
or Mass of \(\mathrm{CO}_2\) formed \(=\frac{69 \times 44 \times 0.2}{12 \times 100}=0.506 \mathrm{~g}\)
Similarly, \(\% \mathrm{H}=2 / 18 \times\left(\right.\) Mass of \(\mathrm{H}_2 \mathrm{O}\) formed \() /\) mass of substance taken \(\times 100\)
Substituting the values of \(\% \mathrm{H}\) and the mass of substance taken, we have,
\(4.8=\frac{2 \times \text { mass of water formed }}{18 \times 0.2} \times 100\) or mass of water formed \(=\frac{4.8 \times 18 \times 0.2}{2 \times 100}=0.0864 \mathrm{~g}\)

Topic: Chemistry, General Organic Chemistry
An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.

Options:
A 0.0864 g of CO2and 0.506 g of water is produced
B 0.505 g of CO2and 0.0864 g of water is produced
C 0.415 g of CO2and 0.0245 g of water is produced
D 0.668 g of CO2and 0.0339 g of water is produced

The correct answer is:
0.505 g of CO2and 0.0864 g of water is produced

Solution:
Correct Option is : (B)
0.505 g of CO2and 0.0864 g of water is produced

Topic: Chemistry, General Organic Chemistry, NEET
An organic compound contains 80% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is: [Atomic wt. of C is 12 and of H is 1]

Options:
A CH
B CH2
C CH3
D CH4

The correct answer is:
CH3

Solution:
% by weight CH8020

% by moles 80122016.6720

Simplest ratio 6.676.67206.6713

Hence, the empirical formula of this compound is CH3.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
An organic compound contains 80% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is: [Atomic wt. of C is 12 and of H is 1]

Options:
A CH
B CH2
C CH3
D CH4

The correct answer is:
CH3

Solution:
% by weight CH8020

% by moles 80122016.6720

Simplest ratio 6.676.67206.6713

Hence, the empirical formula of this compound is CH3.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2021
An organic compound contains C, H and S. The minimum molecular weight of the compound containing 8% Sulphur is

Options:
A 600 g mol-1
B 200 g mol-1
C 400 g mol-1
D 300 g mol-1

The correct answer is:
400 g mol-1

Solution:
% of Sulphur =weight of sulphurweight of compound×100=8
∴ 32weight of compound=8100
∴ weight of compound=400 g mol-1

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2016 (09 Apr Online)
An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, $38.71 \%$ and $\mathrm{H}, 9.67 \%$. The empirical formula of the compound would be

Options:
A $\mathrm{CH}_4 \mathrm{O}$
B $\mathrm{CH}_3 \mathrm{O}$
C $\mathrm{CH}_2 \mathrm{O}$
D $\mathrm{CHO}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{O}$

Solution:

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2008 (Mains)
An organic compound exhibits optical isomerism when

Options:
A Four groups linked to carbon atom are different
B Three groups linked to carbon atom are different
C Two groups linked to carbon atom are different
D All the groups linked to carbon atom are same

The correct answer is:
Four groups linked to carbon atom are different

Solution:
Four groups linked to carbon atom are different

Topic: Chemistry, General Organic Chemistry, JEE Main
An organic compound has \(42.1 \%\) carbon, \(6.4 \%\) hydrogen and remainder is oxygen. If its molecular weight is 342, then its molecular formula is :

Options:
A \(\mathrm{C}_{11} \mathrm{H}_{18} \mathrm{O}_{12}\)
B \(\mathrm{C}_{12} \mathrm{H}_{20} \mathrm{O}_{12}\)
C \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)
D \(\mathrm{C}_{14} \mathrm{H}_{20} \mathrm{O}_{10}\)

The correct answer is:
\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)

Solution:
only $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$ has $42.1 \%$ carbon, $6.4 \%$ hydrogen $\& 51.5$ percent oxygen.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2024 (05 Apr Shift 1)
An organic compound having molecular mass 60 is found to contain $\mathrm{C}=20 \%, \mathrm{H}=$ $6.67 \%$ and $\mathrm{N}=46.67 \%$ while rest is oxygen. On heating it gives $\mathrm{NH}_3$ alongwith a solid residue. The solid residue give violet colour with alkaline copper sulphate solution. The compound is

Options:
A
$\mathrm{CH}_3 \mathrm{NCO}$

B
$\mathrm{CH}_3 \mathrm{CONH}_2$

C
$\left(\mathrm{NH}_2\right)_2 \mathrm{CO}$

D
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CONH}_2$

The correct answer is:
$\left(\mathrm{NH}_2\right)_2 \mathrm{CO}$

Solution:
No Solution Available

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2005
An organic compound is boiled with alcoholic potash. The product is cooled and acidified with HCl. A white solid separates out. The starting compound may be

Options:
A ethyl acetate
B methyl acetate
C ethyl benzoate
D ethyl formate

The correct answer is:
ethyl benzoate

Solution:
Ethyl benzoate is boiled with alc. potash to give ethyl alcohol and potassium benzoate.
Potassium benzoate is converted into benzoic acid, a white solid which separates out.
$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOK}+\mathrm{HCl} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}+\mathrm{KCl}$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
An organic compound is estimated through Dumas method and was found to evolve 6 moles of $\mathrm{CO}_{2}, 4$ moles of $\mathrm{H}_{2} \mathrm{O}$ and 1 mole of nitrogen gas. The formula of the compound is:

Options:
A $\mathrm{C}_{12} \mathrm{H}_{8} \mathrm{~N}$
B $\mathrm{C}_{12} \mathrm{H}_{8} \mathrm{~N}_{2}$
C $\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{~N}_{2}$
D $\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{~N}$

The correct answer is:
$\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{~N}_{2}$

Solution:
$\left[\mathrm{C}_{x} \mathrm{H}_{y} \mathrm{~N}_{z}\right] \frac{\text { Dumas }}{\text { method }} \rightarrow 6 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O}+\mathrm{N}_{2}$

Mol of $\mathrm{CO}_{2}=6,$ so $\mathrm{mol}$ of $\mathrm{C}$ is $=6$

Mol of $\mathrm{H}_{2} \mathrm{O}=4$, so $\mathrm{mol}$ of $\mathrm{H}$ is $=8$

Mol of $\mathrm{N}_{2}=1,$ so $\mathrm{mol}$ of $\mathrm{N}$ is $=2$

$\therefore$ Formula is $\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{~N}_{2}$

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (11 Jan Shift 1)
An organic compound is subjected to chlorination to get compound A using 5.0 g of chlorine. When 0.5 g of compound A is reacted with AgNO3 [Carius Method], the percentage of chlorine in compound A is when it forms 0.3849 g of AgCl. (Round off to the Nearest Integer)
(Atomic masses of Ag and Cl are 107.87 and 35.5 respectively)

Options:

The correct answer is:
19

Solution:
nCl in compound =nAgCl=0.3849 g(107.87+35.5)g/mol

⇒ mass of chlorine =nCl×35.5=0.0953gm

⇒% wt. of chlorine =0.09530.5×100

=19.06%

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (27 Jul Shift 1)
An organic compound (P) having molecular mass 180 is acylated with CH3COCl to get a new compound with molecular mass 348. The number of amino groups present per molecule of compound (P) is

Options:

The correct answer is:
4

Solution:
R−NH2M+CH3COCl→RNHCOCH3M+42+HCl
If the increase in molecular mass is 42, the reactant has one NH2 group.
Actual increase in molecular mass =348-180=168
Number of NH2 groups =16842=4

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
An organic compound undergoes first order decomposition. The time taken for its decomposition to 18 and 110 of its initial concentration are t18 and t110 respectively. What is the value of t18t110×10 ?
(take log102=0.3 )

Options:

The correct answer is:
9

Solution:
As we know
kt=ln⁡A0At
t18t110=ln⁡8ln⁡10=0.9
t18t110×10=9

Topic: Chemistry, Chemical Kinetics, JEE Main
An organic compound undergoes first-order decomposition. The time taken for its decomposition to $1 /$ 8 and $1 / 10$ of its initial concentration are $t_{1 / 8}$ and $t_{1 / 10}$ respectively. What is the value of $\left[\frac{t_{1 / 8}}{t_{1 / 10}}\right] \times 10$ ? $\left(\log _{10} 2=0.3\right)$

Options:

The correct answer is:
9

Solution:
$\begin{array}{l}

t_{1 / 8}=\frac{2.303 \log 8}{k}=\frac{2.303 \times 3 \log 2}{k} \\

t_{1 / 10}=\frac{2.303}{k} \log 10=\frac{2.303}{k} \\

{\left[\frac{t_{1 / 8}}{t_{1 / 10}}\right] \times 10=\frac{\left(\frac{2.303 \times 3 \log 2}{k}\right)}{\left(\frac{2.303}{k}\right)} \times 10=9}

\end{array}$

Topic: Chemistry, Chemical Kinetics, JEE Main
An organic compound which produces a bluish green coloured flame on heating in presence of copper is

Options:
A chlorobenzene
B benzaldehyde
C aniline
D benzoic acid

The correct answer is:
chlorobenzene

Solution:
An organic compound which produces a bluish green flame, when heated in presence of copper is chlorobenzene. This is called Beilstein's test used to detect presence of halogen in an organic compound.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
An organic compound with molecular formula, C7H8O , dissolves in NaOH and gives characteristics colour with neutral FeCl3 . On treatment with bromine-water, it gives a tribromo derivative, C7H5OHBr3. The compound is

Options:
A benzyl alcohol
B o-cresol
C p-cresol
D m-cresol

The correct answer is:
m-cresol

Solution:
The compound may be a phenol, because it gives characteristic colour with FeCl3 .
Only compound (II) can give tribromo-derivative.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
An organic compound $X$ gives a red precipitate on heating with Fehling's solution. Which one of the following reactions yields $X$ as a major product?

Options:
A

B $\mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}+\mathrm{AgOH} \stackrel{\Delta}{\longrightarrow}$
C $2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{Br}+\mathrm{Ag}_2 \mathrm{O} \stackrel{\Delta}{\longrightarrow}$
D $\mathrm{C}_2 \mathrm{H}_2+\mathrm{H}_2 \mathrm{O} \underset{1 \% \mathrm{HgSO}_4 60^{\circ} \mathrm{C}}{\stackrel{40 \% \mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow}}$

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_2+\mathrm{H}_2 \mathrm{O} \underset{1 \% \mathrm{HgSO}_4 60^{\circ} \mathrm{C}}{\stackrel{40 \% \mathrm{H}_2 \mathrm{SO}_4}{\longrightarrow}}$

Solution:
Acetaldehyde + Fehling's solution $\stackrel{\Delta}{\longrightarrow}$ Cuprous oxide (Red ppt.)

Topic: Chemistry, Hydrocarbons, JEE Main
An organic compound X is oxidized by using acidified K2Cr2O7 . The product obtained reacts with phenyl hydrazine, but does not answer silver mtextrror test. The possible structure of X is

Options:
A CH3COCH3
B CH32CHOH
C CH3CHO
D CH3CH2OH

The correct answer is:
CH32CHOH

Solution:
The oxidation product of X reacts with phenyl hydrazine, thus it contains >C=0 group. The same product does not give silver mirror test. Thus, it is a ketone, because only aldehydes give silver mirror test.
Thus, the compound X must be 2o alcohol, as only secondary alcohols give ketones on oxidation and hence, X is CH32CHOH.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
An organic compound ‘X’ showing the following solubility profile is:

Options:
A m-Cresol
B o-Toluidine
C Oleic acid
D Benzamide

The correct answer is:
m-Cresol

Solution:
Meta-cresol

the group should be base sensitivity means it should have acidic hydrogen because it is soluble in 10% NaOH and it should be less acidic than H2CO3 because it is insoluble in NaHCO3.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
An organic compound ‘X’ showing the following solubility profile is:

Options:
A m-Cresol
B o-Toluidine
C Oleic acid
D Benzamide

The correct answer is:
m-Cresol

Solution:
Meta-cresol

the group should be base sensitivity means it should have acidic hydrogen because it is soluble in 10% NaOH and it should be less acidic than H2CO3 because it is insoluble in NaHCO3.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2019 (08 Apr Shift 1)
An organic compound X with molecular formula C11H14 gives an optically active compound on hydrogenation. Upon ozonolysis, X produces a mixture of compounds -P and Q. Compound P gives a yellow precipitate when treated with I2 and NaOH but does not reduce Tollen's reagent. Compound Q does not give any yellow precipitate with I2 and NaOH but gives Fehling's test. The compound X is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
An organic halide is shaken with aqueous $\mathrm{NaOH}$ followed by the addition of dil. $\mathrm{HNO}_3$ and silver । solution gave white ppt. The substance can be [

Options:
A $\mathrm{C}_6 \mathrm{H}_4\left(\mathrm{CH}_3\right) \mathrm{Br}$
B $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Cl}$
C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}$
D None of these

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Cl}$

Solution:
$\mathrm{C}_6 \mathrm{H}_5-\mathrm{CH}_2-\mathrm{Cl} \xrightarrow[\text { (aq) }]{\mathrm{NaOH}} \mathrm{C}_6 \mathrm{H}_5-\mathrm{CH}_2-\mathrm{OH}$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
An organic liquid $A$ containing $C, H$ and $O$ has a pleasant odour with a boiling point of $78^{\circ} \mathrm{C}$ On boiling $A$ with conc. $\mathrm{H}_2 \mathrm{SO}_4$ a colourless gas is produced which decolourises bromine water and alkaline $\mathrm{KMnO}_4$. One mole of this gas also takes one mole of $H_2$. The organic liquid $A$ is

Options:
A $\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}$
B $\mathrm{C}_2 \mathrm{H}_5 \mathrm{CHO}$
C $\mathrm{C}_2 \mathrm{H}_6$
D $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$

Solution:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ on heating with conc. $\mathrm{H}_2 \mathrm{SO}_4$ produces $\mathrm{C}_2 \mathrm{H}_4$. This is ethene, which decolourises bromine water and alkaline $\mathrm{KMnO}_4$ solution due to unsaturation.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
An organization awarded 48 medals in event A'', 25 in event B'' and 18 in event C''. If these medals went to total 60 men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?

Options:
A 15
B 21
C 10
D 9

The correct answer is:
21

Solution:
Let A, B and C denote the set of men who received medals in even A, event B and event C respectively.

Then, nA=48, nB=25, nC=18 , nA∪B∪C=60 and nA∩B∩C=5,

Now we know that, nA∪B∪C=nA+nB+nC-nA∩B-nA∩C-nB∩C+nA∩B∩C

⇒60=48+25+18-n(A∩B)-n(A∩C)-n(B∩C)+5

⇒n(A∩B)+n(A∩C)+n(A∩C)=48+25+18+5-60=36

Therefore, the number of people who received medals in exactly two of the three sports will be 36-3A∩B∩C=36-15=21.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2023 (11 Apr Shift 1)
An orthogonal matrix is

Options:
A $\left[\begin{array}{cc}\cos \alpha & 2 \sin \alpha \\ -2 \sin \alpha & \cos \alpha\end{array}\right]$
B $\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$
C $\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
D $\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$

The correct answer is:
$\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$

Solution:
A square matrix is to be orthogonal matrix if
$A^{\prime} A=I=A A^{\prime}$
$\Rightarrow A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right], A^{\prime}=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
$\Rightarrow A A^{\prime}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right], A^{\prime} A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\therefore A A^{\prime}=A^{\prime} A=I$.

Topic: Mathematics, Matrices, JEE Main
An oscillating circuit consisting of a capacitor with capacitance $C=10 \mu \mathrm{F}$, a coil with inductance $L=6.0 \mu \mathrm{H}$ and active resistance $R=10 \Omega$. The mean power that should be fed to the circuit to maintain undamped harmonic oscillations with an external driving power with $50 \mathrm{~Hz}$ and a $V_m$ of $280 \mathrm{~V}$ is

Options:
A $3.8 \mathrm{~W}$
B $48 \mathrm{~W}$
C $3 \mathrm{~mW}$
D $48 \mathrm{~mW}$

The correct answer is:
$3.8 \mathrm{~W}$

Solution:
$$
\text { } R=10 \Omega L=6 \mu \mathrm{H}, C=10 \mu \mathrm{H}
$$
Impedence of the circuit,
$$
Z=\sqrt{R^2+\left(X_C-X_L\right)^2}
$$
Capacitive reactance, $X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}$
$$
X_C=\frac{1}{2 \times 314 \times 50 \times 10 \times 10^{-6}}
$$
$$
=318.47 \Omega
$$
Inductive reactance $X_L=\omega L=2 \pi f L$
$$
\begin{aligned}
X_L & =2 \times 314 \times 50 \times 6 \times 10^{-6} \\
& =18.84 \times 10^{-4} \Omega
\end{aligned}
$$
Resistance, $R=10 \Omega$
$$
\begin{gathered}
Z=\sqrt{10^2+\left(318.4-18.84 \times 10^{-4}\right)^2} \\
Z=\sqrt{100+(318.39)^2}=\sqrt{101477.36} \\
Z=318.55 \Omega \\
\text { rms current, } i_{\text {rms }}=\frac{V_{\text {rms }}}{Z}=\frac{V_0 / \sqrt{2}}{Z} \\
i_{\text {rms }}=\frac{280}{\sqrt{2} \times 318.55}=0.621 \mathrm{~A}
\end{gathered}
$$
As energy is only dissipated in resistance as heat and no power dissipates in capacitor and inductor, so average power dissipates as heat
$$
P=i_{\mathrm{rms}}^2 \cdot R
$$
$$
\begin{aligned}
& P=(0.621)^2 \times 10 \\
& P=3.8 \mathrm{~W}
\end{aligned}
$$
$3.8 \mathrm{~W}$ power should be fed to maintain an undamped oscillation.

Topic: Physics, Alternating Current, TS EAMCET, TS EAMCET 2018 (04 May Shift 2)
An oscillator is essentially

Options:
A an amplifier with proper negative feedback network circuits
B converts alternating current into direct current
C an amplifier with no feedback network
D an amplifier with proper positive feedback network circuits

The correct answer is:
an amplifier with proper positive feedback network circuits

Solution:
In an oscillator, $\mathrm{L}-\mathrm{C}$ circuit is coupled with transistor amplifier in such a way that there is a positive feed back to the LC circuit i.e., proper energy supply to $\mathrm{LC}$ at proper timings. So that total energy of LC circuit remains same.

Topic: Physics, Semiconductors, VITEEE, VITEEE 2006
An oscillator is nothing but an amplifier with

Options:
A Positive feed back
B Large gain
C No feedback
D Negative feedback

The correct answer is:
Positive feed back

Solution:
In oscillator, a portion of the output power is returned back (feed back) to the input in phase with the starting power. This process is termed as positive feedback.

Topic: Physics, Semiconductors, JEE Main
An oscillator of mass M is at rest in its equilibrium position in a potential, V=12kx –X2 . A particle of mass m comes from the right with speed u and collides completely inelastic with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is: M=10, m=5, u=1, k=1

Options:
A 23
B 13
C 35
D 12

The correct answer is:
13

Solution:
In the first collision mu momentum will be imparted to the system. In the second collision when the momentum of (M + m) is in the opposite direction mu momentum of the particle will make its momentum zero. On 13th collision,
Applying momentum conservation
mu=(M+13m)v
⇒v=muM+13m=u15 ⇒M=2m
⇒v= ωA
⇒u15= KM+13m×A
⇒A=115751=13

Topic: Physics, Oscillations, JEE Main, JEE Main 2018 (16 Apr Online)