datavorous/entrance-exam-dataset huggingface dataset:
Kepler's of aerial velocity of a planet is a consequence of the law of conservation of

Options:
A Energy
B Angular momentum
C Linear momentum
D None of these

The correct answer is:
Angular momentum

Solution:
Areal velocity $\frac{\mathrm{dA}}{\mathrm{dt}} \&$ angular momentum $L$ of a planet are related by equation $\frac{d A}{d t}=\frac{L}{2 M}$, where $\mathrm{M}$ is the mass of planet. Since in planetary motion $\overline{\mathrm{L}}$ is constant $\left(\bar{\tau}_{\text {ext. }}=0\right)$, hence $\frac{\mathrm{dA}}{\mathrm{dt}}$ is also constant.

Topic: Physics, Gravitation, JEE Main
Kepler's second law of planetary motion corresponds to

Options:
A conservation of energy
B conservation of angular momentum
C conservation of linear momentum
D conservation of mass

The correct answer is:
conservation of angular momentum

Solution:
From Kepler's second law of planetary motion,
$$
\frac{d L}{d t}=\frac{L}{2 m}
$$
Thus, angular monentum remains conserved.

Topic: Physics, Gravitation, COMEDK, COMEDK 2021
Kepler's second law regarding constancy of aerial velocity of a planet is a consequence of the law of conservation of

Options:
A Energy
B Angular momentum
C Linear momentum
D None of these

The correct answer is:
Angular momentum

Solution:
Kepler's second law, is a consequence of the fact that the force of gravity is a central force, which in turn implies that angular momentum is constant.

Topic: Physics, Gravitation, JEE Main
Kepler's third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance $\mathrm{r}$

between sun and planet i.e. $\mathrm{T}^{2}=\mathrm{Kr}^{3}$ here $\mathrm{K}$ is constant. If the masses of sun and planet are $\mathrm{M}$ and $\mathrm{m}$ respectively then as per Newton's law of gravitation force of attraction between them is

$\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{r}^{2}}$, here $\mathrm{G}$ is gravitational constant. The relation between $\mathrm{G}$ and $\mathrm{K}$ is described as

Options:
A $\mathrm{GMK}=4 \pi^{2}$
B $\quad \mathrm{K}=\mathrm{G}$
C $\mathrm{K}=\frac{1}{\mathrm{G}}$
D $\quad \mathrm{GK}=4 \pi^{2}$

The correct answer is:
$\mathrm{GMK}=4 \pi^{2}$

Solution:
As we know, orbital speed, $V_{\text {orb }}=\sqrt{\frac{\text { GM }}{\text { r }}}$

Time period $\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{v}_{\text {orb }}}=\frac{2 \pi \mathrm{r}}{\sqrt{\mathrm{GM}}} \sqrt{\mathrm{r}}$

Squarring both sides,

$\mathrm{T}^{2}=\left(\frac{2 \pi \mathrm{r} \sqrt{\mathrm{r}}}{\sqrt{\mathrm{GM}}}\right)^{2}=\frac{4 \pi^{2}}{\mathrm{GM}} \cdot \mathrm{r}^{3}$

$\Rightarrow \quad \frac{\mathrm{T}^{2}}{\mathrm{r}^{3}}=\frac{4 \pi^{2}}{\mathrm{GM}}=\mathrm{K} \Rightarrow \mathrm{GMK}=4 \pi^{2}$

Topic: Physics, Gravitation, JEE Main
Keto-enol tautomerism is found in

Options:
A

B

C

D

The correct answers are:
,

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
Keto-enol tautomerism is not shown by

Options:
A butan-2-one
B 1-phenylbutan-2-one
C p-O2NC6H4CH2COCH2Ph
D PhCOPh

The correct answer is:
PhCOPh

Solution:
The compounds A, B and C show keto-enol tautomerism.
A. CH3-C||O-CH2CH3⇌CH3-C=| OH CH-CH3i ⇌H2C=C-| OH CH2CH3ii
The first enol is more stable, because it has the more substituted double bond.
B. PhCH2-C||O-CH2-CH3⇌PhCH=C-| OH CH2-CH3i ⇌PhCH2-C=| OH CH-CH3ii
The first enol is more stable, because C=C and Ph are conjugated.
C. p-O2NC6H4CH2COCH3⇌p-O2NC6H4CH=C|-OH CH3 (i)⇌p-O2NC6H4CH2C|=OH CHPh (ii)
The first enol is more stable, because the p-O2N extends the delocalization of the conjugated system.
PhCOPh does not show tautomerism as do not have any α -H-atom.

Topic: Chemistry, General Organic Chemistry, JEE Main
Ketones are prepared by

Options:
A Clemmensen's reduction
B Cannizzaro reaction
C Rosenmund's reduction
D Oppenaur's oxidation

The correct answer is:
Oppenaur's oxidation

Solution:
Oppenaur's oxidation is used to prepare Ketone.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Kinetic theory of gases proves

Options:
A
only Boyle’s law

B
only Charles’ law

C
only Avogadro’s law

D
all of these

The correct answer is:
all of these

Solution:
No Solution Available

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2002
Kingdom Animalia is characterised by

Options:
A direct dependence on autotrophs
B indirect dependence on autotrophs
C absence of chlorophyll
D absence of cell wall.

The correct answer is:
absence of cell wall.

Solution:
The main criteria for classification used by Whittaker includes cell structure, thallus organization, mode of nutrition, reproduction and phylogenetic relationships. So, the absence of cell wall is one of the main characteristics of animal kingdom.

Topic: Biology, Animal Kingdom, AIIMS, AIIMS 2011
Kirchhoff's first law i.e. $\Sigma i=0$ at a junction is based on the law of conservation of

Options:
A Charge
B Energy
C Momentum
D Angular momentum

The correct answer is:
Charge

Solution:
Kirchhoff's first law is based on the law of conservation of charge.

Topic: Physics, Current Electricity, JEE Main
Kirchhoff's second law is based on the law of conservation of

Options:
A Charge
B Energy
C Momentum
D Sum of mass and energy

The correct answer is:
Energy

Solution:
Kirchhoff's second law is based on the law of conservation of energy

Topic: Physics, Current Electricity, JEE Main
Kjeldahl’s method cannot be used to estimate nitrogen for which of the following compounds?

Options:
A C6H5NH2
B CH3CH2-C≡N
C C6H5NO2
D

The correct answer is:
C6H5NO2

Solution:
Kjeldahl method is not applicable to nitro or diazo groups present in the ring, as nitrogen atom can't be converted to ammonium sulphate under the reaction conditions.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Kjeldahl's method cannot be used to estimate nitrogen for which of the following compounds?

Options:
A CH3CH2–C≡N
B C6H5NH2
C C6H5NO2
D NH 2 CONH 2

The correct answer is:
C6H5NO2

Solution:
Kjedahl's method cannot be used to test nitrogen in nitro and diazo present in ring because nitrogen in nitro and diazo groups cannot be converted into ammonium sulphate.

Topic: Chemistry, Amines, JEE Main
Kjeldahl's method was used for the estimation of nitrogen in an organic compound. The ammonia evolved from 0.55 g of the compound neutralised 12.5 mL of 1M H2SO4 solution. The percentage of nitrogen in the compound is (Nearest integer)

Options:

The correct answer is:
64

Solution:
Meq of H2SO4 used by NH3=12.5×1×2=25

% of N in the compound =25×10-3×14×1000.55=63.6

or

Meq. of H2SO4= Meq. of NH3

12.5×1×2=25meq of NH3

=25 millimoles of NH3

So Millimoles of N''=25

Moles of N''=25×10-3

wt. of N=14×25×10-3

% N=14×25×10-30.55×100

=63.66

≈64%

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (29 Jun Shift 1)
KMnO4 decomposes on heating at 513 K to form O2 along with

Options:
A MnO2& K2O2
B K2MnO4&Mn
C Mn&KO2
D K2MnO4&MnO2

The correct answer is:
K2MnO4&MnO2

Solution:
Potassium permanganate decomposes when heated, the oxidation state of manganese changes as the oxidation state of KMnO4 is +7 decomposes to potassium manganate which oxidation state is +6 and manganese dioxide its oxidation state is +4.

It is a redox reaction in which Mn which undergo both oxidation and reduction such type of reaction is known as a disproportionation reaction.

2KMnO4→513 KK2MnO4+MnO2+O2

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2024 (29 Jan Shift 1)
KMnO4 is titrated with ferrous ammonium sulphate hexahydrate in presence of dilute H2SO4. Number of water molecules produced for 2 molecules of KMnO4 is _______

Options:

The correct answer is:
68

Solution:
By balancing redox reaction:

10[FeSO4.(NH4)2SO4.6H2O]+2KMNO4+8H2SO4→5Fe2(SO4)3+2MnSO4+10(NH4)2SO4+K2SO4+68H2O

8H++MnO4-+5Fe+2⟶5Fe+3+Mn+2+4H2O

For 2 moles of MnO4-, 8 moles of H2O. are released 5 moles of Mohr's salt contain 30 moles of H2O. So, 10 moles of Mohr's salt contain 60 moles of H2O So, total moles of H2O produced=68

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2023 (13 Apr Shift 1)
KMnO4 oxidises I- in acidic and neutral/faintly alkaline solution, respectively to

Options:
A I2 and IO3-
B IO3- and I2
C IO3- and IO3-
D I2 and I2

The correct answer is:
I2 and IO3-

Solution:
In the acidic medium, MnO4- convert to Mn2+ and I- convert to I2.

2MnO4-+10I-+16H+→2Mn2++5I2+8H2O

In neutral/faintly alkaline solution, MnO4- convert to MnO2 and I- convert to IO3-.

2MnO4-+I-+H2O→2MnO2+2OH-+IO3-

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2023 (30 Jan Shift 2)
$\mathrm{KMnO}_4$ reacts with ferrous sulphate according to the equation $\mathrm{MnO}_4^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_2 \mathrm{O}$ Here $10 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{KMnO}_4$ is equivalent to

Options:
A $20 \mathrm{ml}$ of $0.1 \mathrm{~M~} \mathrm{FeSO}_4$
B $30 \mathrm{ml}$ of $0.1 \mathrm{~M~} \mathrm{FeSO}_4$
C $40 \mathrm{ml}$ of $0.1 \mathrm{~M~} \mathrm{FeSO}_4$
D $50 \mathrm{ml}$ of $0.1 \mathrm{~M~} \mathrm{FeSO}_4$

The correct answer is:
$50 \mathrm{ml}$ of $0.1 \mathrm{~M~} \mathrm{FeSO}_4$

Solution:
In this reaction
$\mathrm{MnO}_4^{-}+5 \mathrm{Fe}_5^{2+}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_2 \mathrm{O}$
5 time quantity of $\mathrm{Fe}^{2+}$ consumed.
So 5 time of $\mathrm{FeSO}_4$ will be equivalent to $50 \mathrm{ml}$

Topic: Chemistry, d and f Block Elements, JEE Main
$\mathrm{KMnO}_4$ reacts with $\mathrm{KI}$, in basic medium to form $\mathrm{I}_2$ and $\mathrm{MnO}_2$. When $250 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{KI}$ solution is mixed with $250 \mathrm{~mL}$ of $0.02 \mathrm{M}$ $\mathrm{KMnO}_4$ in basic medium, what is the number of moles of $\mathrm{I}_2$ formed?

Options:
A $0.015$
B $0.0075$
C $0.005$
D $0.01$

The correct answer is:
$0.0075$

Solution:
$\begin{array}{r}\lceil\text {-factor }=1 \longrightarrow \\ \mathrm{MnO}_4^{-}+\mathrm{I}^{-} \rightleftharpoons \mathrm{MnO}_2+\mathrm{I}_2 \\ \bigsqcup_{n \text {-factor }=3}^{\rightleftharpoons}\end{array}$
Number of milli equivalents of
$\mathrm{MnO}_4^{-}=0.02 \times 3 \times 250=15$
Number of milli equivalents of
$\mathrm{I}_2=0.1 \times 1 \times 250=25$
Thus, here $\mathrm{MnO}_4^{-}$is limiting reagent.
$\therefore$ Number of milli equivalents of $\mathrm{I}_2$ formed $=$
Number of milli equivalent of $\mathrm{MnO}_4^{-}=15$
or number of equivalent of $\mathrm{I}_2$ formed
$=\frac{15}{1000}=0.015$
$\begin{aligned}
\therefore \text { Number of moles of } \mathrm{I}_2 \text { formed } & =\frac{0.015}{2} \\
& =0.0075
\end{aligned}$

Topic: Chemistry, Redox Reactions, JEE Main
Knowing that the Chemistry of lanthanoids $(\operatorname{Ln})$ is dominated by its $+3$ oxidation state, which of the following statements in incorrect?

Options:
A
Because of the large size of the Ln (III) ions the bonding in its compounds is predominantly ionic in character.

B
The ionic sizes of Ln (III) decrease in general with increasing atomic number.

C
Ln (III) compounds are generally colourless.

D
Ln (III) hydroxides are mainly basic in character.

The correct answer is:
Ln (III) compounds are generally colourless.

Solution:
$\mathrm{Ln}^{+3}$ compounds are mostly coloured.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2009
Knowing that the chemistry of lanthanoids $(\mathrm{Ln}$ ) is dominated by its $+3$ oxidation state, which of the following statements is incorrect?

Options:
A The ionic size of Ln (III) decrease in general with increasing atomic number
B Ln (III) compounds are generally colourless.
C Ln (III) hydroxide are mainly basic in character.
D Because of the large size of the Ln (III) ions the bonding in its compounds is predominantly ionic in character.

The correct answer is:
Ln (III) compounds are generally colourless.

Solution:
Most of the $\mathrm{Ln}^{3+}$ compounds except $\mathrm{La}^{3+}$ and $\mathrm{Lu}^{3+}$ are coloured due to the presence of $f$-electrons.

Topic: Chemistry, d and f Block Elements, VITEEE, VITEEE 2017
Knowing that the chemistry of lanthanoids [Ln] is dominated by its +3 oxidation state, which of the following statements is incorrect?

Options:
A The ionic sizes of Ln-III decrease in general with increasing atomic number.
B Ln-III compounds are generally colourless.
C Ln-III hydroxides are mainly basic in character.
D Because of the large size of the Ln-III ions the bonding in its compounds is predominantly ionic in character.

The correct answer is:
Ln-III compounds are generally colourless.

Solution:
As Ln-III compounds are mostly coloured due to presence of unpaired f electrons [f-fe-transitions] are possible

Topic: Chemistry, d and f Block Elements, NEET
Knowing that the chemistry of lanthanoids(Ln) is dominated by its +3 oxidation state, which of the following statements is incorrect?

Options:
A The ionic size of Ln (III) decrease in general with increasing atomic number
B Ln (III) compounds are generally colourless.
C Ln (III) hydroxide are mainly basic in character.
D Because of the large size of the Ln (III) ions the bonding in its compounds is predominantly ionic in character.

The correct answer is:
Ln (III) compounds are generally colourless.

Solution:
Most of the $\mathrm{Ln}^{3+}$ compounds except $\mathrm{La}^{3+}$ and $\mathrm{Lu}^{3+}$ are coloured due to the presence of $f$ -electrons.

Topic: Chemistry, d and f Block Elements, JEE Main
$\mathrm{KO}_2$ (potassium super oxide) is used in oxygen cylinders in space and submarines because it

Options:
A
absorbs $\mathrm{CO}_2$ and increases $\mathrm{O}_2$ content

B
eliminates moisture

C
absorbs $\mathrm{CO}_2$

D
produces ozone.

The correct answer is:
absorbs $\mathrm{CO}_2$ and increases $\mathrm{O}_2$ content

Solution:

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2002
Kohlrausch's law states that at

Options:
A Infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
B Finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
C Infinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte
D Infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte

The correct answer is:
Infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte

Solution:
Kohlrausch's law states, "at infinite dilution each ion contributes its fixed value towards equivalent conductance irrespective of the other ion in combination with it at fixed temperature."

Topic: Chemistry, Electrochemistry, JEE Main
Kohlrausch's law states that at

Options:
A Infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
B Finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte
C Infinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte
D Infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte

The correct answer is:
Infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte

Solution:
Kohlrausch's law states, "at infinite dilution each ion contributes its fixed value towards equivalent conductance irrespective of the other ion in combination with it at fixed temperature."

Topic: Chemistry, Electrochemistry, NEET, NEET 2008 (Mains)
\(K_p\) for the conversion of oxygen to ozone at \(400 \mathrm{~K}\) is \(1.0 \times 10^{-30}\), its standard Gibbs energy change in \(\mathrm{kJ} \mathrm{mol}^{-1}\) is approximately

Options:
A 229.8
B 114.9
C -229.8
D -114.9

The correct answer is:
229.8

Solution:
Given, \(K_p\) for the conversion of oxygen to ozone at \(400 \mathrm{~K}\) is \(1.0 \times 10^{-30}\).
\(\left(\Delta G^{\circ}\right)\) standard Gibbs energy change \(=\) ?
\(\Delta G^{\circ}=-R T \ln K_p=2.303 R T \log _{10} K_p\)
Here,
\(R=8.314 \mathrm{~J} / \mathrm{mol}\)
\(\begin{aligned}
T & =400 \mathrm{~K} \\
\Delta G^{\circ} & =-R T \log K_p \\
& =-8.314 \times 400 \times 2.303 \times \log 10^{-30} \\
& =22976570 \mathrm{~J} \mathrm{~mol}^{-1} \\
& =229.765 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}\)

Topic: Chemistry, Chemical Equilibrium, JEE Main
Kraft temperature is the temperature:

Options:
A below which the aqueous solution of detergents starts freezing.
B below which the formation of micelles takes place.
C above which the aqueous solution of detergents starts boiling.
D above which the formation of micelles takes place.

The correct answer is:
above which the formation of micelles takes place.

Solution:
(i) Micelles formation take place only above a particular temperature called as kraft temperature (TK) (ii) Concentration above which micelle formation become appreciable is critical micells concentration

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2020 (06 Sep Shift 1)
Kranz anatomy is one of the characteristics of the leaves of

Options:
A potato
B wheat
C sugarcane
D mustard

The correct answer is:
sugarcane

Solution:
Sugarcane is a $\mathrm{C}_4$-plant. $\mathrm{C}_4$-plants are characterised by the presence of ' $\mathrm{Kranz}^{\prime}$ (German term meaning halo or wreath) type of anatomy. In this, the leaf mesophyll is undifferentiated and its cells (having normal type of granal chloroplasts) occur in concentric layers around vascular bundles, later is surrounded by large-sized bundle sheath cells (having large-sized agranal chloroplasts) which are arranged in a wreath like manner.

Topic: Biology, Photosynthesis in Higher Plants, NEET, NEET 2010 (Mains)
L2M1T-2 are the dimension of

Options:
A Torque
B Force
C Angular acceleration
D Angular momentum

The correct answer is:
Torque

Solution:
Dimension of torque is L2M1T-2
Torque = Force × Perpendicular distance
i.e., τ=F×r
∴ Dimension formula of τ=MLT-2L
=ML2T-2

Topic: Physics, Units and Dimensions, JEE Main
limn→∞(n+1)1/3n4/3+(n+2)1/3n4/3+.....+(2n)1/3n4/3 is equal to

Options:
A 34(2)4/3-34
B 43(2)3/4
C 43(2)4/3
D 34(2)4/3-43

The correct answer is:
34(2)4/3-34

Solution:
The given limit can be written as
limn→∞1n1+1n13+1+2n13+….+1+nn13
=limn→∞1n∑r=1n1+rn13
=∫01(1+x)1/3dx
=1+x434301
=34(24/3–1)
=34.24/3-34

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (10 Apr Shift 1)
limx→0∫0xtsin10tdtx, is equal to

Options:
A 0
B 110
C -15
D -110

The correct answer is:
0

Solution:
We have, I=limx→0∫0xtsin10tdtx

The given limiting form is the indeterminate form 00.

Thus, using Newton-Leibnitz Rule (differentiation under integral sign) and L' Hospital Rule, we get,

I=limx→0ddx∫0xtsin10tdtddxx

I=limx→0xsin10x1

=0.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2020 (08 Jan Shift 2)
limx→0 sin2x2-1+cosx equals

Options:
A 42
B 22
C 2
D 4

The correct answer is:
42

Solution:
limx→0 sin2x2-1+cosx

On Rationalisation, we get,

=limx→0 sin2x2-1+cosx×2+1+cosx2+1+cosx

=limx→0 2sinx2cosx221-cosx×2+1+cosx

=limx→0 4sin2x2cos2x22sin2x2×2+1+cosx

=limx→0 2cos2x22+1+cosx=2×22

=42.

Topic: Mathematics, Limits, JEE Main, JEE Main 2019 (08 Apr Shift 1)
limx→0x+2sinxx2+2 sin x+1 - sin2x-x+1 is

Options:
A 3
B 1
C 2
D 6

The correct answer is:
2

Solution:
limx→0x+2sinxx2+2sinx+1−sinx−x+1 00form

By using L'Hospital Rule,

=limx→01+2cosx12x+2cosx2x2+2sinx+1−1sin2x−12sin2x−x+1

=1+21+12=332=2.

Topic: Mathematics, Limits, JEE Main, JEE Main 2019 (12 Apr Shift 2)
limx→π21-tanx21-sinx1+tanx2π-2x3 is equal to

Options:
A 18
B 0
C 132
D ∞

The correct answer is:
132

Solution:
limx→π2tanπ4-x21-sinxπ-2x3
Let, x=π2+y

⇒limy→0=tan-y21-cosy-2y3
=limy→0-tany2⋅2sin2y2-8y3=limy→0132tany2y2⋅siny2y22=limy→0132×tany2(y2).limy→0siny2y22
∵ lim x→0 sinx x = lim x→0 tanx x =1
⇒ 1 32 ×1× 1 2
= 1 32

Topic: Mathematics, Limits, JEE Main
limxâ23x+33-x-123-x2-31-x is equal to

Options:

The correct answer is:
36

Solution:
Let, 3x2=t, x→2⇒t→3

limt→3t2+27t2-121t-3t2=limt→3t4+27-12t2t-3

limt→3t2-3t+3t-3t-3=6×6=36

Topic: Mathematics, Limits, JEE Main, JEE Main 2020 (07 Jan Shift 1)
limxâ23x+33-x-123-x2-31-x is equal to

Options:

The correct answer is:
36

Solution:
Let, 3x2=t, xâ2âtâ3

limtâ3t227t2-121t-3t2=limtâ3t4+27-12t2t-3

limtâ3t2-3t+3t-3t-3=6Ã6=36

Topic: Mathematics, Limits, JEE Main
limx→π4cot3x-tanxcosx+π4 is

Options:
A 42
B 82
C 4
D 8

The correct answer is:
8

Solution:
limx→π4cot3x-tanxcosx+π4

=limx→π41tan3x-tanx12cosx-12sinx ∵ cotθ=1tanθ

=limx→π41-tan4xtan3x.12cosx-12sinx

=limx→π41-tan2x1+tan2xtan3x.12cosx-12sinx

=limx→π421-tanx1+tanxsec2xtan3x.cosx-sinx ∵ 1+tan2x=sec2x & a2-b2=a-ba+b

=limx→π421-sinxcosx1+tanxsec2xtan3x.cosx-sinx

=limx→π42cosx-sinx.1+tanx.sec2xcosx.tan3x.cosx-sinx

=limx→π421+tanxcosx.sin3xcos3x.cos2x

=limx→π421+tanxsin3x

=21+tanπ4sin3π4

=8.

Topic: Mathematics, Limits, JEE Main, JEE Main 2019 (12 Jan Shift 1)
$l, m, n$ are the $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ term of a G.P. all positive, then $\left|\begin{array}{ccc}\log l & p & 1 \\ \log m & q & 1 \\ \log n & r & 1\end{array}\right|$ equals

Options:
A
-1

B
2

C
1

D
0

The correct answer is:
0

Solution:

Topic: Mathematics, Determinants, JEE Main, JEE Main 2002
$\int \frac{\sec x}{\sqrt{\log (\sec x+\tan x)}} d x=$

Options:
A $\sqrt{\log (\sec x+\tan x)}+c$
B $\sqrt{\sec x+\tan x)}+c$
C $2 \sqrt{\sec x+\tan x}+c$
D $2 \sqrt{\log (\sec x+\tan x)}+c$

The correct answer is:
$2 \sqrt{\log (\sec x+\tan x)}+c$

Solution:
Let $I=\int \frac{\sec x}{\sqrt{\log (\sec x+\tan x)}} d x$
Put $\log (\sec x+\tan x)=t \Rightarrow \frac{1}{\sec x+\tan x}\left(\sec x \tan x+\sec ^{2} x\right) d x=d t$
$\begin{aligned} & \frac{\sec x(\sec x+\tan x)}{\sec x+\tan x} \mathrm{dx}=\mathrm{dt} \Rightarrow \sec x \mathrm{dx}=\mathrm{dt} \\ \therefore \quad \mathrm{I} &=\int \frac{\mathrm{dt}}{\sqrt{\mathrm{t}}}=\int \mathrm{t}^{-\frac{1}{2}} \mathrm{dt}=\frac{\mathrm{t}^{\frac{1}{2}}}{\left(\frac{1}{2}\right)}+\mathrm{c} \\ &=2 \sqrt{\mathrm{t}}+\mathrm{c}=2 \sqrt{\log (\sec x+\tan x)}+\mathrm{c} \end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
Laboratory method for the preparation of acetyl chloride is

Options:
A $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{SOCl}_2 \rightarrow \mathrm{CH}_3 \mathrm{COCl}$
B $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{PCl}_3 \rightarrow \mathrm{CH}_3 \mathrm{COCl}$
C $\mathrm{CH}_3 \mathrm{COONa}+\mathrm{PCl}_3 \rightarrow \mathrm{CH}_3 \mathrm{COCl}$
D All of these

The correct answer is:
$\mathrm{CH}_3 \mathrm{COOH}+\mathrm{SOCl}_2 \rightarrow \mathrm{CH}_3 \mathrm{COCl}$

Solution:
$\mathrm{CH}_3 \mathrm{COOH}+\mathrm{SOCl}_2 \rightarrow \mathrm{CH}_3 \mathrm{COCl}$: is laboratory method

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Lack of independent assortment of two genes A and B in fruit fly Drosophila is due to:

Options:
A repulsion
B recombination
C linkage
D crossing over.

The correct answer is:
linkage

Solution:
Sex linkage was discovered by Thomas H. Morgan. He discovered it by conducting experiments on eye character in fruit fly, Drosophila. Independent assortment of genes does not occur during cell division.
Related Theory
Coupling and repulsion are two aspects of gene linkages. Coupling is the presence of two dominant alleles of two genes on the same chromosome $(A B)$. The remaining recessive genes of the two genes are present on the other chromosome (ab). Hence, the dominant alleles of genes tend to remain together. Repulsion is the presence of dominant genes on the two homologous chromosomes (Ab/aB).

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2004