datavorous/entrance-exam-dataset huggingface dataset:
The number of enantiomers of the compound $\mathrm{CH}_{3} \mathrm{CHBrCHBrCOOH}$ is

Options:
A 0
B 1
C 3
D 4

The correct answer is:
4

Solution:
The structure $\mathrm{CH}_{3}{ }^{*} \mathrm{CH} \mathrm{Br} \mathrm{C}^{*} \mathrm{H} \mathrm{Br} \mathrm{COOH}$

has two different chiral carbon atoms, hence number of enantiomers (optically active forms) is $2^{n}=2^{2}=4$

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of endothermic process/es from the following is _______.

A. I2( g)→2I(g)

B. HCl(g)→H(g)+Cl(g)

C. H2O(l)→H2O(g)

D. C(s)+O2( g)→CO2( g)

E. Dissolution of ammonium chloride in water

Options:

The correct answer is:
4

Solution:
The decomposition reactions are endothermic reactions. The reactions I2( g)→2I(g) and HCl(g)→H(g)+Cl(g) are decomposition reactions. The heat vaporisation is an endothermic reaction. Hence, the conversion of H2Ol→H2Og is endothermic process. The heat of combustion is an exothermic process. The heat involved in C(s)+O2( g)⟶CO2( g) is heat of combustion. The dissolution of ammonium chloride in water is an endothermic process. NH4Cl(s)→NH4+(aq)+Cl−(aq); ΔH=+15. 1kJ/mol.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (10 Apr Shift 2)
The number of essential amino acids among the following is ________
Histidine, Glycine, Valine, Alanine, Aspartic acid, Lysine, Methionine.

Options:

The correct answer is:
4

Solution:
Histidine, Valine, Lysine, Methionine

Topic: Chemistry, Biomolecules, JEE Main
The number of even numbers of four digits that can be formed using the digits 0, 1, 2, 3, 4 and 5 is

Options:
A 180
B 156
C 144
D 198

The correct answer is:
156

Solution:
(i) When 0 is at unit place
=5Ã4Ã3=60
(ii) When 0 is not at unit place
⇒4Ã4Ã3Ã2=96
(First fill unit place by either 2 or 4 i.e. two ways. Since we can not put â0â at the thousand place it can be filled in 4 ways, remaining as usual)
Total ways =60+96=156

Topic: Mathematics, Permutation Combination, JEE Main
The number of f electrons in the ground state electronic configuration of Np(Z=93) is ______ .(Integer answer)

Options:

The correct answer is:
18

Solution:
1s22 s22p63 s23p64 s23 d104p65 s24 d105p66 s24f145 d106p67 s25f46d1

Np93=Rn865f46 d17 s2

Total no. of electron in f subshell =14 in 4f and 4 in 5f subshell =18e-

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (27 Aug Shift 1)
The number of five digit numbers divisible by 5 that can be formed using the numbers $0,1,2,3,4,5$ without repetition is

Options:
A 240
B 216
C 120
D 96

The correct answer is:
216

Solution:
The five digit number is formed only when unit place is either 0 or 5 .
Case I. When 0 is in unit place, then rest of numbers can be selected in ${ }^5 P_4$ ways.
$=5 !=120$
Case II. When 5 is in unit place, then rest of numbers can be selected in ${ }^5 \mathrm{P}_4$ ways.
$=5 !=120$
And we have to subtract those cases in which 0 is in thousand place.
The number of ways $={ }^4 P_3=24$
$\therefore$ Required number of ways $=120+120-24$ $=216$

Topic: Mathematics, Permutation Combination, JEE Main
The number of five digit numbers greater than 50000 that can be formed by using all the digits 0, 1, 3, 5, 9 only once is

Options:
A 24
B 48
C 150
D 30

The correct answer is:
48

Solution:
The number of arrangements of n objects taken all a time, when repetition of the objects is not allowed is n!.

The numbers formed by 0, 1, 3, 5, 9 greater than 50000 can have 5 or 9 at the first place.

In both the case, we have four digits remaining for the rest of the four places, which can be arranged in 4!=4×3×2×1=24 ways.

Therefore, the total such numbers are 2×24=48.

Topic: Mathematics, Permutation Combination, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 2)
The number of five digit numbers that contain 7 exactly once is equal to

Options:
A 4193
B 3793
C 794
D 4194

The correct answer is:
4193

Solution:
(i) Case Ι→7X_X_X_X_
at places marked X, we can place 0 to 9, except 7, in 9×9×9×9=94 ways
(ii) Case ΙΙ→X_Y_7_Y_Y_
If 7 is at some other place, first place marked X can be filled in 8 ways ( 0 and 7 not allowed) and places marked Y can be filled in 9 ways. 7 can be placed at any of the four places in 4×8×93 ways
⇒ Required number of ways =94+32×93=4193.

Topic: Mathematics, Permutation Combination, JEE Main
The number of five-digit numbers formed with the digits 0, 1, 2, 3, 4 and 5 (without repetition) and divisible by 6 are

Options:
A 72
B 84
C 96
D 108

The correct answer is:
108

Solution:
Divisible by six means divisible by 2 as well as 3.
(i) Case-1: without 0⇒____2/4_
So, we have to put 2 or 4 in ones place.
⇒ Number of ways =4_×3_×2_×1_×2_=2×4!=48
(ii) Case-2: without 3 and 0 at ones place
____0_⇒4×3×2×1×1=4!=24
(ii) Case-2: without 3 and 2 or 4 at ones place
____2/4_⇒3×3×2×1×2=36
⇒ Total number of ways =48+24+36=108

Topic: Mathematics, Permutation Combination, JEE Main
The number of five-digit numbers, greater than 40000 and divisible by 5, which can be formed using the digits 0, 1, 3, 5, 7 and 9 without repetition, is equal to

Options:
A 132
B 120
C 72
D 96

The correct answer is:
120

Solution:
We need to find the number of 5 digit numbers greater than 40000 and divisible by 5.

Case I : Numbers start with 5.

Here if the number starts with 5 it should end with 0 to be divisible by 5.

Case II : Number starts with 7 and the last digit can have 0 or 5.

Case III : The number starts with 9 and the last digit should have 0 or 5.

Total numbers =120.

Hence this is the correct option.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (12 Apr Shift 1)
The number of five-digit telephone numbers having atleast one of their digits repeated is (00000 is also a telephone number)

Options:
A 90000
B 10000
C 30240
D 69760

The correct answer is:
69760

Solution:
Number of all five digit sequences (telephone numbers) is 10Ã10Ã10Ã10Ã10=105
When no digit is repeated, then 10Ã9Ã8Ã7Ã6=30240
â The required number wen atleast one digit is repeated =100000-30240=69760

Topic: Mathematics, Permutation Combination, JEE Main
The number of following factors which affect the percent covalent character of the ionic bond is______

A) Polarising power of cation

B) Extent of distortion of anion

C) Polarisability of the anion

D) Polarising power of anion

Options:

The correct answer is:
3

Solution:
The percentage of covalent character in an ionic bond is determined by the polarization power of the cation, the polarizing ability of the anion, and the degree of distortion of the anion's electron cloud that occurs due to the electric field of the cation. The polarization power of a cation refers to its ability to distort the electron cloud of the anion in the ionic bond. So the options A,B,C are correct.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The number of four letter words that can be formed using the letters of the word BARRACK is

Options:
A
144

B
120

C
264

D
270

The correct answer is:
270

Solution:
If all four letters are different then the number of words ${ }^5 \mathrm{C}_4 \times 4 !=120$
If two letters are $\mathrm{R}$ and other two different letters are chosen from $\mathrm{B}, \mathrm{A}, \mathrm{C}, \mathrm{K}$ then the number of words $={ }^4 \mathrm{C}_2 \times \frac{4 !}{2 !}=72$
If two letters are $\mathrm{A}$ and other two different letters are chosen from $\mathrm{B}, \mathrm{R}, \mathrm{C}, \mathrm{K}$ then the number of words $={ }^4 \mathrm{C}_2 \times \frac{4 !}{2 !}=72$
If word is formed using two $R$ 's and two $A$ 's then the number of words $=\frac{4 !}{2 ! 2 !}=6$
Therefore, the number of four-letter words that can be formed $=120+72+72+6=270$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The number of four-digit numbers formed by using the digits 0, 2, 4, 5 and which are not divisible by 5, is

Options:
A 10
B 8
C 6
D 4

The correct answer is:
8

Solution:
No. of four-digit numbers divisible by 5=6+4=10
_ _ _ 0_ â3!=6
_ _ _ 5_ â 3!-2!=4
Total four-digit numbers =4!-3!=18
â´ Required cases=18-10=8

Topic: Mathematics, Permutation Combination, JEE Main
The number of four-digit numbers strictly greater than 4321 that can be formed using the digit 0,1,2,3,4,5 (repetition of digits is allowed) is:

Options:
A 360
B 288
C 306
D 310

The correct answer is:
310

Solution:
∴ Total Numbers =216+72+18+4=310

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The number of functions $f$ from $\{1,2,3, \ldots, 20\}$ onto $\{1,2,3, \ldots, 20\}$ such that $f(k)$ is a multiple of $3,$ whenever $k$ is a multiple of 4 is:

Options:
A $6^{5} \times(15) !$
B $5 ! \times 6 !$
C $(15) ! \times 6 !$
D $5^{6} \times 15$

The correct answer is:
$(15) ! \times 6 !$

Solution:
Domain and codomain $=\{1,2,3, \ldots, 20\}$.

There are five multiple of 4 as 4,8,12,16 and 20 .

and there are 6 multiple of 3 as 3,6,9,12,15,18 .

Since, when ever $k$ is multiple of 4 then $f(k)$ is multiple of 3

then total number of arrangement

$={ }^{6} c_{5} \times 5 !=6 !$

Remaining 15 elements can be arranged in $15 !$ ways.

Since, for every input, there is an output

$\Rightarrow$ function $f(k)$ in onto

$\therefore \quad$ Total number of arrangement $=15 ! .6 !$

Topic: Mathematics, Functions, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The number of functions f, from the set A=x∈N:x2-10x+9≤0 to the set B=n2:n∈N such that fx≤x-32+1, for every x∈A, is _______.

Options:

The correct answer is:
1440

Solution:
For set A

x2-10x+9≤0⇒x-1x-9≤0

⇒x∈1,9

i.e. A=1,2,3,4,5,6,7,8,9

Given B=n2:n∈N

Now fx≤x-32+1

For x=1: f1≤5⇒B∈12,22

For x=2:f2≤2⇒B∈12

For x=3:f3≤1⇒B∈12

For x=4:f4≤2⇒B∈12

For x=5:f5≤5⇒B∈12,22

For x=6:f6≤10⇒B∈12,22,32

For x=7: f7≤17⇒B∈12,22,32,42

For x=8:f8≤26⇒B∈12,22,32,42,52

For x=9:f9≤37⇒B∈12,22,32,42,52,62

Total number of such function

=2×1×1×1×2×3×4×5×6=2720=1440

Topic: Mathematics, Functions, JEE Main, JEE Main 2022 (27 Jul Shift 2)
The number of functions that can be defined from the set $\mathrm{A}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}\}$ into the set $B=\{1,2,3\}$ is

Options:
A 12
B 24
C 64
D 81

The correct answer is:
81

Solution:
Given, $\mathrm{A}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}\}$ and $\mathrm{B}=\{1,2,3\}$
Element a can be have an image in 3 ways, i.e., ( 1,2 , or 3$)$.
Similarly b, $c$ and $d$ can have an image in 3 ways.
Thus total number of ways $=3 \times 3 \times 3 \times 3 \times=81$.
Let set $A$ have $a$ elements and set $B$ have $b$ elements. Each element in $A$ has $b$ choices to be mapped to. Each such choice gives you a unique function. Since each element has $b$ choices, the total number of functions from $A$ to $B$ is
$\underbrace{b \times b \times b \times \cdots b}_{a \text { times }}=b^a$

Topic: Mathematics, Functions, JEE Main
The number of geometric isomers of the complex CrNH33Cl3 are

Options:

The correct answer is:
2

Solution:
CrNH33Cl3 will form only two geometrical isomers:

Topic: Chemistry, Coordination Compounds, JEE Main
The number of geometric isomers possible for the complex [CoL2Cl2]-

(L=H2NCH2CH2O-) is

Options:

The correct answer is:
5

Solution:

Topic: Chemistry, Coordination Compounds, JEE Main
The number of geometric isomers that can exist for square planar PtClpyNH3NH2OH+ is (py = pyridine):

Options:
A 6
B 2
C 3
D 4

The correct answer is:
3

Solution:
Complex with general formula [Mabcd]+ square planar complex can have three isomers.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2015 (04 Apr)
The number of geometrical isomers found in the metal complexes PtCl2NH32, Ni(CO)4, RuH2O3Cl3 and CoCl2NH34+ respectively, are:

Options:
A 1, 1, 1, 1
B 2, 0, 2, 2
C 2, 1, 2, 2
D 2, 1, 2, 1

The correct answer is:
2, 0, 2, 2

Solution:

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (27 Jul Shift 1)
The number of geometrical isomers of CoNH33NO23 are

Options:
A 4
B 3
C 2
D 0

The correct answer is:
2

Solution:
The complex exhibits two geometrical isomers (i) facial and (ii) meridional.

Topic: Chemistry, Coordination Compounds, JEE Main
The number of geometrical isomers of the compound C6H5-CH=CH-CH=CH-COOH is

Options:

The correct answer is:
4

Solution:
The number of geometrical isomers of the compound C6H5-CH=CH-CH=CH-COOH is given as
n = 2 (number of double bonds)
Total GI=2n=22=4

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of geometrical isomers possible for the complex [Pt(NH3)(NH2OH)(Py)Cl]+ is

Options:

The correct answer is:
3

Solution:
Complex is square planar

Topic: Chemistry, Coordination Compounds, JEE Main
The number of geometrical isomers possible in triamminetrinitrocobalt (III) is X and in trioxalatochromate (III) is Y. Then the value of X+Y is ____.

Options:

The correct answer is:
2

Solution:
Triamminetrinitrocobalt (III) →CoNO23NH33
trioxalatochromate(III) ion →CrC2O433-
CoNO23NH33

CrC2O433-

X+Y=2+0=2.0

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (27 Jul Shift 1)
The number of given orbitals which have electron density along the axis is

px,py,pz,dxy,dyz,dxz,dz2,dx2-y2

Options:

The correct answer is:
5

Solution:
The orbitals oriented along axis have electron density along the axis.

Orientation of 2p orbitals are as follows:

Orientation of dz2& dx2-y2:
,
Thus, px,py,pz,dz2& dx2-y2 are axial orbitals.

The orbitals dxy,dyz and dxz are oriented between the axis.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (25 Jan Shift 2)
The number of given statement/s which is/are correct is_____

(A) The stronger the temperature dependence of the rate constant, the higher is the activation energy.

(B) If a reaction has zero activation energy, its rate is independent of temperature.

(C) The stronger the temperature dependence of the rate constant, the smaller is the activation energy.

(D) If there is no correlation between the temperature and the rate constant then it means that the reaction has negative activation energy.

Options:

The correct answer is:
2

Solution:
Rate constant is given by Arrhenius equation

k=Ae-Ea/RT

Higher the magnitude of activation energy, stronger is the temperature dependence of the rate constant. The pre-exponential factor is a measure of the rate at which collisions occur, irrespective of their energy. If the activation energy of reaction is zero, temperature will have no effect on the rate constant.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The number of greenhouse gases in the following is ________.
CO 2 , O 2 , N 2 O, CH 4 ,CFCs,CO, H 2 O g , O 3

Options:

The correct answer is:
6.00

Solution:
Greenhouse gases that absorbs and emits radiant energy within a thermal infrared range which causes an increase in atmospheric temperature. The list of gases are water H 2 O , CO 2 ,â CH 4 ,â N 2 O,â O 3 â &â CFCs

Topic: Chemistry, Environmental Chemistry, JEE Main
The number of greenhouse gases in the following is \( \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{~N}_{2} \mathrm{O}, \mathrm{CH}_{4} \), CFCs, \( \mathrm{CO}, \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{O}_{3} \) Report your answer to the nearest integer.

Options:

The correct answer is:
6

Solution:
Greenhouse gases that absorbs and emits radiant energy within a thermal infrared range which causes an increase in atmospheric temperature. The list of gases are water H2O, CO2, CH4, N2O,  CFCs and O3

Topic: Chemistry, Environmental Chemistry, JEE Main
The number of groups present in a tripeptide Asp – Glu – Lys is ____

Options:

The correct answer is:
5

Solution:
Asp-Glu-Lys pripeptide is :

No. of CO group =5

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2020 (03 Sep Shift 2)
The number of groups that can be made from 5 different green balls, 4 different blue balls and 3 different red balls, if at least 1 green and 1 blue ball is to be included

Options:
A $3700$
B $3720$
C $4340$
D None of these

The correct answer is:
$3720$

Solution:
Find the required number of groups.
Three groups of 5 different green balls, 4 different blue balls, and 3 different red balls are given.
The total number of groups that can be formed out of given groups are \(2^{12}\).
The total number of groups having no blue balls is \(2^8\).
The total number of groups having no green balls is \(2^7\).
The total number of groups having no green balls and no blue balls is \(2^3\).
So, the total number of groups with at least 1 green and 1 blue balls is given by \(2^{12}-\left(2^7+2^8-2^3\right)=3720\).
Therefore, The number of groups that can be made from 5 different green balls, 4 different blue balls, and 3 different red balls, if at least 1 green and 1 blue balls are to be included is 3720.

Topic: Mathematics, Permutation Combination, JEE Main
The number of halobenzenes from the following that can be prepared by Sandmeyer's reaction is _______

Options:

The correct answer is:
2

Solution:
In Sandmayer reaction only bromobenzene & chlorobenzene are prepared

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2024 (05 Apr Shift 1)
The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorous acid is

Options:
A
zero

B
two

C
one

D
three

The correct answer is:
two

Solution:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2005
The number of hydrogen atoms present in $25.6$ $\mathrm{g}$ of sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ which has a molar mass of $342.3 \mathrm{~g}$ is

Options:
A $22 \times 10^{23}$
B $9.91 \times 10^{23}$
C $11 \times 10^{23}$
D $44 \times 10^{23}$

The correct answer is:
$9.91 \times 10^{23}$

Solution:
No. of moles of a compound
$=\frac{\text { given mass }(\mathrm{gm})}{\text { Molar mass }(\mathrm{gm})}$
i.e. $\frac{25.6}{342.3}=0.0748$ moles.
1 mole of sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ contains $6.022$ $\times 10^{23}$ molecules of it.
Hence $0.0748$ moles contains
$\begin{array}{l}
=6.022 \times 10^{23} \times 0.0748 \\
=0.4504 \times 10^{23} \text { molecules. }
\end{array}$
1 molecule of sucrose by formula is having 22 atoms of hydrogen.
$\therefore 0.4504 \times 10^{23} \times 10^{23} \times 22=9.91 \times 10^{23}$ atoms of hydrogen.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The number of hydroxide ions, produced by one molecule of sodium carbonate $\left(\mathrm{Na}_2 \mathrm{CO}_3\right)$ on hydrolysis is

Options:
A 2
B 1
C 3
D 4

The correct answer is:
2

Solution:
$\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{CO}_3$
and $2 \mathrm{NaOH} \rightleftharpoons \mathrm{Na}^{+}+2 \mathrm{OH}^{-}$
ionisation
Hence, it is clear that $2 \mathrm{OH}^{-}$ions will be formed on hydrolysis of one molecule of sodium corbonate.

Topic: Chemistry, Practical Chemistry, JEE Main
The number of hydroxyl group(s) in Q is

Options:

The correct answer is:
4

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The number of hydroxyl ions in $10 \mathrm{~cm}^{3}$ of $0.2 \mathrm{M}$ HCl solution is

Options:
A $5 \times 10^{-14}$
B $3 \times 10^{9}$
C $3 \times 10^{12}$
D $5 \times 10^{-12}$

The correct answer is:
$3 \times 10^{12}$

Solution:
$\left[\mathrm{H}^{+}\right]=0.2 \mathrm{M}$ (given, $\mathrm{HCl}=0.2 \mathrm{M}$ )
$\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}$
$\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{0.2}=5 \times 10^{-14} \mathrm{M}$
Concentration of hydroxyl ions in $0.2 \mathrm{M} \mathrm{HCl}$
$$
=5 \times 10^{-14} \mathrm{M}
$$
$\therefore$ The number of moles of hydroxyl ions
$$
=5 \times 10^{-14} \times 1000
$$
The number of $\mathrm{OH}^{-}$ in $10 \mathrm{~cm}^{3}$
$$
\begin{array}{l}
=\frac{5 \times 10^{-14} \times 1000 \times 6.023 \times 10^{23}}{10} \\
=3 \times 10^{12}
\end{array}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The number of hyperconjugating structures shown by the carbocations are given below. Which one is not correctly matched?

Options:
A ; 9 hyperconjugating structures
B CH3-C+H-CH3 ; 8 hyperconjugating structures
C CH3-C+H2 ; 3 hyperconjugating structures
D C+H3 ; No hyperconjugating structures

The correct answer is:
CH3-C+H-CH3 ; 8 hyperconjugating structures

Solution:
As CH3-C+H-CH3 shows 6 hyperconjugating structures. So, option B is incorrect. Rest all are correct.

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of hyperconjugation structures involved to stabilize carbocation formed in the above reaction is __________.

Options:

The correct answer is:
7

Solution:
In the first step of the reaction, protonation of alcohol takes place and carbocation is formed after removal of water. The carbocation formed undergo rearrangement to get stability. In the given, methyl shift takes place to convert secondary carbocation to tertiary carbocation. The alpha hydrogen to the positive charged carbon are involved in hyperconjugation.

So, number of hyperconjugation structures in most stable carbocation

= 6 + 1 = 7

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The number of immigration is more than emigration and death are lower than natality. Growth curve of the population will show:

Options:
A Exponential phase
B Lag phase
C Declining phase
D Steady phase

The correct answer is:
Exponential phase

Solution:
In any population, when the sum of immigration and natality is more than the sum of emigration and death rate, then the overall population or population density is increased. Human population show a sigmoid growth curve in which mainly 4 phases are there - a Lag phase - growth is comparatively slow. b Exponential phase - Growth is very high and continuously increasing. c Declining phase - growth is decreased. d Steady phase - growth is zero or stationary.

Topic: Biology, Organisms and Populations, NEET
The number of incorrect statement/s about the black body from the following is______

(A) Emit or absorb energy in the form of electromagnetic radiation.

(B) Frequency distribution of the emitted radiation depends on temperature.

(C) At a given temperature, intensity vs frequency curve passes through a maximum value.

(D) The maximum of the intensity vs frequency curve is at a higher frequency at higher temperature compared to that at lower temperature.

Options:

The correct answer is:
0

Solution:
A black body is an idealised object that absorbs all the radiation that falls on it and emits radiation over a broad range of frequencies.

Black bodies emit and absorb energy in the form of electromagnetic radiation.

The frequency distribution of the emitted radiation from a black body depends on its temperature and is given by Planck's law.

At a given temperature, the intensity vs frequency curve for a black body radiation passes through a maximum value. This frequency is known as the peak frequency or the frequency of maximum emission.

At higher temperatures, the peak frequency of the black body radiation shifts towards higher frequencies (shorter wavelengths). This means that the maximum of the intensity vs frequency curve is at a higher frequency at higher temperature compared to that at lower temperature.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2023 (10 Apr Shift 1)