datavorous/entrance-exam-dataset huggingface dataset:
Total number of elements which do not form hydrides are
Mo, Ca, Fe, Pd, Co, Ru, W, Cr

Options:

The correct answer is:
5

Solution:
Mo, Fe, Co, Ru, W
Metals of group 7,8 and 9 do not form hydride. From group 6 only Cr forms hydride.

Topic: Chemistry, s Block Elements, JEE Main
Total number of enol possible for the compound formed during given reaction will be (including stereoisomer)
CH 3 MgBr (1 equivalent)  +  CH 3 CH 2 COCl→

Options:

The correct answer is:
3

Solution:
2-Butanone will show tautomerism.
Total enol products = 1 + 2 = 3.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Total number of essential amino acid among the given list of amino acids is _______
Arginine, Phenylalanine, Aspartic acid, Cysteine, Histidine, Valine, Proline

Options:

The correct answer is:
4

Solution:
Essential Amino acids are :-
Arginine, Phenylalanine, Histidine, Valine

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2024 (09 Apr Shift 1)
Total number of four digit odd numbers that can be formed using $0,1,2,3,5,7$ are

Options:
A $216$
B $375$
C $400$
D $720$

The correct answer is:
$720$

Solution:
we have the number \(0,1,2,3,5,7\)
Now the digit should be odd and hence last digit should be filled with an odd number
\(\Rightarrow\) Number of way to filled last number \(=4\) (i.e. \(1,3,5,7\) )
\(\Rightarrow\) Number of way to filled third digit \(=6\) (i.e. o,1, 2, 3, 5, 7)
\(\Rightarrow\) Number of way to filled second digit \(=6\) (i.e. o,1, 2, 3, 5, 7)
\(\Rightarrow\) Number of way to filled first digit \(=5\) (i.e.1, \(2,3,5,7\) )
\(\Rightarrow\) Total 4 digit number \(=4 \times 6 \times 6 \times 5\)
\(=720\)

Topic: Mathematics, Permutation Combination, JEE Main
Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are

Options:
A
216

B
375

C
400

D
720

The correct answer is:
720

Solution:
Required number of numbers $=5 \times 6 \times 6 \times 4=36 \times 20=720$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2002
Total number of geometrical isomers for the complex [RhCl(CO)(PPh3)(NH3)] is -

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Rh+ will form a square planar complex.
[M(abcd)] complex is square planar, so it will have three geometrical isomers.

Topic: Chemistry, Coordination Compounds, JEE Main
Total number of inner transition elements are: -
W,Ru,U,Tc,La,Yb,Po,No

Options:

The correct answer is:
3

Solution:
U,Yb,No

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Total number of ionisation isomers of Coen2Br ClNO3 are (including the given compound)

Options:
A 3
B 4
C 5
D 9

The correct answer is:
3

Solution:
Three ionisation isomers
(1) Coen2BrCl NO3
(2) [Coen2BrNO3]Cl
(3) [Coen2(NO3)(Cl)]Br

Topic: Chemistry, Coordination Compounds, JEE Main
Total number of ions from the following with noble gas configuration is

Sr2+(Z=38),Cs+(Z=55),La2+(Z=57)Pb2+(Z=82),Yb2+(Z=70) and Fe2+(Z=26)

Options:

The correct answer is:
2

Solution:
Noble gas configuration= ns2np6

38Sr2+=[36Kr]

Cs+=[54Xe]

Yb702+=[54Xe]4f14

57La2+=[54Xe]5 d1

82Pb2+=[54Xe]4f145 d106 s2

26Fe2+=[18Ar]3d6

Cs+ and Sr2+ have noble gas configuration, which are Xe and Kr, respectively.

Hence, the answer is 2.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Total number of isomeric compounds (including stereoisomers) formed by monochlorination of 2-methylbutane is ______.

Options:

The correct answer is:
6

Solution:
Monochlorination means the substitution of a chlorine. When chlorine atom is added to the below structure, chlorine will substitute at one of the hydrogen.

Mono chlorination of 2-methyl butane gives four structural isomers.

Here only two, i.e., I and III are optically active. Since each optically active compound has two enantiomers, therefore, total six isomeric compounds are possible.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Total number of isomers (including stereoisomers) obtain on monochlorination of methylcyclohexane is

Options:

The correct answer is:
12

Solution:
Compound A and B given above have no chiral centre and doesn't show geometrical isomerism.

Compound C and D given above have two chiral centres each and each of them forms 4 optical isomers.

And compound E has 2 E and Z isomers due to restricted rotation of groups.

Hence, the total number of isomers will be 12.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Total number of lines in Lyman series \( \mathrm{H} \) of spectrum will be (where, \( \mathrm{n}= \) number of orbits).

Options:
A \( \mathrm{n} \)
B \( \mathrm{n}-1 \)
C \( \mathrm{n}-2 \)
D \( \mathrm{n}(\mathrm{n}+1) \)

The correct answer is:
\( \mathrm{n}-1 \)

Solution:
There are many spectral lines formed by the hydrogen spectrum. All the lines are formed in the visible region in the Lyman series. It comes in the ultraviolet region.The total number of lines in Lyman series H of spectrum will be (where, n= number of orbits) n-1.

Topic: Chemistry, Structure of Atom, JEE Main
Total number of lines of the form px+qy+r=0 (where pqâ 0) that intersect the ellipse x22+y2=50 at two integral points (points whose coordinates are integers) is equal to

Options:

The correct answer is:
54

Solution:
If y∈I then y∈-7,-6,....,0,1,...,6,7
Total integral points on x22+y2=50 are 12 which are
2,7,14,1,2,-7-14,1,-2,7,14,-1-2,-7,-14,-1,10,510,-5,-10,5-10,-5
Total number of lines passing through at least two points among these are 12C2=66 in which x=2,x=-2, x=14, x=-14, x=10, x=-10, y=1, y=7, y=-1, y=-7, y=5, y=-5 are rejected.
⇒ Total number of lines are 66-12=54.

Topic: Mathematics, Permutation Combination, JEE Main
Total number of lines touching atleast two circles of the family of four circles x2+y2±3x±8y=0 is

Options:
A 8
B 10
C 12
D 14

The correct answer is:
14

Solution:
Let, C1 be circle x2+y2+8x+8y=0,
C2 be circle x2+y2+8x-8y=0,
C3 be circle x2+y2-8x+8y=0
& C4 be circle x2+y2-8x+8y=0
Now, common tangents between C1 & C2 is 2 ,
common tangents between C1 & C3 is 2 ,
common tangents between C1 & C4 is 3 ,
common tangents between C2 & C3 is 3 ,
common tangent between C2 & C4 is 2
& common tangent between C3 & C4 is 2
Total =14
C1 & C4 touch each other, C2 & C4 touch each other, rest pair for circles intersects each other.

Topic: Mathematics, Circle, JEE Main
Total number of lone pair of electrons in I3- ion is:

Options:
A 12
B 3
C 6
D 9

The correct answer is:
9

Solution:
The structure of I3- ion is:

As we can see central atom has 3 lone pair and two bond pair and outer atoms has 3 lone pair each. So, there are total of 9 lone pair. Central atom will have sp3d hybridisation and linear shape.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2018 (08 Apr)
Total number of low spin complexes are
[Fe(CN)6]3–,[Co(NO2)6]3–,[FeF6]3– , [IrF6]3–,[Co(NH3)6]3+ , [Co(H2O)6]+2,[FeCl4]2–,[Mn(H2O)6]3+

Options:

The correct answer is:
4

Solution:
[FeF 6 ] 3– ,  [Co(H 2 O) 6 ] +2 ,  [FeCl 4 ] 2– ,  [Mn(H 2 O) 6 ] 3+ are high spin complexes and rest of them are low spin complexes.

Topic: Chemistry, Coordination Compounds, JEE Main
Total number of meiotic divisions required for forming 100 zygotes/100 grains of wheat is

Options:
A 50
B 125
C 75
D 100

The correct answer is:
125

Solution:
The zygote is formed by the fusion of a haploid male and female gamete; Thus to determine the number of total meiotic division to produce 100 zygotes the total number of meiotic divisions in males and females should be considered.
In males: One meiotic division forms 4 male gametes, thus 25 meiotic division gives 100 male gametes.
In females: One meiotic division forms 1 egg cell, as the remaining 3 degenerates to provide nutrition to the developing egg cell. Thus 100 meiotic division gives 100 egg cells.
Hence, the total number of meiotic divisions needed to form 100 zygotes = 25 meiotic division in males + 100 meiotic divisions in females = 125.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Total number of molecules with pyramidal shape are:

NH3, ClF3, SO3, PCl3, XeO3, BCl3, ClO3–, SO32–

Options:
A 5
B 4
C 3
D 6

The correct answer is:
5

Solution:
NH 3 ,  PCl 3 ,  XeO 3 – ,  SO 3 2– , ClO 3 –

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Total number of moles of AgCl precipitated on addition of excess of AgNO3 to one mole each of the following complexes CoNH34Cl2Cl, NiH2O6Cl2,PtNH32Cl2 and PdNH34Cl2 is

Options:

The correct answer is:
5

Solution:
Silver nitrate on reaction of chloride ion gives silver chloride as precipitate. Chloride ions outside of coordination sphere will only give precipitates. Thus,
CoNH34Cl2Cl⇒ Gives 1 mole AgCl
NiH2O6Cl2⇒ Gives 2 moles AgCl
PtNH32Cl2⇒ Gives No AgCl
PdNH34Cl2⇒ Gives 2 moles AgCl
Total number of moles of AgCl=5 mole.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Total number of optically active compounds from the following is ______

Options:

The correct answer is:
1

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Total number of optically inactive stereoisomers of succinic acid are

Options:

The correct answer is:
6

Solution:
Only Meso form of succinic acid is optically inactive due to presence of plane of symmetry.

Topic: Chemistry, General Organic Chemistry, JEE Main
Total number of oxygen atom(s) which act(s) as a bridge between any two silicon atoms in a mineral with composition
\( \mathrm{MM}^{\prime} \mathrm{Si}_{3} \mathrm{O}_{\mathrm{x}}\left(\mathrm{M}=\right. \) divalent metal ion and \( \mathrm{M}^{\prime}= \) tetravalent metal ion \( ) . \)

Options:

The correct answer is:
3

Solution:
The (SiO4)4- tetrahedral units are linked together provides a convenient classification of the silicate and Si3O96- is cyclic silicates.

In which, 2 oxygen atoms per tetrahedral are shared, ring structure may be formed of general formula (SiO3)n2n-. Ring containing 3 4, 6 or 8 tetrahedral units are known but 3 & 6 are the most common.

The Cyclic ions Si3O96- occurs in benitoite Ba2+Ti 4+[Si3O9].

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Total number of P-O-P bonds present in P4O10 is

Options:

The correct answer is:
6

Solution:
Structure of P4O10 is
Each phosphorus is attached to 4 oxygen atoms.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Total number of possible stereoisomers of dimethyl cyclopentane is

Options:

The correct answer is:
6

Solution:
Dimethyl cyclopentane

will show stereo isomerism, Its stereo isomers are

will show stereo isomerism, Its stereo isomers are

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (27 Jun Shift 1)
Total number of products form including stereoisomers during monochlorination of 2-methyl butane?

Options:
A 2
B 6
C 4
D 8

The correct answer is:
6

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Total number of relatively more stable isomer(s) possible for octahedral complex Cuen2SCN2 will be____

Options:

The correct answer is:
3

Solution:
This octahedral complex is of M(AA)2b2 type structure in which AA is bidentate ligand and b is monodentate ligand. This can exist in two geometrical isomeric form cis and trans. Out of which trans form is relatively more stable isomer. Now due to ambidentate ligand SCN- it can also shows linkage isomersim.

Cuen2SCN2

So, relatively more stable trans isomers will have three possible isomers.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2022 (28 Jul Shift 1)
Total number of species from the following which can undergo disproportionation reaction ___________.
H2O2,ClO3-,P4,Cl2,Ag,Cu+1, F2,NO2, K+

Options:

The correct answer is:
6

Solution:
The compounds or ions or elements in which central atom has intermediate oxidation state undergo a disproportionation reaction.

Disproportionation reactions are the reactions in which the same element/compound gets oxidised and reduced simultaneously.

1)Cu+ → Cu2+ + Cu

2) 2H2O-12→H2O-2+O02

3) 3 Cl2 + 6 OH− → 5Cl− + ClO3−+ 3 H2O

4) 2NO2(g)+2OH−(aq)→NO2−(aq)+H2O(l)+NO3−

5) P4+3OH-+3H2O→PH3+3H2PO2-

Here, the chlorine reactant is in oxidation state 0, and if we look at the products, the chlorine in the Cl− ion has an oxidation number of −1, where it has undergone reduction. Meanwhile, the oxidation number of the chlorine in the ClO3- ion is +5, which means it has been oxidised.

Hence, H2O2, Cu+, Cl2,ClO3,P4,NO2 shows disproportionation reaction.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Total number of species from the following with central atom utilising \(\mathrm{sp}^2\) hybrid orbitals for bonding is _____.
\(\mathrm{NH}_3, \mathrm{SO}_2, \mathrm{SiO}_2, \mathrm{BeCl}_2, \mathrm{C}_2 \mathrm{H}_2, \mathrm{C}_2 \mathrm{H}_4, \mathrm{BCl}_3, \mathrm{HCHO}, \mathrm{C}_6 \mathrm{H}_6, \mathrm{BF}_3, \mathrm{C}_2 \mathrm{H}_4 \mathrm{Cl}_2\)

Options:

The correct answer is:
6

Solution:
Central atom utilising $\mathrm{sp}^2$ hybrid orbitals
$\mathrm{SO}_2, \mathrm{C}_2 \mathrm{H}_4, \mathrm{BCl}_3, \mathrm{HCHO}, \mathrm{C}_6 \mathrm{H}_6, \mathrm{BF}_3$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Total number of stereo isomers possible for the given structure :

Options:
A 2
B 4
C 3
D 8

The correct answer is:
8

Solution:
There are three stereo center So No of stereoisomer $=2^3=8$

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Total number of stereoisomers of the given compound

will be

Options:

The correct answer is:
4.00

Solution:
Total number of stereo isomers = 2 n−1
=  2 3−1
=  2 2
=4
Here, n = number of chiral centres

Topic: Chemistry, General Organic Chemistry, JEE Main
Total number of substances, which contain hexagonal planar rings in their structures graphite, \( (\mathrm{BN})_{x}, \mathrm{~B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6}, \mathrm{C}_{6} \mathrm{H}_{6} \),
\( \mathrm{B}_{2} \mathrm{H}_{6}, \mathrm{H}_{3} \mathrm{P}_{3} \mathrm{O}_{9} \)

Options:

The correct answer is:
4

Solution:
From the below structures, we can see that all are six member rings, except B2H6 and in case of H3P3O9, hybridisation of P is sp3 , which is not a planar structure. All others are six member planar structures.

sp2 hybridisation with planar structure.

sp2 hybridisation with planar structure.

sp2 hybridisation with planar structure.

sp2 hybridisation with planar structure.

sp3 hybridisation of each phosphorous.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Total number of unpaired electrons in the complex ions \(\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}\) and \(\left[\mathrm{NiCl}_4\right]^{2-}\) is \(\qquad\)

Options:

The correct answer is:
2

Solution:
$\begin{aligned} & \mathrm{Co}^{+3}: 3 \mathrm{~d}^6 \quad \mathrm{t}_{2 \mathrm{~g}}^{2,2,2} \mathrm{e}_{\mathrm{g}}^{0,0} \\ & \text { Unpaired e } \mathrm{e}^{-}=0 \\ & \mathrm{Ni}^{+2}: 3 \mathrm{~d}^8 \quad \mathrm{e}^{2,2} \mathrm{t}_2^{2,1,1} \\ & \text { Unpaired } \mathrm{e}^{-}=2\end{aligned}$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Total numbers of 3-digit numbers that are divisible by 6 and can be formed by using the digits 1, 2, 3, 4, 5 with repetition, is ________

Options:

The correct answer is:
16

Solution:
A number is divisible by 6 when it is divisible by 2 and 3 both.

The three digit number can be a¯b¯2¯.

a+b can be 4,7,10

If a+b=4 then (a, b) can be (1,3), (3,1), (2,2)

If a+b=7 then (a, b) can be(2,5),(5,2),(3,4),(4,3)

If a+b=10 then (a, b) can be (5,5)

So, 8 such cases when 2 is at unit's place

Similarly, there exist 8 such cases when 4 is at unit's place.

Total =16 cases.

Hence this is the required answer.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Total volume of atoms present in a face-centre cubic unit cell of a metal is ( $r$ is atomic radius)

Options:
A
$\frac{20}{3} \pi r^3$

B
$\frac{24}{3} \pi r^3$

C
$\frac{12}{3} \pi r^3$

D
$\frac{16}{3} \pi r^3$

The correct answer is:
$\frac{16}{3} \pi r^3$

Solution:
$\begin{aligned} V & =n \times\left(\frac{4}{3} \pi r^3\right) \\ & =4 \times\left(\frac{4}{3} \pi r^3\right) \\ & =\frac{16}{3} \pi r^3\end{aligned}$

Topic: Chemistry, Solid State, JEE Main, JEE Main 2006
Tracheids are type of water-conducting cell in the xylem which lacks perforations in the cell wall found in all the following plants, except

Options:
A a flowerless green plant of the division Pteridophyte, which comprises the ferns and their relatives
B a plant of a group that comprises those that have seeds unprotected by an ovary or fruit, including the conifers, cycads, and ginkgo.
C a small flowerless green plant of the division Bryophyta, which comprises the mosses and liverworts.
D a plant of a large group that comprises those that have flowers and produce seeds enclosed within a carpel

The correct answer is:
a small flowerless green plant of the division Bryophyta, which comprises the mosses and liverworts.

Solution:
The main difference between bryophytes and tracheophytes is that the bryophytes are nonvascular plants while the tracheophytes are the vascular plants. Bryophyte is called atracheophyte.

Topic: Biology, Plant Kingdom, NEET
Tracheids differ from the tracheary elements in:

Options:
A Having casparian strips
B Being imperforate
C Lacking nucleus
D Being lignified

The correct answer is:
Being imperforate

Solution:
The tracheary elements comprise tracheids and vessels. The main function is the conduction of sap. Tracheids differ from other tracheary elements in being imperforate, comparatively short, and single-celled. While the vessels are longer and multi celled.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2014
Train A is moving along two parallel rail tracks towards north with 72 km h-1 and train B is moving towards south with speed 108 km h-1. Velocity of train B with respect to A and velocity of ground with respect to B are (in m s-1):

Options:
A -30 and 50
B -50 and -30
C -50 and 30
D 50 and -30

The correct answer is:
-50 and 30

Solution:
VA=72 km h-1=72×10003600=20 m s-1 and

VB=-108 km h-1=-108×10003600=-30 m s-1

Velocity of B w.r.t. A:

VB/A=-30-20=– 50 m s-1

Velocity of ground w.r.t. B:

VG/B=0--30=30 m s-1

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Transfer characteristic [output voltage $\left(V_0\right)$ $v s$ input voltage $\left.\left(V_i\right)\right]$ for a base biased transistor in CE configuration is as shown in the figure. For using transistor as a switch, it is used

Options:
A in region III
B both in region (I) and (III)
C in region II
D in region I

The correct answer is:
both in region (I) and (III)

Solution:
In the given graph,
Region (l) Cutoff region
Region (II) - Active region
Region (III)-Saturation region
Using transistor as a switch it is used in cutoff region or saturation region.
Using transistor as a amplifier it is used in active region.

Topic: Physics, Semiconductors, JEE Main
Transgenic plants are the ones -

Options:
A Grown in artificial medium after hybridization in the field
B Produced by a somatic embryo in artificial medium
C Generated by introducing foreign DNA in to a cell and regenerating a plant from that cell
D Produced after protoplast fusion in artifical medium

The correct answer is:
Generated by introducing foreign DNA in to a cell and regenerating a plant from that cell

Solution:
No Solution Available

Topic: Biology, Strategies for Enhancement in Food Production, NEET, NEET 2009 (Mains)
Transition metal complex with highest value of crystal field splitting Δ0 will be

Options:
A MoH2O63+
B CrH2O63+
C OsH2O63+
D FeH2O63+

The correct answer is:
OsH2O63+

Solution:
H2O is a weak field ligand. Crystal field stabilisation energy is maximum for d3 and d8 cases with weak field ligands among the 3d-series elements.But CFSE increases from 3d to 5d series elements due to the increase in nuclear charge. Hence, OsH2O63+ complex case, the CFSE is maximum.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2022 (24 Jun Shift 2)
Transition metals form the most stable complex ions as

Options:
A they exhibit multiple oxidation states
B they have large charge to radius ratio
C vacant orbitals to accommodate electrons are present
D all of these

The correct answer is:
all of these

Solution:
The transition metal form the most stable complex because,

(1) Their charge to size ratio is very high, so they can attract lone pairs from the ligands.

(2) They have vacant orbitals which can accommodate lone pairs of the ligands.

(3) They can show variable oxidation states.

Topic: Chemistry, Coordination Compounds, JEE Main
Transition metals make the most efficient catalysts because of their ability to

Options:
A adopt multiple oxidation states and to form complexes
B form coloured ions
C show paramagnetism due to unpaired electrons
D form a large number of oxides

The correct answer is:
adopt multiple oxidation states and to form complexes

Solution:
The transition metals and their compounds are known for their catalytic activity because of their ability to adopt multiple oxidation states and to form complexes.

Topic: Chemistry, d and f Block Elements, JEE Main
Transitional elements are named transition elements because their characters are

Options:
A In between $s$ and $p$-block elements
B Like that of $p$ and $d$-block elements
C They are members of $/-A$ group
D They are like inactive elements

The correct answer is:
In between $s$ and $p$-block elements

Solution:
In between $s$ and $p$-block elements.

Topic: Chemistry, d and f Block Elements, JEE Main
Transport of food material in higher plants takes place through

Options:
A companion cells
B transfusion tissue
C tracheids
D sieve elements

The correct answer is:
sieve elements

Solution:
Sieve elements are the food conducting cells of phloem tissue. Sieve element in angiosperms are sieve tubes, while in gymnosperms and pteridophytes, these are sieve cells.

Topic: Biology, Transport in Plants, NEET, NEET 2010 (Mains)
Transverse waves of the same frequency are generated in two steel wires $A$ and $B$. The diameter of $A$ is twice that of $B$ and the tension in $A$ is half that in $B$. The ratio of the velocities of the waves in $A$ and $B$ is

Options:
A
$1: 2$

B
$1: \sqrt{2}$

C
$1: 2 \sqrt{2}$

D
$3: 2 \sqrt{2}$

The correct answer is:

$1: 2 \sqrt{2}$

Solution:
The velocity of transverse waves is given by $v=\sqrt{\frac{T}{m}}$ where, $T=$ Tension and $m=$ mass per unit length of the wire. If $r$ is the radius of the wire and $\rho$ its density then,
$m=\pi r^2 \rho$
$\therefore \quad v=\sqrt{\frac{T}{m}}=\sqrt{\frac{T}{\pi r^2 \rho}}$
$\therefore \quad v_A=\frac{\sqrt{T_A}}{r_A \sqrt{\pi \rho}}$
and $\quad v_B=\frac{\sqrt{T_B}}{r_B \sqrt{\pi \rho}}$
Now, $\quad \frac{v_A}{v_B}=\sqrt{\frac{T_A}{T_B}} \cdot \frac{r_B}{r_A}$
$\because \quad r_A=2 r_B$ and $T_A=\frac{1}{2} T_B$
Hence, $\quad \frac{v_A}{v_B}=\frac{1}{2 \sqrt{2}}$

Topic: Physics, Waves and Sound, JEE Main
Treatment of ammonia with excess of ethyl chloride will yield

Options:
A Diethyl amine
B Ethane
C Tetraethyl ammonium chloride
D Methyl amine

The correct answer is:
Tetraethyl ammonium chloride

Solution:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl} \xrightarrow{\mathrm{NH}_3} \mathrm{C}_2 \mathrm{H}_5-\mathrm{NH}_2 \xrightarrow{\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}}\left(\mathrm{C}_2 \mathrm{H}_5\right)_2-\mathrm{NH}$
If $\mathrm{NH}_3$ is in excess, then $1^{\circ}$ amine will be the main product, if $\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}$ is in excess then mixture of $1^{\circ}, 2^{\circ}, 3^{\circ}$ and quaternary amine is obtained.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Treatment of $\text { with } \mathrm{NaNH}_{2} \text { /liq. } \mathrm{NH}_{3}$ gives

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Trichloroacetaldehyde, CCl3CHO reacts with chlorobenzene in presence of sulphuric acid and produces

Options:
A
B
C
D

The correct answer is:

Solution:
It gives D.D.T. (p, p-dichlorodiphenyl-ltrichloroethane)

Topic: Chemistry, Chemistry in Everyday Life, JEE Main