datavorous/entrance-exam-dataset huggingface dataset:
Consider the molecules CH4, NH3 and H2O . Which of the given statements is false?

Options:
A The H - C - H bond angle in CH4 , the H - N - H bond angle in NH3 , and the H - O - H bond angle in H2O are all greater than 90o .
B The H - O - H bond angle in H2O is larger than the H - C - H bond angle in CH4 .
C The H - O - H bond angle in H2O is smaller than the H - N - H bond angle in NH3 .
D The H - C - H bond angle in CH4 is larger than the H - N - H bond angle in NH3 .

The correct answer is:
The H - O - H bond angle in H2O is larger than the H - C - H bond angle in CH4 .

Solution:
Bond Angle	Molecule

104.5o	H2O

107o	NH3

109o28′	CH4

All the molecules are sp3 hybridized and Bond angle of H2O is smaller than NH3 because in H2O two lone pairs are present due to which there will be more lone pair-lone pair repulsion while inNH3,only one lone pair is present, so lone pair-bond pair repulsion is present only.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2016 (Phase 1)
Consider the nitration of benzene using mixed conc. H2SO4 and HNO3 . If a large amount of KHSO4 is added to the mixture, the rate of nitration will be:

Options:
A Faster
B Slower
C Unchanged
D Doubled

The correct answer is:
Slower

Solution:
Q= Product Reactant as product increase Q>K
Q has became equal to K product has to decrease so equilibrium will shift backward direction.
If we add KHS O 4 − ,  conc.HS O 4 − increases, equilibrium shifts backward.

Topic: Chemistry, Hydrocarbons, NEET, NEET 2016 (Phase 1)
Consider the nuclear reaction X200→A110+B80+10n1. If the binding energy per nucleon for X, A and B are 7.4 MeV, 8.2 MeV and 8.1 MeV respectively, then the energy released in the reaction -

Options:
A 70 MeV
B 200 MeV
C 190 MeV
D 10 MeV

The correct answer is:
70 MeV

Solution:
For X: energy E1=200×7.4=1480MeV
For A: energy E2=110×8.2=902MeV
For B: energy E3=80×8.1=648MeV
Energy Released E=E1+E2+E3
E=902+648-1480=70MeV

Topic: Physics, Nuclear Physics, NEET
Consider the nuclear reaction X200âA110 +B80. If the binding energy per nucleon for X, A and B are 7.4 MeV, 8.2 MeV and 8.1 MeV respectively, then the energy released in the reaction is

Options:
A 70 MeV

B 200 MeV
C 190 MeV
D 10 MeV

The correct answer is:
70 MeV

Solution:
For X, energy = 200× 7.4 = 1480 MeV
For A, energy = 110× 8.2 = 902 MeV
For B, energy = 80×8.1 = 648 MeV
Therefore energy released
=(902+648)-1480
= 1550–1480=70 MeV

Topic: Physics, Nuclear Physics, NEET
Consider the nuclear reaction X200âA110+B80. If the binding energy per nucleon for X,A and B are 7.4 MeV,8.2 MeV and 8.1 MeV, respectively, then the energy released in the reaction is

Options:
A 70 MeV
B 200 MeV
C 190 MeV
D 10 MeV

The correct answer is:
70 MeV

Solution:
For X : binding energy =200Ã7.4=1480 MeV
For A : binding energy =110Ã8.2=902 MeV
For B : binding energy =80Ã8.1=648 MeV
â´ Energy released =902+648-1480
=1550-1480
=70 MeV

Topic: Physics, Nuclear Physics, NEET
Consider the pyramid of energy of an ecosystem given below:
If $T_4$ is equivalent to 1000 J , what is the value at $T_1$ ?

Options:
A $\frac{10000}{10} \mathrm{~J}$
B $\frac{10000}{10} \times 4 \mathrm{~J}$
C $10,000 \mathrm{~J}$
D $10,00,000 \mathrm{~J}$

The correct answer is:
$10,00,000 \mathrm{~J}$

Solution:
According to the given pyramid of energy, if $\mathrm{T}_4$ is equivalent to 1000 J , then according to the 10 percent law, which states that only 10 percent of energy is transferred to each trophic level from the lower trophic level, the value at $\mathrm{T}_1$ is $10,00,000 \mathrm{~J}$.
$\begin{aligned} & T_1=10,00,000 \mathrm{~J} \\ & T_2=\frac{10}{100} \times 10,00,000=1,00,000 \mathrm{~J} \\ & T_3=\frac{10}{100} \times 100000=10,000 \mathrm{~J} \\ & T_4=\frac{10}{100} \times 10000=1000 \mathrm{~J}\end{aligned}$

Topic: Biology, Ecosystem, NEET, NEET 2024 (Re-NEET)
Consider the reaction
2NO2(g)+O3(g)âN2O5(g)+O2(g)
The reaction of nitrogen dioxide and ozone represents is first order in NO2(g) and in O3(g) . Which of these possible reaction mechanism is consistent with the rate law?

Mechanism I:	NO 2 (g) + O 3 (g)â NO 3 (g) +â O 2 (g)â(slow)

NO 3 (g) + NO 2 (g)â N 2 O 5 (g)â(fast)

Mechanism II:	O 3 (g)â O 2 (g) + [O]â(fast)

NO 2 (g) + [O] â NO 3 â(slow)

NO 3 (g) + NO 2 (g)â N 2 O 5 â(fast)

Options:
A Only I
B Only II
C Both I and II
D Neither I nor II

The correct answer is:
Both I and II

Solution:
Mechanism I: Rate=k[NO2][O3] Slow step is rate determining
Mechanism II: Rate = k â² [ N O 2 ][ O ] â¦(i)
From 1st step K=[O2][O][O3] (k = eq. constant)
[O]â=âk O 3 O 2 ...(ii)
From equation (i) & (ii)
Rate= k ′ [ N O 2 ][ O 3 ] [ O 2 ] −1
So both mechanism show that reaction is first order with respect to NO2 and O3 .

Topic: Chemistry, Chemical Kinetics, NEET
Consider the reaction 2 NO(g) +  O 2 (g)→2  NO 2 (g) ; Predict whether the reaction is spontaneous at 298 K. (Given:   Δ f G NO  = 86.69 kJ/mol,   Δ f G  NO 2  = 51.84 kJ/mol )

Options:
A Yes, spontaneous
B No, the reaction is Non-spontaneous
C Equilibrium
D Cannot predict

The correct answer is:
Yes, spontaneous

Solution:
For the reaction :
2âNO(g)â+â O 2 (g)âââ2â NO 2 (g)
ΔG =  2Δ f G NO 2 −  2Δ f G NO  +  Δ f G O 2
= 2 Ã 5 1 . 8 4 - 2 Ã 8 6 . 6 9 + 0
= 1 0 3 . 6 8 - 1 7 3 . 3 8
= - 6 9 . 7 kJ/mol
âµÎG is negative
so, the reaction is spontaneous

Topic: Chemistry, Thermodynamics (C), NEET
Consider the reaction A â 2 B + C, Î H = - 1 5 kcal . The energy of activation of backward reaction is 20 kcal mol-1. In presence of catalyst, the energy of activation of forward reaction is 3 kcal mol-1. At 400 K the catalyst causes the rate of the forward reaction to increase by the number of times equal to-

Options:
A e3.5
B e2.5
C e-2.5
D e2.303

The correct answer is:
e2.5

Solution:
E a f - E a b = Î H = - 1 5 kcal
â E a f = - 1 5 + 2 0 = 5 kcal
k catalyst k a = e E a - E a catalyst / RT
= e 5 - 3 Ã 1 0 3 / 2 Ã 4 0 0
= e 2 . 5

Topic: Chemistry, Chemical Kinetics, NEET
Consider the reaction

CH3CH2CH2Br+NaCN→CH3CH2CH2CN+NaBr

This reaction will be the fastest in

Options:
A Ethanol
B Methanol
C N, N' - dimethylformamde (DMF)
D Water

The correct answer is:
N, N' - dimethylformamde (DMF)

Solution:
The reaction proceeds via SN2 mechanism. It is a unimolecular nucleophilic substitution reaction. The reaction is hence favoured in the presence of a polar aprotic solvent. DMF is a polar aprotic solvent. So, the reaction is the fastest with DMF solvent.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET, NEET 2016 (Phase 2)
Consider the reaction

CH3CH2CH2Br+NaCN→CH3CH2CH2CN+NaBr

This reaction will be the fastest in

Options:
A Ethanol
B Methanol
C N, N' - dimethylformamde (DMF)
D Water

The correct answer is:
N, N' - dimethylformamde (DMF)

Solution:
∵ The reaction is SN2, most suitable solvent should be polar aprotic i.e., DMF.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
Consider the reaction,

Cr 2 O 7 2 - + 1 4 H + + 6 e - ⟶ 2 Cr 3 + + 7 H 2 O

What is the quantity of electricity in coulombs needed to reduce 1 mole of Cr 2 O 7 2 - ?
(Given: 1F ≈ 96500 C)

Options:
A 5.79 x 105
B 5.69 x 105
C 5.59 x 105
D 5.49 x 105

The correct answer is:
5.79 x 105

Solution:
1 mole Cr 2 O 7 2 - require 6 Faraday electricity (6 F)
= 6 x 96500 C = 5.79 x 105 C

Topic: Chemistry, Electrochemistry, NEET
Consider the reaction Cr2O72-+ 14H+6e-→ 2Cr3++ 7H2O. What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr2O72-?

Options:
A 960000 C
B 579000 C
C 192000 C
D 1920000 C

The correct answer is:
579000 C

Solution:
Correct Option is : (B)
579000 C

Topic: Chemistry, Electrochemistry, NEET
Consider the reaction,
(i) PCl5(g)⇌PCl3(g)+Cl2(g)
(ii) N2O4(g)⇌2NO2(g)
The addition of an inert gas at constant volume

Options:
A will increase the dissociation of PCl5 as well as N2O4
B will reduce the dissociation of PCl5 as well as N2O4
C will increase the dissociation of PCl5 and step up the formation of NO2
D will not disturb the equilibrium of the reactions

The correct answer is:
will not disturb the equilibrium of the reactions

Solution:
The addition of an inert gas at constant volume will not disturb the equilibrium of the reactions.

Topic: Chemistry, Chemical Equilibrium, NEET
Consider the reaction mechanism

A 2 ⇌ K eq 2A fast (where A is the intermediate)
A+B→k1P (slow)

The rate law for the reaction is

Options:
A k1[A][B]
B k 1 K 1/2 A 2 1/2 B
C k 1 K 1/2 A B
D k1K1/2[A]2[B]

The correct answer is:
k 1 K 1/2 A 2 1/2 B

Solution:
Over all reaction will be
A 2 +2B→2P
here 'A' is intermediate
A 2 ⇌ K eq 2A fast
Keq=[A]2/[A2]
[A]=Keq[A2]
A+B→k1(slow)→RDS
RateRDS=k1[A][B]
=k1Keq[A2].[B]
Note: Overall reaction is A2+2B→2P .

Topic: Chemistry, Chemical Kinetics, NEET
Consider the reaction: $\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightarrow$ $2 \mathrm{NH}_{3(g)}$
The equality relationship between $\frac{d\left[\mathrm{NH}_3\right]}{d t}$ and $-\frac{d\left[\mathrm{H}_2\right]}{d t}$ is:

Options:
A $\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{d\left[\mathrm{H}_2\right]}{d t}$
B $\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{1}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}$
C $+\frac{d[\mathrm{NH}]_3}{d t}=-\frac{2}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}$
D $+\frac{d\left[\mathrm{NH}_3\right]}{d t}=-\frac{3}{2} \frac{d\left[\mathrm{H}_2\right]}{d t}$

The correct answer is:
$+\frac{d[\mathrm{NH}]_3}{d t}=-\frac{2}{3} \frac{d\left[\mathrm{H}_2\right]}{d t}$

Solution:
For the reaction
$$
\begin{aligned}
\mathrm{N}_2(g)+3 \mathrm{H}_2(g) & \rightarrow 2 \mathrm{NH}_3(g) \\
\frac{-d\left[\mathrm{~N}_2\right]}{d t} & =-\frac{1}{3} \frac{d}{d t}\left[\mathrm{H}_2\right] \\
& =\frac{1}{2} \frac{d}{d t}\left(\left[\mathrm{NH}_3\right]\right) \\
\therefore \frac{d}{d t}\left(\left[\mathrm{NH}_3\right]\right) & =-\frac{2}{3} \frac{d}{d t}\left[\mathrm{H}_2\right]
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2006
Consider the reaction of water with F2and suggest, in terms of oxidation and reduction, which species are oxidised/reduced in this process?

Options:
A Oxygen in water is oxidized, and fluorine in F2is reduced.
B Fluorine in F2is oxidized, and oxygen in water is reduced.
C Both oxygen in water and fluorine in F2are oxidized.
D Both oxygen in water and fluorine in F2are reduced.

The correct answer is:
Oxygen in water is oxidized, and fluorine in F2is reduced.

Solution:
Correct Option is : (A)
Oxygen in water is oxidized, and fluorine in F2is reduced.

Topic: Chemistry, Hydrogen, NEET
Consider the reaction

The alkene formed in major amount is

Options:
A
B
C
D

The correct answer is:

Solution:
It is dehydration of alcohol by E1 mechanism through carbocation intermediates. Carbocation formed undergoes rearrangement and produces more substituted alkene as major product.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Consider the reactions
12N2+O2⇌NO2 K1
2NO2⇌N2O4 K2
Using above equations, write down expression for K of the following reaction
N2O4⇌N2+2O2 K

Options:
A K1K2
B K22K1
C 1K1K22
D 1K12K2

The correct answer is:
1K12K2

Solution:
K1=[NO2][N2]1/2[O2] …(i)
K2=[N2O4][NO2]2 …(ii)
K=[N2][O2]2[N2O4] …(iii)
Considering equation (i), (ii) and (iii)
K=1K12K2 .

Topic: Chemistry, Chemical Equilibrium, NEET
Consider the reactions: (a) 6 CO2(g) + 6H2O (l) → C6H12O6(aq) + 6O2(g) (b) O3(g) + H2O2(l) → H2O (l) + 2O2(g) Why is it more appropriate to write these reactions as: (i) 6CO2(g) + 12H2O (l) → C6H12O6(aq) + 6H2O (l) + 6O2(g) (ii) O3(g) + H2O2(l) → H2O (l) + O2(g) + O2(g)

Options:
A Because the modified equations conserve mass more accurately.
B Because the modified equations involve fewer reactants.
C Because the modified equations increase the yield of products.
D Because the modified equations reduce the number of chemical species involved.

The correct answer is:
Because the modified equations conserve mass more accurately.

Solution:
Correct Option is : (A)
Because the modified equations conserve mass more accurately.

Topic: Chemistry, Redox Reactions, NEET
Consider the reactions and arrange the correct sequence with respect to glycolysis

i. Formation of fructose-1,6-bisphosphate
ii. Dehydrogenation
iii. Cleavage to form triose-3-phosphate
iv. Second substrate-level phosphorylation
v. Dehydration

Options:
A i → ii → iii → iv → v
B i → iii → ii → v → iv
C i → ii → iii → v → iv
D i → iii → ii → iv → v

The correct answer is:
i → iii → ii → v → iv

Solution:
Glycolysis is the metabolic pathway that converts glucose, C6H12O6, into pyruvate, CH3COCOO− + H+. This pathway can be thought of as comprising three stages. Stage 1, which is the conversion of glucose into fructose-1,6-bisphosphate, consists of three steps: phosphorylation, isomerization and a second phosphorylation reaction. The strategy of these initial steps in glycolysis is to trap the glucose in the cell and form a compound that can be readily cleaved into phosphorylated three-carbon units. Stage 2 is the cleavage of the fructose-1,6-bisphosphate into two three-carbon fragments. These resulting three-carbon units are readily interconvertible. In stage 3, ATP is harvested when the three-carbon fragments are oxidized to pyruvate.

Topic: Biology, Respiration in Plants, NEET
Consider the reactions:
âââââââ
Identify A, X, Y and Z.

Options:
A A - Methoxyymethane, X - Ethanoic acid, Y - Acetate ion, Z - hydrazine.
B A - Methoxymethane, X - Ethanol, Y - Ethanoic acid, Z - Semicarbazide.
C A - Ethanal, X - Ethanol, Y - But-2-enal, Z - Semicarbazone.
D A - Ethanol, X - Acetaldehyde, Y - Butanone, Z - Hydrazone.

The correct answer is:
A - Ethanal, X - Ethanol, Y - But-2-enal, Z - Semicarbazone.

Solution:
Since 'A' gives positive silver mirror test therefore, it must be an aldehyde or α -Hydroxyketone.
Reaction with semicarbazide indicates that A can be an aldehyde or ketone.
Reaction with OH- i.e., aldol condensation (by assuming alkali to be dilute) indicates that A is aldehyde as aldol reaction of ketones is reversible and carried out in special apparatus.
These indicate option (iii).
âââââââ

Topic: Chemistry, Aldehydes and Ketones, NEET, NEET 2017
Consider the reactions

Identify A, X, Y and Z

Options:
A A-Methoxymethane, X-Ethanol, Y-Ethanoic acid, Z-Semicarbazone
B A-Ethanal, X-Ethanol, Y-But-2-enal, Z-Semicarbazone
C A-Ethanol, X-Acetaldehyde, Y-Butanone, Z-Hydrazone
D A-Methoxymethane, X-Ethanoic Acid, Y-Acetate ion, Z-Hydrazine

The correct answer is:
A-Ethanal, X-Ethanol, Y-But-2-enal, Z-Semicarbazone

Solution:

Topic: Chemistry, Aldehydes and Ketones, NEET
Consider the shown glass slab having a thin biconvex cavity of air centred at O. A black spot at P will have image ( Î¼ glass = 3 2 , the radius of curvature of each surface of the cavity = 10 cm)

Options:
A 7.5 cm on the right side of O
B 7.5 cm on the left side of O
C 3.75 cm on the right side of O
D 3.75 cm on the left side of O

The correct answer is:
3.75 cm on the left side of O

Solution:
1f=μ-1110+110
= 2 3 - 1 2 1 0
=-115 ⇒ f=-15
This lens will behave like a biconcave lens.
So, u = -5 cm
1v-1u=1f
⇒ 1 v - 1 - 5 = 1 - 1 5
-1v=115+15=415
⇒ v=-3.75 cm

Topic: Physics, Ray Optics, NEET
Consider the situation shown in figure. If the current I in the long straight wire xy is increased at a steady rate the induced current in loop A and B will be

Options:
A Clockwise in A and anticlockwise in B
B Anticlockwise in A and clockwise in B
C Clockwise in both A and B
D Anticlockwise in both A and B

The correct answer is:
Clockwise in A and anticlockwise in B

Solution:
Magnetic field in A will be outward and in B it will be inward, given by right hand thumb rule. So to oppose increase in magnetic field in loop in respective direction current induced will be clockwise and anticlockwise respectively.

Topic: Physics, Electromagnetic Induction, NEET
Consider the situation shown in figure. If the switch is closed and after some time it is opened again, the closed loop will show

Options:
A an anticlockwise current-pulse
B a clockwise current-pulse
C an anticlockwise current-pulse and then a clockwise current-pulse
D a clockwise current-pulse and then an anticlockwise current-pulse

The correct answer is:
a clockwise current-pulse and then an anticlockwise current-pulse

Solution:
When switch is closed current in circuit increases suddenly and hence flux in the loop increases in outward direction so clockwise current-pulse is induced.
When switch is opened current in circuit decreases suddenly and hence flux in the loop decreases in outward direction so anticlockwise current-pulse is induced.

Topic: Physics, Electromagnetic Induction, NEET
Consider the situation shown in the figure. If the current I in the long straight wire XY is increased at a steady rate then the induced EMFs in loops A and B will be

Options:
A clockwise in A, anticlockwise in B
B anticlockwise in A, clockwise in B
C clockwise in both A and B
D anticlockwise in both A and B

The correct answer is:
clockwise in A, anticlockwise in B

Solution:
Both coils oppose the increase in current in straight wire, due to which change in magnetic flux is increased through the two coils. Hence current in coil A is clockwise and current in coil B is anticlockwise

Topic: Physics, Electromagnetic Induction, NEET
Consider the statement- “The electronegativity of Nitrogen on the Pauling scale is 3.0 in all the nitrogen compounds” This statement is-

Options:
A Correct
B Incorrect
C The Pauling scale is inaccurate
D Cannot be decided without further information

The correct answer is:
Incorrect

Solution:
Correct Option is : (B)
Incorrect

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Consider the statements for given reaction

1. In the first step, there is breaking of C-O bond
2. The second step is SN2 reaction
3. There is retention of configuration in the first step
4. There is inversion of configuration in the second step

Options:
A 1, 2 and 3 are correct
B 2, 3 and 4 are correct
C 1, 3 and 4 are correct
D 1, 2, 3 and 4 are correct

The correct answer is:
2, 3 and 4 are correct

Solution:
Alcohol forms ester with TsCl in which there is breaking of O-H bond.
Second step is a S N 2reaction I − as nucleophile

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Consider the statements given below regarding contraception and answer as directed thereafter:

(1) Vasectomy involves cutting down of vas deferens in females.
(2) Generally, chances of conception are nil until the mother breast-feeds the infant for a period of maximum up to six months.
(3) Intrauterine devices like copper-T are very effective contraceptives.
(4) Emergency contraceptive pills may be taken up to one week after coitus to prevent conception.

Which of the following two statements are correct?

Options:
A 1,3
B 1,2
C 2,3
D 3,4

The correct answer is:
2,3

Solution:
In a vasectomy, a small part of the vas deferens is removed or tied up through a small incision on the scrotum, whereas in tubectomy, a small part of the fallopian tube is removed or tied up through a small incision in the abdomen or through the vagina.
Lactational amenorrhea (absence of menstruation) method is based on the fact that ovulation and therefore the menstrual cycle does not occur during the period of intense lactation following parturition. Therefore, as long as the mother breast-feeds the child fully, chances of conception are almost nil. However, this method has been reported to be effective only up to a maximum period of six months following parturition.
Intra-uterine devices are presently available as the non-medicated IUDs (e.g., Lippes loop), copper releasing IUDs (CuT, Cu7, Multiload 375) and the hormone releasing IUDs (Progestasert, LNG-20). IUDs increase phagocytosis of sperms within the uterus and the Cu ions released suppress sperm motility and the fertilizing capacity of sperms.
Administration of progestogens or progestogen-estrogen combinations or IUDs within 72 hours of coitus has been found to be very effective as emergency contraceptives as they could be used to avoid possible pregnancy due to rape or casual unprotected intercourse.

Topic: Biology, Reproductive Health, NEET
Consider the statements given below regarding contraception and answer as directed thereafter
A. Medical Termination of Pregnancy (MTP) during first trimester is generally safe
B. Generally chances of conception are nil until mother breast-feeds the infantupto two year
C. Intrauterine devices like copper-T are effective contraceptives
D. Contraception pills may be taken upto one week after coitus to prevent conception
Which two of the above statements are correct?

Options:
A B, C
B C, D
C A, C
D A, B

The correct answer is:
A, C

Solution:
The most used contraceptive product in the united states is the birth control pill. Many combination birth control pills (marketed under various names including ovral, tetragynon and triphasil) are also being used for post coital contraception. Mifepristone is an antihormone that when taken during the first seven weak of pregnancy in conjunction with a tiny amount of prostaglandin to stimulate uterine contraction induces miscarriage by blocking progesterone's quieting effect on the uterus.
For several years the second most used contraceptive method was the intrauterine device a plastic or metal device inserted into the uterus that prevented the fertilized egg from implanting in the endometrial lining.

Topic: Biology, Reproductive Health, NEET, NEET 2008 (Screening)
Consider the three solutions of 1 M concentration
1. Sodium acetate (CH3COONa)
2. Acetic acid + Sodium acetate (CH3COOH+CH3COONa)
3. Acetic acid (CH3COOH)
The pH of these solutions will be lie in the following sequence

Options:
A 3 < 2 < 1
B 2 < 1 < 3
C 1 < 2 < 3
D 3 < 1 < 2

The correct answer is:
3 < 2 < 1

Solution:
The pH of these solutions will be lie in the following sequence

Topic: Chemistry, Ionic Equilibrium, NEET
Consider the vehicle emitting sound wave of frequency 700 Hz moving towards an observer at a speed 22 m s-1. Assuming the observer, as well as the medium to be at rest and velocity of sound in the medium to be 330 m s-1, the frequency of sound as measured by the observer is

Options:
A 25254 Hz
B 19603 Hz
C 22403 Hz
D 750 Hz

The correct answer is:
750 Hz

Solution:
Given,
Speed of sound source =22 m s-1
Frequency of emitted sound =700 Hz
Velocity of sound =330 m s-1
When the sound source is moving the apparent frequency of the sound heard by the observer
n′=nvv-vs
n′=700330330-22
=700330308
=749.99
≃750 Hz

Topic: Physics, Waves and Sound, NEET
Consider third orbit of He+ (helium), using non-relativistic approach, the speed of electron in this orbit will be given constant K=9×109, Z=2 and h (Planck's Constant)= 6.6 × 10−34 J s

Options:
A 2.92×106 m s-1
B 1.46×106 m s-1
C 0.73×106 m s-1
D 3.0×108 m s-1

The correct answer is:
1.46×106 m s-1

Solution:
Energy of electron He+
$E_n$=-13.6Z2n2 eV; Z=2, n=3
E3= -13.6 232×1.6×10-19 J
In Bohr's model energy, E=- KE
E3=-KE3
∴ 9.7×10-19=12mv2
v= 2×9.7×10-199.1×10-31
Velocity of electron in third orbit of He+ is 1.46×106 m s-1.

Topic: Physics, Atomic Physics, NEET, NEET 2015 (Phase 1)
Consider this flow chart and a list of properties given after that.

1. Molecular weight ranging from 18-800 daltons approx.
2. Proteins, nucleic acids, polysaccharides, and lipids.
3. Chemicals more than 800 daltons molecular weight.
4. Has monomers.
5. It generally has polymers.

Which of these are applicable to A and B?

Options:
A A:Â 1,2,3Â andÂ B:Â 4,5
B A:Â 2,4Â andÂ B:Â 1,3,5
C A: 1,4 and B: 2,3,5
D A:Â 1,3,5Â andÂ B:Â 2,4

The correct answer is:
A: 1,4 and B: 2,3,5

Solution:
The given procedure is about chemical analysis of the living matter. The filtrate is the acid-soluble pool while the retentate is the acid-insoluble pool. The acid-soluble pool has biomicromolecules containing low molecular weight molecules having a molecular weight between 18–800 daltons. They are generally monomers such as amino acids, monosaccharides, etc. The acid-insoluble pool has biomacromolecules containing high molecular weight molecules having a molecular weight above 800 daltons. They are generally polymers such as carbohydrates, proteins, nucleic acids, etc.

Topic: Chemistry, Biomolecules, NEET