datavorous/entrance-exam-dataset huggingface dataset:
$\varepsilon_0$ and $\mu_0$ are the electric permittivity and magnetic permeability of free space respectively. If the corresponding quantities of a medium are $2 \varepsilon_0$ and $1.5 \mu_0$ respectively, the refractive index of the medium will nearly be

Options:
A $3$
B $2$
C $\sqrt{2}$
D $\sqrt{3}$

The correct answer is:
$\sqrt{3}$

Solution:
We know
$\begin{aligned} & \text { Refractive index }=\frac{\sqrt{\mu \varepsilon}}{\sqrt{\mu_0 \varepsilon_0}} \\ & =\frac{\sqrt{(1.5) \mu_0(2) \varepsilon_0}}{\sqrt{\mu_0 \varepsilon_0}} \\ & =\sqrt{3}\end{aligned}$

Topic: Physics, Ray Optics, NEET, NEET 2023 (Manipur)
1 : 1 : 1 : 1 ratio of progenies can be obtained if the plants employed for crossing are

A. TTRR Ã ttRR
B. TtRr Ã ttrr
C. TtRR Ã ttrr
D. Ttrr Ã ttRr

Options:
A A,Â CÂ andÂ D
B A,Â B,Â CÂ andÂ D
C B and D
D AÂ andÂ B

The correct answer is:
B and D

Solution:
1 : 1 : 1 : 1 is a dihybrid test cross ratio. It is obtained when crossing is done between a hybrid and a recessive genotype, when two gene loci are considered simultaneously.

Topic: Biology, Principles of Inheritance and Variation, NEET
1, 2-dibromopropane, when heated with Zn dust in ethanol, gives

Options:
A propane
B propene
C propyne
D ethyne

The correct answer is:
propene

Solution:
1, 2-dibromopropane, when heated with Zn dust in ethanol, gives propene

Topic: Chemistry, Hydrocarbons, NEET
1 amu (atomic mass unit) mass is equal to energy of

Options:
A 931 keV
B 93.1 eV
C 931âMeV
D 9.31 MeV

The correct answer is:
931 MeV

Solution:
1 amu=1.66×10-27 kg

Equivalent energy can be calculated by using mass-energy equivalence equation
E=mc2
E=1.66×10-27×3×1082 J

1 MeV=106×1.6×10-19 J

Using above equations we can get,

E= 931 MeV

Topic: Physics, Atomic Physics, NEET
Ï1 and Ï2 are susceptibility of a paramagnetic material at temperatures T1 K and T2 K respectively, then

Options:
A χ1=χ2
B χ1T1=χ2T2
C χ1T2=χ2T1
D χ1T1 =χ2T2

The correct answer is:
χ1T1=χ2T2

Solution:
The susceptibility is inversely proportional to the temperature of substance.
∴X∝ 1 T
ÏT=constant
â´ Ï1â1T1,Ï2â1T2
Ï1T1=Ï2T2

Topic: Physics, Magnetic Properties of Matter, NEET
α -1 antitrypsin is:

Options:
A An antacid
B An enzyme
C Used to treat arthritis
D Used to treat emphysema

The correct answer is:
Used to treat emphysema

Solution:
α -1 antitrypsin is used to treat emphysema.

Topic: Biology, Biotechnology and Its Applications, NEET
1 g of water, of volume $1 \mathrm{~cm}^3$ at $100^{\circ} \mathrm{C}$ is converted into steam at same temperature under normal atmospheric pressure $=\left(\simeq 1 \times 10^5 \mathrm{~Pa}\right)$. The volume of steam formed equals $1671 \mathrm{~cm}^3$. If the specific latent heat of vaporisation of water is $2256 \mathrm{~J} / \mathrm{g}$, the change in internal energy is

Options:
A $2423 \mathrm{~J}$
B $2089 \mathrm{~J}$
C $167 \mathrm{~J}$
D $2256 \mathrm{~J}$

The correct answer is:
$2089 \mathrm{~J}$

Solution:
Given, mass of water, $\mathrm{m}=\mathrm{lg}$
Volume of $1 \mathrm{~g}$ of water $=1 \mathrm{~cm}^3=10^{-6} \mathrm{~m}^3$
Volume of $1 \mathrm{~g}$ of steam $=1671 \mathrm{~cm}^3=1671 \times 10^{-6} \mathrm{~m}^3$
Pressure, $\mathrm{p}=1 \times 10^5 \mathrm{~Pa}$
Latent heat of vaporization of water, $\mathrm{L}=2256 \mathrm{~J} / \mathrm{g}$
Change in volume, $\Delta V=(1671-1) \times 10^{-6} \mathrm{~m}^3$
$$
=1670 \times 10^{-6} \mathrm{~m}^3
$$
Heat supplied, $\Delta \mathrm{Q}=\mathrm{mL}=1 \times 2256=2256 \mathrm{~J}$
As the steam expands, so the work done in expansion is
$$
\begin{array}{rlr}
\Delta \mathrm{W} & =\mathrm{p} \Delta \mathrm{V}=1 \times 10^5 \times 1670 \times 10^{-6} \quad \text { [from Eq. (i)] } \\
& =167 \mathrm{~J} & \ldots \text {.(iii) }
\end{array}
$$
According to first law of thermodynamics,
$$
\begin{array}{rlrl}
\Delta \mathrm{Q} & =\Delta \mathrm{U}+\Delta \mathrm{W} \Rightarrow \Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W} \\
& =2256-167 & & \text { [from Eq. (ii) and (iii)] } \\
& =2089 \mathrm{~J} &
\end{array}
$$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2019 (Odisha)
1 gm of water at 100oC is completely converted into steam at 100oC . 1 gm steam occupies a volume of 1650 cc. (Neglect the volume of 1 gm of water at 100oC). At the pressure of 105 Nm2, latent heat of steam is 540 calgm 1 calorie=4.2 Joule. The increase in the internal energy in Joules is

Options:
A 2310
B 2103
C 1650
D 2150

The correct answer is:
2103

Solution:
Q=540 cal
W=Po V2-V1
V 1 =0
â´ W=105 165Ã10-6=165 J
â´ âU=540Ã4.2-165=2103 J

Topic: Physics, Thermodynamics, NEET
1 gram of hydrated oxalic acid [(COOH)2.2H2O] undergoes combustion it produces 2.2 kcal of heat. What will be its enthalpy of combustion (âHc)?

Options:
A -2.2kcal
B -126 kcal
C -277.2 kcal
D -423 kcal

The correct answer is:
-277.2 kcal

Solution:
Combustion enthalpy is the amount of energy released during complete combustion of one mole of the reactant in the presence of the oxygen.

It is always negative as per the thermodynamic sign convention.

The molar mass of the hydrated oxalic acid is =126 g mol-1

1g⟶-2.2 kcal

126g⟶-2.2×126 kcal=-277.2 kcal

Topic: Chemistry, Thermodynamics (C), NEET
1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to

Options:
A 250 mg
B Zero mg
C 200 mg
D 750 mg

The correct answer is:
250 mg

Solution:
$\begin{aligned}
& \mathrm{M}=\frac{\mathrm{W} \times 1000}{\left.\mathrm{M}_2 \times \mathrm{V} \text { (in } \mathrm{mL}\right)} \\
& \mathrm{W}=\frac{\left.\mathrm{M} \times \mathrm{M}_2 \times \mathrm{V} \text { (in } \mathrm{mL}\right)}{1000}=\frac{0.75 \times 36.5 \times 25}{1000} \\
&=0.684 \mathrm{~g} \text { (Mass of } \mathrm{HCl}) \\
& \underset{36.5 \mathrm{~g}}{\mathrm{HCl}}+\underset{40 \mathrm{~g}}{\mathrm{NaOH}} \longrightarrow \mathrm{HCl}+\mathrm{NaOH}
\end{aligned}$
36.5 g HCl reacts with $\mathrm{NaOH}=40 \mathrm{~g}$
0.684 g HCl reacts with $\mathrm{NaOH}=\frac{40}{36.5} \times 0.684 \simeq 0.750 \mathrm{~g}$
Amount of NaOH left $=1 \mathrm{~g}-0.750 \mathrm{~g}=0.250 \mathrm{~g}=250 \mathrm{mg}$

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2024
1 α -ketoglutaric acid +NH4++NADPH→Xglutamate+H2O+NADP

2

What are X and Z as shown on the arrows?

Options:
A X-Glutamate dehydrogenase, Z-Transaminase
B X-Transaminase, Z-Glutamate dehydrogenase
C X-Nitrogenase, Z-Transaminase
D X-Glumate aminase, Z-Dehydrogenase

The correct answer is:
X-Glutamate dehydrogenase, Z-Transaminase

Solution:
In the process of obtaining nitrogen for making proteins, plants get nitrogen either as nitrates or ammonia. Ammonium form is highly toxic for the plant and hence, it is directly used in the amino acid production in two steps.
First one is called the reactive amination, where the ammonia reacts with α -ketoglutaric acid to form glutamate. This reaction is catalyzed by the enzyme glutamate dehydrogenase. Next part is the transamination where the amino group of amino acid is transferred to keto group of keto acid. Transaminase catalyses this process.

Topic: Biology, Mineral Nutrition, NEET
1 litre of aqueous solution of HCl has pH=1. How many litres of water must be added to create an aqueous solution with pH of 2?

Options:
A 2.0
B 9.0
C 0.1
D 0.9

The correct answer is:
9.0

Solution:
pH=1=-logH+
H+=10-1 M
Final pH=2; H+=10-2 M
M1V1=M2V2
0.1 M× 1 Litre=0.01 V2
V2=10 L
Final volume=10 L
Water added =10-1=9L

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
$0.01 \mathrm{M}$ acetic acid solution is $1 \%$ ionised, then $\mathrm{pH}$ of this acetic acid solution is :

Options:
A 3
B 2
C 4
D 1

The correct answer is:
4

Solution:
Ionisation of $\mathrm{CH}_3 \mathrm{COOH}$ is $1 \%$ in aqueous solution.
So, $\left[\mathrm{H}^{+}\right]=\frac{1}{100} \times$ concentration of $\mathrm{CH}_3 \mathrm{COOH}$
$\begin{aligned}
& =\frac{1}{100} \times 10^{-2}=10^{-4} \\
\mathrm{pH} & =-\log \left[\mathrm{H}^{+}\right]=-\log \left(1 \times 10^{-4}\right) \\
\mathrm{pH} & =4
\end{aligned}$

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2022 (Phase 2)
0.1 M KMnO4 is used for the following titration. What volume of the solution in ml will be required to react with 0.158 g Na2S2O3.5H2O ?

3S2O32-+MnO4-+H2O→8MnO2+6SO42-+2OH-

Options:
A 1.7 ml
B 0.17 ml
C 17 ml
D 1.07 ml

The correct answer is:
17 ml

Solution:
As per as the balanced equation given below
3S2O32-+8MnO4-+H2O→8MnO2+6SO42-+2OH-
3 moles of Na2S2O3=8 moles of KMnO4
3×248 g of Na2S2O3=8×158 g of KMnO4
0.158 g of Na2S2O3=8×158×0.1583×248
=0.27 g KMnO4
Volume =Wt.×1000Molarity×Mol.Wt.
=0.27×10000.1×158=17 ml.

Topic: Chemistry, Redox Reactions, NEET
0.1 M NaCl and 0.05 M BaCl2 solutions are separated by a semi- permeable membrane in container. For this system, choose the correct answer

Options:
A Water flows form BaCl2 solution towards NaCl solution
B There is no movement of any solution across the membrane
C Osmotic pressure of 0.1 M NaCl is lower than the osmotic pressure of BaCl2 (Assume complete dissociation)
D Water flows from NaCl solution towards BaCl2 solution

The correct answer is:
Water flows form BaCl2 solution towards NaCl solution

Solution:
Using πV=iCRT
i = 2 for NaCl
i = 3 for BaCl2
πVNaCl=2×0.1RT=0.2RT
πVBaCl2=3×0.05RT=0.15RT
As πVNaCl>πVBaCl2 . Here water flows from BaCl2 towards NaCl solution.

Topic: Chemistry, Solutions, NEET
1 mol each of AgNO3, CuSO4 and AlCl3 is electrolysed, Then ratio of number of faraday's required are.

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

The correct answer is:
1 : 2 : 3

Solution:
Ag++e– ⟶Ag
Cu2++2e–⟶Cu
Al3++3e– ⟶Al
Thus, number of faradays required by 1 mol each=1 : 2: 3.

Topic: Chemistry, Electrochemistry, NEET
0.01 M solution of KCI and CaCl2 are separately prepared in water. The freezing point of KCI is found to be -2oC . What is the freezing point of CaCl2 aq. solution if it is completely ionized?

Options:
A -3oC
B + 3oC
C -2oC
D -4oC

The correct answer is:
-3oC

Solution:
i for KCl=2, i for for CaCl2=3
âTfâi
âTf(KCl)âTfCaCl2=23
âTfCaCl 2=32Ã2=3oC
Freezing point of CaCl2=-3oC

Topic: Chemistry, Solutions, NEET
0.1 M solution of which of the following compounds will have the lowest freezing point?

Options:
A FeSO4.(NH4)2SO4.6H2O
B [CrCl2(NH3)]Cl
C K3[Fe(CN)6]
D [Co(Cl)(NH3)5]Cl2

The correct answer is:
FeSO4.(NH4)2SO4.6H2O

Solution:
FeSO4.(NH4)2SO4.6H2O (Mohr's Salt) is an example of double salt and have maximum number of ions in the solution, so it will have least freezing point.

Topic: Chemistry, Solutions, NEET
1 mg radium has 2.68Ã108 atoms. Its half-life is 1620 years. How many radium atoms will disintegrate from 1 mg of pure radium in 3240 years?

Options:
A 2.01×109
B 2.01×108
C 1.01×109
D 1.01×108

The correct answer is:
2.01×108

Solution:
Given that, the initial amount of radium
N0=1 mg
and number of atoms in 1 mg radium =2.68Ã108 atoms
T1/2=1620 years
Number of half-lives
n=TT1/2
=32401620
n=2
â´ Amount after 3240 years
N=N012n
=1122
N=14 mg
N=14Ã2.68Ã108 atoms
=0.67Ã108 atoms
â´ Number of disintegrate atoms
=2.68Ã108-0.67Ã108
=2.01Ã108

Topic: Physics, Atomic Physics, NEET
1 mol of AgI/Ag+ is coagulated by

Options:
A 1 mol of KI
B 500 mL of 1 M  K 2 SO 4
C 300 mL of 1 M  Na 3 PO 4
D 1 mol of AgI

The correct answer is:
1 mol of KI

Solution:
1 mol of AgI/Ag+ is coagulated by 1 mol of KI. Here 1 mol Ag+ will combine with 1 mol I- to form precipitate of 1 mol AgI.

Topic: Chemistry, Surface Chemistry, NEET
1 mol of ice is converted to liquid at 273 K; H2O (s) and H2O (l) have entropies 38.20 and 60.03 Jmol-1deg. Enthalpy changes in the conversion will be

Options:
A 59.59 J/mol
B 5959.59 J/mol
C 595.959 J/mol
D 5.959 J/mol

The correct answer is:
5959.59 J/mol

Solution:
ÎS=Sfinal-Sinitial
=(60.03-38.20) Jmol-1K-1
=21.83 Jmol-1K-1
Also ÎS=ÎHfusionT
or ÎHfusion=TÎS=273 KÃ21.83 Jmol-1K-1
=5959.59 Jmol-1 .

Topic: Chemistry, Thermodynamics (C), NEET
0.1 molal aqueous solution of NaBr freezes at -0.335â at atmospheric pressure, kf for water is 1.86 â. The percentage of dissociation of the salt in solution is

Options:
A 90
B 80
C 58
D 98

The correct answer is:
80

Solution:
âTf=ikfm
0.335=iÃ1.86Ã0.1
i=1.80
For NaBr, i=1+Î±
Î±=0.80=80%

Topic: Chemistry, Solutions, NEET
1 mole of a gas with Î³=75 is mixed with 1 mole of a gas Î³=53, then the value of Î³ for the resulting mixture is

Options:
A 75
B 25
C 2416
D 127

The correct answer is:
2416

Solution:
For mixture of gases
n 1 + n 2 Î³ m - 1 = n 1 Î³ 1 - 1 + n 2 Î³ 2 - 1
1 + 1 Î³ m - 1 = 1 7 5 - 1 + 1 5 3 - 1
2 Î³ m - 1 = 5 2 + 3 2
or 2 Î³ m - 1 = 8 2
or 8 Î³ m - 8 = 4
or 8 Î³ m = 12
or Î³ m = 12 8 = 24 16

Topic: Physics, Thermodynamics, NEET
1 mole of FeSO4 (atomic weight of Fe is 55.84 g  mol −1 ) is oxidized to Fe2(SO4)3 . Calculate the equivalent weight of ferrous ion.

Options:
A 55.84
B 27.92
C 18.61
D 111.68

The correct answer is:
55.84

Solution:
2 Fe +2 SO 4 →2 Fe 2 +3 (SO 4 ) 3
Change in oxidation state per Fe atom is 1. So, equivalent weight = atomic weight.

Topic: Chemistry, Redox Reactions, NEET
0.1% of the composition of wastewater is contributed by three different types of materials. Choose the set of substances from the following which depicts all the three types of impurities found in wastewater.

Options:
A Clay, faecal matter and glass
B Cement, cloth fibres and calcium
C Clay, paper fibres and ammonia
D Silt, sand and faecal matter

The correct answer is:
Clay, paper fibres and ammonia

Solution:
Wastewater comprises of 99% water and 0.1% of impurities. These impurities are of three types:
1. Suspended solids, e.g., sand, silt and clay.
2. Colloidal material, e.g., Faecal matter, bacteria, cloth and paper fibres.
3. Dissolved materials, e.g., nutrients (nitrate, ammonia, phosphate, calcium, sodium).

Topic: Biology, Environmental Issues, NEET
1. pH of blood: (P)

2. Blood forms (W)% of the total body weight

3. Proteins form (Pr)% of plasma

4. The thickness of RBC: (R) micrometer

The CORRECT arrangement of the above values is

Options:
A W > P > Pr > R
B W > Pr > P > R
C R > P > Pr > W
D W > R > P > Pr

The correct answer is:
W > P > Pr > R

Solution:
pH of blood is 7.4. Blood forms 8 % of body weight. Proteins form 7% of plasma. Thickness of RBC is 2.5 m

Topic: Biology, Body Fluids and Circulation, NEET
1g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation: C (graphite) + O2(g) → CO2(g) During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the Bomb Calorimeter is 20.7kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm?

Options:
A –1.86 ×102kJ mol–1
B –4.52 ×102kJ mol–1
C –2.48 ×102kJ mol–1
D –3.68 ×102kJ mol–1

The correct answer is:
–2.48 ×102kJ mol–1

Solution:
Correct Option is : (C)
–2.48 ×102kJ mol–1

Topic: Chemistry, Thermodynamics (C), NEET
1-Penten-4-yne reacts with bromine at -80° C to produce

Options:
A 4,4,5,5-Tetrabromopentene
B 1,2-Dibromo-1,4-pentadiene
C 1,1,2,2,4,5-Hexabromopentane
D 4,5-dibromopentyne

The correct answer is:
4,5-dibromopentyne

Solution:
1-Penten-4-yne reacts with bromine at -80° C to produce 4,5-dibromopentyne as follows
CH2=CH-CH2-C≡CH+Br2→-80°CH2Br-CHBr-CH2-C≡CH

Topic: Chemistry, Hydrocarbons, NEET
1-Phenylethanol can be prepared by reaction of benzaldehyde with

Options:
A CH3-Br
B CH3I and Mg
C CH3-Br and AlBr3
D C2H5-I and Mg

The correct answer is:
CH3I and Mg

Solution:
1-Phenylethanol can be prepared by reaction of benzaldehyde with CH3I and Mg as follows

Topic: Chemistry, Aldehydes and Ketones, NEET
2, 2-dichloro propane on hydrolysis yields

Options:
A Acetone
B 2, 2-Propane diol
C Isopropyl alcohol
D Acetaldehyde

The correct answer is:
Acetone

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
2 ×108atoms of carbon are arranged side by side.What is the radius of carbon atom if the length of this arrangement is 2.4 cm?

Options:
A 0.06 nm
B 0.12 nm
C 0.01 nm
D 0.23 nm

The correct answer is:
0.06 nm

Solution:
Correct Option is : (A)
0.06 nm

Topic: Chemistry, Structure of Atom, NEET
2CuI→Cu+CuI2 , the given reaction is an example of

Options:
A Disproprtion
B Neutralisation
C Oxidation
D Reduction

The correct answer is:
Disproprtion

Solution:
2CuI+1→Cu0+CuI2+2 . Oxidation and Reduction both occur for same species so the reaction is Disproportionation (redox).

Topic: Chemistry, Redox Reactions, NEET
2DCDO â OH â X and Y are

Options:
A DCOO-,Â D2CHOH
B HCOO-,Â CH3OH
C HCOO-,Â CD3OH
D DCOO-, CD3OH

The correct answer is:
DCOO-, CD3OH

Solution:
Cannizaro reaction takes place as,

Topic: Chemistry, Aldehydes and Ketones, NEET
2 g molecule of PCl5 are heated in a closed vessel of two litre capacity. When the equilibrium is attained, PCl5 is 40% dissociated into PCl3 and Cl2 . The equilibrium constant is

Options:
A 0.534
B 0.267
C 26.7
D 2.67

The correct answer is:
0.267

Solution:
40% PCl5 is dissociated
x=2×40100=0.8
KC=[PCl3][Cl2][PCl5]=0.4×0.40.6=0.267 .

Topic: Chemistry, Chemical Equilibrium, NEET
0.2 g of an organic compound contains C, H and O. On combustion, it yields 0.15 g CO2 and 0.12 g H2O . The percentage of C, H and O respectively is

Options:
A C = 15%, H = 20%, O = 65%
B C = 10%, H = 8.2%, O = 81.8%
C C = 12.2%, H = 8.8%, O = 79%
D C = 20%, H = 6.66%, O = 73.34%

The correct answer is:
C = 20%, H = 6.66%, O = 73.34%

Solution:
% of C=1244×0.150.2×100=20%
% of H=218×0.120.2×100=6.66%
% of O=100-(20+6.66)=73.34%

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
2 g of benzoic acid (C6H5COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol–1. What is the percentage association of acid if it forms dimer in solution?

Options:
A Degree of association of benzoic acid in benzene is 86.4 %
B Degree of association of benzoic acid in benzene is 99.2 %
C Degree of association of benzoic acid in benzene is 35.4 %
D Degree of association of benzoic acid in benzene is 91.4 %

The correct answer is:
Degree of association of benzoic acid in benzene is 99.2 %

Solution:
Correct Option is : (B)
Degree of association of benzoic acid in benzene is 99.2 %

Topic: Chemistry, Solutions, NEET
2 kg of ice at -20 Â°C is mixed with 5 kg of water at 20 Â°C. The water content of the final mixture is (Latent heat of ice =80 kcal kg-1, the specific heat of water =1 kcal kg-1 Â°C-1 and specific heat of ice =0.5 kcal kgâ1 Â°Câ1 )

Options:
A 7 kg
B 6 kg
C 4 kg
D 3 kg

The correct answer is:
6 kg

Solution:
We have, the heat lost by water =msθ, heat gained by ice =MSθ+ML.
From the principle of calorimetry, we have, the heat lost by water=heat gained by ice.

Hence, msθ1=MSθ2+ML.

5×1×20-0=2×0.5×20+x×80
x=1 kg
1 kg of ice melts to form water.
Hence, final amount of water 5 kg+1 kg=6 kg.

Here, m, s and θ1 are the mass, specific heat and temperature difference for water respectively. Similarly, M, S and θ2 are the corresponding quantities for ice. Here, L is the latent heat of fusion.

Topic: Physics, Thermal Properties of Matter, NEET
2 L of water in a container is heated with a coil of 1 kW at 27 â . The lid of the container is open and energy dissipates at rate of 160 Js-1. Find the time after which temperature will rise from 27 â to 77 â?

[Given specific heat of water is 4.2 kJ/Kg]

Options:
A 8 min 20 sec
B 6 min 2 sec
C 7 min
D 14 min

The correct answer is:
8 min 20 sec

Solution:
Energy gained by water (in 1 s)
= Energy supplied-energy lost
= (1000 J-160 J)=840 J
Total heat required to raise the temperature of water from 27℃ to 77℃ is ms∆θ. Hence, the required time,
t=ms∆θrate by which energy is gained by water
= 24.2×103(50)840=500 s
= 8 min 20 s.

Topic: Physics, Thermal Properties of Matter, NEET
2 mole of an ideal gas at 27oC temperature is expanded reversibly from 2 L to 20 L. Find the entropy change R=2 Calmol −1 K -1

Options:
A 92.1
B 0
C 4
D 9.2

The correct answer is:
9.2

Solution:
For isothermal reversible expansion
-w=q=nRTÃ2.303logâ¡v2v1
=2RTÃ2.303logâ¡202
=2Ã2ÃTÃ2.303Ã1=9.2T
Entropy change, S= q T = 9.2T T =9.2 Cal K â1

Topic: Chemistry, Thermodynamics (C), NEET
2 moles of FeSO4 in acid medium are oxidized by x moles of KMnO4 , whereas 2 moles of FeC2O4 in acid medium are oxidized by y moles of KMnO4 . The ratio of x and y is

Options:
A 13
B 12
C 14
D 15

The correct answer is:
13

Solution:
Fe2+→Fe3++e
Mn7++5e→Mn2+
5moleFeSO4≡1moleMn7+
2moleFeSO4≡25=x
Also
Fe2+→Fe3++e
C3+2→2C4++2e
Mn7++5e→Mn2+
5moleFeC2O4=3moleMn7+
2moleFeC2O4=35×2=y
xy=13

Topic: Chemistry, Redox Reactions, NEET
2 moles of non-volatile solute is added to 1 kg water at -8Â°C . Kf of water is 2 K kg mol-1. Mass of ice that separates out is (Ignoring the effect of change in volume)

Options:
A 250 g
B 500 g
C 750 g
D 100 g

The correct answer is:
500 g

Solution:
For no freezing to take place at -8°C , 4 moles solute should be added to 1 kg water but, since 2 mole solute only is available it can stop the freezing of 500 g water only. 500 g ice is formed.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
$2 \mathrm{~moles}$ of the ideal gas at $27^{\circ} \mathrm{C}$ temperature is expanded reversibly from $2 \mathrm{~L}$ to $20 \mathrm{~L}$. Find the entropy change $\left(\mathrm{R}=2 \mathrm{cal} \mathrm{mol}^{-1}\right.$ $\left.\mathrm{K}^{-1}\right)$ :

Options:
A 92.1
B 0
C 4
D 9.2

The correct answer is:
9.2

Solution:
$\begin{aligned}
& \Delta \mathrm{S}=2.303 \mathrm{n} \mathrm{R} \log _{10} \frac{\mathrm{V}_2}{\mathrm{~V}_1} \\
& =2.303 \times 2 \times 2 \times \log _{10} \frac{20}{2} \\
& =2.303 \times 2 \times 2 \times 1 \\
& =9.212 \mathrm{cal}
\end{aligned}$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2002
2NO2-+O2→2NO3-

This step of nitrification is done by ________bacteria.

Options:
A Pseudomonas
B Thiobacillus
C Nitrobacter
D Nitrococcus

The correct answer is:
Nitrobacter

Solution:
The process of conversion of nitrogen N2 to ammonia is termed as nitrogen fixation. Most of the ammonia is converted to nitrate by soil bacteria, where ammonia is first oxidized to nitrite by the bacteria Nitrosomonas or Nitrococcus. The nitrite is further oxidized to nitrate with the help of Nitrobacter. These steps are called nitrification and the nitrifying bacteria are called chemoautotrophs.

Topic: Biology, Mineral Nutrition, NEET
2 Na (s) + H₂ (g) → 2 NaH (s) Identify the species undergoing oxidation and reduction:

Options:
A Oxidation: H₂, Reduction: Na
B Oxidation: Na, Reduction: H₂
C Oxidation: Na, Reduction: NaH
D Oxidation: H₂, Reduction: NaH

The correct answer is:
Oxidation: H₂, Reduction: Na

Solution:
Correct Option is : (A)
Oxidation: H₂, Reduction: Na

Topic: Chemistry, Redox Reactions, NEET
2 ppm of a pesticide entered in a food chain at the producer level. However, in the apex consumers, 200 ppm of pesticide was identified. The phenomenon that occurred here is:

Options:
A Biodeterioration
B Eutrophication
C Biogeochemical cycling
D Biomagnification

The correct answer is:
Biomagnification

Solution:
Sewage from homes as well as industries is disposed of in the water bodies. Sewage from certain industries such as chemical manufacturing, paper manufacturing, etc. contains toxic substances such as heavy metals. These heavy metals, when taken by aquatic organisms, cannot be metabolised or excreted and hence, they pass it to higher trophic levels. This leads to an increase in the accumulation of toxic substances at each successive trophic level. This is known as biomagnification.

Topic: Biology, Environmental Issues, NEET
2-Butyne can be prepared from which of the following compounds?

Options:
A
B
C CH 2 − CH 2 − CH 2 − CH 2  |                           | Cl                         Cl
D Both A and B

The correct answer is:
Both A and B

Solution:

Topic: Chemistry, Hydrocarbons, NEET
2-Methylbutan-2-ol can be obtained by the acid catalyzed hydration of

Options:
A CH3CH2CH=CH2
B CH3CH=CHCH2
C CH32C=CHCH3
D Either of the three

The correct answer is:
CH32C=CHCH3

Solution:
Proceed backward i.e., remove a molecule of water to get the alkene.

Topic: Chemistry, Hydrocarbons, NEET