datavorous/entrance-exam-dataset huggingface dataset:
On a hilly region, water boils at 95 Â°C. The temperature expressed in Fahrenheit scale is

Options:
A 100 °F

B 200 °F

C 150 °F

D 203 °F

The correct answer is:
203 °F

Solution:
If the temperature of a body on Celsius and Fahrenheit scales are recorded as C and F respectively, then
C-0100-0 = F-32212-32
or C5 =F-329
Here, C=95 °C
∴ 955 = F-329
Or 5F = 1015
∴ F=10155 =203 °F

Topic: Physics, Thermal Properties of Matter, NEET
On a large scale, how is ammonia primarily manufactured?

Options:
A By the reaction of urea and ammonia salts
B By the direct synthesis of nitrogen and hydrogen gases
C By the decomposition of organic matter
D By the decay of nitrogenous organic matter

The correct answer is:
By the direct synthesis of nitrogen and hydrogen gases

Solution:
The paragraph mentions that on a large scale, ammonia is manufactured by Haber's process, which involves the direct synthesis of nitrogen and hydrogen gases.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
On a particle, moving on a circular path with a constant speed v, light is thrown from a projector placed at the centre of the circular path. The shadow of the particle is formed on the wall. The velocity of shadow, up the wall, is

Options:
A v sec 2 Ï
B v cos 2 Ï
C v cos Ï
D None of these

The correct answer is:
v sec 2 Ï

Solution:
v ′ cos Ï R sec Ï = v R
v ′ = v sec 2 Ï

Topic: Physics, Motion In Two Dimensions, NEET
On a plane mirror, a ray of light is incident at an angle of 30° with horizontal. To make the reflected ray vertical, at what angle with horizontal must a plane mirror be placed?

Options:
A 30Â°
B 60°
C 45°
D 54°

The correct answer is:
30°

Solution:
To make the reflected ray vertical, the mirror should make an angle 30Â° with the horizontal.

Topic: Physics, Ray Optics, NEET
On a ship sailing in Pacific Ocean where temperature is 23.4°C, a balloon is filled with 2 L air. What will be the volume of the balloon when the ship reaches Indian ocean, where temperature is 26.1°C?

Options:
A 2.018 L
B 2.654 L
C 1.875 L
D 1.658 L

The correct answer is:
2.018 L

Solution:
Correct Option is : (A)
2.018 L

Topic: Chemistry, States of Matter, NEET
On a smooth plane surface (figure) two block A and B are accelerated rightwards by applying a force 15 N on A. If the mass of B is twice that of A, the force on B is

Options:
A 30 N

B 15 N

C 10 N

D 5 N

The correct answer is:
10 N

Solution:
The acceleration of both the blocks =
153x=5x
∴Force on B=5x×2x=10 N

Topic: Physics, Laws of Motion, NEET
On account of the earth rotating about its axis

Options:
A the linear velocity of objects at equator is greater than at other places
B the angular velocity of objects at equator is more than that of objects at poles
C the linear velocity of objects at all places at the earth is equal, but angular velocity is different
D at all places the angular velocity and linear velocity are uniform

The correct answer is:
the linear velocity of objects at equator is greater than at other places

Solution:
(1) as distance of surface is maximum from axis at equator, hence v is maximum as v=ÏÃr
(2) angular velocity is same everywhere
(3) as distance of surface is maximum from axis at equator, hence v is maximum as v=ÏÃr
(4) as distance of surface is maximum from axis at equator, hence v is maximum as v=ÏÃr

Topic: Physics, Rotational Motion, NEET
On adding excess of AgNO3 solution into 0.01 mole complex compound PtBr4 . xNH3,0.03 moles yellow precipitate was obtained, then value of ‘ x ’ is

Options:
A 2
B 3
C 4
D 5

The correct answer is:
5

Solution:
PtBr4.xNH30.01mole+AgNO3SolutionExcess→AgBr↓0.03moles
As 3 AgBr molecules are formed it means 3Br- are outside the coordination sphere and only one is inside so the value of x=5
i.e., complex is [PtBrNH3)5Br3

Topic: Chemistry, Coordination Compounds, NEET
On adding excess of NH4OH to copper sulphate solution

Options:
A A deep blue solution is obtained
B A blue precipitate of Cu(OH)2 is obtained
C A black precipitate of CuO is obtained
D No change takes place

The correct answer is:
A deep blue solution is obtained

Solution:
With ammonium solution, CuSO4 forms a stable blue complex. First it forms a precipitate of Cu(OH)2 which dissolves in excess of ammonia solution forming a deep blue coloured complex.
CuSO4+2NH4OHâCu(OH)2+(NH4)2SO4
Cu (OH) 2 +2 NH 4 OH+ ( NH 4 ) 2 SO 4 → Cu ( NH 3 ) 4 SO 4 +4 H 2 O Tetraamminecupricsulphate

Topic: Chemistry, d and f Block Elements, NEET
On adding KMnO4 to cold conc. H2SO4 , it gives

Options:
A MnO2
B Mn2O3
C Mn2O7
D MnO3

The correct answer is:
Mn2O7

Solution:
2KMnO 4  +  2H 2 SO 4 → 2KHSO 4  +  Mn 2 O 7 (Green oily liquid)  +  H 2 O
Mn2O7 explodes on heating.
2Mn2O7âÎ4MnO2+3O2

Topic: Chemistry, d and f Block Elements, NEET
On an average, female cockroach produce ______P_____oothecae, each containing _____Q_____ eggs. Select the appropriate option for P and Q.

Options:
A P-9-10; Q-16-18
B P- 9-10; Q-14-16
C P- 14-16; Q-9-10
D P-14-16; Q-10-12

The correct answer is:
P- 9-10; Q-14-16

Solution:
The fertilized are encased in capsules called ootheca. On an average, female cockroach produce 9-10 oothecae, each containing 14-16 fertilized eggs. The development of P. americana is paurometabolous through nymphal stage.

Topic: Biology, Structural Organisation in Animals, NEET
On balancing the given redox reaction

aCr2O72-+bSO32-aq+cH+aq→2aCr3+aq+bSO42-aq+c2H2Ol

The coefficients a,b,c are found to be respectively

Options:
A 8,1,3
B 1,3,8
C 3,8,1
D 1,8,3

The correct answer is:
1,3,8

Solution:
Reduction Half reaction : Cr2O72−+6e-→2Cr3+

Oxidation Half reaction :

SO32−→SO42−+2e-×3

Overall reaction :

Cr2O72−+3SO32−→2Cr3++3SO42-

To balance ‘O’ atoms, adding H2O on LHS

Cr2O72−+3SO32−→2Cr3++3SO42-+4H2O

To balance ‘H’ atoms, adding H+ on RHS

Cr2O72−+3SO32−+8H+→2Cr3++3SO42-+4H2O

Therefore, a=1, b=3, c=8.

Topic: Chemistry, Redox Reactions, NEET, NEET 2023 (All India)
On comparing the proportion of elements found in human tissue and Earth's crust, which one of the following is found to be negligible?

Options:
A Silicon
B Calcium
C Magnesium
D Sulphur

The correct answer is:
Silicon

Solution:
In a human tissue (living tissue), the composition of the elements (percent weight of the human body) given in the question was found to be Sulphur – 0.3 , Magnesium – 0.1 , Calcium – 1.6 and Silicon - negligible.

Topic: Chemistry, Biomolecules, NEET
On complete combustion, 0.3 g of an organic compound gave 0.2 g of $\mathrm{CO}_2$ and 0.1 g of $\mathrm{H}_2 \mathrm{O}$. The percentage composition of carbon and hydrogen in the compound, respectively is:

Options:
A $4.07 \%$ and $15.02 \%$
B $18.18 \%$ and $3.70 \%$
C $15.02 \%$ and $4.07 \%$
D $3.70 \%$ and $18.18 \%$

The correct answer is:
$18.18 \%$ and $3.70 \%$

Solution:
Percentage of carbon $=\frac{12 \times \mathrm{m}_2 \times 100}{44 \times \mathrm{m}}$
$\mathrm{m}=$ mass of organic compound $=0.3 \mathrm{~g}$
$\mathrm{m}_2=$ mass of carbon dioxilde $=0.2 \mathrm{~g}$
$\therefore \% C=\frac{12 \times 0.2 \times 100}{44 \times 0.3}=18.18 \%$
Percentage of hydrogen $=\frac{2 \times \mathrm{m}_1 \times 100}{18 \times \mathrm{m}}$
$\mathrm{m}_1=$ mass of water $=0.1 \mathrm{~g}$
$\therefore \% \mathrm{H}=\frac{2 \times 0.1 \times 100}{18 \times 0.3}=3.70 \%$

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2024 (Re-NEET)
On complete combustion, 0.246 g of an organic compound gave 0.198g of carbon dioxide and 0.1014g of water. Determine the percentage composition of carbon and hydrogen in the compound.

Options:
A Percentage of Carbon=36.78% and Percentage of Hydrogen=7.89%
B Percentage of Carbon=36.78% and Percentage of Hydrogen=4.58%
C Percentage of Carbon=21.95% and Percentage of Hydrogen=7.89%
D Percentage of Carbon=21.95% and Percentage of Hydrogen=4.58%

The correct answer is:
Percentage of Carbon=21.95% and Percentage of Hydrogen=4.58%

Solution:
Correct Option is : (D)
Percentage of Carbon=21.95% and Percentage of Hydrogen=4.58%

Topic: Chemistry, General Organic Chemistry, NEET
On decreasing the wavelength of incident light from 8000 Aâ to 4000 Aâ, the intensity of the scattered light in Rayleigh scattering will become _____ times the initial scattered intensity.

Options:
A 2
B 4
C 16
D 8

The correct answer is:
16

Solution:
Iâ1Î»4
â IfIi=Î»iÎ»f4
=844=24=16

Topic: Physics, Wave Optics, NEET
On electrolysis of dil. sulphuric acid using Platinum Pt electrode, the product obtained at anode will be

Options:
A Oxygen gas
B H2S gas
C SO2 gas
D Hydrogen gas

The correct answer is:
Oxygen gas

Solution:
Dissociation of sulfuric acid,
\(\mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{SO}_4^{-2}(\mathrm{aq})\)
Disssociation of water,
\(\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\)
At Anode: \(4 \mathrm{OH}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O}^{+}(\mathrm{l})+\mathrm{O}_2(\mathrm{~g})+4 \mathrm{e}^{-}\)
At Cathode: \(2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_2(\mathrm{~g})\)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET, NEET 2020 (Phase 1)
On Fe2O3(s)+32C(s)→32CO2(g)+2Fe(s);ΔH°=+234.1kJ
C(s)+O2(g)→CO2(g);ΔH°=-393.5 kJ
Use these equations and ΔH° values to calculate ΔH° for this reaction
4Fe(s)+3O2(g)→2Fe2O3(s)

Options:
A -1648.7 kJ
B -1255.3 kJ
C -1021.2 kJ
D -129.4 kJ

The correct answer is:
-1648.7 kJ

Solution:
2Fe(s)+32CO2(g)→Fe2O3(s)+32C(s);ΔH°=-234.1kJ …(i)
C(s)+O2(g)→CO2(g)(ΔH°=-393.5kJ) …(ii)
Multiplying eq. (i) by 2 and eq. (ii) by 3 and on adding both equations, we get
4Fe(s)+3O2(g)→2Fe2O3(s);ΔH°=(-234.1×2)+(-3×393.5)
=-1648.7 kJ

Topic: Chemistry, Thermodynamics (C), NEET
On heating a mixture containing 1 mole each of Li2CO3 and K2CO3 ..... is/are formed.

Options:
A 2 moles of CO2
B 1 mole of CO2
C 1.5 mole of CO2
D No carbon dioxide

The correct answer is:
1 mole of CO2

Solution:
Li2CO3 decomposes while K2CO3 is stable and does not decompose.
Li 2 CO 3 → Li 2 O + CO 2 1 mole 1 mole

Topic: Chemistry, s Block Elements, NEET
On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 mol of hydrogen (H2) in the presence of a catalyst gives another gas (C) which is basic in nature. Gas C on further oxidation in moist conditions gives a compound (D) which is a part of acid rain. Identify D

Options:
A NH3
B NH2
C HNO2
D HNO3

The correct answer is:
HNO3

Solution:
Correct Option is : (D)
HNO3

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 mol of hydrogen (H2) in the presence of a catalyst gives another gas (C) which is basic in nature. Gas C on further oxidation in moist conditions gives a compound (D) which is a part of acid rain. Identify A

Options:
A NH4NO2
B NH4NO3
C HNO2
D HNO3

The correct answer is:
NH4NO2

Solution:
Correct Option is : (A)
NH4NO2

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 mol of hydrogen (H2) in the presence of a catalyst gives another gas (C) which is basic in nature. Gas C on further oxidation in moist conditions gives a compound (D) which is a part of acid rain. Identify B

Options:
A N3
B N2
C NO2
D NO3

The correct answer is:
N2

Solution:
Correct Option is : (B)
N2

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 mol of hydrogen (H2) in the presence of a catalyst gives another gas (C) which is basic in nature. Gas C on further oxidation in moist conditions gives a compound (D) which is a part of acid rain. Identify C

Options:
A NH3
B NH2
C NO2
D NO3

The correct answer is:
NH3

Solution:
Correct Option is : (A)
NH3

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
On heating NaX with H2SO4 and MnO2 , the halogen that cannot be prepared is

Options:
A I2
B F2
C Cl2
D Br2

The correct answer is:
F2

Solution:
On heating NaX with H2SO4 and MnO2 , F2 cannot be prepared.
2NaX+H2SO4→ΔNa2SO4+2HX
MnO2+4HX→MnX2+X2(g)+2H2O
MnO2 can oxidize HCl, HBr and HI and not HF.
Hence, F2 cannot be prepared by this method.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
On heating one end of a piece of a metal, the other end becomes hot because of

Options:
A energised electrons moving to the other end
B minor perturbation in the energy of atoms
C resistance of the metal
D mobility of atoms in the metal

The correct answer is:
energised electrons moving to the other end

Solution:
Conductivity of heat in metals is due to the presence of free electrons, which move due to increase in temperature.

Topic: Chemistry, Electrochemistry, NEET
On heating one end of a rod, the temperature of whole rod will be uniform when

Options:
A K=1
B K=0
C K=100
D K=∞

The correct answer is:
K=∞

Solution:
dQdt=-KAdÎ¸dx
When K=â, dÎ¸dx=0
i.e., Î¸ is independent of x i.e. constant or uniform.

Topic: Physics, Thermal Properties of Matter, NEET
On heating, some solid substances change from solid to vapour state without passing through liquid state. The technique used for the purification of such solid substances based on the above principle is known as

Options:
A Sublimation
B Distillation
C Chromatography
D Crystallization

The correct answer is:
Sublimation

Solution:
(1) Sublimation : It is the purification technique based on principle that on heating, some solid substances change from solid to vapour state without passing through liquid state.
(2) Distillation : It is used to separate volatile liquids from non-volatile impurities and the liquids having sufficient difference in their boiling point.
(3) Chromatography : It is based on separation by using stationary and mobile phase.
(4) Crystallization : It is based on difference in the solubilities of the compound and impurities in a suitable solvent.

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2024
On heating which of the following releases CO2 most easily?

Options:
A K2CO3
B Na2CO3
C MgCO3
D CaCO3

The correct answer is:
MgCO3

Solution:
Thermal stability order
K2CO3>Na2CO3>CaCO3>MgCO3
Therefore MgCO3 releases CO2 most easily.
MgCO3→Δ MgO+CO2

Topic: Chemistry, s Block Elements, NEET, NEET 2015 (Phase 2)
On heating with dilute sulphuric acid, naphthalene-1 sulphonic acid gives predominantly

Options:
A naphthalene
B naphthalene-2-sulphonic acid
C 1-naphthol
D 2-naphthol

The correct answer is:
naphthalene-2-sulphonic acid

Solution:
Naphthalene-1 sulphonic acid when treated with dilute sulphuric acid at high temperature, gives its isomer naphthalene-2-sulphonic acid.

Topic: Chemistry, Hydrocarbons, NEET
On hydrolysis of a "compound", two compounds are obtained, One of which on treatment with sodium nitrite and hydrochloric acid gives a product which does not respond to iodoform test. The second one reduces Tollen's reagent and Fehling's solution. The "compound" is:

Options:
A $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NC}$
B $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CN}$
C $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{ON}=\mathrm{O}$
D $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CON}\left(\mathrm{CH}_3\right)_2$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NC}$

Solution:
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NC}+2 \mathrm{H}_2 \mathrm{O} \rightarrow$ $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{NH}_2+\mathrm{HCOOH}$

Topic: Chemistry, Aldehydes and Ketones, NEET, NEET 2013 (Karnataka)
On increasing the temperature, the specific resistance of

Options:
A both conductor and semiconductor increases
B both conductor and semiconductor decreases
C conductor increases and semiconductor decreases
D conductor decreases and semiconductor increases

The correct answer is:
conductor increases and semiconductor decreases

Solution:
For conductor, resistivity increases as temperature rises. For semiconductor, resistivity decreases as temperature rises.

Topic: Physics, Semiconductors, NEET
On introducing a thin mica sheet of thickness 2Ã10-6 m and refractive index 1.5 in the path of one of the waves, central bright maxima shifts by n fringes. The wavelength of the wave used is 5000 Aâ , then n is

Options:
A 1
B 2
C 5
D 10

The correct answer is:
2

Solution:
The shift in no. of fringes is given by
nÎ»=Î¼-1t
n=Î¼-1tÎ»=(1.5-1)Ã2Ã10-65000Ã10-10 fringes=2

Topic: Physics, Wave Optics, NEET
On mixing 10 mL of acetone with 40 mL of chloroform, the total volume of the solution is

Options:
A < 50 mL
B > 50 mL
C = 50 mL
D cannot be predicted

The correct answer is:
< 50 mL

Solution:
On mixing 10 mL of acetone with 40 mL of chloroform, the total volume of the solution is < 50 mL as the interparticle forces in between CHCl3 and acetone increases due to H-bonding and thus,
ΔV mixing = -ve.
Chloroform and acetone hence form solution with negative deviation.

Topic: Chemistry, Solutions, NEET
On monochlorination of 2-methylbutane, the total number of compounds including stereo isomers formed are

Options:
A 2
B 4
C 6
D 8

The correct answer is:
6

Solution:
Mono chlorination of 2-methylbutane gives four isomers (I, II, III, IV)

Here only two, i.e., II and IV are optically active. Since each optically active compound has two enantiomers, therefore, in all there are four chiral compounds and two achiral compound (I, III) so total 6 compounds are possible.

Topic: Chemistry, General Organic Chemistry, NEET
On moving a charge of 20 C by 2 cm, 2 J of work is done. The potential difference between the points is

Options:
A 0.1 V
B 8 V
C 2 V
D 0.5 V

The correct answer is:
0.1 V

Solution:
Potential difference between two points in a electric fields is,
VA-VB=Wq0
Where W is work done by moving charge q0 from point A to B.
So, VA-VB=220=0.1 V
(Here W=2 J, q0=20 C)

Topic: Physics, Electrostatics, NEET
On observing light from three different stars P , Q and R , it was found that intensity of violet colour is maximum in the spectrum of P , the intensity of green colour is maximum in the spectrum of R and the intensity of red colour is maximum in the spectrum of Q . If Tp, TQ and TR are the respective absolute temperatures of P , Q and R , then it can be concluded from the above observations that :

Options:
A TP>TQ>TR
B TP>TR>TQ
C TP<TR<TQ
D TP<TQ<TR

The correct answer is:
TP>TR>TQ

Solution:
From Wein's Displacement law,
>λmT=constant ⇒λm∝1T
>From the light spectrum,
>λmP<λmR<λmQ (as λ Red > λ Green > λ Violet )
>Hence,
>TP>TR>TQ

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2015 (Phase 1)
On observing light from three different stars P , Q and R , it was found that intensity of violet colour is maximum in the spectrum of P , the intensity of green colour is maximum in the spectrum of R and the intensity of red colour is maximum in the spectrum of Q . If Tp, TQ and TR are the respective absolute temperatures of P , Q and R , then it can be concluded from the above observations that :

Options:
A TP>TQ>TR
B TP>TR>TQ
C TP<TR<TQ
D TP<TQ<TR

The correct answer is:
TP>TR>TQ

Solution:
From Wein's Displacement law,
Î»mT=constant âÎ»mâ1T
From the light spectrum,
Î»mP<Î»mR<Î»mQ (as Î» Red > Î» Green > Î» Violet )
Hence,
TP>TR>TQ

Topic: Physics, Thermal Properties of Matter, NEET
On oxidation of S 2 O 3 2â by MnO 4 â in neutral aqueous medium, the oxidation state of S would change from:

Options:
A +6 to -2
B -2 to +2
C -2 to +6
D +4 to +6

The correct answer is:
-2 to +6

Solution:
In neutral aqueous medium, thiosulphate is oxidised almost quantitatively to sulphate: 8âMn O 4 â +3â S 2 O 3 2â + H 2 Oâ8â MnO 2 +6â SO 4 2â +2 OH â

Topic: Chemistry, d and f Block Elements, NEET
On passing H2S through acidified KMnO4 , we get

Options:
A S
B MnO2
C KHSO3
D K2SO3

The correct answer is:
S

Solution:
On passing H2S through acidified KMnO4 , we get Sulphur.
2KMnO4+3H2SO4→K2SO4+2MnSO4+3H2O+5O
5H2S+5O→5H2O+5S
2KMnO4+3H2SO4+5H2S→K2SO4+2MnSO4+8H2O+5S

Topic: Chemistry, d and f Block Elements, NEET
On producing the waves of frequency 1000 Hz in a Kundt's tube, the total distance between 6 successive nodes is 85 cm. Speed of sound in the gas-filled in the tube is

Options:
A 300 m s-1
B 350 m s-1
C 340 m s-1
D 330 m s-1

The correct answer is:
340 m s-1

Solution:
Frequency of wave, ν=1000 Hz
Number of successive nodes = 6
The total distance between six successive nodes = 85 cm
From the relation between the distance of nodes and wavelength of the wave
s = 8 5 = n - 1 λ 2 = 6 - 1 λ 2 = 5 λ 2
where, λ = wavelength of the wave
∴ λ = 2 × 8 5 5 = 3 4 cm = 0 . 3 4 m
Hence, the speed of sound in the gas = v λ
= 1000×0.34 = 340 m s-1

Topic: Physics, Waves and Sound, NEET
On prolonged exposure to air, sodium finally changes to

Options:
A Na2CO3
B Na2O
C NaOH
D NaHCO3

The correct answer is:
Na2CO3

Solution:
On prolonged exposure to air, sodium finally changes to Na2CO3
4Na+O2→2Na2O
Na2O+H2O→2NaOH
2NaOH+CO2→Na2CO3+H2O

Topic: Chemistry, s Block Elements, NEET
On reacting with neutral ferric chloride, phenol gives

Options:
A red colour
B blue colour
C violet colour
D green colour

The correct answer is:
violet colour

Solution:
Phenol reacts with neutral FeCl3 solution to give violet colour complex which is soluble in water.
6 C 6 H 5 OH + FeCl 3 → Fe O C 6 H 5 6 3 - + 3 H + + 3 HCl Violet colour complex

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
On reaction of ozone with hydrogen peroxide if we start with one volume of ozone. How many volumes of oxygen will form?

Options:
A 1 volume
B 0.5 volume
C 1.5 volume
D 2 volume

The correct answer is:
2 volume

Solution:
The reaction between ozone and hydrogen peroxide is as follows
H2O2+O31vol.→H2O+2O22vol.
Hence 1 volume of ozone will give 2 volume of O2

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
On reaction with Cl2, phosphorus forms two types of halides ‘A’ and ‘B’. Halide A is yellowish-white powder but halide ‘B’ is colourless oily liquid. Identify A and B

Options:
A A: PCl3B: PCl3
B A: PCl5, B: PCl5
C A: PCl3, B: PCl5
D A: PCl5, B: PCl3

The correct answer is:
A: PCl5, B: PCl3

Solution:
Correct Option is : (D)
A: PCl5, B: PCl3

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
On reduction with hydrogen, 3.6 g of an oxide of metal left 3.2 g of metal. If the atomic weight of metal is 64, the simplest formula of the oxide is

Options:
A MO
B M2O3
C M2O
D M2O5

The correct answer is:
M2O

Solution:
Let the metal oxide by MxOy
MxOy+yH2→xM+yH2O
Moles of metal =3.264=0.05
Moles of metal oxides =3.664x+16y
1 mole of metal oxides gives x mole metal M
0.05x=3.664x+16y
64x+16y=72x
8x=16y
x=2y
Simplest formula is for x = 2, y = 1
The formula is M2O

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
On the application of a constant torque, a wheel is turned from rest through 400 radians in 10 s. The angular acceleration is :

Options:
A 8rad s-2
B 5 rad s-2
C 6 rad s-2
D 7 rad s-2

The correct answer is:
8rad s-2

Solution:
Here, ω0=0;θ=400rad;t=10 s
Now,θ=ω0t+12αt2…i⁡
∴ 4 0 0 = 0 × 1 0 + 1 2 α × 1 0 2
or α = 4 0 0 × 2 1 0 2
orα=8 rad s-2

Topic: Physics, Rotational Motion, NEET
On the application of a constant torque, a wheel is turned from rest through 400 radians in 10 s. The angular acceleration is :

Options:
A 8rad s-2
B 5 rad s-2
C 6 rad s-2
D 7 rad s-2

The correct answer is:
8rad s-2

Solution:
Here, Ï 0 = 0 ; Î¸ = 4 0 0 rad ; t = 1 0 sâ¡
Now, Î¸ = Ï 0 t + 1 2 Î± t 2 â¦ iâ¡
â´ 4 0 0 = 0 Ã 1 0 + 1 2 Î± Ã 1 0 2
or Î± = 4 0 0 Ã 2 1 0 2
or Î± = 8 rad s - 2

Topic: Physics, Rotational Motion, NEET
On the application of a constant torque, a wheel is turned from rest to 400 rad in 10 s. If same torque continues to act, the angular velocity of the wheel after 20 s from the start is

Options:
A 160 rad s-1
B 150 rad s-1
C 120 rad s-1
D 130 rad s-1

The correct answer is:
160 rad s-1

Solution:
By using the following equation,
θ=ω0t+12αt2

α=8 rad s-2
Also, ω=ω0+αt
Here ω0=0, α=8 rad s-2 and t=20 s
∴ω=0+8×20=160rad s-1

Topic: Physics, Rotational Motion, NEET