datavorous/entrance-exam-dataset huggingface dataset:
1.44 g of Titanium (Ti) reacted with an excess of O2 and produced x g of a non-stoichiometric compound Ti1.44O1. The value of x in g is :

Options:
A 1.44
B 2
C 1.77
D None of these

The correct answer is:
1.77

Solution:
Ti + O2 â¶ Ti1.44 O
(1.44 g) (x) g
Apply POAC to Ti
( 1.44g 48 )Ã1={ xg ( 1.44Ã48 )+16 }1.44
x = 4 8 Ã 1 . 4 4 + 1 6 4 8
= 1 . 4 4 + 1 6 4 8
= 1.44 + 0.33
= 1.77 g.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
150 J of energy is incident on area 2 m2. If Qr= 15 J, coefficient of absorption is 0.6, then the amount of energy transmitted is

Options:
A 50 J
B 45 J
C 40 J
D 30 J

The correct answer is:
45 J

Solution:
When thermal radiations (Q) fall on a body, they are partly reflected, partly absorbed and partly transmitted.
Q = Qa+Qr+Qt
And QaQ +QrQ+QtQ= a + r+ t =1
â 15150 + 0.6 + x = 1
or 0.1 + 0.6 + x =1
or x=0.3
Transmitting power, t = QtQ
Or 0.3 = Qt150
â Qt= 45 J

Topic: Physics, Thermal Properties of Matter, NEET
1.56Ã105J of heat is conducted through is 2 m2 wall of 12 cm thick in one hour. Temperature difference between the two sides of the wall is 20 â. The thermal conductivity of the material of the wall is (in Wm-1 K-1)

Options:
A 0.11
B 0.13
C 0.15
D 1.2

The correct answer is:
0.13

Solution:
dQdt=KA∆Tx
1.56×1053600=K×2×2012×10-2
K=1.56×105×12×10-23600×2×10
=1.5612=0.13

Topic: Physics, Thermal Properties of Matter, NEET
190 countries pledged to achieve a significant reduction in the current rate of biodiversity loss at global levels by 2010 in _________ on sustainable development held in 2002 in Johannesburg, South Africa

Options:
A World summit
B Earth summit
C Kyoto Protocol
D Montreal Protocol

The correct answer is:
World summit

Solution:
The earth summit was held in Rio de Janerio in 1992, called upon all nations to take appropriate measures for conservation of biodiversity and sustainable utilization of its benefits. In a follow up the world summit, 190 countries pledge their commitment to achieve by 2010.

Topic: Biology, Biodiversity and its Conservation, NEET
19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00 C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.

Options:
A i = 2.1352 , Ka= 3.07 × 10-3
B i = 1.0753 , Ka= 3.07 × 10-3
C i = 1.0753 , Ka= 4.56 × 10-3
D i = 2.1352 , Ka= 4.56 × 10-3

The correct answer is:
i = 1.0753 , Ka= 3.07 × 10-3

Solution:
Correct Option is : (B)
i = 1.0753 , Ka= 3.07 × 10-3

Topic: Chemistry, Solutions, NEET
200 cm3of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10-3bar. Calculate the molar mass of the protein.

Options:
A 65,986 g mol-1
B 61,022 g mol-1
C 43,231 g mol-1
D 28,346 g mol-1

The correct answer is:
61,022 g mol-1

Solution:
Correct Option is : (B)
61,022 g mol-1

Topic: Chemistry, Solutions, NEET
20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample?

(Atomic weight: Mg = 24)

Options:
A 75
B 96
C 60
D 84

The correct answer is:
84

Solution:
MgCO3s⟶MgOs+CO2g
Moles of MgCO3=2084=0.238 mol
From above equation.
1 mole MgCO3 gives 1 mole MgO
∴ 0.238 mole MgCO3 will give 0.238 mole MgO
=0.238×40 g=9.523 g MgO
Practical yield of MgO=8 g MgO
∴% Purity=89.523×100=84%

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET, NEET 2015 (Phase 2)
200 ml of water is added to 500 ml of 0.2 M solution. What is the molarity of this diluted solution?

Options:
A 0.7093 M
B 0.1428 M
C 0.4005 M
D 0.2897 M

The correct answer is:
0.1428 M

Solution:
No. of millimoles =500×0.2=100
Total volume in mililitre = 500 + 200 = 700
Thus, molarity of diluted Solution: =100700=0.1428 M

Topic: Chemistry, Electrochemistry, NEET
20.8 g of BaCl2 on reaction with 9.8 g of H2SO4 produces 7.3 g of HCl and some amount of BaSO4. The amount of BaSO4 formed is

Options:
A 23.3 g
B 20.8 g
C 9.8 g
D 10.4 g

The correct answer is:
23.3 g

Solution:
According to law of conservation of mass
Mass of reactants = Mass of products
20.8+9.8=7.3+Mass of BaSO4
∴ Mass of BaSO4=23.3 g

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
25.3 g of sodium carbonate, Na2CO3 is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, Na+ and carbonate ions, CO32- are respectively (Molar mass of Na2CO3=106 gmol-1 )

Options:
A 0.477 M and 0.477 M
B 0.955 M and 1.910 M
C 1.910 M and 0.955 M
D 1.90 M and 1.910 M

The correct answer is:
1.910 M and 0.955 M

Solution:
Molarity (M) of Na2CO3 =wtmol.wt.Ã1000vol(ml)=25.3106Ã1000250=0.955 mol/L
And Na2CO3â2Na++CO3-2
Therefore [Na+]=2Ã0.955=1.910 mol/L
[CO3-2]=0.955 mole/L

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
2.56×10-3 equivalent of KOH is required to neutralise 0.12544 g  H 2 XO 4 . The atomic mass of X (in g/mol ) is:
[Given: H 2 XO 4 is a dibasic acid]

Options:
A 16
B 8
C 7
D 32

The correct answer is:
32

Solution:
Let the atomic weight of X = y
No. of equivalents of KOH = No. of equivalents of H 2 XO 4
i.e. 0.1254466 + y2 =2.56×10-3
∴ y = 32

Topic: Chemistry, Redox Reactions, NEET
300 g of water at 25 Â°C is added to 100 g of ice at 0 Â°C. The final temperature of mixture is

Options:
A -53 °C
B -54 °C
C -5 Â°C
D 0 Â°C

The correct answer is:
0 Â°C

Solution:
Heat required for 100 g ice to melt is
Q1=mL=10080=8000 cal
Heat given by 300 g water when it cools to 0 Â°C.
Q2=msÎÎ¸=300125-0
Q2=7500 cal
As Q2<Q1
Therefore ice cannot melt completely and final temperature of mixture is 0 Â°C.

Topic: Physics, Thermal Properties of Matter, NEET
$300 \mathrm{~J}$ of work is done in sliding a $2 \mathrm{~kg}$ block up an inclined plane of height 10 $\mathrm{m}$. Taking $g=10 \mathrm{~m} / \mathrm{s}^2$, work done against friction is:

Options:
A $1000 \mathrm{~J}$
B $200 \mathrm{~J}$
C $100 \mathrm{~J}$
D zero

The correct answer is:
$100 \mathrm{~J}$

Solution:
Potential energy
$$
\begin{aligned}
& =2 \times 10 \times 10 \\
& =200 \mathrm{~J}
\end{aligned}
$$
and work done $=300 \mathrm{~J}$
$\therefore$ Work done against friction
$$
=300-200=100 \mathrm{~J} .
$$

Topic: Physics, Work Power Energy, NEET, NEET 2006
0.365 g of an organic compound containing nitrogen gave 56 ml of nitrogen at S.T.P. The percentage nitrogen in the given compound is

Options:
A 19.18
B 38.36
C 9.18
D 29.18

The correct answer is:
19.18

Solution:
It is Duma's method used to find % of nitrogen
%N=28ÃV at STPÃ10022400ÃW=28Ã56Ã10022400Ã0.236=19.18%

Topic: Chemistry, General Organic Chemistry, NEET
400 cc volume of a gas having Î³=52 is suddenly compressed to 100 cc through an adiabatic process. If the initial pressure is P, then the final pressure will be

Options:
A P/32
B 8P
C 32P
D 16P

The correct answer is:
32P

Solution:
Here V 1 = 400 cc , V 2 = 100 cc ; Î³ = 5 2
Sudden compression is an adiabatic process.
For an adiabatic process, PV Î³ = constant
As P 2 V 2 Î³ = P 1 V 1 Î³
or P 2 = P 1 V 1 V 2 Î³ = P 400 100 5 / 2 = P Ã 4 Ã 4 4 = 3 2 P

Topic: Physics, Thermodynamics, NEET
4.88 g of KClO3 when heated produced 1.92 g of O2 and 2.96 g of KCl. Which of the following statements regarding the experiments is correct?

Options:
A The result illustrates the law of conservation of mass
B The result illustrates the law of multiple properties
C The result illustrates the law of constant proportion
D None of the above laws is followed

The correct answer is:
The result illustrates the law of conservation of mass

Solution:
2KClO34.88g→2KCl2.96g+3O21.92g
Since mass of the products (2.96 + 1.92) is equal to the mass of the reactant, this illustrates the law of conservation of mass.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
50.0 kg of N2(g) and 10.0 kg of H2(g) are mixed to produce NH3(g). What is the amount of NH3(g) formed and the limiting reagent in the production of NH3in this situation?

Options:
A 64.2 kg NH3is formed and Nitrogen is the limiting reagent
B 64.2 kg NH3is formed and Dihydrogen is the limiting reagent
C 56.1 kg NH3is formed and Nitrogen is the limiting reagent
D 56.1 kg NH3is formed and Dihydrogen is the limiting reagent

The correct answer is:
56.1 kg NH3is formed and Dihydrogen is the limiting reagent

Solution:
Correct Option is : (D)
56.1 kg NH3is formed and Dihydrogen is the limiting reagent

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
500 ml of a hydrocarbon gas burnt in excess of oxygen yields 2500 ml of CO2 and 3 litres of water vapour, all the volumes being measured at the same temperature and pressure. The formula of hydrocarbon is

Options:
A C4H10
B C5H12
C C2H2
D C5H10

The correct answer is:
C5H12

Solution:
Suppose the formula of hydrocarbon =CxHy
x=volume of CO2 Volume of hydrocarbon=2500500=5
CxHy1 vol.500 ml+x+y4O2=xCO2x vol.500 xml+y2H2Oy2 vol.500y2 ml
500y2=3000
So y = 12
Molecular formula of hydrocarbon =C5H12 .

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
$6.02 \times 10^{20}$ molecules of urea are present in $100 \mathrm{~mL}$ of its solution. The concentration of solution is

Options:
A $0.02 \mathrm{M}$
B $0.01 \mathrm{M}$
C $0.001 \mathrm{M}$
D $0.1 \mathrm{M}$

The correct answer is:
$0.01 \mathrm{M}$

Solution:
Given, number of molecules of urea $=6.02 \times 10^{20}$
$\begin{aligned}
& \therefore \text { Number of moles }=\frac{6.02 \times 10^{20}}{N_A} \\
& =\frac{6.02 \times 10^{20}}{6.02 \times 10^{23}}=1 \times 10^{-3} \mathrm{~mol}
\end{aligned}$
Volume of the solution
$=100 \mathrm{~mL}=\frac{100}{1000} \mathrm{~L}=0.1 \mathrm{~L}$
Concentration of urea solution
$\begin{aligned}
\left(\text { in } \mathrm{mol} \mathrm{L}^{-1}\right) & =\frac{1 \times 10^{-3}}{0.1} \mathrm{mol} \mathrm{L}^{-1} \\
& =1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}=0.01 \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}$

Topic: Chemistry, Solutions, NEET, NEET 2013 (All India)
6.02×1020 molecules of urea are present in 100 mL of its solution. The concentration of urea Solution: is

Options:
A 0.02 M
B 0.001 M
C 0.01 M
D 0.1 M

The correct answer is:
0.01 M

Solution:
NA=6.02Ã1023=1 mol
6.02Ã1020 molecules=0.001 mol in 100 ml
(0.1 L) solution
So molar concentration =mole volume in L
0.0010.1=0.01 M

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
6.02×1020 molecules urea are present in 100 mL of its solution. The molarity of solution is

Options:
A 0.01 M
B 0.001 M
C 0.1 M
D 0.02 M

The correct answer is:
0.01 M

Solution:
n=6.02×10206.02×1023=0.001
M=nV×1000=0.001100×1000=0.01

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
896 mL vapour of a hydrocarbon ‘A’ having carbon 87.80% and hydrogen 12.19% weighs 3.28g at STP. Hydrogenation of ‘A’ gives 2-methylpentane. Also ‘A’ on hydration in the presence of H2SO4and HgSO4gives a ketone ‘B’ having molecular formula C6H12O. The ketone ‘B’ gives a positive iodoform test. Find the structure of ‘A’ .

Options:
A (CH3)2CH—C≡CH
B CH3—CH2—C≡CH
C (CH3)CH—CH2—C≡CH
D (CH3)2CH—CH2—C≡CH

The correct answer is:
(CH3)2CH—CH2—C≡CH

Solution:
Correct Option is : (D)
(CH3)2CH—CH2—C≡CH

Topic: Chemistry, Hydrocarbons, NEET
965 C of electricity is used for plating of a silver cup by silver. The amount of Ag deposited is:

Options:
A 107.89 g
B 9.89 g
C 1.0002 g
D 1.08 g

The correct answer is:
1.08 g

Solution:
Ag++ e- ⟶ Ag
∵ 96500 C are required to deposite Ag = 108 g
∴ 965 C are required to deposite Ag
= 10896500 × 965 = 1.08 g

Topic: Chemistry, Electrochemistry, NEET
1000 small water drops each of radius r and charge q coalesce together to form one spherical drop. The potential of the bigger drop is larger than that of the smaller one by a factor

Options:
A 1000
B 100
C 10
D 1

The correct answer is:
100

Solution:
Ratio of potential of the bigger drop to smaller drop is given by,
VV0=n23=100023=100

Topic: Physics, Electrostatics, NEET
10.30 mg of O2 is dissolved into a liter of sea water of density 1.03 g/mL. The concentration of O2 in ppm is

Options:
A 8
B 12
C 15
D 10

The correct answer is:
10

Solution:
Mass of O2=10.30 mg
Mass of sea water =1.03 gm=1030 mg
PPM=Mass of O2 Mass of waterÃ106
=10.30Ã10-31030Ã106=10

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
109.5 is the B–A–B angle of which of the following shapes?

Options:
A Trigonal planar
B Tetrahedral
C Trigonal bipyramidal
D Octahedral

The correct answer is:
Tetrahedral

Solution:
Correct Option is : 2
Tetrahedral

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
1.245 g sample of CuSO4.xH2O dissolved in water and excess of H2S was passed. The filtrate contains liberated H2SO4 which requires 20 ml N/2NaOH solutions for neutralization. Calculate the value of x?

Options:
A 3
B 2
C 4
D 5

The correct answer is:
5

Solution:
M eq. âofâ CuSO 4 âÃâ xH 2 Oâ=â M eq. H 2 SO 4 â=â M eq. NaOH
M eq. â CuSO 4 â=â M eq. NaOH
10 3 âÃâwt. Mol.âmass Ãv.f.=NÃV
1.245 M Ã2Ã 10 3 = 4 20 Ã2
M=1.245Ã2Ã 10 3 Ã10
M=249
Molecular weight of CuSO4.xH2O=63.5+32+64+18x
=159.5+18x
249 = 159.5 + 18x
So, x = 5.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus closest to?

Options:
A 1247.5 g mol-1
B 1298.4 g mol-1
C 1285 g mol-1
D 1156.8 g mol-1

The correct answer is:
1247.5 g mol-1

Solution:
Correct Option is : (A)
1247.5 g mol-1

Topic: Chemistry, States of Matter, NEET
0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.

Options:
A % of chlorine = 12.76
B % of chlorine = 37.57
C % of chlorine = 68.25
D % of chlorine = 76.22

The correct answer is:
% of chlorine = 37.57

Solution:
Correct Option is : (B)
% of chlorine = 37.57

Topic: Chemistry, General Organic Chemistry, NEET
A 0.2 mL sample of a solution containing 1.0Ã10-7 curie of 1H3 is injected to the bloodstream of an animal. After sufficient time for circulatory equilibrium to be established, 0.10 mL of blood is found to have an activity of 20 dpm. Calculate the volume of blood in animal, assuming no change in the activity of the sample during circulatory equilibrium.

Options:
A 11110 mL

B 1110 mL

C 11010 mL

D 10110 mL

The correct answer is:
1110 mL

Solution:
Let the volume of blood is V mL.
Thus total volume = V + 0.2 mL after injection of sample
1.1 mL sample has rate = 2060 dps
∴ V+0.2mL sample has rate=20× V+0.2 60×0.1 dps
Since the rate is constant, thus
20×V+0.2 60×0.1=1.0×10-7×3.7×1010 dps
V= 1110 mL

Topic: Physics, Nuclear Physics, NEET
A 0.5 m long metal rod PQ completes the circuit as shown in the figure. The area of the circuit is perpendicular to the magnetic field of flux density 0.15 T. If the resistance of the total circuit is 3âÎ©, calculate the force needed to move the rod in the direction as indicated with a constant speed of 2â m s â1.

Options:
A 5.78  mN

B 3.75ââmN
C 4.95ââmN
D 11.23ââmN

The correct answer is:
3.75  mN

Solution:
F= B 2 â l 2 âv R = ( 0.15 ) 2 ( 0.5 ) 2 Ã2 3
=0.00375âN
=3.75âmN

Topic: Physics, Electromagnetic Induction, NEET
A 1 kg stone at the end of 1 m long string is whirled in a vertical circle at constant speed of 4 m s-1. The tension in the string is 6 N when the stone is at (g=10 m s-2)

Options:
A Top of the circle
B Bottom of the circle
C Half way down
D None of the above

The correct answer is:
Top of the circle

Solution:
Tension at top point T + mg = mu 2 r
â´ T = 1 Ã 4 2 1 - 1 Ã 1 0 = 6 N

Topic: Physics, Laws of Motion, NEET
A 0.01 M ammonia solution is 5 % ionised, its pH will be

Options:
A 11.80
B 10.69
C 7.22
D 12.24

The correct answer is:
10.69

Solution:
Given, aqueous solution of NH3weak base C=0.01 M
Î±=5%=5100
OH-=CÎ±=0.01Ã5100=5Ã10-4
pOH=-logâ¡[OH-]
=logâ¡5Ã10-4
=4logâ¡10-logâ¡5
=4-0.6989
=3.3010
â´ pH=14-pOH=14-3.3010=10.6990

Topic: Chemistry, Ionic Equilibrium, NEET
A 0.001 molal aqueous solution of a complex [MA8] has the freezing point of -0.0054Â°C . If the primary valency of the salt undergoes 100% ionization and Kf for water =1.8 Kmolal-1 the correct representation of complex is

Options:
A [MA8]
B [MA6]A2
C [MA4]A4
D [MA5]A3

The correct answer is:
[MA6]A2

Solution:
âTf=iKfÃm
0.0054=iÃ1.8Ã0.001⇒i=3

Topic: Chemistry, Solutions, NEET
A+2B→C , the rate equation for the reaction is given as

Rate =k[ A ][ B ]

If the concentration of A is kept the same but that of B is doubled what will happen to the rate itself?

Options:
A Doubled
B Halved
C The same
D Quadrupled

The correct answer is:
Doubled

Solution:
R1=k AB
R2=k A2B
∴R2R1=2

Topic: Chemistry, Chemical Kinetics, NEET
A+2B+H2O→C+2D

If A=HNO2 , B=H2SO3 and C=NH2OH , then geometry of D will be

Options:
A trigonal planar
B bent
C tetrahedral
D linear

The correct answer is:
tetrahedral

Solution:
HNO2+2H2SO3+H2O→NH2OH+2H2SO4 , Compound D is H2SO4 in which sulphur shows sp3 hybridisation with tetrahedral geometry.

Topic: Chemistry, Coordination Compounds, NEET
A 2 μF capacitor is charged as shown in the figure. The percentage of its stored energy dissipated as heat after the switch S is turned to position 2 is

Options:
A 0 %
B 20 %
C 75 %
D 80 %

The correct answer is:
80 %

Solution:
Einitial=Q022C=12CV02=V02
Final energy E(f)=Q022Ceff
E(f)=12×2v010×2v0=v025
ΔE(loss)=4v025,
% loss = 45v02v02×100
ΔE(loss) %=80%

Topic: Physics, Electrostatics, NEET
A 2 kg copper block is heated to 500 oC and then it is placed on a large block of ice at 0 oC . If the specific heat capacity of copper is 400 âJ k g â1 Â° C â1 and latent heat of fusion of water is 3.5âÃ 10 5 âJ k g â1 , the amount of ice that can melt is

Options:
A 78 kg
B 75 kg
C 87 kg
D 57 kg

The correct answer is:
87 kg

Solution:
Heat emitted by copper = Heat gained by ice
mcÎÎ¸=mâ²L
â mâ²=mcÎÎ¸L
Given: m=2âkg,âc=400âJ kg-1Â°C-1 ,
ÎÎ¸=500,âL=3.5Ã105 J kg-1
â´ mâ²=2Ã400Ã5003.5Ã105=87 kg

Topic: Physics, Thermal Properties of Matter, NEET
A 0.2 kg object at rest is subjected to a force 0.3 iâ¡ ^ - 0.4 jâ¡ ^ newton. The velocity vector after 6s is

Options:
A 9 i⁡ ^ - 1 2 j⁡ ^
B 8 i⁡ ^ - 1 6 j⁡ ^
C 1 2 i⁡ ^ - 9 j⁡ ^
D 1 6 i⁡ ^ - 8 j⁡ ^

The correct answer is:
9 i⁡ ^ - 1 2 j⁡ ^

Solution:
Here, m = 0.2 kg ; u = 0
F → = 0.3 i ^ - 0.4 j ^ N
v → = ? , t = 6 s
a → = F → m = 0.3 i ^ - 0.4 j ^ 0.2 = 3 2 i ^ - 2 j ^
From v → = u → + a → t , v → = 0 + 3 2 i ^ - 2 j ^ × 6
= 9 i ^ - 1 2 j ^

Topic: Physics, Laws of Motion, NEET
A 2ââkg stone tied at the end of a string 1 m long is whirled along a vertical circle at a constant speed of 4 m s-1. The tension in the string has a value of 52 N when the stone is

Options:
A At the top of the circle
B Half way down
C At the bottom of the circle
D Half way up

The correct answer is:
At the bottom of the circle

Solution:
Here, m=2 kg,r=1 m,v=4 m s-1
Tension at the bottom of the circle,
TL=mg+mv2r
=20+2Ã421=52 N

Topic: Physics, Laws of Motion, NEET
A 2 year old child shows stunted growth and slow brain development. He has thin limbs and also shows wasting of muscles. There is great oedema and swelling present over his body parts. He has very less amount of fat present beneath his skin. Which of the following could be a reason for the development of this condition in him?

Options:
A Replacement of mother's milk by foods less in total proteins as well as calories at an early age
B Replacement of mother's milk by foods less in total proteins, but having high calories at an early age
C Replacement of mother's milk by foods less in total fats, but having high calories at an early age
D Replacement of mother's milk by foods less in vitamins and minerals, but having high calories at an early age

The correct answer is:
Replacement of mother's milk by foods less in total proteins, but having high calories at an early age

Solution:
The child is suffering from 'Kwashiorkor'. Kwashiorkor is a form of protein-energy malnutrition (PEM). It occurs as a result of the replacement of mother's milk by foods having low protein, but a high caloric value, in children more than one year of age. It is characterized by growth failure, failure of brain development, thinning of limbs, wasting of muscles. There is little fat present beneath the skin. Extensive oedema and swelling over the body parts are seen.

Topic: Biology, Digestion and Absorption, NEET
A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

Options:
A Ka = 1.5 × 10-9
B Ka = 1.2 × 10-5
C Ka = 1.5 × 10-5
D Ka = 1.8 × 10-5

The correct answer is:
Ka = 1.5 × 10-9

Solution:
Correct Option is : (A)
Ka = 1.5 × 10-9

Topic: Chemistry, Ionic Equilibrium, NEET
A 3 kg ball strikes a heavy rigid wall with a speed of 10 m s-1 at an angle of 60Â° with the wall. It gets reflected with the same speed at 60Â° with the wall. If the ball is in contact with the wall for 0.2 s, the average force exerted on the ball by the wall is

Options:
A 300 N
B zero

C 1503 N
D 150 N

The correct answer is:
1503 N

Solution:
F=2mv sinθt=2×3×10sin⁡60°0.2=1503 N

Topic: Physics, Center of Mass Momentum and Collision, NEET
A 3 V battery with negligible internal resistance is connected in a circuit as shown in the figure. The current I in the circuit will be

Options:
A 1 A
B 1.5 A
C 2 A
D 13 A

The correct answer is:
1.5 A

Solution:
Resistance in the arms AC and BC are in series,

â´ R â² = 3 + 3 = 6 Î©
Now, R' and 3âÎ© are in parallel,
â´ Req=6Ã36+3=2 Î©
Now, V=IR
â¹ I=32=1.5 A

Topic: Physics, Current Electricity, NEET
A 4 Î¼F capacitor is charged to 400 V and then its plates are joined through a resistance of 1 kÎ©. The heat produced in the resistance is

Options:
A 0.16 J
B 1.28 J
C 0.64 J
D 0.32 J

The correct answer is:
0.32 J

Solution:
The energy stored in the capacitor
=12CV2=12×4×10-6×400×400=0.32 J
This energy will be converted into heat in the resistor.

Topic: Physics, Current Electricity, NEET
A 0.004 M solution of Na2SO4 is isotonic with a 0.010 M solution of glucose at same temperature. The apparent degree of dissociation of Na2SO4 is

Options:
A 25%
B 50%
C 75%
D 85%

The correct answer is:
75%

Solution:
P(Na2SO4)=iCRT=i(0.004)RT
P (glucose) = CRT = 0.010 RT
As solutions are isotonic, i (0.004) RT = 0.01 RT
this gives i = 2.5
Na2SO4→2Na++SO42-
α=i-1n-1
Here n=3
α=2.5-12=1.52=0.75=75%

Topic: Chemistry, Solutions, NEET
A 5.5 m long string has a mass of 0.035 kg. If the tension in the string is 77 N, the speed of a wave on the string is

Options:
A 110 m s-1
B 165 m s-1
C 77 m s-1
D 102 m s-1

The correct answer is:
110 m s-1

Solution:
μ=0.0355.5 kg m-1, T=77 N
Velocity of transverse wave along string
v=Tμ=77×5.50.035=110 m s-1

Topic: Physics, Waves and Sound, NEET
A 5 A current is set up in an external circuit by a 6.0 V storage battery for 6.0 min. The chemical energy of the battery is reduced by

Options:
A 1.08 Ã 10 4 J
B 1.08 × 10 - 4 J
C 1.8 × 10 4 J
D 1.8 × 10 - 4 J

The correct answer is:
1.08 × 10 4 J

Solution:
Chemical energy-reduced
=VIt
= 6 × 5 × 6 × 6 0
= 10800
=1.08×104 J

Topic: Physics, Current Electricity, NEET
A 0.5 kg block slides from A on horizontal track with an initial speed of 3 m sâ1 towards a weightless horizontal spring of length 1 m and force constant 2 N mâ1. The part AB of the track is frictionless and the part BC has the coefficient of static and kinetic friction as 0.22 and 0.20 respectively. If the distance AB and BD are 2 m and 2.14 m respectively, the total distance through which the block moves before it comes to rest completely is -

Options:
A 2.5Â m
B 4.42Â m
C 4.24 m
D 2.44Â m

The correct answer is:
4.24 m

Solution:
At D,KE=12 mv2 â Î¼kmg BD
= 12 Ã 0.5 Ã 32 â 0.2 Ã 0.5 Ã 10 Ã 2.14
= 2.25 â 2.14=0.11 J
0.11=12 kx2 + Î¼Rmg xâx2 + x â 0.11=0
âx=0.1 m
Total distance covered=2 + 2.14 + 0.1 =4.24 m

Topic: Physics, Work Power Energy, NEET