datavorous/entrance-exam-dataset huggingface dataset:
A gas such as carbon monoxide would be most likely to obey the ideal gas law at:

Options:
A High temperatures and low pressures
B Low temperatures and high pressures
C High temperatures and high pressures
D Low temperatures and low pressures

The correct answer is:
High temperatures and low pressures

Solution:
Real gases show ideal gas behaviour at high temperatures and low pressures. This is because at high temperatures and due to low pressure, the molecules will move with a very high speed and the distance between them will be very large and thus the force of attraction between them will be negligible.

Topic: Chemistry, States of Matter, NEET, NEET 2015 (Phase 2)
A gas undergoes change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A:

Options:
A 6 J of the work will be done by the gas
B 6 J of the work will be done by the surrounding on gas
C 10 J of the work will be done by the surrounding on gas
D 10 J of the work will be done by the gas

The correct answer is:
6 J of the work will be done by the surrounding on gas

Solution:
for A to B q=5 W= â8 J, Î U AB =dq+dw â 5â8 Î U AB =â3
for B to A q=5 W= â8 J, Î U BA =dq+dw Î U BA = +3 dw=+3+3 W BA =6 J

Topic: Chemistry, Thermodynamics (C), NEET
A gas undergoes the cyclic process shown in the figure. The cycle is repeated 100 times per minute. The power generated is

Options:
A 240  W
B 100  W
C 60  W
D 120  W

The correct answer is:
100  W

Solution:
Work done per cycle is given by area of the PV curve
=12Ã(30-10)(8-2) [from given]
=12Ã20Ã6=60
The cycle is repeated 100 times, so work done
=60Ã100=6000
As we know,
Power=Work doneTime taken
=600060=100 W

Topic: Physics, Thermodynamics, NEET
A gaseous alkane on complete combustion gives CO2 and H2O . If the ratio of moles of O2 needed for combustion and moles of CO2 formed is 5 : 3. find out the formula of alkane.

Options:
A C4H10
B C5H12
C C3H8
D C2H6

The correct answer is:
C3H8

Solution:
Let the alkane be CnH2n+2
CnH2n+2+n+n+1 2O2→nCO2+(n+1)H2O
Mole of O2Mole ofCO2 =n+n+1 2n=53
n=3
Thus the alkane is C3H8

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2 . The empirical formula of the hydrocarbon is

Options:
A C6H5
B C3H4
C C2H4
D C7H8

The correct answer is:
C7H8

Solution:
CxHy+x+y4O2→xCO2+y2H2O
Number of moles of CO2=3.0844=0.07
So x=0.07
Number of moles of H2O=0.7218=0.04
So y=0.08
x:y=7:8
Therefore, empirical formula of hydrocarbon is C7H8 .

Topic: Chemistry, General Organic Chemistry, NEET
A gaseous mixture consists of 16 g of helium and 16 g of oxygen. The ratio CPCV of the mixture is

Options:
A 1.4
B 1.54
C 1.59
D 1.62

The correct answer is:
1.62

Solution:
μ1=164=4
μ2=1632=12
CPmix=μ1CP1+μ2CP2μ1+μ2
CVmix=μ1CV1+μ2CV2μ1+μ2
γmix=CPmixCVmix

Topic: Physics, Thermodynamics, NEET
A gaseous mixture containing He, CH4 and SO2 was allowed to effuse through a fine hole then find what molar ratio of gases coming out initially? (Given mixture contains He,CH4 and SO2 in 1 : 2 : 3 mole ratio)

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

The correct answer is:
4:4:3

Solution:
nHenCH4=12 164=11
nHenSO2=13 644=43
So, n He : n CH 4 : n SO 2 =4:4:3

Topic: Chemistry, States of Matter, NEET
A gaseous mixture contains an equal number of hydrogen and nitrogen molecules. Specific heat measurements on this mixture at a temperature below 150 K would indicate the value of the ratio of specific heats for the mixture as

Options:
A 32
B 43
C 53
D 75

The correct answer is:
32

Solution:
Below 150 K, hydrogen behaves as monoatomic gas
∴ For the mixture, γ=12γmono+γdi=1253+75=32

Topic: Physics, Kinetic Theory of Gases, NEET
A gaseous mixture contains oxygen and nitrogen in the ratio 1:4 by weight. The ratio of their number of molecules is

Options:
A 1:4
B 1:8
C 7:32
D 3:16

The correct answer is:
7:32

Solution:
Number of moles of O2=132
and number of mol of N2=428
So Number of molecules of  O 2 Number of molecules  N 2 = Number of mol of  O 2 Number of mol of  N 2
=1/324/28=132Ã284
=732=7:32

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A gaseous mixture of 2 moles of A, 3 moles of B, 5 moles of C and 10 moles of D is contained in a vessel. Assuming that gases are ideal and the partial pressure of C is 1.5 atm, total pressure is

Options:
A 3 atm
B 6 atm
C 9 atm
D 15 atm

The correct answer is:
6 atm

Solution:
Partial pressure = no. of moles of gasÃ p total total no.of moles
1.5= 5Ã p total 2+3+5+10
1.5Ã20 5 = p total
p total =6 atm

Topic: Chemistry, States of Matter, NEET
A gaseous mixture was prepared by taking equal moles of $\mathrm{CO}$ and $\mathrm{N}_2$. If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen $\left(\mathrm{N}_2\right)$ in the mixture is

Options:
A $0.8 \mathrm{~atm}$
B $0.9 \mathrm{~atm}$
C $1 \mathrm{~atm}$
D $0.5 \mathrm{~atm}$

The correct answer is:
$0.5 \mathrm{~atm}$

Solution:
$$
\begin{array}{rlrl}
\because & { }^n \mathrm{CO} & ={ }^n \mathrm{~N}_2 \\
& \therefore & p_{\mathrm{CO}} & =p_{\mathrm{N}_2} \\
\text { Given, } & p_{\mathrm{CO}}+p_{\mathrm{N}_2} & =1 \mathrm{~atm} \\
& \text { or } & 2 p_{\mathrm{N}_2} & =1 \mathrm{~atm} \\
\text { or } & p_{\mathrm{N}_2} & =0.5 \mathrm{~atm}
\end{array}
$$

Topic: Chemistry, States of Matter, NEET, NEET 2011 (Screening)
A gaseous organic compound decolourised by alk. KMnO4 solution but gives no precipitate with ammoniacal cuprous chloride is

Options:
A Ethane
B Methane
C Ethene
D Acetylene

The correct answer is:
Ethene

Solution:
alk.  KMnO 4
decolourised by alkene of alkyne

Topic: Chemistry, Hydrocarbons, NEET
A gaseous reaction X2(g)→Z(g)+1/2y(g) shows increase in pressure from 150 mm to 170 mm in 10 minutes. The rate of disappearance of X2 is

Options:
A 4 mm min-1
B 16 mm min-1
C 8 mm min-1
D 2 mm min-1

The correct answer is:
4 mm min-1

Solution:
Pressure α concentration
Rate of appearance of y=170-15010=2 mm min-1
Rate of disappearance of X2=2× rate of appearance of y
=2×2=4 mm min-1.

Topic: Chemistry, Chemical Kinetics, NEET
A Ge specimen is doped with Al. The concentration of acceptor atom is ~1021 atoms m-3. Given that the intrinsic concentration of electron-hole pairs is ~1019 m-3. The concentration of electrons in the specimen is

Options:
A 1017 m-3
B 1015 m-3
C 104 m-3
D 102 m-3

The correct answer is:
1017 m-3

Solution:
∴ ni2=nenn
10192=ne×1021
given, ni=~1019 m-3, nh=~1021atom m-3
ne=1017 m-3

Topic: Physics, Semiconductors, NEET
A gene has two alleles B and b. B is responsible for the expression of character x. b is responsible for the expression of character y. B is completely dominant over b. Another gene A is epistatic on gene B. When A is in the dominant condition it suppresses the expression of B and b and gives rise to character z. In recessive condition (allele a) allows expression of B and b. When an organism with genotype AaBb is crossed with another organism with genotype AbBb, the ratio of characters x, y and z seen in its offsprings is

Options:
A 3:4 :9
B 3:1:12
C 1:12: 3
D 9:4 :3

The correct answer is:
3:1:12

Solution:
According to the data given,
Character x will be expressed in organisms having genotypes aaBB or aaBb. Character y will be expressed in organisms having genotypes aabb. Character z will be expressed in organisms having genotypes AABB, AABb, AAbb, AaBB, AaBb and Aabb
The cross AaBb x AaBb will produce,
AABB, AAbb, aaBB, aabb, AaBB (2), AABb (2), Aabb (2), aab (2) and AaBb (4).
Thus, Character x will be expressed in aaBB (1) and aaBb (2) that is 3 out of 16 possibilities
Thus, Character y will be expressed in aabb (1) that is 1 out of 16 possibilities
Thus, Character z will be expressed in AABB (1), AABb (2), AAbb (1), AaBB (2), AaBb (4) and Aabb (2) that is 12 out of 16 possibilities
The ratio of x : y : z = 3 : 1 : 12

Topic: Biology, Principles of Inheritance and Variation, NEET
A gene is said to be dominant if:

Options:
A it expresses its effect only in the homozygous stage
B it expresses its effect only under heterozygous condition
C it expresses its effect both under homozygous and heterozygous conditions
D it never expresses its effect under any condition

The correct answer is:
it expresses its effect both under homozygous and heterozygous conditions

Solution:
A dominant allele is expressed in homozygous as well as heterozygous condition. A recessive trait is only expressed in the homozygous condition.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2002
A gene locus has two alleles A, a. If the frequency of dominant allele A is 0.4, then what will be the frequency of homozygous dominant, heterozygous and homozygous recessive individuals in the Population?

Options:
A 0.36(AA); 0.48(Aa); 0.16(aa)
B 0.16(AA); 0.24(Aa); 0.36(aa)
C 0.16(AA); 048(Aa); 0.36(aa)
D 0.16(AA); 0.36(Aa); 0.48(aa)

The correct answer is:
0.16(AA); 048(Aa); 0.36(aa)

Solution:
According to the Hardy-Weinberg principle:

Frequency of dominant allele (A) = p =0.4

Frequency of recessive allele (a) = q

Sum total of all the allelic frequencies is 1, i.e. p+q = 1

⇒ 0.4 + q = 1

⇒ q = 0.6

Frequency of AA individuals = p × p = p2 = (0.4)2 = 0.16

Frequency of aa individuals = q × q = q2 = (0.6)2 = 0.36

Frequency of Aa individuals = 2pq = 2(0.4)(0.6) = 0.48

Topic: Biology, Evolution, NEET, NEET 2019 (All India)
A gene showing codominance has

Options:
A alleles tightly linked on the same chromosome.
B alleles that are recessive to each other.
C both alleles independently expressed in the heterozygote.
D one allele dominant on the other.

The correct answer is:
both alleles independently expressed in the heterozygote.

Solution:
Both alleles are independently expressed in heterozygote during codominance. In codominance, the alleles of a gene pair are fully expressed resulting in offspring with a phenotype that is neither dominant nor recessive. Example: ABO blood group system.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2015 (Phase 2)
A gene showing codominance has

Options:
A Alleles tightly linked on the same chromosome
B Alleles that are recessive to each other
C Both alleles independently expressed in the heterozygote
D One allele dominant on the other

The correct answer is:
Both alleles independently expressed in the heterozygote

Solution:
Both alleles are independently expressed in heterozygote during codominance. In codominance, the alleles of a gene pair are fully expressed resulting in offspring with a phenotype that is neither dominant nor recessive. Example: ABO blood group system.

Topic: Biology, Principles of Inheritance and Variation, NEET
A gene whose expression helps to identify transformed cell is known as

Options:
A Selectable Marker
B Vector
C Plasmid
D Structural gene

The correct answer is:
Selectable Marker

Solution:
In recombinant DNA technology, selectable markers help in identifying and eliminating non-transformants and selectively permitting the growth of the transformants.

In 1977, Bolivar and Rodriguez constructed the pBR322 plasmid vector. pBR322 is one of the most widely used E. coli cloning vectors. It contains ampicillin and tetracycline resistance genes that act as selectable markers.

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2017
A gene whose expression helps to identify transformed cell is known as

Options:
A vector.
B restriction enzymes.
C structural gene.
D selectable marker.

The correct answer is:
selectable marker.

Solution:
The selectable marker helps in identifying and eliminating, non-transformants, and selectively permitting the growth of transformants. Normally gene encoding antibiotic resistance are used as selectable markers.

Topic: Biology, Biotechnology - Principles and Processes, NEET
A genetic engineer upon the completion of electrophoresis, moves ahead to elution. What is exactly done by him in this step?

Options:
A The separated band from DNA is exposed to UV radiations.
B The separated band from DNA are cut out from the gel and extracted from the gel piece
C Separation of the recombinant protein from recombinant cell.
D The separated band from DNA fragment can be visualized by ethidium bromide.

The correct answer is:
The separated band from DNA are cut out from the gel and extracted from the gel piece

Solution:
After the completion gel electrophoresis, the DNA sequence needs to amplified by PCR. To obtain the DNA sequence from the gel, the separated bands are cut and then the sequence is extracted. This process is called as elution.

Topic: Biology, Biotechnology - Principles and Processes, NEET
A genetic mechanism and prevents self-pollination is

Options:
A Dichogamy
B Self-incompatibility
C Heterostyly
D Homogamy

The correct answer is:
Self-incompatibility

Solution:
The device to prevent inbreeding is self-incompatibility. This is a genetic mechanism and prevents self-pollen (from the same flower or other flowers of the same plant) from fertilizing the ovules by inhibiting pollen germination or pollen tube growth in the pistil.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
A genetic muscular disorder is

Options:
A osteoporosis
B muscular dystrophy
C myasthenia gravis
D botulism

The correct answer is:
muscular dystrophy

Solution:
Muscular dystrophy is a progressive degeneration of skeletal muscles done mostly due to genetic disorder.

Myasthenia gravis is an autoimmune disorder affecting neuromuscular junction leading to fatigue, weakening, and paralysis of skeletal muscles.

Osteoporosis is a bone disease that occurs when the body loses too much bone, makes too little bone, or both and is called as porous bone.
Botulism is an infection caused by the bacteria Clostridium botulinum.

Topic: Biology, Locomotion and Movement, NEET
A genetically controlled mechanism in which pollen grains of a flower are unable to grow completely on the stigma of the same flower is called ___.

Options:
A Dichogamy
B Self-incompatibility
C Bagging
D Emasculation

The correct answer is:
Self-incompatibility

Solution:
A genetically controlled mechanism in which pollen grains of a flower are unable to grow completely on the stigma of the same flower is called Self-incompatibility

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
A genetically modified bacteria producing human insulin can be obtained by adding a human insulin gene inside the genetic material of a bacteria. Choose the next step in using the modified bacteria to treat diabetes among the following.

Options:
A Add the altered bacterium to human food.
B Inject the altered bacterium into the blood of a person with diabetes
C Put the bacterium into a fermenter to multiply rapidly
D Use the altered bacterium in a nasal spray.

The correct answer is:
Put the bacterium into a fermenter to multiply rapidly

Solution:
Human insulin can be produced via genetically engineered bacteria having the human insulin gene added to its plasmid. Once the modified bacteria has been developed, it can be commercially mass-produced in a bioreactor. The insulin produced can be purified after extraction via the various downstream processing. Using the altered bacteria directly may cause undesired reactions to the diabetic patient.

Topic: Biology, Biotechnology and Its Applications, NEET
A geostationary satellite is orbiting the earth at a height of $5 R$ above that surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2 R$ from the surface of the earth is

Options:
A 5
B 10
C $6 \sqrt{2}$
D $6 / \sqrt{2}$

The correct answer is:
$6 \sqrt{2}$

Solution:
From Keplar third's law
$T^2 \propto r^3$
Hence, $T_1^2 \propto r_1^3$
So,
$\begin{aligned}
& T_2^2 \propto r_2^3 \\
& \frac{T_2^2}{T_1^2}=\frac{r_2^3}{r_1^3} \\
&=\frac{(3 R)^3}{(6 R)^3}
\end{aligned}$
$\begin{aligned}
& \text {or } \frac{T_2^2}{T_1^2}=\frac{1}{8} \\
& T_2^2=\frac{1}{8} T_1^2 \\
& T_2=\frac{24}{2 \sqrt{2}}=6 \sqrt{2} \mathrm{~h}
\end{aligned}$

Topic: Physics, Gravitation, NEET, NEET 2012 (Screening)
A geostationary satellite is orbiting the Earth at a height of 6R above the surface of the earth, R being the radius of the Earth. The time period of another satellite, at a height of 2.5R from the surface of the Earth, is

Options:
A 62 hour
B 6 hour
C 52 hour
D 10 hour

The correct answer is:
62 hour

Solution:
A geostationary satellite has a time period of T=24 hr. From Kepler's law,
T2â R3 âT22T12=R23R13 âT22242=R+2.5R3R+6R3
T2=24Ã248=62 hr.

Topic: Physics, Gravitation, NEET
A geo-stationary satellite is orbiting the Earth at a height of 6R above the surface of the earth, R being the radius of the Earth. The time period of another satellite at a height of 2.5R from the surface of Earth is

Options:
A 62 hr
B 6 hr
C 52 hr
D 10 hr

The correct answer is:
62 hr

Solution:
Geo-satellite means whose time period is
T=24 hr & from law of time period,
T2âR2
â T22T12=R23R13
â T22(24)2=(R+2.5R)3(R+6R)3
T2=24Ã248=62 hr

Topic: Physics, Gravitation, NEET
A girl of mass 50 kg stands on a railroad car of mass 75 kg moving with velocity 20 m s-1. Now, the girl begins to run with a velocity of 10 m s-1 with respect to the car in the same direction, as that of the car. The velocity of the car at this instant will be

Options:
A 20 m s-1

B 24 m s-1
C 16 m s-1
D 18 m s-1

The correct answer is:
16 m s-1

Solution:
According to the law of conservation of linear momentum,
Pi=Pf
â¹ 50+7520=5010+v+â75v
â¹ 125Ã20=5010+v+75v
â¹ 5Ã20=210+v+3v
â¹ 100=20+2v+3v
â¹ 5v=80â¹v=16 m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
A girl riding a bicycle with a speed of 5 m/s towards the north direction observes rain falling vertically down. If she increases her speed to 10 m/s, the rain appears to meet her at 45° to the vertical. What is the speed of the rain relative to the ground?

Options:
A 5i
B 5j
C 5i-5j
D 5i+5j

The correct answer is:
5i-5j

Solution:
Correct Option is : (C)
5i-5j

Topic: Chemistry, Motion In Two Dimensions, NEET
A given double stranded DNA molecule is 1,00,000 base pairs long. The length of this DNA molecule will be

Options:
A 10,000 nm
B 3.4 ×104 nm
C 104 nm
D 2,00, 000 nm

The correct answer is:
3.4 ×104 nm

Solution:
The length of this DNA molecule will be 3.4 ×104 nm because one complete spiral of the helix is 3.4nm ( 34A° ) long and has 10 base pairs.

Topic: Biology, Molecular Basis of Inheritance, NEET
A given mass of gas expands from the state A to the state B by three paths 1, 2 and 3 as shown in the figure. If w1,w2 and w3 respectively be the work done by the gas along three paths then

Options:
A w1>w2>w3
B w1<w2<w3
C w1=w2=w3
D w2<w3<w1

The correct answer is:
w1<w2<w3

Solution:
As greater is the area under the curve, more is the work done.
w1<w2<w3

Topic: Chemistry, Thermodynamics (C), NEET
A given metal crystallizes out with a cubic structure having edge length of 361 pm. If there are four metal atoms in one unit cell, what is the radius of one atom?

Options:
A 40 pm
B 127 pm
C 80 pm
D 108 pm

The correct answer is:
127 pm

Solution:
4 atoms in one unit cell is present in FCC
i.e. 4r= 2 ⋅a
r=2 ⋅a4=1.414×3614
=127 pm

Topic: Chemistry, Solid State, NEET, NEET 2015 (Phase 1)
A given metal crystallizes out with a cubic structure having edge length of 363 pm. If there are four metal atoms in one-unit cell, what is the radius of one atom?

Options:
A 64 pm
B 128.35 pm
C 64.18 pm
D 108 pm

The correct answer is:
128.35 pm

Solution:
FCC:r=a22=3632×1.4141=128.35pm

Topic: Chemistry, Solid State, NEET
A given nitrogen-containing aromatic compound A reacts with Sn/HCl, followed by HNO2 to give an unstable compound B. B on treatment with phenol, forms a beautiful coloured compound C with the molecular formula C12H10N2O. The structure of compound A is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, NEET, NEET 2016 (Phase 2)
A given quantity of an ideal gas is at pressure P and absolute temperature T. The isothermal bulk modulus of the gas is

Options:
A 23P
B P
C 32P
D 2P

The correct answer is:
P

Solution:
In the isothermal process
PV = constant
∴ PdV+VdP=0 or dPdV=-PV
∴ Bulk modulus, B=-dPdV/V=-dPdVV
∴ B=--PVV=P
∴ B=P
Note: Adiabatic bulk modulus is given by B = γP

Topic: Physics, Mechanical Properties of Solids, NEET
A given quantity of an ideal gas is at the pressure P and the absolute temperature T. The isothermal bulk modulus of the gas is

Options:
A 23P
B P
C 32P
D 2P

The correct answer is:
P

Solution:
âµ Î²=ÎPÎVV
â´ In the isothermal process P+ÎP V-ÎV=PV
or PV-PÎV+VÎP-ÎPÎV=PV
But ÎP ÎV maybe neglected.
â´ V ÎP=P ÎV
or ÎPÎVV=P
Hence, ( P ) is correct.

Topic: Physics, Thermodynamics, NEET
A given sample of an ideal gas occupies a volume, V at a pressure, P and absolute temperature, T. The mass of each molecule of the gas is m . Which of the following gives the density of the gas?

Options:
A PkT
B PmkT
C PkTV
D mkT

The correct answer is:
PmkT

Solution:
PV=nRT⇒PV=mMRT

P=RMmVT⇒Pρ=RmNAT

⇒Pρ=kTm

ρ=PmkT.

Topic: Physics, Thermodynamics, NEET, NEET 2016 (Phase 2)
A given sample of somatic cell was treated with chemical X. It was found that chemical X inhibits chromatid separation. During somatic cell division process, the treated cell will reach which stage of division before being interrupted?

Options:
A Anaphase
B Metaphase
C Telophase
D Cytokinesis

The correct answer is:
Metaphase

Solution:
X is Colchicine is a mitotic inhibitor. It doesn't allow spindle fibres to be formed. Thus, condensation of chromosomes will occur and prophase will be completed. But metaphase won't be completed. The metaphase stage is characterised by all the chromosomes coming to lie at the equator with one chromatid of each chromosome connected by its kinetochore to spindle fibres from one pole and its sister chromatid connected by its kinetochore to spindle fibres from the opposite pole

Topic: Biology, Cell Cycle and Cell Division, NEET
A glass flask of volume one litre at 0 Â°C is filled level full of mercury at this temperature. The flask and mercury are now heated to 100 Â°C. How much mercury will spill out if the coefficient of volume expansion of mercury is 1.82Ã10-4 Â°C-1 and linear expansion of glass is 0.1Ã10-4 Â°C-1, respectively?

Options:
A 2.48Ã10-2 L
B 1.52Ã10-2 L
C 1.53Ã104 L
D 1.52Ã10-4 L

The correct answer is:
1.52Ã10-2 L

Solution:
Expansion of mercury
∆Vm=Vmγm∆θ
=1×1.82×10-4100°-0°
=1.82×10-2L
Expansion of flask,
∆Vg=​Vγg∆θ=V⋅3αg∆θ
=​ 1×3×0.1×10-4100o-0o
= ​0.3 ×10-2L
Volume of mercury that over flows
=∆Vm-∆Vg
=1.82-0.3×10-2
=1.52 ×10-2L

Topic: Physics, Thermal Properties of Matter, NEET
A glass prism ABC (refractive index 1.5), immersed in water (refractive index 43 ). A ray of light is incident normally one face AB. If it is totally reflected at the face AC, then

Options:
A sinθ≥89
B sinθ≥23
C sinθ≥32
D 23<sinθ<89

The correct answer is:
sinθ≥89

Solution:
For total internal reflection at face AC
sinθ>μwμg
⇒sinθ≥43×1.5
⇒sinθ≥89

Topic: Physics, Ray Optics, NEET
A glass prism ABC (refractive index 1.5), immersed in water (refractive index 4/3). A ray of light is incident normally on face AB. If it is totally reflected at face AC then

Options:
A sin⁡θ≥89
B sin⁡θ≥23
C sin⁡θ=32
D 23<sin⁡θ<89

The correct answer is:
sin⁡θ≥89

Solution:
For total internal reflection at AC face
sin⁡i≥μwug
sin⁡θ≥43×1.5
sin⁡θ≥89

Topic: Physics, Ray Optics, NEET
A golf ball has a mass of 40g, and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate the uncertainty in the position.

Options:
A 1.46×10–33m
B 2.34×10–33m
C 0.84×10–33m
D 3.25×10–33m

The correct answer is:
1.46×10–33m

Solution:
Correct Option is : (A)
1.46×10–33m

Topic: Chemistry, Structure of Atom, NEET