datavorous/entrance-exam-dataset huggingface dataset:
The amount of bromine (in grams) required to convert 2 g of phenol into 2, 4, 6-tribromophenol is

Options:
A 4.00
B 6.00
C 10.22
D 20.44

The correct answer is:
10.22

Solution:
PhOH+3Br2âC6H2Br3OH+3HBr
1 mole 3 moles 1 mole
94 grams of phenol reacts with 480 gms . of Br2
2 gm . of phenol â48094Ã2=10.22 g

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length $100 \mathrm{~cm}$ to stretch it by $1 \mathrm{~mm}$ is (if Young's modulus of the wire $=2.0 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$ )

Options:
A $10^7$
B $10^5$
C $10^{11}$
D $10^{17}$

The correct answer is:
$10^5$

Solution:
We know that, elastic potential energy per unit volume $=\frac{1}{2} \times$ Stress $\times$ Strain
Also, $Y=\frac{\text { Stress }}{\text { Strain }}$
$\Rightarrow \quad$ Elastic potential energy per unit volume $=\frac{1}{2} \times$ Young's modulus $\times(\text { Strain })^2$
$\begin{aligned} & =\frac{1}{2} \times 2 \times 10^{11} \times\left[\frac{1}{1000}\right]^2 \\ & =\frac{1}{2} \times 2 \times 10^{11} \times 10^{-6}=10^5\left(\mathrm{~J} / \mathrm{m}^3\right)\end{aligned}$

Topic: Physics, Mechanical Properties of Solids, NEET, NEET 2023 (Manipur)
The amount of energy emitted if electron falls from n=3 to n=2 , in hydrogen atom is

Options:
A 0.65 eV
B 1.9 eV
C 10.2 eV
D 12.09 eV

The correct answer is:
1.9 eV

Solution:
ΔE=E3-E2=13.61(2)2-1(3)2=1.9 eV

Topic: Chemistry, Structure of Atom, NEET
The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly: (surface tension of soap solution = 0.03 N m-1)

Options:
A 50.1×10-4 J
B 30.16×10-4 J
C 5.06×10-4 J
D 3.01×10-4 J

The correct answer is:
3.01×10-4 J

Solution:
Soap bubble has two free surface.

Therefore, amount of energy required =2S×A=20.03×4π2×10-22=3.01×10-4 J

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2023 (All India)
The amount of glucose required to prepare 250 mL of $\frac{\mathrm{M}}{20}$ aquepus solution is :
(Molar mass of glucose : $180 \mathrm{~g} \mathrm{~mol}^{-1}$ )

Options:
A 2.25 g
B 4.5 g
C 0.44 g
D 1.125 g

The correct answer is:
2.25 g

Solution:
Molarity, $\mathrm{M}=\frac{\mathrm{w}_2 \times 1000}{\mathrm{M}_2 \times(\mathrm{V})}$
$\mathrm{w}_2=$ Amount of glucose
Given molarity $=\frac{\mathrm{M}}{20}$
$\frac{1}{20}=\frac{w_2 \times 1000}{180 \times 250}$
$w_2=\frac{180 \times 250}{20 \times 1000}$
$=2.25 \mathrm{~g}$

Topic: Chemistry, Solutions, NEET, NEET 2024 (Re-NEET)
The amount of heat energy radiated by a metal at temperature T is E. When the temperature is increased to 3T, energy radiated is

Options:
A 81E
B 9E
C 3E
D 27E

The correct answer is:
81E

Solution:
According to Stefan's law
EâT4
Eâ²E=3TT4or Eâ²=81E

Topic: Physics, Thermal Properties of Matter, NEET
The amount of heat energy required to raise the temperature of 1 g of helium at NTP, from $T_1 \mathrm{~K}$ to $T_2 \mathrm{~K}$ is

Options:
A $\frac{3}{8} N_2 K_B\left(T_2-T_1\right)$
B $\frac{3}{2} N_2 K_B\left(T_2-T_1\right)$
C $\frac{3}{4} N_2 K_B\left(T_2-T_1\right)$
D $\frac{3}{4} N_2 K_B\left(\frac{T_2}{T_1}\right)$

The correct answer is:
$\frac{3}{8} N_2 K_B\left(T_2-T_1\right)$

Solution:
We know that $Q=\frac{F}{2} n R \Delta T$
$\therefore$ Amount of heat required,
$\begin{aligned}
Q & =\frac{3}{2} \times \frac{1}{4} \times K_B N_B \Delta T \\
& =\frac{3}{8} N_B K_B\left(T_2-T_1\right)
\end{aligned}$

Topic: Physics, Kinetic Theory of Gases, NEET, NEET 2013 (All India)
The amount of heat (in calories) required to convert 5 g of ice at 0 °C to steam at 100 °C is [Lfusion=80 cal g-1, Lvaporization=540 cal g-1]

Options:
A 3100
B 3200
C 3600
D 4200

The correct answer is:
3600

Solution:
Ice 0 Â°C converts into steam 100 Â°C in following three steps.
Total heat required Q=Q1+Q2+Q3
Q=5Ã80+5Ã1Ã100-0+5Ã540
Q=3600 cal

Topic: Physics, Thermal Properties of Matter, NEET
The amount of living biomass present in a unit area of an ecosystem at a given time is referred to as:

Options:
A Standing state
B Standing crop
C Gross primary productivity
D Net primary productivity

The correct answer is:
Standing crop

Solution:
Standing crop is the total amount of living material in a specified population at a particular time, generally expressed as biomass. The standing crop may show variation at different times of the year.
Net Primary Productivity (NPP) is the rate at which plants incorporate atmospheric carbon through photosynthesis. (Definition)
Gross primary production (GPP) is the rate at which an ecosystem's producers capture and store a given amount of chemical energy as biomass in a given length of time. (Definition)

Topic: Biology, Ecosystem, NEET
The amount of nutrients, such as carbon, nitrogen, phosphorus and calcium present in the soil at any given time, is referred as:

Options:
A Climax
B Climax community
C Standing state
D Standing crop

The correct answer is:
Standing state

Solution:
The amount of nutrients, such as carbon, nitrogen, phosphorus, calcium, etc., present in the soil at any given time, is referred to as the standing state. It varies in different kinds of ecosystems and also on a seasonal basis.

Topic: Biology, Ecosystem, NEET, NEET 2021
The amount of nutrients, such as carbon, nitrogen, phosphorus, calcium, etc., present in the soil at any given time is referred to as the

Options:
A standing state
B standing crop
C standing soil
D standing earth

The correct answer is:
standing state

Solution:
The amount of nutrients, such as carbon, nitrogen, phosphorus, calcium, etc., present in the soil at any given time, is referred to as the standing state. It varies in different kinds of ecosystems and also on a seasonal basis

Topic: Biology, Ecosystem, NEET
The amount of nutrients such as carbon, nitrogen, phosphorus, calcium etc. present in the soil at any given time is referred to as

Options:
A Standing crop
B Standing mass
C Standing state
D Mineralization

The correct answer is:
Standing state

Solution:
A definite amount of living material present at each trophic level at a specific time is called a standing crop. It can be measured as biomass or a number in a unit area. The nutrients which are necessary for growth such as carbon, phosphorous, calcium, etc. are present in abiotic components such as soil and are known as standing state. Mineralisation is the process by which inorganic nutrients are released into the soil by the degradation of humus by microbes. Hence, the answer is standing state.

Topic: Biology, Ecosystem, NEET
The amount of nutrients such as carbon, nitrogen, potassium and calcium present in the soil at any given time is referred to as :

Options:
A Humus
B Detritus
C Standing state
D Standing crop

The correct answer is:
Standing state

Solution:
The amount of nutrients such as carbon, nitrogen, phosphorus, calcium, etc., present in the soil at any given time is referred to as the standing state. Standing crop is the amount of living material present in different trophic levels at a given time.

Topic: Biology, Ecosystem, NEET, NEET 2023 (Manipur)
The amount of oxygen required by the microbes in the decomposition of organic matter is called:

Options:
A Chemical oxygen demand
B Biochemical oxygen demand
C Total oxygen demand
D Dissolve oxygen

The correct answer is:
Biochemical oxygen demand

Solution:
BOD refers to the amount of oxygen that would consume if all the organic matter in one litre of water is to be oxidized by bacteria. The sewage water is treated till BOD is reduced. The greater the BOD, greater is the polluting potential. Once the BOD is reduced the effluent is then passed into a settling tank where the bacterial flocs are allowed to settle.

Topic: Biology, Microbes in Human Welfare, NEET
The amount of silver deposited by passing 241.25 C of charge through silver nitrate solution is

Options:
A 2.7 g
B 2.7 mg
C 0.27 g
D 0.54 g

The correct answer is:
0.27 g

Solution:
Given, current = 241.25 C
We know that 1 C electricity will deposit 1.118 × 10-3g of silver.
∴ 241.25 C electricity will deposit
= (1.118 × 10-3) × 241.25
= 0.27 g of silver.

Topic: Chemistry, Electrochemistry, NEET
The amount of solute (molar mass 60 g mol-1 ) that must be added to 180 g of water so that the vapour pressure of water is lowered by 10 % is

Options:
A 60 g
B 120 g
C 30 g
D 24 g

The correct answer is:
60 g

Solution:
Relative lowering of vapour pressure is given as
po-pspo=wAmAÃmBwB
As relative lowering of vapour pressure
=po-pspo=10100
mA=60,wB=180,wA=x
So 10100=x/60180/18
110=x/6010
It mean 60 g of the solute must be added to 180 g of water so that the vapour pressure of water is lowered by 10 %.

Topic: Chemistry, Redox Reactions, NEET
The amount of the Golgi bodies will be highest in

Options:
A Ciliated epithelial cell
B Red blood cell
C Smooth muscle cell
D Goblet cell

The correct answer is:
Goblet cell

Solution:
Goblet cell been secretory in function will have more amount of Golgi bodies and vacuoles as compared to other cells from the options. Golgi bodies are involved in packaging of proteins.

Topic: Biology, Structural Organisation in Animals, NEET
The amount of urea which is excreted out in the urine per day of an adult human is

Options:
A 25 to 30 grams
B 1 to 1.5 grams
C 18 grams
D 100 grams

The correct answer is:
25 to 30 grams

Solution:
The animals must detoxify ammonia by converting it into a relatively nontoxic form such as urea or uric acid. Mammals, including humans, produce urea, whereas reptiles and many terrestrial invertebrates produce uric acid. An adult human excretes, on an average, 1 to 1.5 litres of urine per day. The urine formed is a light yellow coloured watery fluid which is slightly acidic (pH-6.0) and has a characterestic odour. On an average, 25-30 gm of urea is excreted out per day.

Topic: Biology, Excretory Products and their Elimination, NEET
The amount of urea which is excreted out in the urine per day of an adult human is

Options:
A 25 to 30 grams
B 1 to 1.5 grams
C 18 grams
D 100 grams

The correct answer is:
25 to 30 grams

Solution:
Terrestrial organisms have evolved other mechanisms to excrete nitrogenous wastes. The animals must detoxify ammonia by converting it into a relatively nontoxic form such as urea or uric acid. Mammals, including humans, produce urea, whereas reptiles and many terrestrial invertebrates produce uric acid. Animals that secrete urea as the primary nitrogenous waste material are called ureotelic animals. An adult human excretes, on average, 1 to 1.5 litres of urine per day with 25â30gm of urea.

Topic: Biology, Excretory Products and their Elimination, NEET
The amount of urine released by humans in a day is

Options:
A 1 to 1.5 litres of slightly acidic (pH-6.0) urine having 45-60 gm of urea.
B 1 to 1.5 litres of slightly acidic (pH-6.0) urine having 25-30 gm of urea.
C 0 to 1 litres of slightly alkaline (pH-7.3) urine having 25-30 gm of urea.
D 1 to 1.5 litres of slightly acidic (pH-6.0) urine having 45-60 gm of urea.

The correct answer is:
1 to 1.5 litres of slightly acidic (pH-6.0) urine having 25-30 gm of urea.

Solution:
Urine
Daily excretion : 1 to 1.5 litres of urine per day.
Colour : Light yellow coloured watery fluid.
pH = 6.0 (slightly acidic).
It is having a characteristic odour.
Approximately 25-30 gm of urea is excreted per day.
Analysis of urine helps in clinical diagnosis of disorders of the kidney.

Topic: Biology, Excretory Products and their Elimination, NEET
The amount of water required to form one molecule of glucose in a photosynthetic reaction is

Options:
A 6 molecules
B 10 molecules
C 1 molecules
D 12 molecules

The correct answer is:
12 molecules

Solution:
Summary of photosynthesis is 6CO2 + 12H2O + Light energy
â Chlorophyll C 6 H 1 2 O 6 + 6O 2 â + 6H 2 O
Twelve molecules of water are required to form one molecule of glucose during photosynthesis.

Topic: Biology, Photosynthesis in Higher Plants, NEET
The amount of work done in increasing the voltage across the plates of the capacitor from 5 V to 10 V is W. The work done in increasing it from 10 V to 15 V will be

Options:
A W
B 0.6Â W
C 1.25Â W
D 1.67 W

The correct answer is:
1.67 W

Solution:
The work done in increasing the potential is given by
W=dUâ(V22-V12)
â´ W1W2=(V22-V12)(V32-V22) here V1=5v, V2=10V and V3=15V
â´ WW2+100-25225-100=75125
â´ W2=1.67 W

Topic: Physics, Electrostatics, NEET
The amount of work done in moving a charge of 5 C across two points having a potential difference of 15 Volts is equal to

Options:
A 0.333Â J
B 3Â J
C 6Â J
D 75 J

The correct answer is:
75 J

Solution:
Work done = (charge) × (potential difference)
W=q(ΔV)=5×15=75 J

Topic: Physics, Electrostatics, NEET
The amplitude and time period in SHM are 0.8 cm and 0.2 secâ¡ respectively. If the initial phase is Ï/2 radian, then the equation representing SHM is -

Options:
A y=0.8cos⁡10πt
B y=0.8sin⁡πt
C y=3×0.8sin⁡πt
D y=0.8sin⁡10πt

The correct answer is:
y=0.8cos⁡10πt

Solution:
Use general equation of SHM
y=asinâ¡Ït+Ï
=0.8sinâ¡2Ït0.2+Ï2
=0.8sinâ¡10Ï t+Ï2
=0.8cosâ¡Â 10Ït

Topic: Physics, Oscillations, NEET
The amplitude and time period in SHM are 0.8 cm and 0.2 sec respectively. If the initial phase is Ï2 radian, then the equation representing SHM is

Options:
A y=0.8cos⁡10πt
B y=0.8sin⁡πt
C y=3×0.8sin⁡πt
D y=0.8sin⁡10πt

The correct answer is:
y=0.8cos⁡10πt

Solution:
y=asin⁡ωt+ϕ
y=0.8sin⁡2πt0.2+π2
y=0.8cos⁡10πt

Topic: Physics, Oscillations, NEET
The amplitude of a simple pendulum is 10 cm. When the pendulum is at a displacement of 4 cm from the mean position, the ratio of kinetic and potential energies at that point is

Options:
A 5.25
B 2.5
C 4.5
D 7.5

The correct answer is:
5.25

Solution:
K.EP.E=12mw2A2-x212mw2 x2
K.EP.E=A2-x2x2
Given A=10âcm and x=4 cm
KE PE = 10 2 â 4 2 4 2 = 84 16 = 21 4 =5.25

Topic: Physics, Oscillations, NEET
The amplitude of a simple pendulum is 10 cm. When the pendulum is at a displacement of 4 cm from the mean position, the ratio of kinetic and potential energies at that point is

Options:
A 5.25
B 2.5
C 4.5
D 7.5

The correct answer is:
5.25

Solution:
K.EP.E=12mw2A2-x212mw2 x2
K.EP.E=A2-x2x2
Given A=10 cm and x=4 cm
KE PE = 10 2 − 4 2 4 2 = 84 16 = 21 4 =5.25

Topic: Physics, Oscillations, NEET
The amplitude of the charge oscillating in a circuit decreases exponentially as $Q=Q_0 e^{-R t 2 L}$, where $Q_0$ is the charge at $t=0 \mathrm{~s}$. The time at which charge amplitude decreases to $0.50 Q_0$ is nearly:
[Given that $R=1.5 \Omega, L=12 \mathrm{mH}, \ln (2)=0.693$ ]

Options:
A 19.01 ms
B 11.09 ms
C 19.01 s
D 11.09 s

The correct answer is:
11.09 ms

Solution:
Given $Q=Q_0 e^{-R t / 2 L}$
$R=1.5 \Omega, L=12 \mathrm{mH}, \operatorname{In}(2)=0.693$
$Q=0.5 Q_0, t=$ ?
$0.5 Q_0=Q_0 e^{-R t / 2 L}$
$\Rightarrow \quad \frac{1}{2}=e^{-R t / 2 L}$
Taking log on both sides
$\ln \left(\frac{1}{2}\right)=\ln e^{-R t / 2 L}$
$\Rightarrow \ln 2=\frac{R t}{2 L}$
$t=\frac{2 L \operatorname{In} 2}{R}=\frac{2 \times 12 \times 10^{-3} \times 0.693}{1.5}$
$t=11.09 \mathrm{~ms}$

Topic: Physics, Alternating Current, NEET, NEET 2024 (Re-NEET)
The amplitude of the two waves is in ratio 5 : 2. If all other conditions for the two waves are same, then what is the ratio of their energy densities?

Options:
A 5 : 2
B 5 : 4
C 4 : 5
D 25 : 4

The correct answer is:
25 : 4

Solution:
Energy density of wave is given by
u=2Ï2n2Ïa2
Or uâa2 as n and Ï are consants
â´u1u2=a12a22=5222
So, u1:u2=25:4

Topic: Physics, Waves and Sound, NEET
The anaphase-promoting complex is activated by

Options:
A M cdk cyclin
B G1 cdk cyclin
C S cdk cyclin
D Transcription factor

The correct answer is:
M cdk cyclin

Solution:
The cell cycle is controlled by the enzymes like cyclin-dependent kinases (Cdks). The Cdks phosphorylate amino acids like serine and threonine which initiate or block activities related to cell cycle. In eukaryotes, multiple cyclins are present and each of which acts during a specific checkpoint in cell cycle regulation. Common classes of cyclins includes

G1-phase cyclins

G1/S-phase cyclins

S-phase cyclins

M-phase cyclins

M-phase cyclins form M-CDK complexes and drive the cell's entry into mitosis and acts on the anaphase-promoting complex.

Topic: Biology, Cell Cycle and Cell Division, NEET
The anatomical unit of muscle is

Options:
A muscle fibre
B sarcomere
C filaments
D sutures

The correct answer is:
muscle fibre

Solution:
Myofibrils are assembled of repeated structures called sarcomeres (the smallest functional unit of muscle fibre). However, a single muscle cell has many myofibril (muscle fibre) units. Each sarcomere is composed of actin (called thin) and myosin (called thick) filaments, arranged in a precise order, and protein complexes that support the filament structures.

Topic: Biology, Structural Organisation in Animals, NEET
The anatomy of spring wood shows some peculiar features. Identify the correct set of statements about spring wood.

(a) It is also called the early wood.

(b) In spring, cambium produces xylem elements with narrow vessels.

(c) It is lighter.

(d) The spring wood along with autumn wood shows alternate concentric rings forming annual rings.

(e) It has a lower density.

Choose the correct answer from the options given below:

Options:
A (a), (c), (d) and (e) only
B (a), (b) and (d) only
C (c), (d) and (e) only
D (a), (b), (d) and (e) only

The correct answer is:
(a), (c), (d) and (e) only

Solution:
The true statements are:

1. It is also called early wood.

2. In spring cambium produces xylem elements with wider vessels.

3. Spring wood is lighter.

4. The spring wood along with autumn wood shows alternate concentric rings forming annual rings.

5. It has a lower density.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2022 (Phase 1)
The angle at which the intensity of transmitted light becomes maximum when a polarising sheet is rotated between two crossed polaroids is
[θ is the angle between transmission axes of first and middle polaroid]

Options:
A θ=90°
B θ=0°
C θ=45°
D it is independent of angle

The correct answer is:
θ=45°

Solution:
If the polarising sheet is making an angle θ with the first polarid, then energy of transmitted light, according to law of Malus, will become
I=I02×cos2θ×cos2(90−θ)
=I02×cos2θ×sin2θ
The intensity of light transmitted from the analyser will be maximum when θ=45°.

Topic: Physics, Wave Optics, NEET
The angle between pass axis of polarizer and analyser is 45o. The percentage of polarized light passing through analyser is

Options:
A 75%
B 25%
C 50%
D 100%

The correct answer is:
50%

Solution:
Given,
θ=45o
If I0 be the intensity of light before passing through the analyser, then intensity of light after passing through analyser,
I=I0 cos2⁡θ
=I0cos2⁡45o
=I0⋅122
I=I02
Hence, the percentage of polarized light passing through analyser
=I0-I02I0×100
=I0/2I0×100
=12×100
=50%

Topic: Physics, Wave Optics, NEET
The angle between the electric lines of force and the equipotential surface is

Options:
A 45°
B 90°
C 180°
D 0°

The correct answer is:
90°

Solution:
Electric lines of force are the electric field lines and electric field lines are perpendicular to equipotential surfaces

Topic: Physics, Electrostatics, NEET, NEET 2022 (Phase 1)
The angle between the overlapping of one s - orbital and one p - orbital is

Options:
A 180o
B 120o
C 190oÂ 28′
D 120o 60'

The correct answer is:
180o

Solution:
The overlap between s - and p - orbitals occurs along internuclear axis and hence the angle is 180o

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
The angle between vectors A â and B â is 60Â°. What is the ratio of A â Â· B â and |AâÃBâ| ?

Options:
A 1 2
B 1 3
C 1 4
D 1 5

The correct answer is:
1 3

Solution:
Now, = A â Â· B â | A â Ã B â | = AB cos Î¸ AB sin Î¸ = cot Î¸
Since Î¸ = 60Â°,
AâÂ·Bâ|AâÃBâ|=cot 60Â°=13

Topic: Physics, Rotational Motion, NEET
The angle of 1' (minute of arc) in radian is nearly equal to

Options:
A 2.91×10-4 rad
B 4.85×10-4 rad
C 4.80×10-6 rad
D 1.75×10-2 rad

The correct answer is:
2.91×10-4 rad

Solution:
First, convert one minute to a second, 1'=160° and then convert it into radian, 160×π180 rad, so we can say that in one minute there is, 2.91×10-4 rad.

Topic: Physics, Experimental Physics, NEET, NEET 2020 (Phase 2)
The angle of a prism is 60o and its refractive index is 2. The angle of minimum deviation suffered by a ray of light in passing through it is

Options:
A About 20o
B 30o
C 60o
D 45o

The correct answer is:
30o

Solution:
Refractive index, μ=sin⁡A+δm2sin⁡A2
⇒ 2=sin⁡60o+δm2sin⁡60o2
⇒ 2×sin 30o=sin⁡60o+δm2
⇒ sin⁡45o=sin⁡60o+δm2
⇒ δm=30o

Topic: Physics, Ray Optics, NEET
The angle of a prism is ‘ A ’. One of its refracting surfaces is silvered. Light rays falling at an angle of incidence 2A on the first surface returns back through the same path after suffering reflection at the silvered surface. The refractive index μ, of the prism is:

Options:
A 2sin⁡A
B 2cos⁡A
C 12cos⁡A
D tan⁡A

The correct answer is:
2cos⁡A

Solution:
To retrace its path light rays should fall normally on the reflecting surface. So
r2=0
⇒r1=A-r2⇒r1=A
Now applying snell rule between incident ray and refracted ray.
1sin⁡2A=nsin⁡A⇒2sin⁡Acos⁡A=nsin⁡A
⇒n=2cos⁡A

Topic: Physics, Ray Optics, NEET, NEET 2014
The angle of dip at a certain place on earth is 60Â° and the magnitude of the earth's horizontal component of the magnetic field is 0.26 G. The magnetic field at the place on earth is

Options:
A 0.13 G

B 0.26 G
C 0.52 G
D 0.65 G

The correct answer is:
0.52 G

Solution:
Horizontal component of earth's magnetic field is the component of earth's magnetic field along the horizontal direction.
BH=B cos â¡Î´
0.26=B cos â¡60Â°
â B=0.26cos â¡60Â°=0.52 G

Topic: Physics, Magnetic Properties of Matter, NEET
The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index Î¼) is

Options:
A sin-1μ
B sin-11μ
C tan-11μ
D tan-1μ

The correct answer is:
tan-1μ

Solution:
According to Brewster's law of polarization,
μ=tanip where ip is angle of incidence
∴ip=tan-1μ

Topic: Physics, Wave Optics, NEET
The angle of incidence for a ray of light at a refracting surface of a prism is 45o. The angle of prism is 60o. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are:

Options:
A 45o;12
B 30o; 2
C 45o;2
D 30o;12

The correct answer is:
30o; 2

Solution:
At minimum deviation δmin=2i-A
δmin=245-60
δmin=30o
Refractive index of material is
μ=sin⁡δmin+A2sin⁡A2=sin⁡30+602sin⁡30o
μ=sin⁡45osin⁡30o=1212=2

Topic: Physics, Ray Optics, NEET, NEET 2016 (Phase 1)
The angle of minimum deviation produced by an equilateral prism is 46o. The refractive index of material of the prism.

Options:
A 1.6
B 1.5
C 1.4
D 1.8

The correct answer is:
1.6

Solution:
δmin=2 i–A⇒i=53o
and 2r=A
μ=sin⁡isin⁡r=sin⁡53sin⁡30
μ=1.6

Topic: Physics, Ray Optics, NEET
The angle which the velocity vector of a projectile, thrown with a velocity v, at an angle θ to the horizontal, will make with the horizontal, after time t of its being thrown up is

Options:
A Î¸
B tan-1⁡θt
C tan-1⁡v cosθv sinθ-gt
D tan-1⁡v sinθ-gtv cosθ

The correct answer is:
tan-1⁡v sinθ-gtv cosθ

Solution:
velocity at any time t, during the motion of a projectile is given as
v→=vcos θi^+vsin⁡θ-gtj^
∴ tanα=vyvx=v sinθ-gtv cosθ

Topic: Physics, Motion In Two Dimensions, NEET
The angle which the velocity vector of a projectile, thrown with a velocity v, at an angle Î¸ to the horizontal, will make with the horizontal after time t of its being thrown up is

Options:
A Î¸
B tan-1⁡θt
C tan-1⁡v cosθv sinθ-gt
D tan-1⁡v sinθ-gtv cosθ

The correct answer is:
tan-1⁡v sinθ-gtv cosθ

Solution:
velocity at any time t, during the motion of a projectile is given as
v→=vcos θi^+vsin⁡θ-gtj^
∴ tanα=vyvx=v sinθ-gtv cosθ

Topic: Physics, Motion In Two Dimensions, NEET