datavorous/entrance-exam-dataset huggingface dataset:
The absolute growth rate is

Options:
A measurement and the comparison of total growth rate per unit time.
B growth rate in whole life.
C growth from birth to death.
D per unit initial parameters

The correct answer is:
measurement and the comparison of total growth rate per unit time.

Solution:
Quantitative comparisons between the growth of living system can also be made in two ways : (i) measurement and the comparison of total growth per unit time is called the absolute growth rate. (ii) The growth of the given system per unit time is expressed on a common basis, e.g., per unit initial parameter is called the relative growth rate.

Topic: Biology, Plant - Growth and Development, NEET
The absorption of monosaccharide such as glucose in the small intestine occurs by:

Options:
A Simple diffusion method
B Facilitated transport method
C Active transport method
D All of the above mention methods

The correct answer is:
All of the above mention methods

Solution:
The absorption of some monosaccharide such as glucose and some aminoacids may occur by all three possible ways. Glucose is absorbed from lumen of small intestine by simple diffusion, facilitated transport as well as by an active transport method.

Topic: Biology, Digestion and Absorption, NEET
The acceleration due to gravity at a depth R2 below the surface of the earth is

Options:
A g
B 2g
C g2
D g4

The correct answer is:
g2

Solution:
The acceleration due to gravity at a distance x(x<R) from centre of earth

g'=gxR⇒g'=gR2R=g2

Topic: Physics, Gravitation, NEET
The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then

Options:
A d=12km
B d=1 km
C d=32km
D d=2 km

The correct answer is:
d=2 km

Solution:
Acceleration due to gravity in both cases are

Above earth surfaceg′=g1-2hRe

Below earth surfaceg′=g1-dRe
As per the problem, g( 1− 2h R e )=g( 1− d R e )

Simplifying the above relations, we get
d=2h
The depth is d=2×1 km

Topic: Physics, Gravitation, NEET, NEET 2017
The acceleration due to gravity on the planet $\mathrm{A}$ is 9 times the acceleration due to gravity on planet B. A man jumps to a height of $2 \mathrm{~m}$ on the surface of A. What is the height of jump by the same person on the plane $\mathrm{B}$ ?

Options:
A $2 / 9 \mathrm{~m}$
B $18 \mathrm{~m}$
C $6 \mathrm{~m}$
D $2 / 3 \mathrm{~m}$

The correct answer is:
$18 \mathrm{~m}$

Solution:
Height of jump on the planet B $=\frac{g_{\mathrm{A}}}{g_{\mathrm{B}}} \times$ height of jump on the planet A
$=2 \times 9=18 \mathrm{~m}$

Topic: Physics, Gravitation, NEET, NEET 2003
The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2 m on the surface of A. What is the height of jump by the same on the planet B?

Options:
A 6 m

B 23 m
C 29 m
D 18 m

The correct answer is:
18 m

Solution:
It is given that, acceleration due to gravity on plane A is 9 times the acceleration due to gravity on the planet B , i.e.,
gA=9gB .... (i)
From the 3rd equation of motion, we have
v2=2gh
At planet A ,
hA=v22gA .... (ii)
At planet B ,
hB=v22gB .... (iii)
Dividing equation (ii) by equation (iii), we get
hAhB=gBgA
From equation (i), gA=9gB
hAhB=gB9gA=19⇒hB=9hA=9×2=18 m

Topic: Physics, Laws of Motion, NEET
The acceleration due to gravity on the surface of a planet is one-fourth of the value on Earth. When a brass ball is brought to this planet, its

Options:
A Mass is halved
B Weight is halved
C Mass becomes one-fourth
D Weight becomes one-fourth

The correct answer is:
Weight becomes one-fourth

Solution:
We know that,
Weight w=mg
So, w∝g
According to question,
w2w1=g4g
w2w1=14

Topic: Physics, Gravitation, NEET
The acceleration due to gravity on the surface of moon is 1.7 m${{s}^{-2}}$ . What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m${{s}^{-2}}$)

Options:
A 9.3s
B 8.4s
C 7.3s
D 7.47s

The correct answer is:
8.4s

Solution:
Correct Option is : (B)
8.4s

Topic: Chemistry, Oscillations, NEET
The acceleration due to the gravity on the moon is one-sixth of the acceleration due to gravity on the earth. If the average density of the moon is three fifth that of earth, the moon's radius in terms of earth's radius Re is

Options:
A 0.16 Re
B 0.27 Re
C 0.32 Re
D 0.36 Re

The correct answer is:
0.27 Re

Solution:
We know
g=43πGRρ . so
g Moongearth=43πGRMoon ρMoon43πGReρe
⇒ g/6g=43πGRmρm/43πGReρe
⇒ Rm=518Re=0.27 Re

Topic: Physics, Gravitation, NEET
The acceleration of 3 kg block in the arrangement shown in the diagram, when the given system is released from rest is

Options:
A 4.3 m s-2

B 3.7 m s-2

C 6 m s-2

D 4 m s-2

The correct answer is:
4.3 m s-2

Solution:
The maximum static friction on 5 kg block will be = 0.8 × 50 = 40 N

But the pulling force by 3 kg block is only 30 N. Hence 5 kg block will not move.
Let a1kg = 2a (from constraint relation)
For 3 kg block : 30 - 2T = 3a and for 1 kg block : T = 2a
Solving the above two equations, we get
a = (30/7) m s-2 = 4.3 m s-2

Topic: Physics, Laws of Motion, NEET
The acceleration of a body due to attraction of the earth (radius R ) at a distance 2R from the surface of the earth is ( g= acceleration due to gravity at the surface of the earth)

Options:
A g9
B g3
C g4
D g

The correct answer is:
g9

Solution:
g′=gR2R+h2
g′=gR2R+2R2
g′=gR23R2
g′=g9

Topic: Physics, Gravitation, NEET
The acceleration of a particle increases linearly with time t as 6t. If the initial velocity of the particle is zero and the particle starts from the origin, then the distance travelled by the particle in time t will be

Options:
A t
B t2
C t3
D t4

The correct answer is:
t3

Solution:
We know that a=dvdt
Then according to the question
dvdt=6t or dv=6tdt,v=6t22=3t2,
Also, dx=3t2dt â x=3t33=t3

Topic: Physics, Motion In One Dimension, NEET
The acceleration of an electron due to the mutual attraction between the electron and a proton when they are 1.6 Å apart is,
me≃9×10-31 kg, e=1.6×10-19 C (Take 14πε0=9×109 N m2 C-2)

Options:
A 1024 m s-2
B 1023 m s-2
C 1022 m s-2
D 1025 m s-2

The correct answer is:
1022 m s-2

Solution:
Recall the formula of electrostatic force interns of charge on electron and septation between them, F=Ke2r2, now use Newton's second law, ma=Ke2r2

⇒a=Ke2mr2,

so, the acceleration, a=9×1091.6×10-1921.6×10-1029×10-31

⇒a=10-29×1051=1022 m s-2

Topic: Physics, Electrostatics, NEET, NEET 2020 (Phase 2)
The acceleration of electron in the first orbit of hydrogen atom is (where the symbols have their usual meanings)

Options:
A 4π2mh3
B h24π2mr
C h24π2m2r3
D m2h24π2r3

The correct answer is:
h24π2m2r3

Solution:
From mvr=nh2π,v=nh2πmr
Acceleration, a=v2r=n2h24π2m2r2(r)=h24π2m2r3

Topic: Physics, Atomic Physics, NEET
The acceleration of the 10 kg mass block shown in the figure is:-

Options:
A 3.517.5g
B 7.517.5g
C 14.517.5g
D g7

The correct answer is:
g7

Solution:
Refer to the free-body diagrams
T-10gsinâ¡30Â°=10a or T-5g=10a
Again , 7.5gâT=7.5âa
Adding, 2.5g=17.5a
or a= 25g 175 = g 7

Topic: Physics, Laws of Motion, NEET
The acceleration of the 500 g block in figure is

Options:
A 6g13 downwards
B 7g13 downwards
C 8g13 downwards
D 9g13 upwards

The correct answer is:
8g13 downwards

Solution:
500g -T=500a -(1)
T-100gsinâ¡30Â°-Tâ²â²=100a
or T-Tâ²â²-50g=100a -(2)
Again, Tâ²â²-50g=50a -(3)
From Eqs. (ii) and (iii)
T-100g=150a -(4)
Adding Eqs. (i) and (iv), 400g = 650 a or a=400g650=8g13
This acceleration is downwards

Topic: Physics, Laws of Motion, NEET
The acceleration to gravity at a height 1/20th of the radius of the earth above the earth surface is 9ms-2. Its value at a point at an equal distance below the surface of the earth in ms-2 is about

Options:
A 8.5
B 9.5
C 9.8
D 11.5

The correct answer is:
9.5

Solution:
At a height h,

gh=g1-2hR

for h=R20 we get

9=g1-110=910g

At a depth d,

gd=g1-dR

for d=R20

gd=g1-120=1920g

⇒gd9=1918⇒gd=9.5 m s-2

Topic: Physics, Gravitation, NEET
The acceleration-displacement graph of a particle executing simple harmonic motion is shown in the figure. The frequency of oscillation is

Options:
A 2.5Ï s-1
B 2π10 s-1
C 12π5 s-1
D 12π20 s-1

The correct answer is:
2.5Ï s-1

Solution:
Using slope of the curve we can write equation
a=-10 x
Comparing it with a=âÏ2 x
Ï2=10
Ï=10
frequency f=Ï2Ï=12Ï10 s-1

Topic: Physics, Oscillations, NEET
The acellular forms of life could have originated around

Options:
A 1000 mya
B 1500 mya
C 2000 mya
D 3000 mya

The correct answer is:
3000 mya

Solution:
The first non-cellular forms of life could have originated
3 billion years back. They would have been giant molecules (RNA, protein, polysaccharides, etc.). These capsules reproduced their molecules perhaps. The first cellular form of life did not possibly originate until about 2000 million years ago. These were probably single-cells. All life forms were in a water environment only.

Topic: Biology, Evolution, NEET
The acid concentration in CAM plants is more at

Options:
A night
B daytime
C dawn
D dusk

The correct answer is:
night

Solution:
CAM is the crassulacean acid metabolism which is used by some plants to adapt in arid conditions with the help of the carbon fixation pathway. The peculiarity of these plants is that their stomata are closed during the day and opens at night when the temperature is low. This characteristic feature enables the CAM plants to minimize photorespiration and helps them survive in dry conditions or when exposed to high temperatures. The opening of stomata at night helps them to diffuse carbon dioxide in their leaves which is then further converted to malate and other organic acids by the enzyme PEP carboxylase.

Topic: Biology, Photosynthesis in Higher Plants, NEET
The acid concentration in CAM plants is more at

Options:
A day time
B night
C dawn
D dusk

The correct answer is:
night

Solution:
Crassulacean Acid Metabolism (CAM pathway) occurs in succulent plants. CO2 is fixed during the night (dark) to malic acid via PEP carboxylase. This CO2 comes from respiration (the breakdown of starch) as also from the atmosphere. Malic acid gets stored in vacuoles during the night which is decarboxylated to release during the day. Formation of malic acid during dark is called acidification or phase-I. Release of for actual photosynthesis during day is called deacidification or phase-II.
The CAM plants also contain the enzyme of Calvin cycle. Thus during day time, malic acid breaks into pyruvate and CO2. While CO2, enters the calvin cycle, pyruvate is used up to regenerate PEP. The succulents, therefore synthesize plenty of organic acids from CO2 during night (when stomata are open) and plenty of carbohydrates during the day (when stomata are closed).

Topic: Biology, Photosynthesis in Higher Plants, NEET
The acid form of an acid base indicator is yellow in acid and red in basic form. What is the change in pH in order to change the indicator from 80% yellow to 80% red.

Options:
A 1.2
B 2.1
C 0.12
D 12

The correct answer is:
1.2

Solution:
First case Second case HIn Acid form 80% yellow 20% yellow ⇌ H ⊕ + In − Base form 30% red 80% red
First case
pH1=pKa+log[In-][HIn]
=pKa+log2080
=pKa+log1/4 …(i)
Second case
pH2=pKa+log8020
=pKa+log4 …(ii)
From equations (i) and (ii), we get
pH2-pH1=2log4=4log2
=4×0.3=1.2

Topic: Chemistry, Ionic Equilibrium, NEET
The acid having O – O bond is

Options:
A H2S4O6
B H2S2O6
C H2S2O3
D H2S2O8

The correct answer is:
H2S2O8

Solution:
Only H2S2O8 has O – O bond here

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
The acid used in separating biomolecules is

Options:
A trichloroacetic acid
B hydrochloric acid
C sulphuric acid
D nitric acid

The correct answer is:
trichloroacetic acid

Solution:
Chemical analysis is done to separate various biomolecules from the tissues. The living tissue is first grinded with trichloroacetic acid in mortar and pestle and slurry is prepared. This slurry is filtered and thus two parts: retentate and filtrate is obtained. Each part is studied further to find out the biomolecules present in them. Trichloroacetic acid is used because it will not alter the molecules present in the tissues when we grind them.

Topic: Chemistry, Biomolecules, NEET
The acidic, basic and amphoteric oxides, respectivley, are

Options:
A Cl2O,CaO,P4O10
B MgO,Cl2O,Al2O3
C Na2O,SO3,Al2O3
D N2O3,Li2O,Al2O3

The correct answer is:
N2O3,Li2O,Al2O3

Solution:
Non metal oxide is acidic (N2O3)
Metal oxide is basic (Li2O)
Al2O3 is amphoteric.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
the acidic group in aqueous solution of glycine is

Options:
A -COOH
B -COO-
C -NH2
D − NH 3 +

The correct answer is:
− NH 3 +

Solution:
In aqueous solution, the α -amino acids form mostly Zwitter ions.
NH2CH2COOH⇌H3N+-CH2-COO-
If a base is added, H+ ion is furnished by -N+H3 . Hence, the acidic group is -N+H3.

Topic: Chemistry, Biomolecules, NEET
The acromion process can be observed in

Options:
A pelvic girdle
B pectoral girdle
C carpals
D tarsals

The correct answer is:
pectoral girdle

Solution:
The scapula (shoulder blade) consists of a sharp ridge, the spine, and a triangular body. The end of the spine projects as a flattened and expanded process called the acromion process. This process articulates with the clavicle. The scapula is a large triangular flat bone that is located in the dorsal part of the thorax between the second and the seventh ribs.

Topic: Biology, Locomotion and Movement, NEET
The acrosome of sperm is derived from:

Options:
A Golgi body
B Mitochondria
C Ribosomes
D Centriole

The correct answer is:
Golgi body

Solution:
Acrosome is the inverted cap like structure which is present on the sperm head. It is derived from the Golgi apparatus. Acrosome contains digestive enzymes like hyaluronidase. When the sperm comes in contact with the zonapellucida of the ovum, the acrosome releases these enzymes. They help in digestion of the zonapellucida and enable the sperm to penetrate the ovum. This is known as acrosomal reaction.
Mitochondria are present in the middle piece of the sperm. They provide the energy required by the sperm for fertilizing the ovum.

Topic: Biology, Human Reproduction, NEET
The action of jelly and cream in copulation is:

Options:
A Spermicidal and immobilizing the sperms also
B Entangles the sperms
C Preventing the ova to be released
D Enables the sperms to reach towards ovum speedly

The correct answer is:
Spermicidal and immobilizing the sperms also

Solution:
They act as spermicidal and immobilize the sperm.
Spermicide is a kind of birth control that has chemicals that stop sperm from reaching an egg.

Topic: Biology, Reproductive Health, NEET
The action of the vaginal diaphragm is

Options:
A To prevent the ova to come in the uterus
B To prevent the sperm to come in contact with ova
C Spermicidal
D Anti–implantational

The correct answer is:
To prevent the sperm to come in contact with ova

Solution:
The vaginal diaphragm, cervical caps and vaults are dome-shaped rubber plastic covers or barriers inserted into the female reproductive tract to cover the cervix during coitus. They prevent conception by blocking the entry of sperms through the cervix.

Topic: Biology, Reproductive Health, NEET
The activation energy for a simple chemical reaction $\mathrm{A} \rightarrow \mathrm{B}$ is $\mathrm{E}_a$ in forward direction. The activation energy for reverse reaction:

Options:
A is negative of $\mathrm{E}_a$
B is always less than $\mathrm{E}_a$
C can be less than or more $\mathrm{E}_a$
D is always double of $\mathrm{E}_a$

The correct answer is:
can be less than or more $\mathrm{E}_a$

Solution:
The activation energy for the reverse reaction, $\mathrm{E}_{\mathrm{a}(\mathrm{rev})}$, is the difference between the product energy and transition state at the peak of the diagram. $\mathrm{H}_r$ is the difference between the potential energy of the reactant and the potential energy of the product.
Related Theory
In a reversible reaction of bi-molecular type the activation energy for forward reaction is positive. Whereas, for the backward reaction activation energy is negative.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2003
The activation energy for the reaction: 2 HI(g) → H2+ I2(g) is 209.5 kJ mol–1at 581K. What is the fraction of molecules of reactants having energy equal to or greater than activation energy?

Options:
A 6.658 × 10–19
B 3.124 × 10–19
C 1.471 × 10–19
D 7.324 × 10–19

The correct answer is:
1.471 × 10–19

Solution:
Correct Option is : (C)
1.471 × 10–19

Topic: Chemistry, Chemical Kinetics, NEET
The activation energy of a reaction can be determined from the slope of which of the following graphs?

Options:
A lnÂ KÂ vs.Â T
B lnÂ KTÂ vs.Â T
C ln K vs.1T
D T ( ln  K ) vs. 1 T

The correct answer is:
ln K vs.1T

Solution:
As per Arrhenius equation
K=A e-EaRT
ln K= In A-EaRT
Plot of ln K v/s 1T will give
Slop=-EaR
∴Ea=-R(slope)

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2015 (Phase 1)
The activation energy of a reaction can be determined from the slope of which of the following graphs?

Options:
A In K vs. T
B In K vs. 1T
C <InKT vs. 1T
D InKT vs. T

The correct answer is:
In K vs. 1T

Solution:
Arrhenius equation K=A.e-Ea/RT
lnK=lnA-EaRT
So, activation energy of reaction can be determined from the slope of lnK vs 1T .
Using y=mx+c we can find the value of Ea from slope value.

Topic: Chemistry, Chemical Kinetics, NEET
The activation energy of a reaction is zero. The rate constant (k) of the reaction

Options:
A increases with the increase in temperature
B decreases with increase in temperature
C decreases with the decrease in temperature
D is independent of temperature

The correct answer is:
is independent of temperature

Solution:
When the activation energy of a reaction is zero, the rate constant (k) of the reaction is nearly independent of temperature.
Ea=0
So k=Ae-Ea/RT
=Ae-0=A
Hence the rate constant is independent of temperature.

Topic: Chemistry, Chemical Kinetics, NEET
The active chemical drug reserpine is obtained from

Options:
A Datura
B Rauwolfia
C Atropa
D Papaver

The correct answer is:
Rauwolfia

Solution:
The genetic variation shown by the medicinal plant Rauwolfia vomitoria growing in different Himalayan ranges might be in terms of potency and concentration of the active chemical (Reserpine) obtained from roots of plants.

Topic: Biology, Biodiversity and its Conservation, NEET
The active form of Entamoeba histolytica is a protozoan parasite in the

Options:
A large intestine
B stomach
C liver
D blood only

The correct answer is:
large intestine

Solution:
Entamoeba histolytica is a protozoan parasite in the large intestine of human which causes amoebiasis (amoebic dysentery). Symptoms of this disease include constipation, abdominal pain, and cramps, stools with excess mucous, and blood clots. The active form of Entamoeba histolytica feeds upon erythrocytes, mucosa, and submucosa of the colon as it causes damage by lysis of epithelial cells, neutrophils, and also red blood cells.

Topic: Biology, Human Health and Diseases, NEET, NEET 2015 (Phase 1)
The active mass of solid is generally taken as

Options:
A > 1
B = 1
C < 1
D 0

The correct answer is:
= 1

Solution:
The active mass of solids is usually taken as unity.

Topic: Chemistry, Chemical Equilibrium, NEET
The activity of a radioactive element decreased to one-third of original activity I0 in 9 yr. After further 9 yr, its activity will be

Options:
A I0
B 23I0
C I0/9
D I0/3

The correct answer is:
I0/9

Solution:
In 9 years, activity becomes I = I 0 3
In further 9 years, activity would becomes
I ′ = I 3 = I 0 3 × 3 = I 0 9

Topic: Physics, Nuclear Physics, NEET
The activity of a radioactive element decreased to one-third of the original activity I0 in a period of nine years. After a further lapse of nine years its activity will be

Options:
A I0
B 23I0
C I09
D I06

The correct answer is:
I09

Solution:
II0=12tT
⇒13=129T
⇒I′I0=1218T=129T2
⇒I′=I09

Topic: Physics, Atomic Physics, NEET
The activity of a radioactive element decreases to one-third of the original activity I0 in a period of nine years. After a further lapse of nine years its activity will be

Options:
A I0
B 23I0
C I09
D I06

The correct answer is:
I09

Solution:
I=I0e-Î»t
â I03=I0e-9Î»
e-9Î»=13
After a further lapse of nine years, total time is eighteen years I=I0eâÎ»Ã18
âI=I0eâÎ»Ã92
âI=I0132
âI=I09

Topic: Physics, Nuclear Physics, NEET
The activity of a radioactive sample decreases to 13rd of its original value in 3 days. Then, in 9 days its activity will become

Options:
A (127) of the original value
B (19) of the original value
C (118) of the original value
D (13) of the original value

The correct answer is:
(127) of the original value

Solution:
R=R0e-Î»t
â 13=e-Î» Ã 3=e-3Î» â¦(i)
Again R'R0=e-Î» Ã 9=e-9Î»=(e-3Î»)3
=133 [From Eq.i]
=127
â Râ = R027
Hence, in 9 days activity will become 127 of the original value.

Topic: Physics, Nuclear Physics, NEET
The activity of a radioactive sample is measured as 9750 counts per minute at t=0 and as 975 counts per minute at t=5 minutes. The decay constant is approximately

Options:
A 0.922 per minute
B 0.691 per minute
C 0.461 per minute
D 0.230 per minute

The correct answer is:
0.461 per minute

Solution:
dN dt =λN
9750=λ N 0 …. (i)
975=λN … (ii)
Dividing (i) by (ii)
N N 0 = 1 10
λ= 2.303 t log N 0 N = 2.303 5 log10
=0.4606=0.461 per minute

Topic: Physics, Nuclear Physics, NEET
The activity of a radioactive substance becomes from 8000 Bq to 1000 Bq in 12 days. What is the half-life of the radioactive substance?

Options:
A 4 days
B 6 days
C 2 days
D 3 days

The correct answer is:
4 days

Solution:
The value of initial sample is 8000. It decays in 4 days to half of the initial value i.e. 4000 atoms. In next four days, it further days to 2000 atoms. In next four days, it will again decay to 1000 atom i.e. half of the present initial value.
Hence,
So, half-life of sample T12=4 days

Topic: Physics, Nuclear Physics, NEET
The activity of a sample of radioactive material is A1 at time t1 and A2 at time t2 t2>t1. If its mean life is T then which of the following is correct.

Options:
A A1t1=A2t2
B A1+A2t2-t1= constant
C A2=A1et1-t2/T
D A2=A1et1/Tt2

The correct answer is:
A2=A1et1-t2/T

Solution:
A=A0e-tT
In time t1 activity is A1
⇒A1=A0e-t1T ....... (i)
In time t2 activity is A2
⇒A2=A0e-t2T ...... (ii)
Dividing equation (ii) by (i)
⇒A2A1=e-t2Te-t1T
⇒A2=A1e-t2-t1T
⇒A2=A1et1-t2T

Topic: Physics, Nuclear Physics, NEET
The activity of a sample of radioactive material is A1 at time t1 and A2 at time t2t2>t1 . Its mean life is T. Which of the following is correct?

Options:
A A1t1=A2t2
B A1+A2t2–t1= constant
C A2=A1et1–t2T
D A2=A1et1Tt2

The correct answer is:
A2=A1et1–t2T

Solution:
We know that for time t1 activity is A1 , therefore A1=A0eât1T .....(1)
Similarly for time t2 activity is A2 , Therefore A2=A0eât2T ......(2)
Now, dividing equation (1) by (2) A1A2=et2ât1T
Therefore, A2=A1et1ât2T

Topic: Physics, Nuclear Physics, NEET
The activity of a sample reduces from A0 to A03 in one hour. The activity after 3 hours more will be

Options:
A A033
B A 0 9
C A 0 9 3
D A 0 27

The correct answer is:
A 0 9

Solution:
After one hour A 0 3 A 0 = 1 3 = 1 2 1/T
After next 3 hours means four hours from initial
A A 0 = 1 2 4/T = 1 2 1/T 4 = 1 3 4 = 1 9

Topic: Physics, Nuclear Physics, NEET
The addition of a catalyst during a chemical reaction alters which of the following quantities?

Options:
A Entropy
B Internal energy
C Enthalpy
D Activation energy

The correct answer is:
Activation energy

Solution:
Catalyst is going to affect the activation energy of a chemical reaction. Activation energy is the minimum energy required to from activated complex or Transition state.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2016 (Phase 1)
The addition of Br2 to (E)-but-2-ene gives

Options:
A R, R-2-3- dibromobutane
B S,Â R-2-3- dibromobutane
C R, S-2-3- dibromobutane
D A mtextxture of R,Â R and S,Â SÂ 2,Â 3 dibromobutane

The correct answer is:
R, S-2-3- dibromobutane

Solution:
The addition of bromtextne to E- but-2-ene gives R, S-2-3- dibromobutane, and it is always anti as bromtextne atoms added are on opposite sides.
In E-but-2-ene addition of Br2 occurs as:
R, S-2-3- dibromobutane.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET