datavorous/entrance-exam-dataset huggingface dataset:
If g is the acceleration due to gravity on the surface of the earth, its value at a height equal to double the radius of the earth is

Options:
A g
B g2
C g3
D g9

The correct answer is:
g9

Solution:
Acceleration due to gravity at a height h form the surface of the earth
Â Â Â Â Â Â Â Â Â Â Â Â g′=g11+hR2
Given,Â Â h=2R
â´Â Â Â Â Â Â Â g′=g11+22
orÂ Â Â Â Â Â g′=g9

Topic: Physics, Gravitation, NEET
If gynoecium is situated in the centre and other parts of the flower are located on the rim of the thalamus almost at the same level, it is called ___.

Options:
A Epigynous
B Perigynous
C Hypogynous
D None of the above

The correct answer is:
Perigynous

Solution:
If gynoecium is situated in the centre and other parts of the flower are located on the rim of the thalamus almost at the same level, it is called Perigynous.

Topic: Biology, Morphology of Flowering Plants, NEET
If ΔH<0,ΔS>0,ΔG<0 then the reaction is

Options:
A Spontaneous reaction at high temperature
B Spontaneous reaction at all temperatures
C Non spontaneous reaction at low temperature
D Non spontaneous reaction at high temperature

The correct answer is:
Spontaneous reaction at all temperatures

Solution:
Correct Option is : 2
Spontaneous reaction at all temperatures

Topic: Chemistry, Thermodynamics (C), NEET
If H2SO4 ionises as H2SO4+2H2Oâ2H3O++SO42-, then total number of ions produced by 0.1 Molar and 1 L aqueous H2SO4 will be:

Options:
A 9.03Ã1021
B 3.01Ã1022
C 6.02Ã1022
D 1.8× 10 23

The correct answer is:
1.8× 10 23

Solution:
1 mole of H2SO4 gives = 3 moles of ions or 3Ã6.023Ã1023 ions
â´0.1 mole of H2SO4 will give =0.1Ã3Ã6.023Ã1023 ions =1.8Ã1023 ions

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
If ∆Hfo for H2O2 and H2O are -188 kJ/mol and -286 kJ/mol, what will be the enthalpy change of the reaction:

2H2O2l→2H2Ol+O2(g)

Options:
A -196 kJ
B -494 kJ
C 146 kJ
D -98 kJ

The correct answer is:
-196 kJ

Solution:
Given reaction is 2H2O2lâ2H2Ol+O2(g)
ΔH Reaction o = ΔH f o Products − ΔH f o Reactants
Enthalpy change =2Ã-286-2Ã-188
=-196 kJ

Topic: Chemistry, Thermodynamics (C), NEET
If ∆H is the change in enthalpy and ∆E, the change in internal energy accompanying a gaseous reaction, then

Options:
A ∆H is always greater than ∆E
B ∆H<∆E only if the number of moles of the products is greater than the number of moles of the reactants
C ∆H is always less than ∆E
D ∆H<∆E only if the number of moles of products is less than the number of moles of the reactants

The correct answer is:
∆H<∆E only if the number of moles of products is less than the number of moles of the reactants

Solution:
As ∆H=∆E+∆ngRT
If np<nr ;∆ng=np-nr=-ve
Hence ∆H<∆E

Topic: Chemistry, Thermodynamics (C), NEET
If ΔH is the change in enthalpy and ΔE the change in internal energy for a gaseous reaction then

Options:
A ΔH is always greater than ΔE
B ΔH<ΔE only if the number of moles of the products is greater than the number of the reactants
C ΔH is always less than ΔE
D ΔH<ΔE only if the number of moles of the products is less than the number of moles of the reactants

The correct answer is:
ΔH<ΔE only if the number of moles of the products is less than the number of moles of the reactants

Solution:
The change in enthalpy and the change in internal energy for a gaseous reaction is given by
ΔH=ΔE+ΔngRT
If Δn=-ve than ΔH<ΔE .

Topic: Chemistry, Thermodynamics (C), NEET
If Henle's loop were absent from mammalian nephron, which one of the following is to be expected ?

Options:
A there will be no urine formation
B there will be hardly any change in the quality and quantity of urine formed
C the urine will be more concentrated
D the urine will be more dilute

The correct answer is:
the urine will be more dilute

Solution:
The principal function of the loop of Henle appears to be the recovery of water and sodium chloride from the urine. This function allows production of urine, that is more concentrated than blood, limiting the amount of water needed as intake for survival. Thus, if Henle's loop were absent from mammalian nephron, urine will be more dilute.
Related Theory
Mammalian kidneys have the loop of Henle and length of the loop in the kidney determines how concentrated the final urine is. Birds, and reptiles don't have long loops of Henle in their kidneys. They use their cloaca to pull water from their faeces and urine.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2003
If hydrogen and oxygen are mixed and kept in the same vessel at room temperature, the reaction does not take place to from water because

Options:
A activation energy for the reaction is very high at room temperature
B molecules have no proper orientation to react to form water
C the frequency of collisions is not high enough for the reaction to take place
D no catalyst is present in the reaction mixture

The correct answer is:
activation energy for the reaction is very high at room temperature

Solution:
The activation energy for the reaction is very high at room temperature hence reaction will not take place.

Topic: Chemistry, Chemical Kinetics, NEET
If IO4- is converted into I2 in acid medium, calculate the equivalent mass of IO4-.

Options:
A M/14
B M/7
C M/5
D M/3

The correct answer is:
M/7

Solution:
Equivalent mass =Molecular weight Change in oxidation number per mole
Suppose molecular weight is M
Oxidation number of I2 in IO4- in Acidic medium i.e., IÃ(-8)+1e-=+7
So eq. wt. =M/7 .

Topic: Chemistry, Redox Reactions, NEET
If in a circular coil A of radius R, current i is flowing and in another coil B of radius 2R, a current 2i is flowing, then the ratio of the magnetic fields, BA and BB produced by them will be

Options:
A 1
B 2
C 12
D 4

The correct answer is:
1

Solution:
Magnetic field in circular coil A is
Similarly, BA=Î¼0Ni2R
R is radius and i is current flowing in coil.
BB=Î¼0N(2i)2 .(2R)
=Î¼0Ni2R
BABB=11=1

Topic: Physics, Magnetic Effects of Current, NEET
If in a double stranded DNA, 15 pitches in a helix are present, then the number of total nucleotides in it will be

Options:
A 150
B 300
C 600
D Infinite

The correct answer is:
300

Solution:
In 1953, J.D. Watson (an American biologist) and F.H.C. Crick (a British physicist) proposed a three-dimensional model of physiological DNA (i.e., B-DNA) based on X-ray diffraction data of DNA obtained by Franklin and Wilkins. For this epoch-making discovery, Watson, Crick and Wilkins got Nobel Prize in medicine in 1962. According to them, the diameter of DNA is 20 nm or 20 Angstroms. Adjacent bases are separated 0.34 nm or by 3.4 Angstroms along the axis. The length of a complete turn of helix is 3.4 nm or 34 Angstroms, i.e., there are 10 bp per turn/pitch. Thus, 15 pitches will have 150 bp, i.e., 300 nucleotides.

Topic: Biology, Molecular Basis of Inheritance, NEET
If in a p-n junction diode, a square input signal of 10 V is applied as shown

Then the output signal across RL will be

Options:
A
B
C
D

The correct answer is:

Solution:
The current will flow through RL when the diode is forward biased.

Topic: Physics, Semiconductors, NEET
If in a photosynthetic system H2O18 type is provided, what will be the type of oxygen released in the process?

Options:
A O18
B O16
C normal O2
D O14

The correct answer is:
O18

Solution:
Photosynthesis is the process of converting light to chemical energy and using that to produce glucose.
The general word formula for photosynthesis is -
6CO2+12H2O C6H12O6+6H2O+6O2 in presence of light and chlorophyll. The whole photosynthesis process is divided into two broad parts -
Light reaction where the light energy is used to generate ATP and NADPH2 which is later used in the Dark reaction to reduce carbon dioxide into glucose (carbohydrate) molecule. The free oxygen released as a byproduct in the photosynthesis occurs by splitting water. The oxygen of the carbon dioxide is used in making the glucose molecule through the Calvin cycle.
Splitting of water molecule into protons and oxygen in presence of sunlight is called as photolysis of water that occurs in the non-cyclic photophosphorylation. Thus, if H2O18 is used in the photosynthesis process, the oxygen released during photosynthesis will be O18 type.

Topic: Biology, Photosynthesis in Higher Plants, NEET
If in a p-n junction, a square input signal of 10 V is applied as shown, then the output across RL will be

Options:
A
B
C
D

The correct answer is:

Solution:
In the given circuit the P-N junction is acting as a half wave rectifier (as it is forward-bias) output will be +Ve

Topic: Physics, Semiconductors, NEET, NEET 2015 (Phase 1)
If in a p-n junction, a square input signal of 10 V is applied as shown, then the output across RL will be

Options:
A
B
C
D

The correct answer is:

Solution:
In the given circuit the P-N junction is acting as a half wave rectifier (as it is forward-bias) output will be +Ve

Topic: Physics, Semiconductors, NEET
If in a p-n junction diode, a square input signal of 10 V is applied as shown

Then the output signal across R L will be

Options:
A

B

C

D

The correct answer is:

Solution:
The current will flow through R L when the diode is forward biased.

Topic: Physics, Semiconductors, NEET
If in a pond there are 20 lotus plants last year and through reproduction 8 new plants are added, taking the current population to 28, calculate the birth rate.

Options:
A 4 offsprings per year
B 0.4 offspring per lotus per year
C 2.5 offsprings per lotus per year
D 25 offsprings per year

The correct answer is:
0.4 offspring per lotus per year

Solution:
A population has certain attributes that an individual organism does not. An individual may have births and deaths, but a population has birth rates and death rates. In a population these rates refer to per capita births and deaths, respectively. The rates, hence, are expressed is change in numbers (increase or decrease) with respect to members of the population. Here is an example, If in a pond there are 20 lotus plants last year and through reproduction 8 new plants are added, taking the current population to 28, then the birth rate calculate as 8/20 = 0.4 offspring per lotus per year.

Topic: Biology, Organisms and Populations, NEET
If in nuclear fission, a piece of uranium of mass 5.0 g is lost, the energy obtained in kWh is

Options:
A 1.25×107
B 2.25×107
C 3.25×107
D 0.25×107

The correct answer is:
1.25×107

Solution:
E=Δmc2
=0.5×10-3×3×1082
=4.5×1013
E=4.5×10133.6×106
=1.25×107 kWh

Topic: Physics, Nuclear Physics, NEET
If in the circuit, power dissipation is 150 W, then R is

Options:
A 2 Ω
B 6 Ω
C 5 Ω
D 4 Ω

The correct answer is:
6 Ω

Solution:
Power =V2R
150=V2Rnet
=152Rnet
1Rnet=1R+12
∴ Rnet=2RR+2
∴ 150 =152R+22R
∴ R=6 Ω

Topic: Physics, Current Electricity, NEET
If initial concentration is doubled, the half-life period of reaction is also doubled. The order of reaction is

Options:
A zero
B third
C second
D first

The correct answer is:
zero

Solution:
For nth order reaction
t1/2∝1([A0])n-1
for zero order reaction
t1/2∝[A0] .

Topic: Chemistry, Chemical Kinetics, NEET
If input in a full-wave rectifier is e=50sinâ¡314t volt, diode resistance is 100 Î© and load resistance is 1 kÎ© then, pulse frequency of output voltage is

Options:
A 50Â Hz
B 100 Hz
C 150Â Hz
D 200Â Hz

The correct answer is:
100 Hz

Solution:
Input supply is e=50sin⁡314t volt
So frequency is f=ω2π=3142π f=50 Hz
For full wave rectifier frequency of output wave is double the input frequency. So it will be 100 Hz.

Topic: Physics, Semiconductors, NEET
If θ is the polarising angle for two optical media whose critical angles are C1 and C2, then the correct relation is

Options:
A sin⁡θ=sin⁡C2sin⁡C1
B θ=sin⁡C2sin⁡C1
C tan⁡θ=sin⁡C1sin⁡C2
D sin⁡θ=sin⁡C1sin⁡C2

The correct answer is:
tan⁡θ=sin⁡C1sin⁡C2

Solution:
According to the Brewster's Law
âtanâ¡Î¸=Î¼denserÎ¼rarer
Or tanâ¡Î¸=1sinâ¡C2Ãsinâ¡C11
[ âµÎ¼=1sinâ¡C, where, C is the critical angle]
â tanâ¡Î¸=sinâ¡C1sinâ¡C2

Topic: Physics, Wave Optics, NEET
If $\phi$ is the work function of photosensitive material in eV and light of wavelength of numerical value $\lambda=\frac{h c}{e}$ metre, is incident on it with energy above its threshold value at an instant then the maximum kinetic energy of the photo-electron ejected by it at that instant (Take $h$-Plank's constant, $c$-velocity of light in free space) is (in SI units):

Options:
A $e+2 \phi$
B $2 e-\phi$
C $e-\phi$
D $e+\phi$

The correct answer is:
$e-\phi$

Solution:
$(\mathrm{K} . \mathrm{E})_{\max }=\frac{h c}{\lambda}-\phi_0$
and $\lambda=\frac{h c}{e}$
$\Rightarrow(\mathrm{K} . \mathrm{E})_{\max }=\frac{h c}{\frac{h c}{e}}-\phi_0$
$\Rightarrow(K . E)_{\max }=e-\phi_0 \Rightarrow(K . E .)_{\max }=e \dashv \phi$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2024 (Re-NEET)
If K1 and K2 are maximum kinetic energies of photoelectrons emitted when lights of wavelength Î»1 and Î»2 respectively incident on a metallic surface. If Î»1=3Î»2, then

Options:
A K1>K23
B K1<K23
C K1=3K2
D K2=3K1

The correct answer is:
K1<K23

Solution:
Eienstein's photoelectric equation is given as,
K1=hc1λ1-1λ0 ...(i)
and K2=hc1λ2-1λ0 ...(ii)
Dividing Equation (i) by Equation (ii), we get
K1K2=1λ1-1λ01λ2-1λ0
Since, λ1=3λ2
K1K2=13λ2-1λ01λ2-1λ0
=13λ2 1-3λ0/λ21λ21-1λ0
K1K2=13 1-3λ0/λ21-1λ0/λ2
3K1K2=1-3/λ0/λ21-1/λ0λ2<1
∴ 3K1K2<1⇒K1<K23

Topic: Physics, Dual Nature of Matter, NEET
If K1 and K2 are the maximum kinetic energies of photo electrons emitted when lights of wavelength λ1 and λ2 respectively are incident on a metallic surface and λ1=3λ2. Then

Options:
A K1>K2/3
B K1<K2 /3
C K1=3K2
D K2=3K1

The correct answer is:
K1<K2 /3

Solution:
Maximum kinetic energies of photo electrons for light of wavelength K1=hcλ1-f ,
Maximum kinetic energies of photo electrons for light of wavelength K2=hcλ2-f
Also given, λ1=3λ2
K2K1=3hcλ1-ϕhcλ1-ϕ=3hcλ1-ϕ3hcλ1-ϕ
Therefore K2K1>3

Topic: Physics, Dual Nature of Matter, NEET
If ks and kp respectively are effective spring constant in series and parallel combination of springs as shown in the figure, find kâ¡ sâ¡ kâ¡ pâ¡

Options:
A 9 2
B 3 7
C 2 9
D 7 3

The correct answer is:
2 9

Solution:
The effective spring constant ks of this arrangement is
1 kâ¡ sâ¡ = 1 kâ¡ + 1 2 kâ¡
1 kâ¡ sâ¡ = 2 + 1 2 kâ¡ = 3 2 kâ¡
kâ¡ sâ¡ = 2 kâ¡ 3
The effective spring constant kp of this arrangement is
kâ¡ pâ¡ = kâ¡ 1 + kâ¡ 2
= kâ¡ + 2 kâ¡ = 3 kâ¡
∴ kâ¡ sâ¡ kâ¡ pâ¡ = 2 kâ¡ / 3 3 kâ¡ = 2 9

Topic: Physics, Oscillations, NEET
If kinetic energy of a body is increased by $300 \%$ than percentage change in momentum will be:

Options:
A $100 \%$
B $150 \%$
C $265 \%$
D $73.2 \%$

The correct answer is:
$100 \%$

Solution:
Here $P_1=\sqrt{2 m E_1}$ and $P_2=\sqrt{2 m E_2}$
$\begin{aligned}
P^{\prime} & =\sqrt{2 m\left(E+\frac{300}{100} E_1\right)} \\
& =\sqrt{2 m 4 E_1} \\
& =2 P_1
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2002
If l1,l2,l3 are the lengths of the emitter, base and collector of a transistor, then

Options:
A l1=l2=l3
B l3<l2>l1
C l3<l1<l2
D l3>l1>l2

The correct answer is:
l3>l1>l2

Solution:
In a transistor base is thinnest and emitter is largest therefore l2 is least

Topic: Physics, Semiconductors, NEET
If Levitt's hypothesis is to be believed, then the opening and closing of stomata take place as a result of ______ in and out of guard cells.

Options:
A Active transport of calcium ions
B Active transport of potassium ions
C Diffusion of potassium ions
D Diffusion of calcium ions

The correct answer is:
Active transport of potassium ions

Solution:
According to Levitt's proton pump theory entry of potassium ions in the guard cells from surrounding epidermal cells in exchange of protons, causes opening of stomata. This exchange of protons and potassium ions requires energy and thus is an active process. Increase in potassium ions, increases the osmotic concentration of guard cells causing endo-osmosis. This makes the guard cells turgid causing opening of stomata

Topic: Biology, Transport in Plants, NEET
If light is available for a given duration and then again made unavailable, the biosynthetic process will

Options:
A continue for long duration.
B continue for unlimited time.
C continue for short duration.
D will stop immediately.

The correct answer is:
continue for short duration.

Solution:
One of the steps in the photosynthesis process is dark reaction where the carbondioxide is reduced to glucose via the Calvin cycle and this process is independent of light. This phase is called as biosynthetic phase. The process is dependent on the products of light reaction but not on light itself. If the light becomes unavailable to the plant, the dark reaction phase (biosynthetic process) will continue until the products are available. After the depletion of the products, this process will stop until light is available again. Hence, dark reaction will continue for a short duration even if light is not available.

Topic: Biology, Photosynthesis in Higher Plants, NEET
If linkage was known at the time of Mendel then which of the following laws, he would NOT have been able to explain?

Options:
A Law of dominance
B Law of independent assortment
C Law of segregation
D Law of purity of gametes

The correct answer is:
Law of independent assortment

Solution:
Mendel's Law of Independent Assortment states that "when two or more characteristics are inherited, individual hereditary factors assort independently during gamete production, giving different traits an equal opportunity of occurring together."
When two or more different genes are located to close to each other in a chromosome are said to be linked. Linked gene are found to avoid recombination and are not assorted during gamete production.
Linked genes are in contradiction to the Law of independent assortment thus, if Mendel had known about linkage during his experiments, he would have not been able to explain the inheritance of two genes without one affecting the other.

Topic: Biology, Principles of Inheritance and Variation, NEET
If $M(A, Z), M_P$ and $M_n$ denote the masses of the nucleus ${ }_Z^A X$, proton and neutron respectively in units of $u\left(1 \mathrm{u}=931.5 \mathrm{MeV} / c^2\right)$ and $\mathrm{BE}$ represents its binding energy in $\mathrm{MeV}$, then

Options:
A $M(A, Z)=Z M_p+(A-Z) M_n-\mathrm{BE} / c^2$
B $M(A, Z)=Z M_p+(A-Z) M_n+\mathrm{BE}$
C $M(A, Z)=Z M_p+(A-Z) M_n-B E$
D $M(A, Z)=Z M_p^P+(A-Z) M_n^n+B E / c^2$

The correct answer is:
$M(A, Z)=Z M_p+(A-Z) M_n-\mathrm{BE} / c^2$

Solution:
Binding energy of a nucleus containing $N$ neutrons and $Z$ protons is
$\begin{aligned}
& \mathrm{BE}=\left[N M_n+Z M_p-M(A, Z)\right] c^2 \\
\Rightarrow & \frac{\mathrm{BE}}{c^2}=N M_n+Z M_p-M(A, Z) \\
\Rightarrow & \frac{\mathrm{BE}}{c^2}=(A-Z) M_n+Z M_p-M(A, Z) \\
\Rightarrow & M(A, Z)=Z M_p+(A-Z) M_n-\mathrm{BE} / c^2
\end{aligned}$

Topic: Physics, Nuclear Physics, NEET, NEET 2008 (Screening)
If $\lambda_m$ denotes the wavelength at which the radioactive emission from a black body at a temperature $T K$ is maximum, then:

Options:
A $\lambda_m \propto T^4$
B $\lambda_m$ is independent of $T$
C $\lambda_m \propto T$
D $\lambda_m \propto T^{-1}$

The correct answer is:
$\lambda_m \propto T^{-1}$

Solution:
According to Wein's displacement law
$$
\begin{aligned}
\lambda_m T & =\text { constant } \\
\Rightarrow \quad \lambda_n & \propto T^{-1}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2004
If M is the mass of the earth and R its radius, the ratio of the gravitational acceleration and the gravitational constant is

Options:
A R2M
B MR2
C MR2
D MR

The correct answer is:
MR2

Solution:
Gravitational acceleration is given by
g=GMR2
Where G = gravitational constant
â´gG=MR2

Topic: Physics, Gravitation, NEET
If MO is the mass of an oxygen isotope 8O17, Mp and Mn are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is

Options:
A MOc2
B MO-17MnÂ c2
C MO-8MpÂ c2
D ( 8 M p +9 M n − M O ) C 2

The correct answer is:
( 8 M p +9 M n − M O ) C 2

Solution:
The binding energy is given as
BE=Îmc2
Where, Îm= mass defect
And c= speed of light
â BE=8 Mp+9Mn-MOc2

Topic: Physics, Nuclear Physics, NEET
If MO is the mass of an oxygen isotope O178, MP and MN are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is

Options:
A (MO – 17 MN) c2

B (MO– 8 MP) c2

C (8 MP+9MN – MO) c2

D MOc2

The correct answer is:
(8 MP+9MN – MO) c2

Solution:
Binding energy = [ZMP + (A â Z) MN â MO] c2
= [8MP + (17 â 8) MN â MO]c2 = (8 MP + 9MN â MO) c2

Topic: Physics, Nuclear Physics, NEET
If magnetic field is along +z-axis and a potential difference is applied across an ionized gas chamber placed in the x-y plane, then

(i) Magnetic force on positive ions will act along +y axis

(ii) Magnetic force on electrons will act along +y axis

Options:
A both (i) and (ii) are correct
B both (i) and (ii) are wrong
C only (i) is correct
D only (ii) is correct

The correct answer is:
both (i) and (ii) are wrong

Solution:
Force on +ve ions =qvÃB
=qvi^ÃBk^=qvB-j^
Force on electrons =-qvÃB
=-qv-i^ÃBk^=qvB-j^

Topic: Physics, Magnetic Effects of Current, NEET
If male plant is hexaploid and female plant is tetraploid, the endosperm will be:

Options:
A 6n
B 8n
C 12n
D 7n

The correct answer is:
7n

Solution:
In hexaploid male plant 6n chromosomes are present. It produces triploid gametes. Diploid female gamete and diploid polar nuclei are produced by tetraploid female plant. Endosperm is a triploid tissue. It is produced by triple fusion between one male gamete and two polar nuclei.
Triple fusion= 2 polar nuclei +1 male gamete = 2(2n) + 3n =4n+3n =7n

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
If masses of the plates are 2m and m and their edges are a and 2a respectively, then the distance between the centre of mass of the system and the point O is

Options:
A a2
B a
C 3a2
D 2a3

The correct answer is:
a

Solution:
xcm=2m×a2+m×2a3m=a
ycm=2m×a2-ma3m=0
So distance of center of mass of the system is a .

Topic: Physics, Center of Mass Momentum and Collision, NEET
If \(\mathrm{NaCl}\) is doped with \(10^{-3} \mathrm{~mol} \%\) of \(\mathrm{SrCl}_2\), what is the concentration of cation vacancies?

Options:
A \(10^{-5}\)
B \(10^{-6}\)
C \(10^{-7}\)
D \(10^{-4}\)

The correct answer is:
\(10^{-6}\)

Solution:
Correct Option is : (B)
\(10^{-6}\)

Topic: Chemistry, Solid State, NEET
If molality of the dilute solution is doubled, the value of molal depression constant Kf will be

Options:
A Doubled
B Halved
C Tripled
D Unchanged

The correct answer is:
Unchanged

Solution:
Kf (molal depression constant) is a characteristic of solvent and is independent of molality.

Topic: Chemistry, Solutions, NEET, NEET 2017
If moment of inertia of disc about the diameter is given as I, then the moment of inertia of the same disc about an axis perpendicular to its plane and passing through its rim is

Options:
A 6 I
B 4 I
C 2 I
D 8 I

The correct answer is:
6 I

Solution:
Moment of inertia of a disc about diameter is
14 MR2=I (given)
â´ MR2=4I
Now, required moment of inertia =32MR2
=324I=6I

Topic: Physics, Rotational Motion, NEET
If mother's blood group is A and father's blood group is AB, then child's blood group cannot be

Options:
A A
B B
C O
D AB

The correct answer is:
O

Solution:
Father has AB blood group, so his genotype is IAIB . Mother has A blood group, so her genotype could be IAIA or IAIO. For getting all the possibilities we have to assume that mother is heterozygous. Father will produce two types of gametes, one with IA while the other has IB . The mother will also produce two types of gametes, one with IA while the other has IO.

Topic: Biology, Principles of Inheritance and Variation, NEET
If mother's blood group is A and father's blood group is B, child's blood group could be

Options:
A A, B, AB
B A, B
C A, B, AB, O
D A, AB

The correct answer is:
A, B, AB, O

Solution:
ABO blood group is controlled by gene I and its types are IA, IB and IO.
As mother is A and the father is B, the possible genotypes for parents are - Mother - IAIA/IAIO and Father - IBIB/IBIO .
Therefore, the child's blood group genotype possibilities are -
IAIA/IAIO-A
IBIO/IBIO-B
IoIO-O
IAIB-AB

Topic: Biology, Principles of Inheritance and Variation, NEET