datavorous/entrance-exam-dataset huggingface dataset:
If the dipole moment of HCl is 1.08 D and the bond distance is 1.27 Å , the partial charge on hydrogen and chlorine, respectively are

Options:
A +1.0 and -1.0
B +0.85 and -0.85
C +0.356 and -0.356
D +0.178 and -0.178

The correct answer is:
+0.85 and -0.85

Solution:
Dipole moment = charge × bond distance
1.08= charge ×1.27
Charge =1.08/1.27=0.85
Since, HCl is electrically neutral, H+ and Cl- will carry the same charge with opposite sign.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
If the dipole moment of Toluene and Nitro-benzene are 0.43 D and 3.93 D respectively, then what is the expected dipole moment of p-Nitrotoluene?

Options:
A 3.50 D
B 2.18 D
C 4.36 D
D 5.30 D

The correct answer is:
4.36 D

Solution:
Methyl group has +I effect and -NO2 group has -I effect. Therefore, in p-nitro toluene the dipole moments of âCH3 and -NO2 groups act in the same direction. So, the resultant dipole moment is additive.
i.e., 3.93 + 0.43 = 4.36 debye

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
If the directions of electric and magnetic field vectors of a plane electromagnetic wave are along positive y-direction and positive z-direction respectively, then the direction of propagation of the wave is along

Options:
A Positive z -direction
B Negative z -direction
C Negative y -direction
D Positive x -direction

The correct answer is:
Positive x -direction

Solution:
e^=E^×B^, direction of propagation is always perpendicular to plane of E and B . It will be positive in x - direction.

Topic: Physics, Electromagnetic Waves, NEET
If the distance between Na+ and Cl- ions in NaCl crystal is 'a' pm, the edge length of the unit cell is

Options:
A 4 a pm
B a4pm
C a2pm
D 2a pm

The correct answer is:
2a pm

Solution:
The edge length = 2 × distance between Na+ and Cl- =2×a=2a

Topic: Chemistry, Solid State, NEET
If the distance between the earth and the sun becomes four times, then the time period will become

Options:
A 4 times
B 8 times
C 14 times
D 18 times

The correct answer is:
8 times

Solution:
According to Kelper's third law (law of periods),
T2âR3
where T is time taken by the planet to go once around the sun and R is the semi-major axis (distance) of the elliptical orbit.
â´ T3=k R3 â¦(i)
Where k is a constant of proportionality.
When R becomes 4 times, let the time period be Tâ².
â´ Tâ²2=k(4R)3 â¦(ii)
â´ T2Tâ²2=164
or TTâ²=18
or Tâ²=8T
So, time period becomes 8 times of previous value.

Topic: Physics, Gravitation, NEET
If the distance between the earth and the sun were half its present value, the number of days in a year would have been

Options:
A 64.5
B 129
C 182.5
D 730

The correct answer is:
129

Solution:
T T â² = ( R R/2 ) 3 2 â T â² =T ( 1 2 ) 3 2
T â² =123/2Ã 365 days=122 Ã 365 days
=182.52Ã22days=91.25 Ã 1.414 days=129 days

Topic: Physics, Gravitation, NEET
If the distance between the virtual image from its real object is 60 cm in case of a concave mirror. The focal length of the concave mirror if the image formed is 3 times magnified image is

Options:
A +Â 15Â cm
B -Â 15Â cm
C +Â 22.5Â cm
D -22.5 cm

The correct answer is:
-22.5 cm

Solution:
x + 3x=60 ⇒x=15 cm
u=-15 cm
m=+ 3
∴ m=ff-u⇒ + 3=ff-(-15)
∴ 3f+ 45=f ⇒f=- 22.5 cm

Topic: Physics, Ray Optics, NEET
If the distance between two consecutive base pairs is 0.34 nm and the total number of base pairs of a DNA double helix in a typical mammalian cell is 6.6×109bp bp, then the length of the DNA is approximately

Options:
A 2.5 meters
B 2.2 meters
C 2.7 meters
D 2.0 meters

The correct answer is:
2.2 meters

Solution:
Given the distance between two consecutive base pairs is 0.34 nm=0.34×10-9 m.

A total number of bases =6.6×109 bp.

Length of DNA = Total no. of bp×Distance between two consecutivebp

= 6.6×109×0.34×10-9

=2.24 m

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2020 (Phase 1)
If the Ecello for a given reaction has a negative value, which of the following gives the correct relationship for the values of ∆Go and Keq?

Options:
A âGo>0; Keq<1
B âGo>0;Keq>1Â
C âGo<0;Keq>1
D âGo<0;Keq<1

The correct answer is:
âGo>0; Keq<1

Solution:
ΔG∘=-2.303RT log k
ΔG∘=−nFEo
∵Ecello=-ve Ecello<0, then,

ΔG∘=−nF-Eo=+ve

From the equation, ΔG∘=-2.303RT log k

∴∆G∘=+ve & Keq=-ve
i.e., ∆G∘>0 & Keq<1.

Topic: Chemistry, Electrochemistry, NEET, NEET 2016 (Phase 2)
If the $E_{\text {cell }}^{\circ}$ for a given reaction has a negative value then which of the following gives the correct relationships for the values of $\Delta G^{\circ}$ and $K_{\mathrm{eq}}$ ?

Options:
A $\Delta G^{\circ} < 0 ; K_{\mathrm{eq}}>1$
B $\Delta G^{\circ} < 0 ; K_{\text {eq }} < 1$
C $\Delta G^{\circ}>0 ; K_{\mathrm{eq}} < 1$
D $\Delta G^{\circ}>0 ; K_{\text {eq }}>1$

The correct answer is:
$\Delta G^{\circ}>0 ; K_{\mathrm{eq}} < 1$

Solution:
$\Delta G^{\circ}=-n F E^{\circ}$
When $E^{\circ}$ is negative, then $\Delta G^{\circ}>0$
$\Delta G^{\circ}=-R T \ln K_{e q}$
When $\Delta G^{\circ}>0, K_{e q}=10^{-x}$ which is less than one, $i e, K_{e q} < 1$.

Topic: Chemistry, Electrochemistry, NEET, NEET 2011 (Screening)
If the earth shrinks such that its mass does not change but radius decreases to one-quarter of its original value then one complete day will take

Options:
A 96 h

B 48 h
C 6 h
D 1.5 h

The correct answer is:
1.5 h

Solution:
We know that angular momentum of the spin =Iω
By the conservation of angular momentum
25MR2.2πT=25MR42.2πT′
T′=T16=2416=1.5 h

Topic: Physics, Rotational Motion, NEET
If the egg of sexually reproducing organisms contains 16 chromosomes, what will be the number of dyads in anaphase I?

Options:
A 16
B 8
C 32
D 64

The correct answer is:
32

Solution:
Egg contains a haploid number of chromosomes; thus, the diploid number will be 32 in the oocyte. In anaphase I, homologous chromosomes segregate in the form of a dyad; thus, the number of dyads will be 32.

Topic: Biology, Cell Cycle and Cell Division, NEET
If the electric lines of force are shaped like arcs of concentric circles with their centre at point O in a certain section of an electric field, then

Options:
A The intensity of the field in this section should at each point be inversely proportional to its distance from point O
B The intensity of the field in this section should at each point be inversely proportional to square of its distance from point O
C The intensity of the field in this section should at each point be inversely proportional to cube of its distance from point O
D Nothing can be said

The correct answer is:
The intensity of the field in this section should at each point be inversely proportional to its distance from point O

Solution:
To prove this we will calculate work done by electric force in a close loop ABCD.

During displacement AD and BC work done by electric force = 0 as electrostatic is perpendicular to displacement.
Work done during AB=R α E1
Work done during CD=-r α E2
Total work done =R α E1-r α E2
Close loop work done should be zero
∴E1E2=rR

Topic: Physics, Magnetic Effects of Current, NEET
If the electron in hydrogen atom jumps from the third orbit to second orbit, the wavelength of the emitted radiation is given by

Options:
A λ=365R
B λ=5R36
C λ=5R
D λ=R6

The correct answer is:
λ=365R

Solution:
1Î»=R1n12-1n22
=R122-132
1Î»=5R36
So Î»=365R

Topic: Chemistry, Structure of Atom, NEET
If the electron in the hydrogen atom jumps from the third orbit to the second orbit, the wavelength of the emitted radiation in term of Rydberg constant is

Options:
A 65R
B 365R
C 647R
D 367R

The correct answer is:
365R

Solution:
1λ=R122-132
=R14-19
1λ=R 9-436
=5R36
λ=365R

Topic: Physics, Atomic Physics, NEET
If the emission rate of a blackbody at 0 Â°C is R , then the rate of emission at 273 Â°C is

Options:
A 2R
B 4R
C 8R
D 16R

The correct answer is:
16R

Solution:
According Stefan's law, emission rate of a ideal blackbody
E∝T4
For T=0 °C=273 K
E1∝2734 ...(i)
For T=273 °C=273+273=546 K
E2∝5434 .....(ii)
From Equations (i) and (ii),
⇒ E1E2=2735464⇒RE2=124 ∵ E1=R
Hence, E2=16R

Topic: Physics, Thermal Properties of Matter, NEET
If the end correction of an open pipe is 0.8 cm then the inner radius of that pipe will be

Options:
A 13 cm
B 23Â cm
C 32Â cm
D 0.2Â cm

The correct answer is:
23 cm

Solution:
For open organ pipe
âL=1.2Ãr
â´r=âLÃ11.2=0.81.2=23 cm.

Topic: Physics, Waves and Sound, NEET
If the endosperm cells of an angiosperm seed are pentaploid, then such a seed may have been formed by which of the following parents?

Options:
A Triploid female and diploid male.
B Diploid male and tetraploid female.
C Both parents are tetraploid.
D Triploid male and diploid female.

The correct answer is:
Diploid male and tetraploid female.

Solution:
If more than two sets of chromosomes are present then that condition is called polyploidy. The correct answer should be B because diploid male undergo meiosis and give rise to haploid male gamete(n). Similarly, the tetraploid female gamete undergoes meiosis that also gives rise to gametes (2n). These female gametes ( 2n+2n polar nuclei) fuse with a male gamete that gives rise to pentaploid endosperm.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
If the endosperm of an angiosperm has 24 chromosomes, what would be the number of chromosomes in the megaspore mother cell of the same plant?

Options:
A 8
B 16
C 24
D 32

The correct answer is:
16

Solution:
Ploidy of endosperm in angiosperm is 3n as it is post fertilization product formed by triple fusion and ploidy of mega spore mother cell is 2n.

Topic: Biology, Plant Kingdom, NEET
If the enthalpy change for the transition of liquid water to steam is $30 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $27^{\circ} \mathrm{C}$, the entropy change for the process would be

Options:
A $1.0 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
B $0.1 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
C $100 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
D $10 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

The correct answer is:
$100 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

Solution:
$$
\begin{aligned}
& \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \\
& \text { Given, } \Delta H_{\text {vap. }}=30 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta G^{\circ}=0 \text { at equilibrium, } \\
& \Delta S_{\text {vap }}=\frac{\Delta H_{\text {vap }}}{T} \\
& =\frac{30 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}}{300 \mathrm{~K}} \\
& =100 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\
&
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2011 (Screening)
If the equilibrium concentration of O2in air at 25°C is 1.6 × 10-2, what is the concentration of O3?

Options:
A 1.3 × 10-66
B 1.0 × 10-32
C 4.0 × 10-25
D 5.0 × 10-53

The correct answer is:
1.3 × 10-66

Solution:
Correct Option is : (A)
1.3 × 10-66

Topic: Chemistry, Ionic Equilibrium, NEET
If the equilibrium constant for N2g+O2(g)⇌2NO(g) is K, the equilibrium constant for 12N2g+12O2(g)⇌NO(g) will be:

Options:
A K12
B 12K
C K
D K2

The correct answer is:
K12

Solution:
\(\begin{aligned}
& \mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) ; \mathrm{K} \\
& \frac{1}{2} \mathrm{~N}_2(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) ; \mathrm{K}^\prime \\
& \mathrm{K}=\frac{\left[\mathrm{NO}^2\right.}{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]} \\
& \mathrm{K}^\prime=\frac{\mathrm{NO}}{\left[\mathrm{N}_2\right]^{1 / 2}\left[\mathrm{O}_2\right]^{1 / 2}} \\
& \therefore \mathrm{K}^\prime=\sqrt{\mathrm{K}}
\end{aligned}\)

Topic: Chemistry, Chemical Equilibrium, NEET, NEET 2015 (Phase 2)
If the equivalent resistance between the terminal points P and Q is 4 Î© in the given circuit, then find out the resistance of R in ohms

Options:
A 7
B 4
C 2
D 5

The correct answer is:
7

Solution:
Equivalent circuit is,
Therefore equivalent resistance is,
Req=R4R3R+4R3
4=4R7
R=7 Î©

Topic: Physics, Current Electricity, NEET
If the error in measuring the radius of the sphere is 2% and that in measuring its mass is 3%, then the error in measuring the density of material of the sphere is

Options:
A 5%
B 7%
C 9%
D 11%

The correct answer is:
9%

Solution:
As density =massvolume
â´ Ï=M43ÏR3=34MÏR3
â´ The percentage error in density is
Taking "log" both sides we get lnâÏ=ln 3 4 +lnâMâlnâÏâ3âlnâR
differentiating, we get Δρ ρ = ΔM M − 3ΔR R
But as error is always added â ÎÏ Ï = Îm m + 3ÎR R
â ÎÏÏÃ100%=ÎMM+3ÎRRÃ100%
=3%+32%=3%+6%=9%

Topic: Physics, Units and Dimensions, NEET
If the error in the measurement of momentum is 20 %, then the error in the calculation of kinetic energy is (assume the error in measurement of m as zero)

Options:
A 20 %
B 44 %
C 40 %
D 200 %

The correct answer is:
44 %

Solution:
KE=12p2m
KE1=12p2m
p1=65 p
KE2=12 3625p2m
ΔKE= KE2-KE1
=12 65p2-p2m
=1125 p2m
Percentage change in kinetic energy =ΔKEKE1×100
=1125×100=44%

Topic: Physics, Units and Dimensions, NEET
If the error in the measurement of radius of a sphere is $2 \%$, then the error in the determination of volume of the sphere will be

Options:
A $4 \%$
B $6 \%$
C $8 \%$
D $2 \%$

The correct answer is:
$6 \%$

Solution:
Percentage error in radius is given as 2% i.e. \(\frac{\Delta r}{r} \times 100=2 \%\)
Volume of sphere \(V=\frac{4 \pi_2}{3}\)
Percentage error in volume \(\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=3 \times \frac{\Delta \mathrm{r}}{\mathrm{r}} \times 100=3 \times 2=6 \%\)

Topic: Physics, Mathematics in Physics, NEET, NEET 2008 (Screening)
If the excess pressure inside a soap bubble is balanced by oil column of height 2 mm, then the surface tension of soap solution will be (r=1 cm and density d=0.8 g cc-1), g=10 m s-2

Options:
A 4 N m-1
B 4Ã10-1 N m-1
C 4Ã10-2 N m-1
D 4Ã10-3 N m-1

The correct answer is:
4Ã10-2 N m-1

Solution:
Excess pressure inside a soap bubble of radius R is
=4TR .....(i)
Where T is surface tension of liquid film.
Pressure due to oil column
=hÏg .....(ii)
Where h is height of column, Ï the density and g the gravity.
From Eqs. (i) and (ii), we get
4TR=hÏg
â¹ T=hÏgR4
Given, h=2 mm, g=980 cm s-2,
Ï=0.8 gcc-1, R=1 cm
â´ T=0.2Ã0.8Ã9804
=2Ã10â3Ã0.8Ã103Ã10Ã1Ã10â24 N m-1
=4Ã10â2 N m-1

Topic: Physics, Mechanical Properties of Fluids, NEET
If the extension in both the springs increases from x to x0 on flowing current I in the rod from B to A then, the value of magnetic field will be -

Options:
A mgIl
B mgxIl
C mgIlx0
D mgx0-xIlx0

The correct answer is:
mgx0-xIlx0

Solution:
mg=2kx ....i
mg+IBl=2kx0 ...... 2
mg+IBlmg=x0x or
IBl=mgx0x- 1 or B=mgIlx0-xx

Topic: Physics, Magnetic Properties of Matter, NEET
If the external pressure (Pex) is greater than the internal pressure (p) of a gas, what sign is assigned to the work done on the system during compression?

Options:
A Positive
B Negative
C Zero
D It depends on the specific conditions

The correct answer is:
Positive

Solution:
If the external pressure (Pex) is greater than the internal pressure (p) during compression, the work done on the system is assigned a positive sign.

Topic: Chemistry, Thermodynamics (C), NEET
If the filiform apparatus has 28 chromosomes, then what will be their number in the nucellus and funicle of the cell, respectively?

Options:
A 56, 28
B 28, 56
C 56, 56
D 84, 56

The correct answer is:
56, 56

Solution:
Filliform apparatus are finger like projection present on synergids present in embryo sac. Synergids are haploid cell as they are part of female gametophyte. Thus n=28 . Nucellus and funicle are diploid theu will have 56 chromosome each.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
If the first one-third of a journey is travelled at 20 km h-1, next one-third at 40 km h-1 and the last one-third at 60 km h-1, then the average speed for the whole journey will be

Options:
A 32.7 km h-1
B 35 km h-1
C 40 km h-1
D 45 km h-1

The correct answer is:
32.7 km h-1

Solution:
Let t1, t2, t3 be the time taken in travelling distances s 3 , s 3 , s 3 . Then

t 1 = s / 3 v 1 = s / 3 2 0 , t 2 = s / 3 v 2 = s / 3 4 0 , t 3 = s / 3 v 3 = s / 3 6 0
â´ Average speed = s s / 3 2 0 + s / 3 4 0 + s / 3 6 0 = 3 1 2 0 + 1 4 0 + 1 6 0
= 3 6 + 3 + 2 1 2 0 = 3 Ã 1 2 0 1 1 = 32.7 km/h

Topic: Physics, Motion In One Dimension, NEET
If the force exerted by an electric dipole on a charge q at a distance of 1 m is F, the force at a point 2 m away in the same direction will be

Options:
A F2
B F4
C F6
D F8

The correct answer is:
F8

Solution:
Force exerted by an electric dipole on a charge is inversely proportional to the cube of distance of the charge from the centre of the dipole
Fâ1r3
â´Â FF′=r′r3
FF′=213
F′=F8

Topic: Physics, Electrostatics, NEET
If the formula of basic Beryllium nitrate is [BexO(NO3)6] . What is the value of 'n' here

Options:
A 2
B 3
C 4
D 5

The correct answer is:
4

Solution:
Basic Beryllium nitrate is [Be4O(NO3)6]
Hence, x=4

Topic: Chemistry, s Block Elements, NEET
If the forward bias voltage in a PN diode is decreased, the length of the depletion region will

Options:
A Increase
B Decrease
C Not change
D Initially increase and then decrease

The correct answer is:
Increase

Solution:
In a PN diode when the forward bias voltage is decreased, then length of depletion region will increase and vice versa.

Topic: Physics, Semiconductors, NEET
If the forward voltage in a diode is increased, the width of the depletion region

Options:
A increases
B decreases
C fluctuates
D no change

The correct answer is:
decreases

Solution:
On increasing the forward bias voltage, the barrier energy decreases. This results in the flow of majority charge carries. Hence, width of depletion region decreases

Topic: Physics, Semiconductors, NEET
If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia complex (which behaves as a strong electrolyte) is -0.0558 °C , the number of chloride(s) in the coordination sphere of the complex is [Kf for water =1.86 Kkg mol-1 ]

Options:
A 1
B 2
C 4
D 3

The correct answer is:
3

Solution:
ΔTf=iKfm
0.0558=i×1.86×0.01
i=3
As Co is in +3 oxidation state so formula should be
[Co(NH3)5Cl]Cl2.[Co3+ forms octahedral complexes]

Topic: Chemistry, Solutions, NEET
If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia complex (which behaves as a strong electrolyte) is -0.0558°C, the number of chloride(s) in the coordination sphere of the complex is
[ Kf of water = 1.86 K kg mol-1 ]

Options:
A 0
B 1
C 2
D 3

The correct answer is:
1

Solution:
ΔTf=iKfm
0.0558=i×1.86×0.01
i=3
i.e. 2 Cl- ions are present outside and one Cl- present inside the co-ordination sphere.

Topic: Chemistry, Solutions, NEET
If the frequency of a recessive allele is 0.4. What will be the frequency of individuals with dominant phenotype in the population?

Options:
A 0.68
B 0.6
C 0.84
D 0.36

The correct answer is:
0.84

Solution:
According to the Hardy-Weinberg principle,
Frequency of dominant allele A=p
Frequency of recessive allele a=q
Frequency of AA individuals =pÃp=p2
Frequency of aa individuals = qÃq=q2
Frequency of Aa individuals =2pq
Here q = 0.4
We know that, p+q=1 , so the value of p comes out to be = 0.6
Total dominant phenotypes = 2pq+p2
= 2Ã0.6Ã0.4+0.6Ã0.6
= 0.48+0.36=0.84

Topic: Biology, Evolution, NEET
If the friction is sufficient to prevent the block from sliding, then the minimum value of F for which the cube begins to topple about an edge is

Options:
A 23mg
B 34mg
C 12mg
D mg

The correct answer is:
34mg

Solution:
Balancing torque about point O
mga2=F2a3
F=34mg

Topic: Physics, Rotational Motion, NEET
If the fringe width obtained in Young's double-slit experiment is 1.33 mm, then on immersing the whole set up in the water with refractive index 1.33, new fringe width will be

Options:
A Does not change
B 1.0 mm
C 2.0 mm
D 2.69 mm

The correct answer is:
1.0 mm

Solution:
Î²â²=Î²Î¼=1.331.33
=1 mm

Topic: Physics, Wave Optics, NEET
If the galvanometer G does not show any deflection in the circuit shown, the value of R is given by:

Options:
A 400 Ω
B 200 Ω
C 50 Ω
D 100 Ω

The correct answer is:
100 Ω

Solution:
As given in the question,

Ig=0

Potential drop across R will be equal to 2 V.

Now, current through the left circuit can be written as,

I=10400+R.

Therefore, potential drop across R will be,

VR=IR⇒2=10R400+R⇒800+2R=10R⇒R=100 Ω

Topic: Physics, Current Electricity, NEET, NEET 2023 (All India)
If the gene of interest is inserted at the BamHI site in pBR322, the recombinant plasmid will

Options:
A Show ampicillin & tetracycline resistance only
B Show tetracycline resistance
C Will grow well on tetracycline containing medium
D Will not grow on tetracycline containing medium

The correct answer is:
Will not grow on tetracycline containing medium

Solution:
pBR322 has genes for resistance against two antibiotics (tetracycline and ampicillin), an origin of replication and a variety of restriction sites for cloning of restriction fragments contained through cleavage with a specific enzyme. Foreign DNA is inserted at a site located in one of the two genes for resistance against antibiotics so that it will inactivate one of the two resistance genes. In plasmid pBR322, the gene conferring resistance to ampicillin (ApR) can be interrupted by the insertion of a DNA fragment into the PstI site, and the gene conferring resistance to tetracycline (TcR) can be interrupted by the insertion of a DNA fragment into the BamHI site. If we ligate a foreign DNA at the BamHI site of tetracycline resistance gene in pBR322 the recombinant plasmid will not grow on tetracycline containing medium.

Topic: Biology, Biotechnology - Principles and Processes, NEET
If the gene of interest is introduced in a tetracycline-resistant gene region of pBR322, then:

Options:
A The host will become resistant for tetracycline
B The host will start producing tetracycline
C The host will start producing ampicillin.
D The host will loose resistance for antibiotic tetracycline.

The correct answer is:
The host will loose resistance for antibiotic tetracycline.

Solution:
The recombinant plasmid will loose tetracycline nature due to insertion of foreign DNA and this is known as insertional inactivation but can still be selected out from non recombinants by plating the transformants on tetracycline medium.

Topic: Biology, Biotechnology - Principles and Processes, NEET
If the half cell reaction A+e-→A- has a large negative reduction potential, it follows that

Options:
A A is readily reduced
B A is readily oxidised
C A- is readily reduced
D A- is readily oxidised

The correct answer is:
A- is readily oxidised

Solution:
If the half cell reaction A+e-→A- has a large negative reduction potential, it follows that A- is readily oxidised as negative ERPo means more EOPo , i.e., more is the tendency to get itself oxidized.

Topic: Chemistry, Electrochemistry, NEET
If the half-cell reactions are given as
(i) Fe 2+ +2 e â âFes; E o =â0.44V

(ii) 2H+(aq)+O2(g)+2eâH2O(l); Eo=+1.23 V

The Eo for the reaction
Fes+2H++O2gâFe2+(aq)+H2Ol

Options:
A – 0.79 V
B – 1.67 V
C 1.67 V
D 0.79 V

The correct answer is:
1.67 V

Solution:
Subtracting equation (i) from (ii) we get equation (iii) as follows
Fe(s)+2H++12O2→Fe2+(aq)+H2O(l) …(iii)
So, Eo=Eo2-Eo1=+1.23-(-0.44)=+1.67V.

Topic: Chemistry, Electrochemistry, NEET
If the half-life of a radioactive element is 10 hours, its average life is

Options:
A 1.44 hours
B 66.93 hours
C 14.4 hours
D 0.693 hours

The correct answer is:
14.4 hours

Solution:
λt12=ln⁡2=0.693
⇒ tavg=1λ =t12ln2=100.693
⇒ tavg=14.4 hours

Topic: Physics, Nuclear Physics, NEET