datavorous/entrance-exam-dataset huggingface dataset:
Two coils have mutual inductance 0.005 H. The current changes in the first coil according to equation I=I0sinâ¡Ït, where I0=10 A and Ï=100Ï rad s-1. The maximum value of E.M.F. in the second coil is

Options:
A 2π
B 5π
C π
D 4π

The correct answer is:
5π

Solution:
e=Mdidt=0.005×ddt(I0sinωt)
=0.005×I0ωcosωt
∴emax=0.005×10×100π=5π [ ∵cosωt=1 ]

Topic: Physics, Electromagnetic Induction, NEET
Two coils of self inductance $2 \mathrm{mH}$ and 8 $\mathrm{mH}$ are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is:

Options:
A $16 \mathrm{mH}$
B $10 \mathrm{mH}$
C $6 \mathrm{mH}$
D $4 \mathrm{mH}$

The correct answer is:
$4 \mathrm{mH}$

Solution:
Mutual inductance between coils is
$$
\begin{aligned}
M & =k \sqrt{I_1 I_2} \\
M & =1 \sqrt{2 \times 10^{-3} \times 8 \times 10^{-3}}\{x=1\} \\
& =4 \times 10^{-3}=4 \mathrm{mH}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, NEET, NEET 2006
Two coils of self inductance 2 mH and 8 mH are placed so close together that the effective flux in one coil is completely linked with the other. The mutual inductance between these coils is

Options:
A 16mH
B 10mH
C 6mH
D 4mH

The correct answer is:
4mH

Solution:
Mutual inductance between coils is
M = K L 1 L 2
or, M = 1 2 Ã 1 0 - 3 Ã 8 Ã 1 0 - 3 âµ K = 1
= 4 Ã 1 0 - 3 = 4 mH

Topic: Physics, Electromagnetic Induction, NEET
Two coils P and Q are kept near each other. When no current flows through coil P and current increases in coil Q at the rate 10 A s-1, the e.m.f. in coil P is 15 mV. When coil Q carries no current and current of 1.8 A flows through coil P, the magnetic flux linked with the coil Q is

Options:
A 1.4 mWb
B Â 2.2 mWb
C 2.7 mWb
D 2.9 mWb

The correct answer is:
2.7 mWb

Solution:
ep=M.dIQdt
15Ã10-3=MÃ10
â´ M=15Ã10-4 H
Φ Q =M I p
=15Ã10-4Ã1.8
=27.0Ã10-4
=2.7Ã10-3
=2.7 mWb

Topic: Physics, Electromagnetic Induction, NEET
Two concentric and coplanar circular coils have radii a and bâ«a as shown in the figure. The resistance of the inner coil is R and the current in the outer coil is increased from 0 to i, then the total charge circulating in the inner coil is

Options:
A 2μ0iπba2R
B 2μ0ibπa2R
C μ0ibπa2R
D μ0i2bπa2R

The correct answer is:
μ0i2bπa2R

Solution:
Initial flux linked with inner coil when i=0 is zero.

Final flux linked with inner coil when i=i is

ϕ=μ0i2bπa2

∆q=∆ϕR

⇒∆q=μ0i2bπa2R

Topic: Physics, Electromagnetic Induction, NEET
Two concentric circular coils, one of small radius r 1 and the other of large radius r 2 , such that r 1 << r 2 , are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement.

Options:
A $\frac{{{\mu }_{0}}\pi {{r}_{1}}}{{{r}_{2}}}$
B $\frac{{{\mu }_{0}}\pi {{r}_{2}}}{2{{r}_{1}}}$
C $\frac{{{\mu }_{0}}\pi {{r}_{2}}}{{{r}_{1}}}$
D $\frac{{{\mu }_{0}}\pi {{r}_{1}}}{2{{r}_{2}}}$

The correct answer is:
$\frac{{{\mu }_{0}}\pi {{r}_{1}}}{2{{r}_{2}}}$

Solution:
Correct Option is : (D)
$\frac{{{\mu }_{0}}\pi {{r}_{1}}}{2{{r}_{2}}}$

Topic: Chemistry, Electromagnetic Induction, NEET
Two concentric conducting spheres of radii r1 and r2r1 < r2 carry electric charges of +âQ and -Q respectively. The region between the sphere is filled with two insulating layers of dielectric constant Îµ1 and Îµ2 and width d1 and d2 respectively. Variation of the potential and electric field with radial distance from O is given. Select the correct one. (assume Vat r2=0 )

Options:
A
B
C
D

The correct answer is:

Solution:
Electric field
r<ââr1, Eâ=â0
r1â<ârâ<âr1â+âd, E=Q4ÏÎµ0r2Îµ1
r1â+âdâ<ârâ<âr2, E=Q4ÏÎµ0Îµ2r2
râ>âr2, E=0
Potential can be find out by integrating
i.e. V=-â«ârE.dr=0

Topic: Physics, Electrostatics, NEET
Two concentric co-planar circular loops of radius R and r<<R are placed as shown in the figure. The mutual inductance of the system will be

Options:
A πμ0r22R
B πμ0r24R
C πμ0R24r
D πμ0R22r

The correct answer is:
πμ0r22R

Solution:
The magnetic flux through the smaller loop due to the bigger loop is

ϕ21=μ0i2R×πr2

⇒M=ϕ21i=πμ0r22R

Topic: Physics, Electromagnetic Induction, NEET
Two concentric spheres of radii R and r have similar charges with equal surface charge densities σ. The electric potential at their common centre is

Options:
A σ/ε0
B σε0R-r
C σε0R+r
D σ4πε0R+r

The correct answer is:
σε0R+r

Solution:
The potential at the centre is the same as that on the surface,
â´ v =ÏrÎµ0+ÏRÎµ0

Topic: Physics, Electrostatics, NEET
Two concentric spherical conducting shells of radii R and 2R carry charges Q and 2Q respectively. Change in electric potential on the outer shell when both are connected by a conducting wire is k=14ÏÎµ0

Options:
A zero
B 3 kQ 2 R
C kQ R
D 2 kQ R

The correct answer is:
zero

Solution:
When two concentric shells are connected by a wire the whole charge on inner shell is transferred to the outer shell.
Potential of the outer shell
V i = kQ 2 R + k 2 Q 2 R
V i = 3 kQ 2 R
V f = k 3 Q 2 R
â´ Î V = V f - V i = 0

Topic: Physics, Electrostatics, NEET
Two concentric spherical conducting shells of radii R and 2 R carry charges Q and 2 Q respectively. Change in electric potential on the outer shell when both are connected by a conducting wire is

Options:
A zero
B 3kQ2R
C kQR
D 2kQR

The correct answer is:
zero

Solution:
The initial potential at the outer shell
V1=kQ2R+k2Q2R= 3kQ2R
The final potential at outer shell is
V2=k3Q2R â´ÎV=V1- V2=0

Topic: Physics, Electrostatics, NEET
Two condensers, one of capacity $C$ and other of capacity $C / 2$ are connected to a $V$-volt battery, as shown. The work done in charging fully both the condensers is

Options:
A $\frac{1}{4} C V^2$
B $\frac{3}{4} C V^2$
C $\frac{1}{2} C V^2$
D $2 \mathrm{CV}^2$

The correct answer is:
$\frac{3}{4} C V^2$

Solution:
Work done in charging fully both the condensers
$\begin{aligned}
& =\frac{1}{2} C V^2+\frac{1}{2}\left(\frac{C}{2}\right) V^2 \\
& =\frac{3}{4} C V^2
\end{aligned}$

Topic: Physics, Capacitance, NEET, NEET 2007
Two conducting circular loops of radii R1 and R2 are placed in the same plane with their centres coinciding. If R1≫R2, the mutual inductance M between them will be directly proportional to:

Options:
A R1R2
B R2R1
C R22R1
D R12R2

The correct answer is:
R22R1

Solution:
M=ϕ12I1=B1A2I1=μ0I12R1πR22I1

M=μ0πR222R1

∴ M∝R22R1

Topic: Physics, Electromagnetic Induction, NEET, NEET 2021
Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1mm. Conductor B is a hollow tube of outer diameter 2mm and inner diameter 1mm. What is the ratio of the resistance of Conductor A (RA) to the resistance of Conductor B (RB)?

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

The correct answer is:
3 : 1

Solution:
Correct Option is : (D)
3 : 1

Topic: Chemistry, Current Electricity, NEET
Two converging lenses have focal length f1 and f2 . The optical axes of the two lenses coincide. This lens system is used to form an image of an object. It is observed that the size of the image does not depend on the distance between the system and object. The distance between lenses is

Options:
A f1f2f1+f2
B f1+f2
C f1-f2
D f1-f22f1

The correct answer is:
f1+f2

Solution:
For 1st lens, m1=f1x-f1
Where x1 is the distance of the object from 1st lens.
For 2nd lens, (Let d be the distance between the lenses)
m2=x-f1d-f1-f2x-df1+f1f2
Overall magnification =m1m2=f1d-f1-f2x-df1+f1f2 is independent of x if d=f1+f2

Topic: Physics, Ray Optics, NEET
Two convex lenses of focal lengths 0.3 m and 0.05 m are used to make a telescope. The distance kept between the two in order to obtain an image at infinity is

Options:
A 0.35 m

B 0.25 m

C 0.175 m

D 0.15 m

The correct answer is:
0.35 m

Solution:
When the final image is formed at infinity by a telescope, it is called in normal adjustment and then the length of the telescope is given by
L=f0+fe
Where f0 is the focal length of objective lens and fe is the focal length of eye-piece
â´L= f0+fe=0.3+0.05=0.35 m

Topic: Physics, Ray Optics, NEET
Two convex lenses of power 2 D and 5 D are separated by a distance 13 m . The power of the optical system formed is

Options:
A + 2 D

B - 2 D
C - 3 D

D + 3 D

The correct answer is:
+ 3 D

Solution:
Given
P1=2 D
P2=3 D
d=13 m
We know that
1F=1f1+1f2-df1 . f2
∴ P=P1+P2-dP1 . P2
=2+3-13×2×3
⇒ P=3 D

Topic: Physics, Ray Optics, NEET
Two coplanar concentric circular coils of radii r and 2âr , have the same number of turns n . The smaller coil carries a clockwise current i, while the larger coil carries an anticlockwise current 2âi . The magnetic field induction at the centre is

Options:
A 3μ0ni4r
B zero
C μ0ni4r
D μ0ni2r

The correct answer is:
zero

Solution:
B 1 = μ 0 ni 2r ,  B 2 = μ 0 n 2i 2 2r
B R = B 1 − B 2
B 1 = B 2 , both are in opposite direction

Topic: Physics, Magnetic Effects of Current, NEET
Two copper vessels $A$ and $B$ have the same base area but of different shapes. A takes twice the volume of water as that $B$ requires to fill upto a particular common height. Then the correct statement among the following is:

Options:
A Pressure on the base area of vessels A and B is same.
B Pressure on the base area of vessels A and B is not same.
C Both vessels A and B weigh the same.
D Vessel B weighs twice that of A.

The correct answer is:
Pressure on the base area of vessels A and B is same.

Solution:
Pressure, $\mathrm{P}=h \rho g$
In above equation, $\rho$ represents density of water, $g$ represents acceleration due to gravity and $h$ is the height of the vessel.
As both vessels have same height and containing water, then densities are also same. Hence, the pressure on the base area of vessels $A$ and $B$ is same.

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2022 (Phase 2)
Two crosses between the same pair of genotypes or phenotypes in which the source of the gametes are reversed in one cross, is known as

Options:
A Reverse cross
B Test cross
C Reciprocal cross
D Dihybrid cross

The correct answer is:
Reciprocal cross

Solution:
Dihybrid cross is made between individuals having contrasting traits in order to study the inheritance of two pairs of alleles. Test cross is a back cross between individual with a dominant trait and its recessive parent to know whether the individual is homozygous or heterozygous for the trait.

Topic: Biology, Principles of Inheritance and Variation, NEET
Two crosses between the same pair of genotypes or phenotypes in which the sources of the gametes arc reversed in one cross, is known as:

Options:
A test cross
B reciprocal cross
C dihybrid cross
D reverse cross

The correct answer is:
reciprocal cross

Solution:
Reciprocal cross is the cross, with the phenotype of each sex reversed as compared with the original cross, to test the role of parental sex on inheritance pattern.
Related Theory
Forward genetics is the examination of the genetic cause of an altered or abnormal phenotype introduced by a chemical mutagenesis or mutation by irradiation (e.g. phenotype $\rightarrow$ genotype). In reverse genetics, a particular gene is altered and the phenotype is investigated (e.g., genotype $\rightarrow$ phenotype).

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2003
Two deuterons undergo nuclear fusion to form a Helium nucleus. The energy released in this process is (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV)

Options:
A 23.6 MeV
B 30.2 MeV
C 25.8 MeV
D 32.4 MeV

The correct answer is:
23.6 MeV

Solution:
∵ Q=∑B.E. of products- ∑B.E. of reactants
H2+1H2→2H4
Energy released=Q=4B.E.2H4 -4B.E.1H2
=4×7−4×1.1=28−4.4=23.6 MeV

Topic: Physics, Nuclear Physics, NEET
Two different electrolytic cells filled with molten Cu(NO3)2 and molten Al(NO3)3 respectively are connected in series. When electricity is passed 2.7âgâAl is deposited on electrode. Calculate the weight of Cu deposited on cathode.

[Cu=63.5;Al=27.0 g mol-1]

Options:
A 190.5 g
B 9.525 g
C 63.5 g
D 31.75 g

The correct answer is:
9.525 g

Solution:
wt.of Al depositedwt.of Cu deposited=eq.wt.of Aleq.wt.of Cu
2.7wt.ofÂ Cu=27/363.5/2
wt. of Cu = 9.525 g

Topic: Chemistry, Electrochemistry, NEET
Two different first order reactions have rate constants k1 and k2 at T1(k1>k2) . If temperature is increased from T1 to T2 , then new constants become k3 and k4 respectively. Which among the following relations is correct?

Options:
A k1>k2=k3=k4
B k1<k3 and k2<k4
C k1=k3=k4
D k1>k2>k3>k4

The correct answer is:
k1<k3 and k2<k4

Solution:
Rate constant increases with increase in temperature. Hence, k1<k3 and k2<k4 .

Topic: Chemistry, Chemical Kinetics, NEET
Two discs A and B are in contact and rotating with angular velocity with angular velocities ω1 and ω2 respectively as shown. If there is no slipping between the discs, then

Options:
A ω1=ω2
B ω1>ω2
C ω1<ω2
D data insufficient

The correct answer is:
ω1<ω2

Solution:
For no slipping the linear velocity at point of contact for both the discs must be same
ω1R1=ω2R2 or ω1ω2=R2R1
But R1>R2
∴ω1<ω2

Topic: Physics, Rotational Motion, NEET
Two discs of moments of inertia I1and I2about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed ω1and ω2are brought into contact face to face with their axes of rotation coincident.Calculate the loss in kinetic energy of the system in the process.

Options:
A $\frac{{{I}_{1}}{{I}_{2}}}{2\left( {{I}_{1}}+{{I}_{2}} \right)}{{\left( {{\omega }_{1}}+{{\omega }_{2}} \right)}^{2}}$
B $-\frac{{{I}_{1}}{{I}_{2}}}{2\left( {{I}_{1}}+{{I}_{2}} \right)}{{\left( {{\omega }_{1}}+{{\omega }_{2}} \right)}^{2}}$
C $-\frac{{{I}_{1}}{{I}_{2}}}{2\left( {{I}_{1}}+{{I}_{2}} \right)}{{\left( {{\omega }_{1}}-{{\omega }_{2}} \right)}^{2}}$
D $\frac{{{I}_{1}}{{I}_{2}}}{2\left( {{I}_{1}}+{{I}_{2}} \right)}{{\left( {{\omega }_{1}}-{{\omega }_{2}} \right)}^{2}}$

The correct answer is:
$-\frac{{{I}_{1}}{{I}_{2}}}{2\left( {{I}_{1}}+{{I}_{2}} \right)}{{\left( {{\omega }_{1}}-{{\omega }_{2}} \right)}^{2}}$

Solution:
Correct Option is : (C)
$-\frac{{{I}_{1}}{{I}_{2}}}{2\left( {{I}_{1}}+{{I}_{2}} \right)}{{\left( {{\omega }_{1}}-{{\omega }_{2}} \right)}^{2}}$

Topic: Chemistry, Rotational Motion, NEET
Two discs of moments of inertia I1and I2about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed ω1and ω2are brought into contact face to face with their axes of rotation coincident. Does the law of conservation of angular momentum apply to the situation?

Options:
A No, because the moments of inertia of the discs are different.
B Yes, because the discs have the same angular speed.
C Yes, because there is no net external torque on the system.
D No, because the axes of rotation are coincident.

The correct answer is:
Yes, because there is no net external torque on the system.

Solution:
Correct Option is : (C)
Yes, because there is no net external torque on the system.

Topic: Chemistry, Rotational Motion, NEET
Two discs of moments of inertia I1and I2about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed ω1and ω2are brought into contact face to face with their axes of rotation coincident.Find the angular speed of the two-disc system.

Options:
A $\frac{{{I}_{1}}{{\omega }_{2}}+{{I}_{2}}{{\omega }_{2}}}{{{\omega }_{1}}+{{\omega }_{2}}}$
B $\frac{{{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}}}{{{I}_{1}}+{{I}_{2}}}$
C $\frac{{{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}}}{{{\omega }_{1}}+{{\omega }_{2}}}$
D $\frac{{{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{1}}}{{{\omega }_{1}}+{{\omega }_{2}}}$

The correct answer is:
$\frac{{{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}}}{{{I}_{1}}+{{I}_{2}}}$

Solution:
Correct Option is : (B)
$\frac{{{I}_{1}}{{\omega }_{1}}+{{I}_{2}}{{\omega }_{2}}}{{{I}_{1}}+{{I}_{2}}}$

Topic: Chemistry, Rotational Motion, NEET
Two discs of same moment of inertia ( I ) are rotating in same sense about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities ω1 and ω2. They are brought into contact face to face coinciding the axis of rotation. The expression for loss in energy during this process is

Options:
A 12Iω1+ω22
B 14Iω1-ω22
C Iω1-ω22
D 18 ω1-ω22

The correct answer is:
14Iω1-ω22

Solution:
The initial angular momentum of two discs is I1ω1+I2ω2=Iω1+ω2.

The final angular momentum of system is I1+I2ω=2Iω

From the conservation of angular momentum, ω=ω1+ω22.

Initial kinetic energy, KEi=12Iω12+ω22

and final kinetic energy, KEf=122Iω1+ω224

Loss in kinetic energy is ΔKE=12I1I2I1+I2 ω1-ω22
=12I22I ω1-ω22
=14Iω1-ω22

Topic: Physics, Rotational Motion, NEET, NEET 2017
Two DNA strands have the same length. Which one will denature at a lower temperature?

Options:
A The one having more number of guanine and cytosine molecules
B The one having more number of adenine and thymine molecules
C The one having least number of guanine and cytosine molecules
D More than one option is correct

The correct answer is:
More than one option is correct

Solution:
In a double helix structure of DNA, the bases in the two complementary strands are paired with the help of hydrogen bonds. There are two hydrogen bonds between one adenine and thymine base pairs whereas there are three hydrogen bonds between one guanine and cytosine base pairs. The energy required to break the bond between G-C will be more as compared to energy required for A-T bonds. So the DNA with a maximum number of A-T pairing will melt faster at lower temperature as compared to others.

Topic: Biology, Molecular Basis of Inheritance, NEET
Two drops, of the same radius, are falling through air with a steady velocity of 5ââcm s-1. If the two drops coalesce, the terminal velocity would be

Options:
A 10  cm s-1
B 2.5  cm s-1
C 5Ã(4)1/3ââcm s-1
D 5×3  cm s-1

The correct answer is:
5Ã(4)1/3ââcm s-1

Solution:
If R is radius of bigger drop formed, then
43ÏR3=2Ã43Ïâr3orR=21/3r
As, v 0 â r 2
â´ ( v 0 ) 1 v0=R2r2=(21/3r)2r2=22/3
or ( v 0 ) 1 = v 0 2 2/3
5Ã ( 4 ) 1/3
=( 5 )Ã ( 2 ) 2/3
=5Ã(4)1/3ââcm s-1

Topic: Physics, Mechanical Properties of Fluids, NEET
Two electron beams having velocities in the ratio 1:2 are subjected to the same transverse magnetic field. The ratio of the radii is

Options:
A 1 : 2

B 2 : 1

C 4 : 1

D 1 : 4

The correct answer is:
1 : 2

Solution:
r= mv qB ⇒ r 1 r 2 = v 1 v 2 ⇒ r 1 r 2 = v 1 v 2

Topic: Physics, Magnetic Effects of Current, NEET
Two electrons occupying the same orbital are distinguished by:

Options:
A Principal quantum number
B Magnetic quantum number
C Azimuthal quantum number
D Spin quantum number

The correct answer is:
Spin quantum number

Solution:
Two electrons occupying the same orbital will have the same principal quantum number, same magnetic quantum number, and same azimuthal quantum number but they always have the different spin quantum numbers. If one electron has the clockwise rotation then other electron will have anticlockwise rotation.

Topic: Chemistry, Structure of Atom, NEET, NEET 2016 (Phase 1)
Two electrons revolving around a nucleus at distance 'r' and '4r'. The ratio of their time period is

Options:
A 1 : 4
B 4 : 1
C 8: 1
D 1 : 8

The correct answer is:
1 : 8

Solution:
T2∝r3
TrT4r=r4r3/2
=18

Topic: Chemistry, Structure of Atom, NEET
Two element X (atomic mass = 75) and Y (atomic mass = 16) combine to give a compound having 75.8% of X. The formula of the compound is

Options:
A XY
B X2Y
C X2Y2
D X2Y3

The correct answer is:
X2Y3

Solution:
Element
Atomic weight
%
Ratio
Simplest ratio

X
75
75.8
75.875=1
2

Y
16
24.2
24.216=1.5
3

X2Y3

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
Two elements A and B form compounds having formula AB2and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2lowers the freezing point by 2.3 K whereas 1.0 g of AB4lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol–1. Calculate atomic masses of A and B.

Options:
A A = 34.25 u and B = 42.64 u
B A = 25.58 u and B = 42.64 u
C A = 25.58 u and B = 78.36 u
D A = 34.25 u and B = 78.36 u

The correct answer is:
A = 25.58 u and B = 42.64 u

Solution:
Correct Option is : (B)
A = 25.58 u and B = 42.64 u

Topic: Chemistry, Solutions, NEET
Two energy lavels of an electron in an atom are separated by 2.3 eV. The frequency of radiation emitted when the electrons go from higher to lower level is

Options:
A 6.95 × 1014 Hz
B 3.68 × 1015 Hz
C 5.6 × 1014 Hz
D 9.11× 1015 Hz

The correct answer is:
5.6 × 1014 Hz

Solution:
Given, E2-E1= 2.3 eV
Or v= E2-E1h=2.3 × 1.6×10-196.6 ×10-34
= 0.55 × 1015
= 5.5 × 1014 Hz

Topic: Physics, Atomic Physics, NEET
Two energy levels of an electron in an atom are separated by 2.3 eV. The frequency of radiation emitted when the electrons go from higher to lower level is

Options:
A 6.95Â ÃÂ 1014Â Hz
B 3.68Â ÃÂ 1015Â Hz
C 5.6 Ã 1014 Hz
D 9.11ÃÂ 1015Â Hz

The correct answer is:
5.6 Ã 1014 Hz

Solution:
Given, E2-E1= 2.3 eV
Or Î½=E2-E1h=2.3 Ã 1.6Ã10-196.6 Ã10-34
= 0.56 Ã 1015
= 5.6 Ã 1014 Hz

Topic: Physics, Atomic Physics, NEET
Two enzymes responsible for restricting the growth of bacteriophages in E.coli were isolated. One was methylase and other was a restriction endonuclease. What is the significance of methylase?

Options:
A Able to cut the DNA of bacteriophage at specific site
B Able to of remove the methyl group and hence prevent the action of restriction of endonuclease on host DNA
C Protection of host DNA from the action of restriction endonuclease by adding methyl group to one or two bases usually within the sequences recognized by restriction enzyme
D Able to ligate the two cohesive ends of DNA molecule

The correct answer is:
Protection of host DNA from the action of restriction endonuclease by adding methyl group to one or two bases usually within the sequences recognized by restriction enzyme

Solution:
The host DNA is protected by other enzymes called methylases, which methylate adenine bases within host recognition sequences. For each restriction endonuclease, the host cell produces a corresponding methylase that marks the host DNA and prevents its degradation. These pairs of enzymes are referred to as restriction-modification systems.

Topic: Biology, Biotechnology - Principles and Processes, NEET
Two equal masses hung from two massless springs of spring constants k1 and k2. They have equal maximum velocity when executing simple harmonic motion. The ratio of their amplitudes is

Options:
A k1k212
B k1k2
C k2k1
D k2k112

The correct answer is:
k2k112

Solution:
As the masses are executing SHM the velocity, v=aω
∵ v1=v2 (given)
∴ a1ω1=a2ω2
a1a2=ω2ω1 .....(i)
We know that the time period,
T=2πmk
T2π=mk⇒1ω=mk
⇒ ω=km
⇒ ω∝k .....(ii)
From Equations (i) and (ii), we get
a1a2=k2k1

Topic: Physics, Oscillations, NEET
Two equally charged small metal balls placed at a fixed distance experience a force F. A similar uncharged metal ball after touching one of them is placed at the middle point between the two balls. The force experienced by this ball is

Options:
A F2
B F
C 2F
D 4F

The correct answer is:
F

Solution:
Initially, FAB=14ÏÎµ0.q.qr2=14ÏÎµ0.q2r2
Finally, force on C
FC=FCB-FCA=14ÏÎµ0.q2qr22-14âÏÎµâââ0.q2q2r22=14ÏâÎµâ0.q2r2
So FC=FAB = F

Topic: Physics, Electrostatics, NEET
Two factories are sounding their sirens at 800 Hz. A man goes from one factory to other at a speed of 2 m/s. The velocity of sound is 320 m/s. The number of beats heard by the person in one second will be

Options:
A 10
B 4
C 2
D 8

The correct answer is:
10

Solution:
n 1 = n 0 vâ v 0 vâ v s =800 320â2 320 = 800Ã318 320 âHz
n 2 = n 0 vâ v 0 vâ v s =800 320+2 320 = 800Ã322 320 âHz
â´ Beat frequency = n2 â n1
= 8 0 0 3 2 2 - 3 1 8 3 2 0
= 800Ã4 320 =10âHz

Topic: Physics, Waves and Sound, NEET