datavorous/entrance-exam-dataset huggingface dataset:
A plant that is heterozygous for two genes was selfed. A total of 1000 seeds were collected. Find out the total number of seeds heterozygous for the first gene and homozygous for the second gene.

Options:
A 125
B 250
C 300
D 500

The correct answer is:
250

Solution:
The plant is heterozygous for two genes. Thus, its genotype is AaBb. Its selfing is actually a dihybrid cross.
The various genotypes of dihybrid cross are 1 AABB, 1 AAbb, 1 aabb, 1 aaBB, 2 AaBB, 2 AABb, 2 Aabb, 2 aaBb and 4 AaBb. Out of these AaBB and Aabb are the only two genotypes which are heterozygous for first gene and homozygous for second gene. Thus out of 16 possibilities 4 possibilities satisfy the given condition. Out of 1000 seeds the number of individuals will be 416×1000=250

Topic: Biology, Principles of Inheritance and Variation, NEET
A plant with genotype AABbCcDD is self-pollinated. Provided that the four genes are independently assorting, what proportion of the progeny will show the genotype AAbbccDD?

Options:
A 1/4
B 1/16
C 1/64
D 1/256

The correct answer is:
1/16

Solution:
Gamete are formed by reductional division having half the chromosomal content of the parent cell. During the meiotic division, crossing over occurs resulting in formation of new combination. For the given genotype the gamete combination will be - ABCD, ABcD, AbcD, AbCD.
Mendel's law of independent assortment states that "the alleles of two (or more) different genes get sorted into gametes independently of one another."
On selfing, the outcome will be

ABCD	ABcD	AbCD	AbcD

ABCD	AABBCCDD	AABBCcDD	AABbCCDD	AABbCcDD

ABcD	AABBCcDD	AABBccDD	AABbCcDD	AABbccDD

AbCD	AABbCCDD	AABbCcDD	AAbbCCDD	AAbbCcDD

AbcD	AABbCcDD	AABBccDD	AAbbCcDD	AAbbccDD

Thus, the genotype AAbbccDD will be expressed in 1/16th of progenies.

Topic: Biology, Principles of Inheritance and Variation, NEET
A plant with racemose inflorescence, zygomorphic flowers, five sepals and petals showing vexillary aestivation, monocarpellary and unilocular superior ovary, androecium consisting of ten stamens.

The above-listed conditions are characteristics of which family of plants?

Options:
A Solanaceae
B Fabaceae
C Liliaceae
D Brassicaceae

The correct answer is:
Fabaceae

Solution:
The given characteristics belong to the members of the family Fabaceae. These are plants with racemose inflorescence, zygomorphic flowers, five sepals and petals showing vexillary aestivation, monocarpellary and unilocular superior ovary, androecium consisting of ten stamens. The above floral formula stands corrected for plants like soybean, grams, pulses and Sesbania.

Topic: Biology, Morphology of Flowering Plants, NEET
A plasmid has accepted the gene of interest, such as gene coding for streptomycin. How can we make clones of this plasmid?

Options:
A Inserting it into a virus to generate multiple copies
B Treating it with a restriction enzyme in order to cut the molecule into small pieces
C Inserting it into a suitable bacterium in order to produce multiple copies
D Running it on a gel electrophoresis in order to determine the size of the gene of interest

The correct answer is:
Inserting it into a suitable bacterium in order to produce multiple copies

Solution:
Plasmid is an extra-chromosomal DNA found in bacteria which can self-replicate. To make multiple copies of a plasmid (a small circular DNA), it should be inserted into a bacterium. A plasmid would not replicate if it were inserted into a virus, as viruses do not contain plasmid and don't have the necessary conditions for its replication. Treating the plasmid with a restriction enzyme will cut into smaller fragments, rendering the plasmid useless. Running the plasmid on a gel (using gel electrophoresis) would only tell us the size of the plasmid and not make clones or copies.

Topic: Biology, Biotechnology - Principles and Processes, NEET
A plate has a length 5Â±0.1 cm and breadth 2Â±0.01 cm. Then the area of the plate is

Options:
A 10 Â±0.25 cm2
B 10Â Â±0.01Â cm2
C 10Â Â±0.001Â cm2
D 10Â Â±1Â cm2

The correct answer is:
10 Â±0.25 cm2

Solution:
A = ab
Δ A A = Δ a a + Δ b b
= 0.1 5 + 0.01 2
=0.2+0.0510

=0.2510
ΔA=0.025×A
=0.25cm2

Topic: Physics, Units and Dimensions, NEET
A player caught a cricket ball of mass 150 g moving at the rate of 20 m s-1. If the catching process be completed in 0.1 s, the force of impact exerted by the ball on the hands of the player is

Options:
A 0.3 N

B 30 N

C 300 N

D 3000 N

The correct answer is:
30 N

Solution:
Force,
F=mu-vt=150Ã10-3(20-0)0.1
=30 N

Topic: Physics, Laws of Motion, NEET
A player throws a ball upwards with an initial speed of 29.4 m/s. What is the maximum height the ball reaches, and how long does it take for the ball to return to the player's hands, considering the acceleration due to gravity as 9.8 m/s² and neglecting air resistance?

Options:
A The maximum height is 14.7 meters, and it takes 6.0 seconds for the ball to return.
B The maximum height is 14.7 meters, and it takes 3.0 seconds for the ball to return.
C The maximum height is 59.4 meters, and it takes 3.0 seconds for the ball to return.
D The maximum height is 59.4 meters, and it takes 6.0 seconds for the ball to return.

The correct answer is:
The maximum height is 14.7 meters, and it takes 3.0 seconds for the ball to return.

Solution:
Correct Option is : (B)
The maximum height is 14.7 meters, and it takes 3.0 seconds for the ball to return.

Topic: Chemistry, Motion In One Dimension, NEET
A pleiotropic gene

Options:
A is a gene evolved during Pliocene.
B controls a trait only in combination with another gene.
C controls multiple traits in an individual.
D is expressed in only primitive plants.

The correct answer is:
controls multiple traits in an individual.

Solution:
It is a single gene that controls multiple traits in an individual. Examples of pleiotropy are phenylketonuria, albinism, sickle cell anemia, autism and schizophrenia.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2015 (Phase 2)
A pleiotropic gene

Options:
A Is a gene evolved during Pliocene
B Controls a trait only in combination with another gene
C Controls multiple traits in an individual
D Is expressed in only primitive plants

The correct answer is:
Controls multiple traits in an individual

Solution:
It is a single gene that controls multiple traits in an individual. Examples of pleiotropy are phenylketonuria, albinism, sickle cell anemia, autism and schizophrenia.

Topic: Biology, Principles of Inheritance and Variation, NEET
A plot of $\log (x / m)$ versus $\log p$ for the adsorption of a gas on a solid gives a straight line with slope equal to:

Options:
A $\log \mathrm{K}$
B $-\log \mathrm{K}$
C $n$
D $1 / n$

The correct answer is:
$1 / n$

Solution:
The empirical relation $\frac{x}{m}=k p^{\frac{1}{n}}$, is known as the Freundlich adsorption isotherm. Taking the logarithm on both sides, $\log \frac{x}{m}=\log k=\frac{1}{n} \log p$
$$
\text { If a following curve is plotted }
$$

Topic: Chemistry, Surface Chemistry, NEET, NEET 2006
A plot of log xmvs log P for the adsorption of a gas on solid gives a straight line with slope equal to:

Options:
A 1n
B logK
C -logK
D n

The correct answer is:
1n

Solution:
According to Freundlich
xm=KP1n
logxm=logK+1nlogP
Slope=1n

Topic: Chemistry, Surface Chemistry, NEET
A p-n junction in series with a resistance of 5 kÎ© is connected across a 50 V DC source. If the forward bias resistance of the junction is 50 Î©, the forward bias current is

Options:
A 1 mA

B 2 mA

C 20 mA

D 9.9 mA

The correct answer is:
9.9 mA

Solution:
Total resistance = 50 + 5000 = 5050 Ω
∴ Forward bias current i=505050=1101=9.9 mA

Topic: Physics, Semiconductors, NEET
A p-n photodiode is made of a material with a band gap of $2.0 \mathrm{eV}$. The minimum frequency of the radiation that can be absorbed by the material is nearly

Options:
A $10 \times 10^{14} \mathrm{~Hz}$
B $5 \times 10^{14} \mathrm{~Hz}$
C $1 \times 10^{14} \mathrm{~Hz}$
D $20 \times 10^{14} \mathrm{~Hz}$

The correct answer is:
$5 \times 10^{14} \mathrm{~Hz}$

Solution:
$p-n$ photodiode is a semiconductor diode that produces a significant current when illuminated. It is reversed biased but is operated below the breakdown voltage.
Energy of radiation $=$ band gap energy ie, $h v=2.0 \mathrm{eV}$
$$
\text { or } v=\frac{2.0 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}} \approx 5 \times 10^{14} \mathrm{~Hz}
$$

Topic: Physics, Semiconductors, NEET, NEET 2008 (Screening)
A point charge +10 mC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. What is the magnitude of the electric flux through the square?

Options:
A 3.1 × ${{10}^{5}}$N ${{m}^{2}}$/C
B 4.2 × ${{10}^{5}}$N ${{m}^{2}}$/C
C 2.9 × ${{10}^{5}}$N ${{m}^{2}}$/C
D 2.2 × N ${{m}^{2}}$/C

The correct answer is:
2.2 × N ${{m}^{2}}$/C

Solution:
Correct Option is : (D)
2.2 × N ${{m}^{2}}$/C

Topic: Chemistry, Electrostatics, NEET
A point charge is surrounded by eight identical charges at a distance r as shown in the figure. How much work is done by the force of electrostatic repulsion when the point charge at the centre is moved to infinity?

Options:
A zero
B 8q24πε0r
C 8q4πε0r
D 64q24πε0r

The correct answer is:
8q24πε0r

Solution:
Work done =-ÎU=Ui-Uf
=8×q×q4πε0r-0=8q24πε0r

Topic: Physics, Electrostatics, NEET
A point charge Q is placed at the corner of a square plate of side a then the flux through the square plate is -

Options:
A Q8∈0
B Q4∈0
C Q∈0
D Zero

The correct answer is:
Zero

Solution:
The electric field due to Q at any point of square plate will be along the plane of square therefore angle between electric field and area vector of plate is 90°.∴ = 0

Topic: Physics, Electrostatics, NEET
A point mass m=20 kg, is suspended by a massless spring of constant 2000 N/M. The point mass is released when elongation in the spring is 15 cm. The equation of displacement of particle as a function of time is (Take g=10 m/s2)

Options:
A y=10sin⁡10t
B y=10cos⁡10t
C y=10sin⁡10t+π6
D None of these

The correct answer is:
y=10sin⁡10t+π6

Solution:
The motion of block is S.H.M.
∴ y=Asin⁡ωt+ ϕ
Here amplitude is A=mgk
=20 ×102000m=10 cm
At t=0, displacement of body with respect to mean position is
y=15-10=5cm
∴ 5=10sin⁡ω×0+ ϕ
or 12=sin⁡ϕ ⇒ ϕ=π6
∴ y=10sin⁡10t+π6

Topic: Physics, Oscillations, NEET
A point mass m is moved in a vertical circle of the radius r with the help of a string. The velocity of the mass is 7gr at the lowest point. The tension in the string at the lowest point is:

Options:
A 6mg
B 7mg
C 8mg
D 1mg

The correct answer is:
8mg

Solution:
First, make the free body diagram when a mass bob is attached with a string of length, r rotating in a vertical circle.

Given, v=7gr, now make the equation between centripetal force and surface tension which are acted in opposite direction,

⇒T-mg=m(7gr)2r⇒T=8 mg.

Topic: Physics, Laws of Motion, NEET, NEET 2020 (Phase 2)
A point mass m is suspended at the end of a massless wire of length L and cross-section area A. If Y is Young's modulus for the wire, then the frequency of oscillations for the SHM along the vertical line is

Options:
A 12πYAmL
B 2πmLYA
C 1πYAmL
D πmLYA

The correct answer is:
12πYAmL

Solution:
In this case,
Stress = mgA
Strain = lL (where l is extension)
Now, Young's modulus Y is given by
Y=stressstrain=mg/Al/L
mg=YAlL
So, kl=YAlL (∵ mg=kl)
(k is force constant)
Now, frequency is given by
n=12πkm
=12πYAmL

Topic: Physics, Oscillations, NEET
A point near the equator has: (H and V are horizontal and vertical components of Earth's magnetic field at a point)

Options:
A Hâ«V
B V=HÂ â 0
C V≫H
D V=H=0

The correct answer is:
H≫V

Solution:
As tan⁡δ=VH
At equator, δ=0,
So, H≫V

Topic: Physics, Magnetic Properties of Matter, NEET
A point object is moving with velocity v 0 = 2 i ^ - 3 j ^ + 4 k ^ in front of a moving plane mirror whose normal is along x - axis. The mirror is moving with velocity v m = i ^ - 4 j ^ + 2 k ^ . Find the velocity vector of image.

Options:
A - 5 j ^
B - 3 j ^ + 4 k ^
C - 4 j ^ + 2 k ^
D 2 i ^ - 3 j ^ + 2 k ^

The correct answer is:
- 3 j ^ + 4 k ^

Solution:
Let velocity of image vi=x i^+y j^+z k^
Along the normal, i.e., along x-axis.
v im = - v om
⇒ x - 1 = - 2 - 1
⇒ x = 0
Along y and z - axes v i = v 0

Topic: Physics, Mathematics in Physics, NEET
A point object is placed at a distance of 25 cm from a convex lens of local length 20 cm. If a glass slab of thickness t and reflective index 1.5 is inserted between the lens and object. The image is formed at infinity. Find the thickness t.

Options:
A 15 cm
B - 15 cm
C 5 cm
D - 5 cm

The correct answer is:
15 cm

Solution:
For final image at infinity image by the slab must be at the focus of convex lense shift
= t 1 - 1 μ ⇒ 5 = t 1 - 2 3
t = 15 cm

Topic: Physics, Ray Optics, NEET
A point object is placed at a distance of 30 cm from a convex mirror of a focal length 30 cm. The image will form at

Options:
A Infinity
B Pole
C 15 cm behind the mirror
D No image will be formed

The correct answer is:
Infinity

Solution:
The image will be formed at infinity. It is because, according to the question the object distance = 30 cm and focal length is 30 cm here, both the object distance and the focal length coincides therefore it refers that the object is placed at the focus. when an object is placed at focus in a convex mirror the image formation will be at ' infinity '.

Topic: Physics, Ray Optics, NEET
A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40 cm from it, the final image would be formed at a distance of :

Options:
A 20 cm from the lens, it would be a real image.
B 30 cm from the lens, it would be a real image.
C 30 cm from the plane mirror, it would be a virtual image.
D 20 cm from the plane mirror, it would be a virtual image.

The correct answer is:
20 cm from the plane mirror, it would be a virtual image.

Solution:
first, for image formation from lens
u=-60 cm

f=+30 cm

⇒v=ufu+f=-60×30-60+30=60 cm

this real image formed by lens acts as virtual object for mirror

Real image from plane mirror is formed 20 cm in front of mirror, hence at 20 cm distance from lens. Now, for second refraction from lens, u=-20 cm

f=+30 cm

v=ufu+f=-20×30-20+30=-60 cm

So, final virtual image is 60 cm from lens, or 20 cm behind mirror

Topic: Physics, Ray Optics, NEET, NEET 2021
A point object O and a mirror M move with velocities of 3 cm s-1 and 4 cm s-1 respectively as shown in the figure. OO â² is the normal to the mirror. What is the velocity of the image?

Options:
A 5 cm s-1
B 7 cm s-1
C 8 cm s-1
D 219 cm s-1

The correct answer is:
7 cm s-1

Solution:
Only velocity of the mirror along OO'=2cm s-1 is relevant
Relative to mirror : velocity = 5 cm s-1
⇒ the velocity of image relative to mirror = 5 cm s-1 towards left.
Relative to laboratory frame : velocity of image = 7 cm s-1.

Topic: Physics, Ray Optics, NEET
A point object O is placed at a distance of 0.3 m from a convex lens of focal length 0.2 m . It is then cut into two halves each of which is displaced by 0.0005 m as shown in figure.

Image will be formed from the lens at a distance of

Options:
A 30 cm
B 40 cm
C 50 cm
D 60 cm

The correct answer is:
60 cm

Solution:
By using lense maker formula
1v-1u=1f
1v=1u+1f
1v=10.2+1-0.3âv=60 cm

Topic: Physics, Ray Optics, NEET
A point object O is placed in front of a glass rod having a spherical end of the radius of curvature 30 cm. The image would be formed at

Options:
A 30 cm left from the pole
B Infinity
C 1 cm to the right from the pole
D 18 cm to the left from the object

The correct answer is:
30 cm left from the pole

Solution:
By using lens maker's formula,
Î¼âv-Î¼âu=Î¼â -Î¼âR
â¹1.5v-1-15=1.5-1+30
â¹v= -30 cm

Topic: Physics, Ray Optics, NEET
A point object O is placed on the principal axis of a convex lens of focal length f=20 cm at a distance of 40 cm to the left of it. The diameter of the lens is 10 cm. An eye is placed 60 cm to the right of the lens and a distance h below the principal axis. The maximum value of h to see the image is. (assume paraxial approximation to be valid here)

Options:
A 0
B 2.5 cm
C 5 cm
D 10 cm

The correct answer is:
2.5 cm

Solution:
1 v = 1 fâ¡ + 1 u
= 1 2 0 - 1 4 0 = 1 4 0
v = 40 cm

To locate field of view ray passing through upper end should be used. At location of eye, its height should not be below 20 tan Î¸ from axis to see the image i.e., 2 0 5 4 0 = 2.5 cm

Topic: Physics, Ray Optics, NEET
A point on the periphery of a rotating disc has its acceleration vector making an angle 30Â° with the velocity vector. Then, the ratio of the magnitude of centripetal acceleration to the tangential acceleration is

Options:
A sin30°
B cos30°
C tan30°
D none of these

The correct answer is:
tan30°

Solution:
tan⁡θ=aNat

Topic: Physics, Laws of Motion, NEET
A point on the periphery of rotating disc has its acceleration vector making on angle 30o with velocity vector, then the ratio of magnitude of centripetal acceleration to tangential acceleration is

Options:
A sin⁡30o
B cos⁡30o
C tan⁡30o
D none of these

The correct answer is:
tan⁡30o

Solution:
From the diagram, aNat=tan⁡θ=tan⁡30o

Topic: Physics, Laws of Motion, NEET
A point P consider at the contact point of a wheel on the ground which roll on the ground without slipping, then the value of the displacement of point $\mathrm{P}$ when wheel completes half of the rotation (if radius of wheel is $1 \mathrm{~m}$ )is:

Options:
A $2 \mathrm{~m}$
B $\sqrt{\pi^2+4} \mathrm{~m}$
C $\pi \mathrm{m}$
D $\sqrt{\pi^2+2} \mathrm{~m}$

The correct answer is:
$\sqrt{\pi^2+4} \mathrm{~m}$

Solution:
Displacement $P Q$
$\begin{aligned}
& =\sqrt{(P M)^2+(Q M)^2} \\
& =\sqrt{(\pi R)^2+(2 R)^2} \\
& =\sqrt{\pi^2+4}\{\because R=1 \mathrm{~m}\}
\end{aligned}$

Topic: Physics, Rotational Motion, NEET, NEET 2002
A point P lies on the axis of a flat coil carrying a current. The magnetic moment of the coil is Î¼. What will be the magnetic field at point P? It is given that the distance of P from the centre of coil is d, which is large compared to the radius of the coil.

Options:
A Î¼02ÏÎ¼d3
B μ04πμd3
C μ06πμd2
D μ08πμd2

The correct answer is:
μ02πμd3

Solution:
The magnetic field at P due to the flat coil of n turns, radius r, carrying current i is

B=Î¼02Â·nir2d2+r23/2âÎ¼02Â·nir2d3(d>>r)

=Î¼02ÏÂ·nÏr2id3=Î¼02ÏÂ·Î¼d3

Topic: Physics, Current Electricity, NEET
A point P lies on the axis of a ring of mass M and radius a at a distance a from its centre O. A small particle of mass m starts from P and reaches O under gravitational attraction only. Its speed when it reaches O is

Options:
A 2GMa
B GMa
C 2GMa2-1
D 2GMa1-12

The correct answer is:
2GMa1-12

Solution:
The gravitational potential at P, VP=-GMAP
VP=-GMa2+a2=-GM2a and gravitational potential at O, VO=-GMa
Let v be the velocity of the particle when it reaches O.
∴ K.E. at O=12mv2
Gravitational P.E. = Gravitational potential × mass
∴ P.E. at P=-GMm2a and P.E. at O=-GMma
By the principle of conservation of mechanical energy
12mv2-GMma=-GMm2a
∴v2=2GMma-GMm2a=2GMa1-12
∴v=2GMa1-12

Topic: Physics, Gravitation, NEET
A point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x=a \sin (\omega t+ \frac{\pi}{6})$. After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?

Options:
A $\frac{T}{8}$
B $\frac{T}{6}$
C $\frac{T}{3}$
D $\frac{T}{12}$

The correct answer is:
$\frac{T}{12}$

Solution:
Key Idea : Velocity is the time derivative of displacement.
Writing the given equation of a point performing SHM
$x=a \sin \left(\omega t+\frac{\pi}{6}\right)$
Differentiating Eq. (i), w.r.t. time, we obtain
$y=\frac{d x}{d t}=a \omega \cos \left(\omega t+\frac{\pi}{6}\right)$
It is given that $y=\frac{a \omega}{2}$, so that
or $\frac{a \omega}{2}=a \omega \cos \left(\omega t+\frac{\pi}{6}\right)$
$\frac{1}{2}=\cos \left(\omega t+\frac{\pi}{6}\right)$
or $\cos \frac{\pi}{3}=\cos \left(\omega t+\frac{\pi}{6}\right)$
$\begin{array}{ll}
\text {or } & \omega t+\frac{\pi}{6}=\frac{\pi}{3} \\
\Rightarrow & \omega t=\frac{\pi}{6} \\
\text {or } & t=\frac{\pi}{6 \omega}=\frac{\pi \times T}{6 \times 2 \pi}=\frac{T}{12}
\end{array}$
Thus, at $\frac{T}{12}$ velocity of the point will be equal to half of its maximum velocity.

Topic: Physics, Oscillations, NEET, NEET 2008 (Screening)
A point performs simple harmonic oscillation of period T and the equation of motion is given by x = a sin ω t + π 6 . After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?

Options:
A T 3
B T 12
C T 8
D T 6

The correct answer is:
T 12

Solution:
x = a sin ω t + π 6
d⁡ x d⁡ t = a ω cos ω t + π 6
Maximum velocity = a ω
∴ aω2=aωcosωt+π6
∴ cosωt+π6=12
ωt+π6=60ο=2π6 rad
ω t = 2 π 6 - π 6 = π 6
∴t=π6ω=π6×T⁡2π=T⁡12∵ ω=2πT⁡

Topic: Physics, Oscillations, NEET
A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q are at a distance of 2 m and 3 m respectively from the source. The ratio of the intensities of the waves at P and Q is

Options:
A 9:4
B 2:3
C 3:2
D 4:9

The correct answer is:
9:4

Solution:
The average power per unit area that is incident perpendicular to the direction of propagation is called the intensity.
Intensity of sound
I=P4πr2
I∝1r2
I1I2=r2r12
Here, r1=2 m, r2=3 m
Substituting the values, we have
I1I2=322=94

Topic: Physics, Waves and Sound, NEET
A point source emits sound equally in all directions in a non-absorbing medium. Two points $\mathrm{P}$ and $\mathrm{Q}$ are at distances of 2 $\mathrm{m}$ and $3 \mathrm{~m}$ respectively from the source. The ratio of the intensities of the wave at $P$ and $Q$ is:

Options:
A $3: 2$
B $2: 3$
C $9: 4$
D $4: 9$

The correct answer is:
$9: 4$

Solution:
$$
\begin{aligned}
& d_1=2 \mathrm{~m}_1 d_2=3 \mathrm{~m} \\
& \text { Intensity } \propto \frac{1}{(\text { distance })^2} \\
& I_1 \propto \frac{I}{2^2} \text { and } I_2 \propto \frac{1}{3^2} \\
& \therefore \quad \frac{I_1}{I_2}=\frac{9}{4}
\end{aligned}
$$

Topic: Physics, Waves and Sound, NEET, NEET 2005
A point source emits sound equally in all directions in a non-absorbing medium. Two points P and Q are at distances of 2 m and 3 m respectively from the source. The ratio of the intensities of the waves at P and Q is

Options:
A 3 : 2

B 2 : 3
C 9 : 4

D 4 : 9

The correct answer is:
9 : 4

Solution:
d 1 =2  m,  d 2 =3  m
Intensity ∝ 1 distance 2
I⁡ 1 ∝ 1 2 2 and I⁡ 2 ∝ 1 3 2
∴ I⁡ 1 I⁡ 2 = 9 4

Topic: Physics, Waves and Sound, NEET
A point source of light is kept at a distance of 15 cm from a converging lens, on its optical axis. The focal length of the lens is 10 cm and its diameter is 3 cm. A screen is placed on the other side of the lens, perpendicular to the axis of the lens, at a distance 20 cm. from it. Then find the area of the illuminated part of the screen?

Options:
A π4 cm2
B π6 cm2
C Ï2 cm2
D Ï3 cm2

The correct answer is:
π4 cm2

Solution:
1 V - 1 u = 1 f â 1 v - 1 - 1 5 = 1 1 0 â 1 v = 1 1 0 - 1 1 5
1 V = 3 - 2 3 0 â v = 3 0 cm
for r , 1 . 5 3 0 = r 1 0 â r = 1 2 cm â A = Ï r 2 â A = Ï 1 2 2
A = Ï 4 cm 2

Topic: Physics, Ray Optics, NEET
A polarized light of intensity I0 is passed through another polarizer whose pass axis makes an angle of 60o with the pass axis of the former. What is the intensity of emerging polarized light from second polarizer?

Options:
A I=I0
B I=I06
C I=I05
D I04

The correct answer is:
I04

Solution:
By Malus law, I=I0cos2⁡θ
(I = Intensity of emergent polarized light)
Where θ=60o, I= ?
=I0×cos2⁡60o
( I0= Intensity passed through polarizer)
=I0×122=I04

Topic: Physics, Wave Optics, NEET
A polarizer - analyser set is adjusted such that the intensity of light coming out of the analyser is just 10 % of the original intensity. Assuming that the polarizer - analyser set does not absorb any light, the angle by which the analyser needs to be rotated further, to reduce the output intensity to zero, is

Options:
A 71.6o
B 18.4o
C 90o
D 45o

The correct answer is:
18.4o

Solution:
I=I0cos2θ
I010=I0cos2θ
Cosθ=110=0.31<12 which is 0.707
sin-1(1/2)=45o and 90o-θ<45o so only one option is correct
i.e. 18.4o
angle rotated should be =90o-71.6o=18.4o

Topic: Physics, Wave Optics, NEET