datavorous/entrance-exam-dataset huggingface dataset:
A pulley fixed with celling carries a string with blocks of mass ′m′ and '3m' attached to its ends. The masses of string and pulley are negligible. When the system is released, its centre of mass moves with acceleration -

Options:
A g
B g/2
C g/4
D zero

The correct answer is:
g/4

Solution:
Acceleration of each mass w.r.t. pulley
=3mg–mg3m+m=g/2
Acceleration of centre of mass
=3mg/2–mg/23m+m=g/4

Topic: Physics, Center of Mass Momentum and Collision, NEET
A pulse of a wave train travels along a stretched string and reaches the fixed end of the string. It will be reflected with

Options:
A A phase change of 180° with speed changed
B The same phase as the incident pulse with no reversal of velocity
C A phase change of 180Â° with no change of speed
D The same phase as the incident pulse but with velocity reversed

The correct answer is:
A phase change of 180° with no change of speed

Solution:
When a pulse of a wave train travels along the stretched string and reaches the fixed end of the string, then it will be reflected back to the same medium. the reflected ray suffers a phase change of π with the incident wave but wave speed after reflection does not change.

Topic: Physics, Waves and Sound, NEET
A pulse or a wave train travels along a stretched string and reaches the fixed end of the string. It will be reflected back with

Options:
A The same phase as the incident pulse but with velocity reversed
B A phase change of 180° with no reversal of velocity
C The same phase as the incident pulse with no reversal of velocity
D A phase change of 180° with velocity reversed

The correct answer is:
A phase change of 180° with velocity reversed

Solution:
On reflection from fixed end (denser medium) a phase difference of π is introduced and velocity is reversed.

Topic: Physics, Waves and Sound, NEET
A pump motor is used to deliver water at a certain rate from a given pipe. To obtain 'n' times water from the same pipe in the same time the amount by which the power of the motor should be increased -

Options:
A n1/2
B n2
C n3
D n

The correct answer is:
n3

Solution:
Volume delivered per sec.=Av
mass delivered per sec=Avd
momentum delivered per sec=Av2d=force
power=force × velocity= Av3d
i.e. power ∝v3

Topic: Physics, Work Power Energy, NEET
A pump motor is used to deliver water at a certain rate from a given pipe. To obtain n times water from the same pipe in the same time the amount by which the power of the motor should be increased

Options:
A n1/2
B n2
C n3
D n

The correct answer is:
n3

Solution:
The volume delivered per second = Av
Mass delivered per second = Avd
Momentum delivered per second = Av2d = Force
Power = Force × Velocity = Av3d
i.e. Power ∝v3

Topic: Physics, Work Power Energy, NEET
A pure homozygous pea plant with yellow round seeds is crossed with a pure homozygous pea plant with green wrinkled seeds. Their progeny produced is allowed to undergo self-pollination. It produces four different phenotypes. Each phenotype is an expression of different types of genotypes. Select the correct option from the table.

Yellow round seeds	Green round seeds	Yellow wrinkled seeds

I.	Types of genotypes:	9	3	3

II.	Types of genotypes:	4	2	2

III.	Types of genotypes:	3	3	3

IV.	Types of genotypes:	5	2	2

Options:
A I.
B II.
C III.
D IV.

The correct answer is:
II.

Solution:
In the dihybrid cross, yellow round seeds are produced because of genotypes YYRR, YyRr, YyRR and YYRr (4 genotypes). Green Round seeds are produced by genotypes yyRR and yyRr (2 genotypes). Yellow wrinkled seeds are produced due to YYrr and Yyrr (2 genotypes).

Topic: Biology, Principles of Inheritance and Variation, NEET
A pure pea plant having yellow and round seeds in an inflated pod was crossed with a pure pea plant having green and wrinkled seeds in a constricted pod. The proportion of plants having green round seeds in an inflated pod in the F2 generation is

Options:
A 2764
B 964
C 364
D 164

The correct answer is:
964

Solution:
In the given question, the cross is between YYRRII and yyrrii. The F1 hybrid is YyRrIi.
For getting, the F2 generation the cross is YyRrIi x YyRrIi. It is a pea plant where all characters are independent assorted so we can consider the crosses separately:
In Yy x Yy , the proportion of yellow seeds produced is 34 and the proportion of green seeds produced is 14
In Rr x Rr , the proportion of round seeds produced is 34 and the proportion of wrinkled seeds produced is 14
In Ii x Ii, the proportion of inflated pods produced is 34 and the proportion of constricted pods produced is 14
Thus the proportion of plants having green round seeds in an inflated pod in the F2 generation is 14×34×34 =964

Topic: Biology, Principles of Inheritance and Variation, NEET
A quantity of PCl5 was heated in a 2 litre vessel at 525 K. It dissociates as

PCl5g â PCl3g + Cl2g

At equilibrium 0.2 mol each of PCl5, PCl3 and Cl2 is found in the reaction mixture. The equilibrium constant Kc for the reaction is -

Options:
A 0.2
B 0.5
C 0.1
D 0.05

The correct answer is:
0.1

Solution:
PCl5at equi. moles 0.2â¡ â PCl30.2â¡ + Cl2 0.2â¡
at equi. conc 0.22=0.1 0.22=0.1 0.22=0.1

K c = PCl 3 Cl 2 PCl 3 = 0.1×0.1 0.1 =0.1

Topic: Chemistry, Chemical Equilibrium, NEET
A racing car travels on a track (without banking) ABCDEFA (Fig). ABC is a circular arc of radius 2 R. CD and FA are straight paths of length R and DEF is a circular arc of radius R = 100 m. The coefficient of friction on the road is µ = 0.1. The maximum speed of the car is 50 m s–1. Find the minimum time for completing one round.

Options:
A 89.7s
B 86.3s
C 81.6s
D 88.97s

The correct answer is:
86.3s

Solution:
Correct Option is : (B)
86.3s

Topic: Chemistry, Laws of Motion, NEET
A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is ( C = Velocity of light)

Options:
A EC
B 2EC
C 2EC2
D EC2

The correct answer is:
2EC

Solution:
E=hCλP=hλ ⇒ P=EC
Initial momentum of light when falls on a perfectly reflecting surface.
Pi=EC
When it is reflected momentum is in opposite direction
Pf=-EC
Net momentum =Pf-Pi
= -EC-EC
=-2EC
ΔP=2EC

Topic: Physics, Electromagnetic Waves, NEET, NEET 2015 (Phase 1)
A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is

Options:
A Ec
B 2Ec
C Ec
D Ec2

The correct answer is:
2Ec

Solution:
Initial momentum of surface
Â Â Â Pi=EC
Where, cÂ = velocity of light (constant).
Since, the surface is perfectly, reflecting, so the same momentum will be reflected completely.
Final momentum
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Pf=Ec (negative value)
â´Change in momentum
Îp=Â pf-pi
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =-Â Ec-Â Ec=Â -2Ec
Thus , momentum transferred to the surface is
Îp′=|âp|=2Ec

Topic: Physics, Dual Nature of Matter, NEET
A radioactive element decays by Î²- emission. If the mass of parent and daughter nuclide are m1 and m2 respectively. Calculate energy liberated during emission.

Options:
A m1-m2- 2mec2
B m2-m1- 2mec2
C me-m2- 2m1c2
D m2-me- 2m1c2

The correct answer is:
m1-m2- 2mec2

Solution:
The masses of parent and daughter elements nuclide are m1and m2 respectively.
Mass of parent element = m1+ Zme (where Z is atomic number of parent element)
Mass of daughter element - m2+ Z+1me (As atomic number of element increases by one due to β – emission)
Mass decay = mass of parent atom - one β loss - [mass of daughter atom along with one electron]
=m1+Zme-me-[m2+ Z+1me]
=[m1-m2-2me]
Energy liberated =m1-m2- 2mec2

Topic: Physics, Nuclear Physics, NEET
A radioactive element has a rate of disintegration 10,000 disintegrations per minute at a particular instant. After four minutes it becomes 2500 disintegrations per minute. The decay constant per minute is

Options:
A 0.2loge2â¡
B Â 0.5loge2â¡Â
C 0.6loge2â¡Â
D 0.8loge2â¡Â

The correct answer is:
0.5loge2⁡

Solution:
NN0=e-Î»t
â´ 250010000=e-Î»Ã4
â´ 14=e-4Î»
â´ e4Î»=4
â´ 4Î»=logeâ¡4
â´ 4Î»=logeâ¡22
â´ 4Î»=2logeâ¡2
â´ Î»=24logeâ¡2
â´ Î»=0.5logeâ¡2

Topic: Physics, Nuclear Physics, NEET
A radioactive nucleus A with a half-life T, decays into a nucleus B. At t=0, there is no nucleus B. At some time t, the ratio of the number of B to that of A is 0.3. Then, t is given by:

Options:
A t=Tlogâ¡1.3
B t=T2logâ¡Â 2logâ¡Â 1.3
C t=Tlogâ¡ 1.3logâ¡ 2
D t=Tlogâ¡Â 1.3

The correct answer is:
t=Tlogâ¡ 1.3logâ¡ 2

Solution:
Let N A and N B be the number of molecules of A and B after time t.
Also, after time t, N B N A =0.3( given ) Also, let N 0 be the total number of nucleus initial
After time t, N A + N B = N 0
N A +0.3 N A = N 0
∴   N A = N 0 1.3
Also, rate of disintegration of A
N A = N 0 e −λt
⇒ N 0 1.3 = N 0 e −λt ;  1 1.3 = e −λt    or  log( 1.3 )=λt
t= log( 1.3 ) λ   ∴  t= Tlog( 1.3 ) log( 2 ) = Tlog( 1.3 ) log2

Topic: Physics, Nuclear Physics, NEET
A radioactive nucleus of mass $M$ emits a photon of frequency $v$ and the nucleus recoils. The recoil energy will be

Options:
A $h^2 v^2 / 2 M c^2$
B zero
C $h \mathrm{~V}$
D $M c^2-h v$

The correct answer is:
$h^2 v^2 / 2 M c^2$

Solution:
Momentum of a photon
$$
p=\frac{h v}{c}
$$
Hence, Recoil energy
$$
E=\frac{p^2}{2 M}
$$
$$
\begin{aligned}
& \therefore \quad E=\frac{\left(\frac{h v}{c}\right)^2}{2 M} \\
& \text { or } \quad E=\frac{h^2 v^2}{2 M c^2} \\
&
\end{aligned}
$$

Topic: Physics, Nuclear Physics, NEET, NEET 2011 (Screening)
A radioactive nucleus of mass number A, initially at rest, emits on α-particle with a speed v. What will be the recoil speed of the daughter nucleus?

Options:
A 2v(A-4)
B 2v(A+4)
C 4v(A-4)
D 4v(A+4)

The correct answer is:
4v(A-4)

Solution:
As per conservation law of momentum 0=4v+A-4vr
∴ Recoil speed vr=4v(A-4)

Topic: Physics, Center of Mass Momentum and Collision, NEET
A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After 5 minutes, the rate is 1250 disintegrations per minute. Then the decay constant per minute is

Options:
A 0.4ln2
B 0.2ln2
C 0.1ln2
D 0.8ln2

The correct answer is:
0.4ln2

Solution:
As rate of disintegration is proportional to the number of atoms actually present, therefore,
N 0 = 5000, N = 1 2 5 0 , t = 5 min
As N = N 0 e - Î» t
â´ ln Nâ¡ Nâ¡ 0 = - Î» t ; ln 1 2 5 0 5 0 0 0 = - Î» t ; ln 1 4 = - Î» Ã 5 ; - 2 ln 2 = - Î» Ã 5 ; Î» = 0 . 4 ln 2

Topic: Physics, Nuclear Physics, NEET
A radioactive sample at any instant has its disintegration rate 5000 dpm. After 5 minutes, the rate is 1250 dpm. Then, the decay constant (per minute) is

Options:
A 0.8 ln2
B 0.4 ln2
C 0.2 ln2
D 0.1 ln2

The correct answer is:
0.4 ln2

Solution:
12505000=12n
n=2
T12=tn=52=2.5 min
decay constant ,
λ=ln2T12=ln22.5=0.4 ln2

Topic: Physics, Nuclear Physics, NEET
A radioactive sample S1 having activity A1 has twice the number of nuclei as another sample S2 activity A2. If A2=2A1, then ratio of the half-life of S1 to the half-life of S2 is

Options:
A 1:2
B 2:1
C 4:1
D 1:4

The correct answer is:
4:1

Solution:
Since, we know,
activity, A=Î»N=0.693t12N
ât12=0.693AN
Thus, ( t 1 2 ) 1 ( t 1 2 ) 2 =N1A1âA2N2
Given, A2=2A1
and N1=2N2
=2A1A1Ã2N2N2=4 :1

Topic: Physics, Nuclear Physics, NEET
A radioactive sample S1 having an activity of 5 μCi has twice the number of nuclei as another sample S2 which has an activity of 10 μCi. The half lives of S1 and S2 can be

Options:
A 20 yr and 5 yr, respectively
B 20 yr and 10 yr, respectively
C 10 yr each
D 5 yr each

The correct answer is:
20 yr and 5 yr, respectively

Solution:
Activity of S1 = 12 (activity of S2)
or λ 1 N 1 = 1 2 λ 2 N 2
Or λ 1 λ 2 = N 2 2 N 1 T 1 T 2 = 2 N 1 N 2 or (T = half - life = ℓ n 2 λ )
Given N1 = 2 N2,
Therefore T 1 T 2 = 4

Topic: Physics, Nuclear Physics, NEET
A radioactive sample S1 having the activity A1 has twice the number of nuclei as another sample S2 of activityA2. If A2=2A1, then the ratio of half-life of S1 to the half-life of S2 is

Options:
A 4
B 2
C 0.25
D 0.75

The correct answer is:
4

Solution:
Activity, A=Î»N=0.693T1/2 N
Where T1/2 is the half-life of a radioactive sample,
â´ A1A2=N1T1 ÃT2N2
T1T2=A2A1 ÃN1N2
=2A1A1 Ã2N2N2=41

Topic: Physics, Nuclear Physics, NEET
A radioactive sample S1 having the activity A1 has twice the number of nuclei as another sample S2 of activityA2. If A2=2A1, then the ratio of the half-life of S1 to the half-life of S2 is

Options:
A 4
B 2
C 0.25
D 0.75

The correct answer is:
4

Solution:
Activity A, Where T is the half-life of a radioactive sample,
∴A1A2=N1T1 ×T2N2
T1T2=A2A1 ×N1N2
=2A1A1 ×2N2N2=41

Topic: Physics, Nuclear Physics, NEET
A radioactive substance X decays into another radioactive substance Y. Initially only X was present. λx and λy are the disintegration constants of X and Y. Nx and Ny are the number of nuclei of X and Y at any time t. Number of nuclei Ny will be maximum when:

Options:
A NyNx-Ny=λyλx-λy
B NxNx-Ny=λxλx-λy
C λyNy=λxNx
D λyNx=λxNy

The correct answer is:
λyNy=λxNx

Solution:
Net rate of formation of Y at any time t is:
dNydt=λxNx-λyNy
Ny is maximum when dNydt=0
or λyNy=λxNx

Topic: Physics, Nuclear Physics, NEET
A rail track made of steel having a length of 10 m is clamped on a railway line at its two ends (Figure). On a summer day, due to a rise in temperature by 20°C, it is deformed as shown in the figure. What is the displacement of the center, denoted as x, if the coefficient of linear expansion for steel is αsteel = 1.2 × 10-5/°C?

Options:
A 10 cm
B 11 cm
C 12 cm
D 13 cm

The correct answer is:
11 cm

Solution:
Correct Option is : (B)
11 cm

Topic: Chemistry, Thermal Properties of Matter, NEET
A railway engine whistling at a constant frequency moves with a constant speed. It goes past a stationary observer standing beside the railway track. The frequency (n′)of the sound heard by the observer is plotted against time (t), which of the following best represents the resulting curve

Options:
A
B
C
D

The correct answer is:

Solution:
It is known that Doppler's effect depends on velocity not on distance. When the source is approaching the stationary observer, the apparent frequency heard by the observer is
nâ²=vÃnv-vs=constant
But nâ²>n.
When the source has crossed the observer, apparent frequency heard by the observer is
nâ²â²=vÃnv+vs=another constant
But nâ²â²<n.option (d) is correct.

Topic: Physics, Waves and Sound, NEET
A raindrop of radius 0.3 mm has a terminal velocity in air =1 m s-1. The viscous force on it is (Given the viscosity of the air is 18 Ã10-5 g cm-1 s-1 )

Options:
A 101.73×10-4 dyne
B 101.73×10-5 dyne
C 16.95×10-4 dyne
D 16.95×10-5 dyne

The correct answer is:
101.73×10-4 dyne

Solution:
F=6 π η r v
=6×3.14×(18×10-5)×0.03×100
=101.73×10-4 dyne

Topic: Physics, Mechanical Properties of Fluids, NEET
A ratio transmitter operates at a frequency 880 kHz and a power of 10 kW. The number of photons emitted per second is

Options:
A 1.72×1031
B 1.327×1025
C 1.327×1037
D 1.327×1045

The correct answer is:
1.72×1031

Solution:
Number of photons emitted per second,
n=PowerEnergy of photon=Phv
=100006.6Ã10-34Ã880Ã103=1.72Ã1031

Topic: Physics, Dual Nature of Matter, NEET
A ratio transmitter operates at a frequency 880 kHz and a power of 10 kW. The number of photons emitted per second is

Options:
A 1.72×1031
B 1.327×1025
C 1.327×1037
D 1.327×1045

The correct answer is:
1.72×1031

Solution:
Number of photons emitted per sec,
n=PowerEnergy of photon=Phv
=100006.6×10-34×880×103=1.72×1031

Topic: Physics, Dual Nature of Matter, NEET
A ray is incident at an angle of incidence i on one surface of a small angle prism (with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is μ, then the angle of incidence is nearly equal to:

Options:
A 2Aμ
B μA
C μA2
D A2μ

The correct answer is:
μA

Solution:
r2=0

r1=A

Apply Snell's law

sin i=μ sin r1

for small angle r1=A

i=μA

Topic: Physics, Ray Optics, NEET, NEET 2020 (Phase 1)
A ray of light from a liquid (Âµ=3) is incident on a system of two right-angled prisms of refractive index 3 and 2 as shown. The ray suffers zero deviation when emerges into the air from CD. The angle of incidence i is

Options:
A 45°
B 60°
C 30°
D 90°

The correct answer is:
45°

Solution:
3 cos i=2cos r1 Snell's law....(1)
Sin i=2 sin r1 Snell law... (2)
12 + 22 will give
3 cos2i + sin2i=2
i=45°

Topic: Physics, Ray Optics, NEET
A ray of light is incident at an angle of incidence, $i$, on one face of a prism of angle $A$ (assumed to be small) and emerges normally from the opposite face. If the refractive index of the prism is $\mu$, the angle of incidence $i$, is nearly equal to

Options:
A $\mu A$
B $\frac{\mu A}{2}$
C $A / \mu$
D $A / 2 \mu$

The correct answer is:
$\mu A$

Solution:
Refractive index
$\begin{gathered}
\mu=\frac{\sin i}{\sin r} \\
\mu=\frac{\sin i}{\sin A}
\end{gathered}$
In case of small angle.
$\begin{aligned}
\mu & =\frac{i}{A} \\
i & =\mu A
\end{aligned}$

Topic: Physics, Ray Optics, NEET, NEET 2012 (Screening)
A ray of light is incident normally on a glass slab of thickness 5 cm and refractive index 1.6. The time taken to travel by a ray from source to surface of the slab is the same as to travel through glass slab. The distance of the source from the surface is

Options:
A 4 cm
B 8 cm
C 12 cm
D 16 cm

The correct answer is:
8 cm

Solution:
Velocity of light in medium,v=5t
Velocity of light in vaccum, c= x t
μ= c v = x 5
x=5μ=8 cm

Topic: Physics, Ray Optics, NEET
A ray of light is incident normally on one face of a right-angled isosceles prism and then it grazes the hypotenuse. The refractive index of the material of the prism is

Options:
A 1.33
B 1.414
C 1.5
D 1.732

The correct answer is:
1.414

Solution:
Refractive index, Î¼=1sinâ¡C,
Where C is the critical angle
Here, C=45Â°
â´ Î¼=1sinâ¡45Â°
or Î¼= 2
or Î¼=1.414

Topic: Physics, Ray Optics, NEET
A ray of light is incident normally on one of the faces of a prism of apex angle 30Â°and refractive index 2. The angle of deviation of the ray is

Options:
A

30°

B

45°

C

15Â°

D none of these

The correct answer is:

15Â°

Solution:
Using Snell’s law for the refraction at AC, we get

μ  sini=(1)  sinr

2sin30°=sinr

sinr=12

⇒     r=45°

∴ Angle of deviation =45°−30°=15°

Topic: Physics, Ray Optics, NEET
A ray of light is incident normally on one of the faces of a prism of apex angle 30Â° and refractive index 2. The angle of deviation of the ray is

Options:
A

30°

B

45°

C

15Â°

D none of these

The correct answer is:

15Â°

Solution:
Using Snell’s law for the refraction at AC, we get

μ  sini=(1)  sinr

2sin30°=sinr

sinr=12

⇒     r=45°

∴ Angle of deviation =45°−30°=15°

Topic: Physics, Ray Optics, NEET
A ray of light is incident on a plane mirror at an angle of 30°. At what angle with the horizontal must a plane mirror be placed so that the reflected ray becomes vertically upwards?

Options:
A 40°
B 20°
C 30°
D 60°

The correct answer is:
30°

Solution:
When a light ray falls on a mirror at an angle 30°, then the reflected ray will make the same angle with the plans as shown in Figure (i)
In order to make the reflected ray vertical, the mirror should be rotated at an angle of 60°,
So, the mirror should be tilted by 602=30° as shown in Figure. (ii)

Topic: Physics, Ray Optics, NEET
A ray of light is incident on a prism ABC of refractive index 3 as shown in the figure.

The angle by which the second identical prism must be rotated so that the final ray suffers net minimum deviation is

Options:
A 60°
B 45°
C 90°
D 30°

The correct answer is:
60°

Solution:
In the position shown net deviation suffered by the ray of light should be minimum. Therefore, the second prism should be rotated by 60° ( anticlockwise).

Topic: Physics, Ray Optics, NEET
A ray of light is incident on a surface of glass slab at an angle 45°. If the lateral shift produced per unit thickness is 1/3, then the angle of refraction produced is

Options:
A tan-1⁡32
B tan-1⁡1-23
C sin-1⁡1-23
D tan-1⁡23-1

The correct answer is:
tan-1⁡1-23

Solution:
Lateral shift, Y=tsinâ¡i-rcosâ¡r
Yt=sinâ¡i-rcosâ¡r
=sinâ¡icosâ¡r-cosâ¡isinâ¡rcosâ¡r
=sinâ¡i-cosâ¡itanâ¡r
â 13=sinâ¡45o-cosâ¡45otanâ¡r=121-tanâ¡r
â 23=1-tanâ¡r
â tanâ¡r=1-23
â´ r=tan-1â¡1-23

Topic: Physics, Ray Optics, NEET
A ray of light is incident on an equilateral glass prism placed on a horizontal table. Which of the following is true for the condition of minimum deviation?

Options:
A PQ is horizontal
B QR is horizontal
C RS is horizontal
D either PQ or RS is horizontal

The correct answer is:
QR is horizontal

Solution:
The refracted ray is parallel to the base of the prism in the case of minimum deviation. Hence, the correct option is (QR is horizontal).

Topic: Physics, Ray Optics, NEET
A ray of light is incident on the plane mirror at rest. The mirror starts turning at a uniform angular acceleration of Ï rad sâ2. The reflected ray at the end of 14 s must have turned through

Options:
A 90°
B 45°
C 22.5°
D 11.25°

The correct answer is:
11.25°

Solution:
As we know, if a mirror is rotated.
Given, angular acceleration, Î±=Ï rad/s2
Initial angular velocity =Ï0=0
Time, t=14 s
Now, as Î¸=Ï0t+12Î±t2
=0Ã14+12ÃÏÃ142
=1802Ã16=5.625o
Since, mirror is turned by 5.625o, therefore reflected ray will be turned by 2Ã5.625=11.25o.
Hence, option 4 is correct.

Topic: Physics, Ray Optics, NEET
A ray of light passes through four transparent media with refractive index n1,n2,n3 and n4 as shown in the figure. The surfaces of all media are parallel.

If the emergent ray DE is parallel to incidence ray AB, then

Options:
A n1=n4
B n2=n4
C n3=n4
D n1=n2+n3+n43

The correct answer is:
n1=n4

Solution:
It is given that the light rays to medium (1) and medium (4) are parallel
â´ n1=n4

Topic: Physics, Ray Optics, NEET
A ray of light passing through a prism of refractive index 2 undergoes minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. The angle of the prism is

Options:
A 60°
B 90°
C 75°
D 30°

The correct answer is:
90°

Solution:
According to given problem, i=2r=A
∵ angle of prism, A=2r
Since, δm=2i-A
⇒ δm=2A-A=A
We have, n=sin⁡A+δm2sin⁡A2 ⇒ 2=sin⁡Asin⁡A2
⇒ 2=2sin⁡A2cos⁡A2sin⁡A2
⇒ cos⁡A2=22 ⇒ A2=cos-1⁡12
⇒ A2=45° ⇒ A=90°

Topic: Physics, Ray Optics, NEET
A ray of light strikes a transparent slab of the refractive index of 2 at an angle of incidence 45°. The angle between the reflected and refracted rays is

Options:
A 95°
B 120°
C 135°
D 105°

The correct answer is:
105°

Solution:
According to the figure,
Given, i=45°
Let the angle between the reflected ray and refracted ray is
Apply Snell's law at the air-glass surface,
sin⁡isin⁡r=n1
⇒sin⁡45°sin⁡r=2
sin⁡r=12⋅2=12
∴ r=30°
From the diagram, θ=180-i+r
=180-45°+30°
=180°-75°=105°

Topic: Physics, Ray Optics, NEET
A ray of light suffers a minimum deviation when incident on an equilateral prism of refractive index 2. The angle of deviation is

Options:
A 30°
B 45°
C 60°
D 50°

The correct answer is:
30°

Solution:
Given, the refractive index =2
Prism angle, A=60°
For a prism of the position minimum deviation,
μ=sin⁡A+δm2sin⁡A2
2=sin⁡60o+δm2sin⁡60o2
=sin⁡60o+δm2sin⁡30o
⁡sin⁡ 60o+δm2=22=12
⇒ 60+δm2=45°⇒δm=30°

Topic: Physics, Ray Optics, NEET
A ray of light travelling in a transparent medium of refractive index $\mu$ falls, on a surface separating the medium from air at an angle of incidence of $45^{\circ}$. For which of the following value of $\mu$ the ray can undergo total internal reflection?

Options:
A $\mu=1.33$
B $\mu=1.40$
C $\mu=1.50$
D $\mu=1.25$

The correct answer is:
$\mu=1.50$

Solution:
For total internal reflection
$$
\begin{aligned}
& \mathrm{i}>\mathrm{c} \\
\Rightarrow & \sin \mathrm{i}>\sin \mathrm{c} \\
\Rightarrow & \sin 45^{\circ}>\frac{1}{\mu} \\
\Rightarrow & \mu>\sqrt{2} \\
\Rightarrow & \mu>1.4
\end{aligned}
$$

Topic: Physics, Ray Optics, NEET, NEET 2010 (Screening)