datavorous/entrance-exam-dataset huggingface dataset:
A radionuclide with half-life 1620 s is produced in a reactor at a constant rate of 1000 nuclei per second. During each decay energy, 200 MeV is released. If the production of radionuclides started at t = 0, then the rate of release of energy at t = 3240 s is

Options:
A 1.5×105 MeV s-1
B 1.5×102 MeV s-1
C 2.5×102 MeV s-1
D 3.5×105 MeV s-1

The correct answer is:
1.5×105 MeV s-1

Solution:
Let N be the number of nuclei at time t, then net rate of increase of nuclei at instant t is,z
dâ¡ Nâ¡ dâ¡ t = Î± - Î» Nâ¡ (where Î± = rate of production of nuclei)
â´ â« 0 Nâ¡ dâ¡ Nâ¡ Î± - Î» Nâ¡ = â« 0 t dâ¡ t
â´ Nâ¡ = Î± Î» 1 - eâ¡ - Î» t ...(i)
Rate of decay at this instant R = Î»N = Î± ( 1 â eâ Î»t)
Hence, rate of release of energy at this time = R (energy released in each decay)
= Î± 1 - eâ¡ - Î» t 2 0 0 MeV / s
Substituting the values, we have
rate of release of energy = 1 0 0 0 1 - eâ¡ - 0 . 6 9 3 1 6 2 0 Ã 3 2 4 0 2 0 0
= 1.5 Ã 105 MeV/s

Topic: Physics, Nuclear Physics, JEE Main
A rain drop of radius $2 \mathrm{~mm}$ falls from a height of 500 $\mathrm{m}$ above the ground. It falls with decreasing acceleration due to viscous resistance of air until half of its original height. It attains its maximum (terminal) speed and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first half and second half of the journey? [Density of $\mathrm{H}_2 \mathrm{O}=10^3 \mathrm{~kg}$ / $\mathrm{m}^3$. What is the work done by the resistance force in the entire journey if its speed on reaching the ground is $10 \mathrm{~m} / \mathbf{s}$ ?

Options:

None

Solution:
Here, $r=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}, \rho=10^3 \mathrm{~kg} / \mathrm{m}^3$
Distance moved in half of the journey $=\mathrm{S}=\frac{500}{2}=250 \mathrm{~m}$.
Mass of the drop $=$ Vol. $\times$ density $=\frac{4}{3} \pi r^3 \rho$
$$
=\frac{4}{3} \times \frac{22}{7} \times\left(2 \times 10^{-3}\right)^3 \times 10^3=3.35 \times 10^{-5} \mathrm{~kg}
$$
Work done $=\mathrm{W}=\mathrm{F} \times \mathrm{S}=3.35 \times 10^{-5} \times 9.8 \times 250=0.082 \mathrm{~J}$
Work done by the gravitational force remains same weather the drop falls under decreasing acceleration or with uniform speed. So, the work done in second half of the journey is also $0.082 \mathrm{~J}$.
Energy of the drop on reaching the ground $=E=m g h=3.35 \times 10^{-5} \times 9.8 \times 500=0.164 \mathrm{~J}$ (considering no resistive forces).
Kinetic energy $=\frac{1}{2} m v^2=\frac{1}{2} \times 3.35 \times 10^{-5} \times 10^2=1.675 \times$ $10^{-3} \mathrm{~J}=0.001675=0.1623 \mathrm{~J}$
$=$ Work done by the resistive force.

Topic: Physics, Work Power Energy
A raindrop falling from a height $h$ above ground, attains a near terminal velocity when it has fallen through a height $(3 / 4) h$. Which of the diagrams shown in figure correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground?

Options:
A

B

C

D

The correct answer is:

Solution:
P.E. is maximum when drop start falling at $t=0$ as it fall is P.E. decrease gradually to zero. So, it rejects the graph (a), (c) and (d).
K.E. at $t=0$ is zero as drop falls with zero velocity, its velocity increases (gradually), hence, first KE also increases. After sometime speed (velocity) is constant this is called terminal velocity, so, KE also become constant. It happens when it falls $\left(\frac{3}{4}\right)$ height or remains at $\left(\frac{4}{4}\right)$ from ground, then PE decreases continuously as the drop is falling continuously. The variation in PE and KE is best represented by (b).

Topic: Physics, Work Power Energy
A raindrop with radius R=0.2 mm falls from a cloud at a height h=2000 m above the ground. Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is : [Density of water fw=1000 kg m-3 and Density of air fa=1.2 kg m-3, g=10 m/s2 Coefficient of viscosity of air =1.8×10-5N s m-2

Options:
A 250.6 m s-1
B 43.56 m s-1
C 4.94 m s-1
D 14.4 m s-1

The correct answer is:
4.94 m s-1

Solution:
At terminal speed

a=0

Fnet=0

mg=Fv=6πηRv

V=mg6πηRV

v=ρw4π3R3 g6πηR

=2ρwR2 g9η

=40081 m/s

=4.94 m/s

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2021 (27 Jul Shift 2)
a. Ramesh Chandra Dagar, a farmer of Sonipat, is associated with integrated organic farming and has recycled agricultural waste.
b. Ahmed Khan in Bangalore is associated with recycling of plastic into polyblend powder used to lay roads.

From given statements,

Options:
A a is correct, b is incorrect
B a is incorrect, b is correct
C both a and b are correct
D both a and b are incorrect

The correct answer is:
both a and b are correct

Solution:
Ramesh Chandra Dagar, a farmer in Sonipat, Haryana, is doing just this. He maintains bee-keeping, dairy management, water harvesting, composting and agriculture in a chain of processes, which support each other and allow an extremely economical and sustainable venture.
Ahmed Khan realised that plastic waste was a real problem. Polyblend, a fine powder of recycled modified plastic, was developed by his company. This mixture is mixed with bitumen that is used to lay roads. In collaboration with R. V. college of engineering and the Bangalore city corporation, Ahmed Khan proved that blends of polyblend and bitumen, when used to lay roads, enhanced the bitumen's water-repellant properties and helped to increase road life by a factor of three.

Topic: Biology, Environmental Issues, NEET
A random variable $\mathrm{X}$ assumes value $1,2,3, \ldots \ldots . n$ with equal probabilities. If the ratio of variance of $=\sum p_i x_i^2-\left(\sum p_i x_i\right)^2$ to expected value of $\mathrm{X}$ is equal to 4 , then the value of $n$ is

Options:
A 35
B 50
C 30
D 25

The correct answer is:
25

Solution:
$\begin{aligned} & \text { Variance }=\sum p_i X_i^2-\left(\sum p_i X_i\right)^2 \\ & =\frac{1}{n} \times \frac{n(n+1)(2 n+1)}{6}-\left\{\frac{1}{n} \times \frac{n(n+1)}{2}\right\}^2 \\ & =\frac{n(n+1)}{2 n}\left\{\frac{2 n+1}{3}-\frac{n(n+1)}{2 n}\right\} \\ & \text { Expected value }=\sum p_i x_i=\frac{1}{n} \cdot \frac{n(n+1)}{2}=\frac{n(n+1)}{2 n} \\ & \text { Ratio }=\frac{2 n+1}{3}-\frac{n(n+1)}{2 n}\end{aligned}$
$\Rightarrow n^2-25 n=0$
$\Rightarrow n(n-25)=0$
$\Rightarrow n=0$ or $n=25$
$\Rightarrow n=25$ as $n=0$ is not possible

Topic: Mathematics, Probability, JEE Main
A random variable $\mathrm{X}$ assumes values 1,2 , $3, \ldots ., \mathrm{n}$ with equal probabilities, if $\operatorname{var}(\mathrm{X})=\mathrm{E}(\mathrm{X})$, then $\mathrm{n}$ is

Options:
A 4
B 5
C 7
D 9

The correct answer is:
7

Solution:
$\begin{aligned} & \mathrm{X}=1,2,3, \ldots \mathrm{n} \\ & \begin{aligned} \mathrm{P}(\mathrm{X}) & =\frac{1}{\mathrm{n}} \\ \mathrm{E}(\mathrm{X}) & =\sum_{\mathrm{i}=1}^{\mathrm{n}} x_{\mathrm{i}} \mathrm{p}_{\mathrm{i}} \\ & =\frac{(1+2+3+\ldots+\mathrm{n})}{\mathrm{n}} \\ & =\frac{\mathrm{n}(\mathrm{n}+1)}{2 \mathrm{n}} \\ \mathrm{E}(\mathrm{X}) & =\frac{\mathrm{n}+1}{2}\end{aligned}\end{aligned}$
$\begin{aligned} & \operatorname{Var}(\mathrm{X})=\sum_{\mathrm{i}=1}^{\mathrm{n}} x_{\mathrm{i}}^2 \mathrm{p}_{\mathrm{i}}-[\mathrm{E}(\mathrm{X})]^2 \\ & =\frac{1^2+2^2+3^2+\ldots+\mathrm{n}^2}{\mathrm{n}}-\left(\frac{\mathrm{n}+1}{2}\right)^2 \\ & =\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6 \mathrm{n}}-\left(\frac{\mathrm{n}+1}{2}\right)^2 \\ & =\frac{(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\left(\frac{\mathrm{n}+1}{2}\right)^2 \\ & \operatorname{Var}(X)=E(X) \\ & \text {...[Given] } \\ & \frac{(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\left(\frac{\mathrm{n}+1}{2}\right)^2=\frac{\mathrm{n}+1}{2} \\ & \frac{2 \mathrm{n}^2+\mathrm{n}+2 \mathrm{n}+1}{6}-\left(\frac{\mathrm{n}^2+2 \mathrm{n}+1}{4}\right)=\frac{\mathrm{n}+1}{2} \\ & \frac{4 n^2+6 n+2-3 n^2-6 n-3}{12}=\frac{n+1}{2} \\ & \mathrm{n}^2-1=6(\mathrm{n}+1) \\ & \mathrm{n}^2-1=6 \mathrm{n}+6 \\ & \mathrm{n}^2-6 \mathrm{n}-7=0 \\ & \therefore \quad \mathrm{n}=-1 \text { or } \mathrm{n}=7 \\ & \text { But } \mathrm{n} \neq-1 \\ & \therefore \quad \mathrm{n}=7 \\ & \end{aligned}$

Topic: Mathematics, Probability, JEE Main
A random variable Xâ½B(n, p). If values of mean and variance of X are 18 and 12 respectively then total number of possible values of X are

Options:
A 54
B 55
C 12
D 18

The correct answer is:
55

Solution:
Mean = np = 18
Variance = npq = 12
â npqnp=1218
â q=23
â p=1-q=1-23
â p=13
Now np=18
â n13=18
â n=54
â â´ Values of X are
0, 1, 2, ........... 54
â´ 55 Values.

Topic: Mathematics, Probability, JEE Main
A random variable $X \sim B(n, p)$, if values of mean and variance of $\mathrm{X}$ are 18,12 respectively, then $\mathrm{n}=$

Options:
A 54
B 18
C 12
D 55

The correct answer is:
54

Solution:
We have $n p=18$ and $n p q=12$
$$
\begin{aligned}
& \therefore \mathrm{q}=\frac{12}{18}=\frac{2}{3} \quad \Rightarrow \mathrm{p}=1-\frac{2}{3}=\frac{1}{3} \\
& \therefore \mathrm{n}\left(\frac{1}{3}\right)=18 \quad \Rightarrow \mathrm{n}=54
\end{aligned}
$$

Topic: Mathematics, Statistics, JEE Main
A random variable X follows binomial probability distribution with probability P(X), with mean as 2, probability of success as p and probability of failure as q such that p+q=1. If ΣX2P(X)=285, then the probability of exactly 2 success is

Options:
A 3×214510
B 32×21659
C 3×2510
D 45×259

The correct answer is:
32×21659

Solution:
Given, x¯=np=2 mean
Variance =npq=ΣX2P(X)-x¯2
npq=285-4⇒npq=85
2×q=85⇒q=45,p=15,n=10
Probability of exactly 2 success is 10C2152458=32×21659

Topic: Mathematics, Probability, JEE Main
A random variable $X$ follows the binomial distribution and $X \sim B(n, 0.3)$. If the mean of $X$ is three times as large as the standard deviation of $X$, then $n=$

Options:
A $9$
B $21$
C $27$
D $3$

The correct answer is:
$21$

Solution:
Given, $p=0.3, q=1-p=0.7$
Also, Mean $(\mu)=3 \times$ Standard deviation $(\sigma)$.
$\Rightarrow \quad \mu=3(\sigma)$...(i)
$\begin{aligned} & \text { Now, } \mu=n p=0.3 n \\ & \text { and } \sigma=\sqrt{n p q}=\sqrt{n(0 \cdot 3)(0 \cdot 7)}=\sqrt{0.21 n} \\ & \text { From Eq. (i), }(0.3) n=3 \sqrt{(0.21) n}\end{aligned}$
$\begin{array}{ll}\Rightarrow & (0.1) n=\sqrt{(0.21) n} \\ \Rightarrow & \left(\frac{n}{10}\right)^2=\left(\frac{21 n}{100}\right) \\ \Rightarrow & n^2-21 n=0 \Rightarrow n(n-21)=0 \\ \Rightarrow & n=0,21\end{array}$
But $n \neq 0$
Thus, $n=21$.

Topic: Mathematics, Statistics, JEE Main
A random variable $X$ has Poisson distribution with mean 2. Then $P(X>1.5)$ equals

Options:
A
$\frac{2}{\mathrm{e}^2}$

B
0

C
$1-\frac{3}{\mathrm{e}^2}$

D
$\frac{3}{\mathrm{e}^2}$

The correct answer is:
$1-\frac{3}{\mathrm{e}^2}$

Solution:
$$
\begin{aligned}
& P(x=k)=e^{-\lambda} \frac{\lambda^k}{k !} \\
& P(x \geq 2)=1-P(x=0)-P(x=1) \\
& =1-e^{-\lambda}-e^{-\lambda}\left(\frac{\lambda}{1 !}\right) \\
& =1-\frac{3}{e^2} .
\end{aligned}
$$

Topic: Mathematics, Probability, JEE Main, JEE Main 2005
A random variable X has the following probability distribution:

X	0	1	2	3	4

PX	k	2k	4k	6k	8k

The value of P1<x<4x≤2is equal to

Options:
A 47
B 23
C 37
D 45

The correct answer is:
47

Solution:
To find P1<x<4x≤2 or PAB

We know that

PAB=PA∩BPB

Given

x	0	1	2	3	4

Px	k	2k	4k	6k	8k

PA=2,3

PB=0,1,2

PA∩B=Px=2

PB=Px=0+Px=1+Px=2

So PAB=Px=2Px=0+Px=1+Px=2

=4kk+2k+4k=4k7k=47

Hence option A is correct.

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (24 Jun Shift 2)
A random variable X has the following probability distribution:
X:12345PX: k22kk2k5k2
Then, PX>2 is equal to:

Options:
A 712
B 136
C 16
D 2336

The correct answer is:
2336

Solution:
W e know ∑Pi=1⇒6k2+5k=1

6k2+5k-1=0

6k2+6k-k-1=0

6k-1k+1=0⇒k=-1 (rejected) ; k=16

PX>2=k+2k+5k2

=16+26+536=6+12+536=2336

Topic: Mathematics, Probability, JEE Main, JEE Main 2020 (09 Jan Shift 2)
A random variable $X$ has the probability distribution:
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline \mathrm{X}: & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline \mathrm{p}(\mathrm{X}): & 0.15 & 0.23 & 0.12 & 0.10 & 0.20 & 0.08 & 0.07 & 0.05 \\
\hline
\end{array}
For the events $E=\{X$ is a prime number $\}$ and $F=\{X < 4\}$, the probability $P(E \cup F)$ is

Options:
A
$0.87$

B
$0.77$

C
$0.35$

D
$0.50$

The correct answer is:
$0.77$

Solution:
$P(E \cup F)=P(E)+P(F)-P(E \cap F)=0.62+0.50-0.35=0.77$

Topic: Mathematics, Probability, JEE Main, JEE Main 2004
A rat is running on ice with speed v=Ïm s-1. Suddenly he decides to turn by 90Â° and want to keep running with the same speed throughout. What is the least amount of time (in seconds) he needs for such a turn? Suppose that ratâs feet can move independently. Coefficient of friction between ratâs feet and ice is 0.125. (Given: Ï2=g )

Options:

The correct answer is:
4

Solution:
v=μRg ;

π2=18R×10 ;

R=4π25 ;

T=πR2v=π2π×4π25=4 s

Topic: Physics, Motion In One Dimension, JEE Main
A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of 213+1231310 is

Options:
A 1:41613
B 43613:1
C 23613:1
D 1:2613

The correct answer is:
43613:1

Solution:
Total number of terms in the expansion of 213+1231310=10+1=11

Now, 5th term from end =11-5+1=7th from beginning.

We need to find a ratio of the 5th term from the beginning to the 5th term from the end (i.e. 7th term from beginning) in the binomial expansion of 213+1231310=T5T7

=10C421310-412313410C621310-6123136 ∵Tr+1=nCran-rbr

=10!4!6!×22×12434310!6!4!×243×12632

=43613:1

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (12 Jan Shift 1)
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, JEE Main
A ray emanating from the point $(0,6)$ is incident on the ellipse $25 x^2+16 y^2=1600$ at the point $P$ with ordinate $S$. After reflection, ray cuts the $Y$-axis at $B$. The length of $P B$ is

Options:
A 5
B 7
C 12
D 13

The correct answer is:
13

Solution:
$\because$ Ellipse is $25 x^2+16 y^2=1600$
$\frac{x^2}{8^2}+\frac{y^2}{10^2}=1$
Coordinating foci an $\left(0, \pm \sqrt{\left(10^2-8^2\right)}\right)$ i.e. $(0, \pm 6)$
Let coordinates of $P$ are $(\lambda, 5)$
$\begin{aligned} & \therefore \quad \frac{\lambda^2}{8^2}+\frac{5^2}{10^4}=1 \\ & \Rightarrow \lambda= \pm 4 \sqrt{3} \\ & \therefore P \equiv( \pm 4 \sqrt{3}, 5) \text { (lie in I or II quadrants) }\end{aligned}$
According to reflection property, a ray passing through focus $B^{\prime}(0,6)$ will passing through $B(0,-6)$ (other focus). If $P$ lies in I quadrant, then
$P B=\sqrt{48+(5+6)^2}=\sqrt{169}=13$

Topic: Mathematics, Ellipse, JEE Main
A ray of laser of a wavelength 630 nm is incident at an angle of 30° at the diamond-air interface. It is going from diamond to air. The refractive index of diamond is 2.42 and that of air is 1. Choose the correct option.

Options:
A Angle of refraction is 24.41°.
B Angle of refraction is 30°.
C Refraction is not possible.
D Angle of refraction is 53.4°.

The correct answer is:
Refraction is not possible.

Solution:
sinθC=1μ=12μ2<sinθC

sinθ>sinθC

θ>θC

Total internal reflection will happen.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (25 Jul Shift 1)
A ray of light AO in vacuum is incident on a glass slab at angle 60° and refracted at angle 30° along OB as shown in the figure. The optical path length of light ray from A to B is:

Options:
A 23a+2b
B 2a+2b
C 2a+2b3
D 2a+2b3

The correct answer is:
2a+2b

Solution:
Optical path length =OA+μOB
=acos 60o+μbcos 30o
=2a+μ2b3
From Snell’s law,
sin 60o=μ sin 30o
μ=3
Optical path length =2a+32b3
=2a+b

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (10 Apr Shift 1)
A ray of light along x+3y=3 gets reflected upon reaching X-axis, the equation of the reflected ray is

Options:
A y=3x-3
B 3y=x-1
C y=x+3
D 3y=x-3

The correct answer is:
3y=x-3

Solution:
Let incident ray intersecting Y-axis at B and X-axis at A.

∆OAB is congruent to ∆OAB'
Hence OB'=OB
Equation of B'A:x3+y-1=1
or x-3y=3.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2013 (07 Apr)
A ray of light coming from the point (2,23) is incident at an angle 30° on the line x=1 at the point A. The ray gets reflected on the line x=1 and meets x -axis at the point B. Then, the line AB passes through the point

Options:
A 3,-13
B (4,-32)
C (3,-3)
D 4,-3

The correct answer is:
(3,-3)

Solution:
Equation of

P'B→y-23=tan120°·(x-0)

⇒ 3x+y=23

Now satisfy options

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2020 (06 Sep Shift 1)
A ray of light entering from air into a denser medium of refractive index 43, as shown in figure.
The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle θ should be equal to:

Options:
A sin-173
B sin-154
C sin-174
D sin-153

The correct answer is:
sin-173

Solution:
At maximum angle θ ray at point B goes in gazing emergence, at all less values of θ, TIR occurs. At point B

43×sinθ''=1×sin90°

θ''=sin-134

θ'=π2-θ''

At point A

1×sinθ=43×sinθ'

sinθ=43×sinπ2-θ''

sinθ=43coscos-174

sinθ=43×74

θ=sin-173

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (25 Jul Shift 2)
A ray of light incidents on a plane mirror at an angle of $30^{\circ}$. The deviation produced in the ray is

Options:
A $30^{\circ}$
B $60^{\circ}$
C $90^{\circ}$
D $120^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
given, incident angle is $30^{\circ}$.
Deviation angle $=d=2 \theta$
where $\mathrm{d}$ is the angle of deviation and $\theta$ is the angle of incidence. $\mathrm{d}=2 \times 30^{\circ} .=60^{\circ}$

Topic: Physics, Ray Optics, JEE Main
A ray of light is incident along a line which meets another line 7x-y+1=0 at the point 0,1. The ray is then reflected from this point along the line y+2x=1. Then the equation of the line of incidence of the ray of light is :

Options:
A 41x-25y+25=0
B 41x+25y-25=0
C 41x-38y+38=0
D 41x+38y-38=0

The correct answer is:
41x-38y+38=0

Solution:
Let slope of incident ray be m.
∴ angle of incidence = angle of reflection
∴ m-71+7m=-2-71-14=913
⇒m-71+7m=913 or m-71+7m=-913
⇒13m-91=9+63m, or13m-91=-9-63m
⇒ m=-2 or m=4138
⇒ y-1=4138x-0
⇒ 38y-38-41x=0

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2016 (10 Apr Online)
A ray of light is incident at an angle of 45 o (relative to the normal) on a thin film of a liquid with an index of refraction of 2. Assuming that the medium on both sides of the film is air, find the minimum thickness of the film required for constructive interference in the reflected light for sodium light of wavelength 600 nm.

Options:
A 122 nm
B 212 nm
C 173 nm
D 320 nm

The correct answer is:
122 nm

Solution:
From Snell's law
1×sin45o=2sinr
sinr=12

r=30°

∆x=2μtcosr
For constructive interference in reflection
2μtcosr+λ2=nλ

tmin=λ4μcosr

tmin=600 nm4×2×cos30o≈122 nm

Topic: Physics, Wave Optics, JEE Main
A ray of light is incident at an angle of 60 ∘ on one face of a prism of angle 30 ∘ . The emergent ray of light makes an angle of 30 ∘ with incident ray. The angle made by the emergent ray with second face of prism will be:

Options:
A 0°
B 90°
C 30°
D 45°

The correct answer is:
90°

Solution:
Incidence angle on one face of a prism, i=60°. Angle of the prism, A=30°,

the mergent ray that makes an angle 30° with incident ray. We know, deviation angle, δ= Angle between the incident ray and emergent ray=30°

⇒δ=i+e-A
⇒30°=60°+e−30∘
⇒e=0°. So, the angle with the face, =90° .

Topic: Physics, Ray Optics, JEE Main, JEE Main 2018 (16 Apr Online)
A ray of light is incident at an angle of incidence 60° on the glass slab of refractive index 3. After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is 43 cm. The thickness of the glass slab is _____ cm.

Options:

The correct answer is:
12

Solution:
From snell's law, sinisinr=3⇒sin60°sinr=3⇒r=30°

From the diagram shown, d=tsini-rcosr⇒t×1232=43⇒t=12 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (24 Jun Shift 2)
A ray of light is incident at the glass-water interface at an angle i. If it finally emerges parallel to the surface of the water, then the value of Î¼g would be

Options:
A 43sini
B 1sini
C 4 3
D 1

The correct answer is:
1sini

Solution:
Applying, Snell's law Î¼sini= constant at 1 and 2, we have
Î¼ 1 sin i 1 = Î¼ 2 sin i 2
Here, Î¼1=Î¼glass, i1=i
â´ Î¼2=Î¼air=1 and i2=90Â°
â´ Î¼gsini=1Ãsin90Â°
â´ Î¼g=1sini

Topic: Physics, Ray Optics, JEE Main
A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is Î¸iC and Brewster's angle of incidence is Î¸iB, such that sinÎ¸iCsinÎ¸iB=Î·=1.28. The relative refractive index of the two media is

Options:
A 0.4
B 0.2
C 0.9
D 0.8

The correct answer is:
0.8

Solution:
μ1: refractive index of rarer medium
μ2: retractive index of denser medium
sinθC=μ1μ2
tanθB=μ1μ2
sin θ C =tan θ B
sin θ C sin θ B = tan θ B sin θ B = 1 cos θ B =1.28
tan θ B = μ 1 μ 2 = 1− ( 1 1.28 ) 2 ( 1 1.28 )
=0.8

Topic: Physics, Ray Optics, JEE Main, JEE Main 2014 (19 Apr Online)
A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is Î¸iC and Brewster's angle of incidence is Î¸iB, such that sinÎ¸iCsinÎ¸iB=Î·=1.28. The relative refractive index of the two media is

Options:
A 0.4
B 0.2
C 0.9
D 0.8

The correct answer is:
0.8

Solution:
μ1: refractive index of rarer medium
μ2: retractive index of denser medium
sinθC=μ1μ2
tanθB=μ1μ2
sin θ C =tan θ B
sin θ C sin θ B = tan θ B sin θ B = 1 cos θ B =1.28
tan θ B = μ 1 μ 2 = 1− ( 1 1.28 ) 2 ( 1 1.28 )
=0.8

Topic: Physics, Ray Optics, JEE Main
A ray of light is incident from air on a glass plate having thickness 3 cm and refractive index 2. The angle of incidence of a ray is equal to the critical angle for glass-air interface. The lateral displacement of the ray when it passes through the plate is ____×10-2cm . (given sin 15°=0.26)

Options:

The correct answer is:
52

Solution:
Let C be the critical angle.

Then, sinC=1μ, where, μ is the refractive index of glass.

So, sinC=12⇒C=45°

As the angle of incidence is equal to the critical angle, so using Snell's law,

sinC=μsinθ

⇒12=2 sinθ

⇒θ=30°

Now, lateral displacement of the ray is

x=t sin(i-r) sec r

⇒x=3sin 45°-30°sec 30°

⇒x=3(0.26)23

⇒x=0.52 cm

⇒x=52×10-2 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (25 Jan Shift 1)
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, JEE Main
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 Snellâs law for the refraction at AC, we get

Î¼ââsini=(1)ââsinr

2sin30Â°=sinr

sinr=12

ââââââr=45Â°

â´ Angle of deviation =45Â°â30Â°=15Â°.

Topic: Physics, Ray Optics, JEE Main
A ray of light is incident on a $60^{\circ}$ prism at the minimum deviation position. The angle of refraction at the first face (ie, incident face) of the prism is

Options:
A zero
B $30^{\circ}$
C $45^{\circ}$
D $60^{\circ}$

The correct answer is:
$30^{\circ}$

Solution:
The refracting angle of prism
$A=r_1+\mathrm{r}_2$
For minimum deviation
$\begin{array}{l}
\mathrm{r}_1 =\mathrm{r}_2=\mathrm{r} \\
\therefore \mathrm{A} =2 \mathrm{r} \\
\text {or } \mathrm{r} =\frac{\mathrm{A}}{2}=\frac{60^{\circ}}{2}=30^{\circ}
\end{array}$

Topic: Physics, Ray Optics, JEE Main