datavorous/entrance-exam-dataset huggingface dataset:
A lead bullet strikes a target with a velocity of 480 ms-1. After the impact, the bullet falls dead and the heat produced in the process gets equally shared between the bullet and the target. If the specific heat capacity of lead is c=0.03 cal/g/Â°C, then the rise in temperature of the bullet is

Options:
A 557℃
B 457℃
C 857℃
D 754℃

The correct answer is:
457℃

Solution:
âQ=12K.E.âmcâT=1212mÂ v2J
âT=v24Jc= 480Ã4804Ã4.2Ã(0.03Ã103)=457â

Topic: Physics, Thermodynamics, JEE Main
A leak proof cylinder of length 1 m, made of a metal which has very low coefficient of expansion is floating vertically in water at 0oC such that its height above the water surface is 20 cm. When the temperature of water is increased to 4oC, the height of the cylinder above the water surface becomes 21 cm. The density of water at T=4oC, relative to the density at T=0oC is close to:

Options:
A 1.26
B 1.04
C 1.01
D 1.03

The correct answer is:
1.01

Solution:
According to Archemedes Principle, weight of the body is balanced by the upward thrust force due to buoyancy.

At 0°C, mg=A(100-20)ρ0°Cg, where A is area of cross-section of cylinder and ρ0°C is density of water at 0°C.
At 4°C, mg=A(100-21)ρ4°Cg, where ρ4°C is density of water at 4°C.
ρ4°Cρ0°C=8079=1.01

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2020 (08 Jan Shift 1)
A lens forms a real image of an object at a distance of 20 cm. It is observed that on placing another identical lens in contact with it, the image gets shifted by 10 cm towards the combination. The power of the lens is

Options:
A 5 D
B 2.5 D
C 10 D
D 2 D

The correct answer is:
5 D

Solution:
As image formed is real therefore lens must be convex, v=20 cm. Let f1, be focal length for this lens.
∴1f1=1v-1u=120-1u
After placing it in contact with another lens, the image is shifted 10 cm towards the combination.
i.e,. v=20-10cm=10 cm
∴110-1u=1f1+1f2=120-1u+1f2⇒f2=20 cm
∴ P=10020m-1=5D

Topic: Physics, Ray Optics, JEE Main
A lens having focal length $f$ and aperture of diameter $\mathrm{d}$ forms an image of intensity $I$. Aperture of diameter $\frac{\mathrm{d}}{2}$ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively

Options:
A $f$ and $\frac{1}{4}$
B $\frac{3 f}{4}$ and $\frac{I}{2}$
C f and $\frac{3 I}{4}$
D $\frac{\mathrm{f}}{2}$ and $\frac{\mathrm{I}}{2}$

The correct answer is:
f and $\frac{3 I}{4}$

Solution:
Intensity, $I \propto A^2$
$\Rightarrow \frac{\mathrm{I}_2}{\mathrm{I}_1}=\left[\frac{\mathrm{A}_2}{\mathrm{~A}_1}\right]^2=\frac{\pi \mathbf{r}^2-\frac{\pi \mathbf{r}^2}{4}}{\pi \mathbf{r}^2}=\frac{3}{4}$
$\Rightarrow \quad \mathrm{I}_2=\frac{3}{4} \mathrm{I}_1$ and focal length remains
unchanged.

Topic: Physics, Wave Optics, NEET, NEET 2010 (Screening)
A lens having focal length $f$ and aperture of diameter $\mathrm{d}$ forms an image of intensity $I$. Aperture of diameter $\frac{\mathrm{d}}{2}$ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively

Options:
A $f$ and $\frac{1}{4}$
B $\frac{3 f}{4}$ and $\frac{I}{2}$
C f and $\frac{3 I}{4}$
D $\frac{\mathrm{f}}{2}$ and $\frac{\mathrm{I}}{2}$

The correct answer is:
f and $\frac{3 I}{4}$

Solution:
Intensity, $I \propto A^2$
$\Rightarrow \frac{\mathrm{I}_2}{\mathrm{I}_1}=\left[\frac{\mathrm{A}_2}{\mathrm{~A}_1}\right]^2=\frac{\pi \mathbf{r}^2-\frac{\pi \mathbf{r}^2}{4}}{\pi \mathbf{r}^2}=\frac{3}{4}$
$\Rightarrow \quad \mathrm{I}_2=\frac{3}{4} \mathrm{I}_1$ and focal length remains
unchanged.

Topic: Physics, Wave Optics, JEE Main
A lens is made of flint glass (refractive index = 1.5). When the lens is immersed in a liquid of refractive index 1.25, the focal length

Options:
A Increases by a factor of 1.25
B Increases by a factor of 2.5
C Increases by a factor of 1.2
D Decreases by a factor of 1.2

The correct answer is:
Increases by a factor of 2.5

Solution:
Lens-maker's formula is given by
1f= aÎ¼g-11R1-1R2 â¦i
If the lens is immersed in a liquid of refractive index Î¼1 then
1f1= lÎ¼g-11R1-1R2 â¦ii
Here, lÎ¼g is refractive index of glass w.r.t liquid
Dividing Eq. (i) by Eq. (ii), we have
f1f= aÎ¼g-1 lÎ¼g-1
âf1f=1.5-11.51.25-1
â f 1 f = 0.5Ã1.25 0.25 =2.5â f 1 =( 2.5 )f
Hence, focal length increases by a factor of 2.5.

Topic: Physics, Ray Optics, JEE Main
A lens of focal power $0.5 D$ is

Options:
A A convex lens of focal length $0.5 \mathrm{~m}$
B A concave lens of focal length $0.5 \mathrm{~m}$
C A convex lens of focal length $2 \mathrm{~m}$
D A concave lens of focal length $2 \mathrm{~m}$

The correct answer is:
A convex lens of focal length $2 \mathrm{~m}$

Solution:
$P=\frac{1}{f} \Rightarrow f=\frac{1}{0.5}=2 m$

Topic: Physics, Ray Optics, JEE Main
A lens of power +2 diopters is placed in contact with a lens of power -1 diopter. The combination will behave like

Options:
A A convergent lens of focal length $50 \mathrm{~cm}$
B A divergent lens of focal length $100 \mathrm{~cm}$
C A convergent lens of focal length $100 \mathrm{~cm}$
D A convergent lens of focal length $200 \mathrm{~cm}$

The correct answer is:
A convergent lens of focal length $100 \mathrm{~cm}$

Solution:
$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{P_1}{100}+\frac{P_2}{100}=\frac{1}{100}$
$\Rightarrow f=100 cm$

Topic: Physics, Ray Optics, JEE Main
A letter ${ }^{\prime} \mathrm{A}^{\prime}$ is constructed of a uniform wire with resistance $1.0 \Omega$ per $\mathrm{cm}$, The sides of the letter are $20 \mathrm{~cm}$ and the cross piece in the middle is $10 \mathrm{~cm}$ long. The apex angle is 60 . The resistance between the ends of the legs is close to:

Options:
A
$50.0 \Omega$

B
$10 \Omega$

C
$36.7 \Omega$

D
$26.7 \Omega$

The correct answer is:
$26.7 \Omega$

Solution:
For $\mathrm{ADE} \frac{1}{\mathrm{R}^{\prime}}=\frac{1}{2 \mathrm{x}}+\frac{1}{10}$
or $\quad R^{\prime}=\frac{20 x}{10+2 x}$
$$
\mathrm{R}_{\mathrm{BC}}=\frac{20 \mathrm{x}}{10+2 \mathrm{x}}+20-\mathrm{x}+20-\mathrm{x}
$$
or $\frac{20 \mathrm{x}}{10+2 \mathrm{x}}+40=2 \mathrm{x}$
Solving we get
$$
\mathrm{x}=10 \Omega
$$
Putting the value of $x=10 \Omega$ in equation (i) We get
$$
\begin{aligned}
\mathrm{R}_{\mathrm{BC}} & =\frac{20 \times 10}{10+2 \times 10}+20-10+20-10 \\
& =\frac{80}{3}=26.7 \Omega
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2013 (09 Apr Online)
A letter is known to have come either from LONDON or CLIFTON; on the postmark only the two consecutive letters $\mathrm{ON}$ are legible. The probability that it came from LONDON is

Options:
A $\frac{5}{17}$
B $\frac{12}{17}$
C $\frac{17}{30}$
D $\frac{3}{5}$

The correct answer is:
$\frac{12}{17}$

Solution:
We define the following events :
$A_1$ : Selecting a pair of consecutive letter from the word LONDON.
$A_2$ : Selecting a pair of consecutive letters from the word CLIFTON.
$\mathrm{E}$ : Selecting a pair of letters 'ON'.
Then $P\left(A_1 \cap E\right)=\frac{2}{5} ;$ as there are 5 pairs of consecutive letters out of which 2 are $\mathrm{ON}$.
$P\left(A_2 \cap E\right)=\frac{1}{6}$; as there are 6 pairs of consecutive letters of which one is ON.
$P\left(\frac{A_1}{E}\right)=\frac{P\left(A_1 \cap E\right)}{P\left(A_1 \cap E\right)+P\left(A_2 \cap E\right)}$

Topic: Mathematics, Probability, JEE Main
A Li7 target is bombarded with a proton beam current of magnitude 10-4 A for an hour to produce Be7 of activity 1.8Ã108 disintegrations per second. If we assume that one radioactive nucleus of beryllium is produced by bombarding 1000 protons, then its half-life is close to

Options:
A 100 days
B 110 days
C 150 days
D 80 days

The correct answer is:
100 days

Solution:
At time t, let say there are N atoms of Be7 (radioactive). Then the net rate of formation of Be7 nuclei at this instant is,
d⁡ N⁡ d⁡ t = 1 0 - 4 1 . 6 × 1 0 - 1 9 × 1 0 0 0 - λ N⁡
or d⁡ N⁡ d⁡ t = 6 . 2 5 × 1 0 1 1 - λ N⁡
or ∫ o⁡ N⁡ 0 d⁡ N⁡ 6 . 2 5 × 1 0 1 1 - λ N⁡ = ∫ 0 3 6 0 0 d⁡ t
where N0 are the number of nuclei at t = 1 hour or 3600 s.
∴ - 1 λ ln 6 . 2 5 × 1 0 1 1 - λ N⁡ 0 6 . 2 5 × 1 0 1 1 = 3 6 0 0
λN0 = activity of Be7 at t = 1 hour = 1.8 × 108 disintegrations per second
∴ - 1 λ ln 6 . 2 5 × 1 0 1 1 - 1 . 8 × 1 0 8 6 . 2 5 × 1 0 1 1 = 3 6 0 0
∴ λ = 8.0 × 10-8 s-1

Therefore, half-life t1/2=0.6938.0×10-8=8.66×106 s

= 100.26 days

Topic: Physics, Nuclear Physics, JEE Main
A lift ascends with constant acceleration a=1 m s-2 , then with constant velocity and finally, it stops under constant retardation a=1 m s-2 . If total distance ascended by the lift is 7 m, in a total time of the journey is 8 s. Find the time (in second) for which lift moves with constant velocity.

Options:

The correct answer is:
6

Solution:
a=vt1
t1=va ...(i)
t0=8-2t1 ...(ii)
s0=7-2s1 ...(iii)
Solving equation (i), (ii) and (iii), we get
t0=6 s

Topic: Physics, Motion In One Dimension, JEE Main
A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the
ball as observed by the man in the lift and a man standing stationary on the ground are respectively

Options:
A
g, g

B
g - a, g - a

C
g - a, g

D
a, g

The correct answer is:
g - a, g

Solution:
Man in the lift is in a non - inertial frame so we have to take into account the pseudo acceleration

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2002
A lift of mass $1000 \mathrm{~kg}$ is moving with an acceleration of $1 \mathrm{~m} / \mathrm{s}^2$ in upward direction. Tension developed in the string, which is connected to the lift, is

Options:
A $9,800 \mathrm{~N}$
B $10,000 \mathrm{~N}$
C $10,800 \mathrm{~N}$
D $11,000 \mathrm{~N}$

The correct answer is:
$10,800 \mathrm{~N}$

Solution:
$T=m(g+a)=1000(9.8+1)=10800 \mathrm{~N}$

Topic: Physics, Laws of Motion, JEE Main
A lift of mass M=500 kg is descending with speed of 2 m s-1. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of 2 m s-2. The kinetic energy of the lift at the end of fall through to a distance of 6 m will be ______ kJ.

Options:

The correct answer is:
7

Solution:
Given, u=2 m s-1, a=2 m s-2, s=6 m

Using kinematics third equation of motion, final speed of lift is v2=u2+2as

⇒v=22+226

=4+24=28 m s-1

Now, kinetic energy of the lift is KE=12mv2

=12(500)28

=7000 J

=7 kJ

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (31 Jan Shift 1)
A light beam is described by E=800sinωt-xc. An electron is allowed to move normal to the propagation of light beam with a speed of 3×107 m s-1. What is the maximum magnetic force exerted on the electron?

Options:
A 1.28×10-21 N
B 12.8×10-18 N
C 1.28×10-18 N
D 12.8×10-17 N

The correct answer is:
12.8×10-18 N

Solution:
E0=800 V

B0=E0c=8003×108

Fm=qvB0=1.6×10-19×3×107×8003×108

=12.8×10-18 N

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2021 (26 Aug Shift 2)
A light beam of diameter 3R is incident symmetrically on a glass hemisphere of radius R and of refractive index n=3. The radius of the beam at the base of the hemisphere is

Options:
A R3
B R23
C 32R
D R2

The correct answer is:
R3

Solution:
From the given ray diagram shown in the figure below μ1sini=μ2sinr
⇒1×sin60°=3×sinr
⇒r=30°
In ΔMNP
tan30°=xR2
⇒x=R23
From the geometry of the figure
⇒r=32R-x=32R-R23
=3-123R=R3

Topic: Physics, Ray Optics, JEE Main
A light beam of wavelength $800 \mathrm{~nm}$ passes through a single slit and projected on a screen lsept at $5 \mathrm{~m}$ away from the slit. What should be the slit width for the ray optics approximation to be valid?

Options:
A $0.5 \mathrm{~mm}$
B $2 \mathrm{~mm}$
C $1.5 \mathrm{~mm}$
D $0.25 \mathrm{~mm}$

The correct answer is:
$2 \mathrm{~mm}$

Solution:
Ray optics approximation can be taken upto FresneIs' distance which is given by
$$
z_f=\frac{a^2}{\lambda}
$$
Given, $\quad Z_f=5 \mathrm{~m}$, wavelength, $\lambda=800 \mathrm{~nm}=800 \times 10^{-9} \mathrm{~m}$, Slit width $a=$ ?
$$
\begin{aligned}
a & =\sqrt{Z_f \cdot \lambda} \\
& =\left(\sqrt{5 \times 800 \times 10^{-9} \times 10^3 \times 10^3}\right) \mathrm{mm} \\
a & =2 \mathrm{~mm}
\end{aligned}
$$

Topic: Physics, Wave Optics, JEE Main
A light bulb, a capacitor and a battery are connected together as shown here, with switch $S$ initially open. When the switch $S$ is closed, which one of the following is true

Options:
A The bulb will light up for an instant when the capacitor starts charging
B The bulb will light up when the capacitor is fully charged
C The bulb will not light up at all
D The bulb will light up and go off at regular intervals

The correct answer is:
The bulb will light up for an instant when the capacitor starts charging

Solution:
Initially when key is closed, the capacitor acts as short-circuit, so bulb will light up. But finally the capacitor becomes fully charged, so it will act as open circuit, so bulb will not glow.

Topic: Physics, Capacitance, JEE Main
A light bulb and an inductor coil are connected to an ac source through a key as shown in the figure below. The key is closed and after sometime an iron rod is inserted into the interior of the inductor. The glow of the light bulb

Options:
A decreases
B remains unchanged
C will fluctuate
D increases

The correct answer is:
decreases

Solution:
We know that the formula of impedance for L-R circuit, Z=R2+XL2, here, R=resistance and XL=ωL, where, L is the inductance when we insert the iron rod in inductor then the value of L increased and then the value of impedance also increased, and due to constant voltage current will be decreased across the bulb and the glow of the bulb will decrease.

Topic: Physics, Alternating Current, NEET, NEET 2020 (Phase 2)
A light bulb of resistance $R=16 \Omega$ is attached in series with an infinite resistor network. with identical resistances $\mathrm{r}$ as shown below. A $10 \mathrm{~V}$ battery derives current in the circuit. What should be the value of $r$ such that the bulb dissipated about $1 \mathrm{~W}$ of power.

Options:
A $14.8 \Omega$
B $29.6 \Omega$
C $7.4 \Omega$
D $3.7 \Omega$

The correct answer is:
$14.8 \Omega$

Solution:
$$
\begin{array}{l}
\mathrm{P}_{\text {bulb }}=\frac{v^{2}}{R}=i^{2} R \\
1=\frac{v^{2}}{16} \\
\mathrm{~V}_{\mathrm{B}}=4 \mathrm{~V} \\
1=i^{2} \times 16 \\
\mathrm{I}_{\mathrm{B}}=\frac{1}{4} \mathrm{Amp} . \\
6=\frac{1}{4} \times r_{e q}(\text { equivalent of groups of } \mathrm{r})
\end{array}
$$
Where
$$
r_{e q}=r+\frac{r_{e q} \cdot r}{r_{e q}+r}
$$

Topic: Physics, Current Electricity, JEE Main
A light cylindrical tube T of length l and radius rcontaining air is inverted in water (density d1 ). One end of the tube is open and the other is closed. A block B of density d2 >d1 is kept on the tube as shown in the figure. The tube stays in equilibrium in the position shown. Find the volume of B. Density of the air is negligible as compared with the density of water :

Options:
A v=d1d2-d1pr2l3
B v=d2d2-d1pr2l3
C v=d1d1-d2pr2l3
D v=d1d2pr2l3

The correct answer is:
v=d1d2-d1pr2l3

Solution:
The free body diagram of block + tube inclusive of water is as shown. Since the block + tube system shown in figure is in equilibrium.
∴ Net weight of system = buoyant fore

Let v be the volume of the block, then
d2vg=v+πr2l3 d1g
⇒ d2-d1v=πr2l3
or v=d1d2-d1πr2l3

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A light emitting diode (LED) is fabricated using GaAs semiconducting material whose band gap is $1.42 \mathrm{eV}$. The wavelength of light emitted from the LED is :

Options:
A $1400 \mathrm{~nm}$
B $650 \mathrm{~nm}$
C $875 \mathrm{~nm}$
D $1243 \mathrm{~nm}$

The correct answer is:
$875 \mathrm{~nm}$

Solution:
$\lambda=\frac{1240}{1.42}=875 \mathrm{~nm}$ (Approx)

Topic: Physics, Semiconductors, JEE Main, JEE Main 2024 (09 Apr Shift 1)
A light from Paschen series of hydrogen atom is able to eject photoelectrons from a metal. Then the work function of the metal is

Options:
A $3: 4 \mathrm{eV}$
B $1.54 \mathrm{eV}$
C None of these
D $1.1 \mathrm{eV}$

The correct answer is:
$1.1 \mathrm{eV}$

Solution:
In Paschen series electron transmits from some higher state to $n=3$ state
So photons of paschen series have energies, ranging from
$E_{\min }=E_{n=4}-E_{n=3}$
$\begin{aligned} & =-0.85-(-1.5 \mathrm{l}) \\ & =0.66 \mathrm{eV}\end{aligned}$
to $E_{\max }=E_{n=\infty}-E_{n=3}$
$\begin{aligned} & =0-(-1.5 \mathrm{l}) \\ & =1.51 \mathrm{eV}\end{aligned}$
So, if a photon of paschen series able to emit a photoelectron from a metal, then work function $\phi_0$ of metal must be such that,
$\phi_0 \leq$ Photon energy
or $\phi_0 \leq 1.51 \mathrm{eV}$
Only matching option is options (d).

Topic: Physics, Atomic Physics, JEE Main
A light inelastic thread passes over a small frictionless pulley. Two blocks of masses m=1 kg and M = 3 kg, respectively, are attached with the thread and heavy block rests on a surface. A particle P of mass 1 kg moving upward with a velocity of 10 m s-1 collides with the lighter block and sticks to it. The speed of the bigger block just after the string is taut will be [g=10 m s-2]

Options:
A 2.5 m s-1
B 4 m s-1
C 5 m s-1
D 2 m s-1

The correct answer is:
2 m s-1

Solution:
Let us assume that just after the collision, the speed of the particle plus 1 kg block is u

2u=1Ã10âu=5 m s-1

After the collision and until the string is taut again, the particle block system is in free fall. So just before the string is taut, the particle plus 1 kg block will be moving downward with a velocity of 5 m s-1 and just after the string is taut, let us assume that the speed of both the blocks and the particle is v, then

2Ã5=5v

âv=2 m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A light of intensity 16 mW and energy of each photon 10 eV incident on a metal plate of work function 5 eV and area 10-4 m2 then find the maximum kinetic energy of emitted electrons and the number of photoelectrons emitted per second if photon efficiency is 10%.

Options:
A 5 eV, 1011
B 10 eV, 1012
C 5 eV, 1013
D 10 eV, 1014

The correct answer is:
5 eV, 1011

Solution:
Maximum KE=hCλ-ϕ
=10-5
=5 eV
Intensity I=NphCλtA⇒Npt=IAhCλ
and the number of emitted photoelectrons per second. =Npt×10%
=IAhCλ×110=16×10-3×10-410×1.6×10-19×10
=1011

Topic: Physics, Dual Nature of Matter, JEE Main
A light of wavelength 3000 Ao falls on a metal surface. Ejected e- is further accelerated by a potential difference of 2V , then final K.E of the e- is found to be 8Ã10-19J . If threshold energy for the metal surface is 'Ï'eV . Then find the numerical value of 8Ï

Options:

The correct answer is:
9.00

Solution:
KE final â=â E photon â+â E potential ââÏ
where Ï = threshold energy for metal surface
E photon â=â hc Î» â=â 6.6Ã 10 â34 Ã3Ã 10 8 3000Ã 10 â10
â6.6Ã 10 â19 J=4.12âeV
E potential â=â2âeV
KE final = 8Ã 10 â19 J 1.6Ã 10 â19 J =5âeV
Hence, ϕ=(4.12+2−5) eV=1.125 eV
Numerical value = 8×1.125=9

Topic: Chemistry, Structure of Atom, JEE Main
A light planet is revolving around a massive star in a circular orbit of radius R with a period of revolution T. If the force of attraction between planet and star is proportional to R-3/2 then choose the correct option :

Options:
A T2∝R5/2
B T2∝R7/2
C T2∝R3/2
D T2∝R3

The correct answer is:
T2∝R5/2

Solution:
Let the gravitational force of attraction be F=GMmR3/2.

Now, this gravitational force will provide the required centripetal force. Therefore,

GMmR3/2=mω2R

⇒ω2∝1R5/2 ∵T=2πω

⇒T2∝R5/2

Topic: Physics, Gravitation, JEE Main, JEE Main 2024 (01 Feb Shift 2)
A light ray emerging from the point source placed at $\mathrm{P}(1,3)$ is reflected at a point $\mathrm{Q}$ in the axis of $x$. If the reflected ray passes through the point $R$ $(6,7)$, then the abscissa of $Q$ is:

Options:
A
1

B
3

C
$\frac{7}{2}$

D
$\frac{5}{2}$

The correct answer is:
$\frac{5}{2}$

Solution:
Let abcissa of $\mathrm{Q}=x$
$\begin{aligned} \therefore \quad & Q=(x, 0) \\ & \tan \theta=\frac{0-7}{x-6}, \tan \left(180^{\circ}-\theta\right)=\frac{0-3}{x-1}\end{aligned}$
Now, $\tan \left(180^{\circ}-\theta\right)=-\tan \theta$
$$
\therefore \quad \frac{-3}{x-1}=\frac{-7}{x-6} \Rightarrow x=\frac{5}{2}
$$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2013 (09 Apr Online)
A light ray emits from the origin making an angle 30° with the positive $x$-axis. After getting reflected by the line x+y=1, if this ray intersects x-axis at Q, then the abscissa of Q is

Options:
A 2(3-1)
B 23+3
C 23-3
D 32(3+1)

The correct answer is:
23+3

Solution:
On plotting the diagram of given data we get,

Let Q(h,0)

∵OP reflected by x+y=1

So, the image of Q lies on y=x3

∴ x-h1=y1=-2(h-1)2

⇒x=1, y=1-h

Since it lies on y=x3

∴ 1-h=13

⇒h=1-13=3-13

On rationalising the numerator we get,

⇒h=3-13×3+13+1

⇒h=23+3

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2023 (29 Jan Shift 1)
A light ray enters a solid glass sphere of refractive index μ=3 at an angle of incidence 60°. The ray is both reflected and refracted at the farther surface of the sphere. The angle (in degrees) between the reflected and refracted rays at this surface is _____________.

Options:

The correct answer is:
90

Solution:
Apply Snell's law at S_
1sin 60°=3sin r
sin r=12
r=30°
from geometry
r'=30°
Again apply snell's law on S2
3sinr'=1sine
32=sine
∴e=60°
from geometry
r'+θ+e=180°
θ=90°

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (02 Sep Shift 2)
A light ray falls on a square glass slab as shown in the diagram. The index of refraction of the glass, if total internal reflection is to occur at the vertical face, is equal to :

Options:
A
$\frac{(\sqrt{2}+1)}{2}$

B
$\sqrt{\frac{5}{2}}$

C
$\frac{3}{2}$

D
$\sqrt{\frac{3}{2}}$

The correct answer is:
$\sqrt{\frac{3}{2}}$

Solution:
At point A by Snell's law
$$
\mu=\frac{\sin 45^{\circ}}{\sin r} \Rightarrow \sin r=\frac{1}{\mu \sqrt{2}}
$$
At point $\mathrm{B}$, for total internal reflection,
$$
\sin i_1=\frac{1}{\mu}
$$
From figure, $i_1=90^{\circ}-r$
$$
\begin{aligned}
& \therefore\left(\sin 90^{\circ}-r\right)=\frac{1}{\mu} \\
& \Rightarrow \cos r=\frac{1}{\mu}
\end{aligned}
$$
Now $\cos r=\sqrt{1-\sin ^2 r}=\sqrt{1-\frac{1}{2 \mu^2}}$
$$
=\sqrt{\frac{2 \mu^2-1}{2 \mu^2}}
$$
From eqs (ii) and (iii)
$$
\frac{1}{\mu}=\sqrt{\frac{2 \mu^2-1}{2 \mu^2}}
$$
Squaring both sides and then solving, we get
$$
\mu=\sqrt{\frac{3}{2}}
$$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2013 (23 Apr Online)
A light ray is incident by grazing one of the face of a prism and after refraction ray does not emerge out, what should be the angle of prism while critical angle is $C$

Options:
A Equal to $2 C$
B Less than $2 C$
C More than $2 C$
D None of the above

The correct answer is:
More than $2 C$

Solution:
From ray diagram
$A=C+\theta$ for $\operatorname{TIR}$ at $A C$
$\theta\gt C_{\text {so }} A\gt2 C$

Topic: Physics, Ray Optics, JEE Main
A light ray is incident on the lower medium boundary at an angle 30° with the normal. Which of the following statements is true?

Options:
A If Î¼ 2 > 2 ,then total deviation is 60Â°
B If μ 2 < 2 , then total deviation is 30°
C If μ 2 > 2 ,then total deviation is 120°
D If μ 2 < 2 ,then total deviation is 180°

The correct answer is:
If Î¼ 2 > 2 ,then total deviation is 60Â°

Solution:
For T.I.R. at A
4 sin 3 0 â = Î¼ 2 sin 9 0 â
Î¼ 2 = 2
Case -I
If Î¼ 2 < 2 , then always T.I.R. takes place and in this situation angle of deviation is 120o.
Case-II
Î¼ 2 > 2 , then for the given angle of incidence no. T.I.R takes place at A, so light strikes on interface B for T.I.R.
Î¼ 2 sin Î± = 2 sin 9 0 â
or Î¼ 2 = 2 sin Î±
sin Î± < 1
Î¼ 2 > 2

Topic: Physics, Ray Optics, JEE Main
A light ray is incident perpendicular to one face of a $90^{\circ}$ prism and is totally internally reflected at the glass-air interface. If the angle of reflection is $45^{\circ}$, we conclude that the refractive index $n$

Options:
A
$\mathrm{n} < \frac{1}{2}$
B
$\mathrm{n}>\sqrt{2}$

C
$n>\frac{1}{\sqrt{2}}$

D
$\mathrm{n} < \sqrt{2}$

The correct answer is:
$\mathrm{n}>\sqrt{2}$

Solution:
Angle of incidence $\mathrm{i}>\mathrm{C}$ for total internal reflection. Here $\mathrm{i}=45^{\circ}$ inside the medium.
$$
\begin{aligned}
& \therefore 45^{\circ}>\sin ^{-1}(1 / n) \\
& \Rightarrow n>\sqrt{ } 2 .
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2004