datavorous/entrance-exam-dataset huggingface dataset:
$A B C$ is a triangle, right angled at $A$. The resultant of the forces acting along $\overrightarrow{A B}, \overrightarrow{A C}$ with magnitudes $\frac{1}{A B}$ and $\frac{1}{A C}$ respectively is the force along $\overrightarrow{A D}$, where $D$ is the foot of the perpendicular from $\mathrm{A}$ onto $\mathrm{BC}$. The magnitude of the resultant is

Options:
A
$\frac{A B^2+A C^2}{(A B)^2(A C)^2}$

B
$\frac{(A B)(A C)}{A B+A C}$

C
$\frac{1}{A B}+\frac{1}{A C}$

D
$\frac{1}{\mathrm{AD}}$

The correct answer is:
$\frac{1}{\mathrm{AD}}$

Solution:
Magnitude of resultant
$$
\begin{aligned}
& =\sqrt{\left(\frac{1}{A B}\right)^2+\left(\frac{1}{A C}\right)^2}=\frac{\sqrt{A B^2+A C^2}}{A B \cdot A C} \\
& =\frac{B C}{A B \cdot A C}=\frac{B C}{A D \cdot B C}=\frac{1}{A D}
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2006
$A B C$ is an equilateral triangle inscribed in a circle of centre $O$ and radius $5 \mathrm{~cm}$. Let the diameter through $C$ meet the circle again at $\mathrm{D}$.
Assertion (A): $A D \cdot B D < O B \cdot O C$
Reason (R): $2\left(A D^{2}+B D^{2}\right)=C D^{2}=100$ sq cm

Options:
A Both A and R are individually true and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$
B Both $\mathrm{A}$ and $\mathrm{R}$ are individually true but $\mathrm{R}$ is not the correct explanation of $\mathrm{A}$
C $\mathrm{A}$ is true but $\mathrm{R}$ is false
D A is false but $\mathrm{R}$ is true

The correct answer is:
A is false but $\mathrm{R}$ is true

Solution:
(A) : Consider $A D \cdot B D < O B \cdot O C$
Now, $\mathrm{OA}=5 \mathrm{~cm}$ (Radius), $\mathrm{OB}=5 \mathrm{~cm}$ (Radius) and $\mathrm{OC}=5 \mathrm{~cm}($ Radius $)$
Since, $\Delta O A D$ and $\triangle O B D$
are congruent by SAS therefore
$$
A D=O A=5 \mathrm{~cm}
$$
and $B D=O B=5 \mathrm{~cm}$
Thus, $A D \cdot B D=5 \times 5=25$
and $\quad O B \cdot O C=5 \times 5=25$
Thus, we have
$A D \cdot B D=25=O B \cdot O C$
$$
\begin{aligned}
\operatorname{Now}(\mathrm{R}): & 2\left(A D^{2}+B D^{2}\right) \\
&=2[25+25]=100 \\
\text { and } \quad & C D^{2}=(10)^{2}=100
\end{aligned}
$$
Thus, $2\left(A D^{2}+B D^{2}\right)=C D^{2}=100$ sq. $\mathrm{cm}$.
Hence, (A) is false and (R) is true.

Topic: Mathematics, Circle, JEE Main
$\triangle A B C$ is formed by $A(1,8,4), B(0,-11,4)$ and $C(2,-3,1)$. If $D$ is the foot of the perpendicular from $A$ to $B C$. Then the coordinates of $D$ are

Options:
A (– 4, 5, 2)
B (4, 5, – 2)
C (4, – 5, 2)
D (4, – 5, – 2)

The correct answer is:
(4, 5, – 2)

Solution:
The vertices of $\triangle A B C$ are given as $A(1,8,4)$, $B(0,-11,4)$ and $C(2,-3,1)$.
Equation of line $B C$,
$$
\frac{x}{2}=\frac{y+11}{8}=\frac{z-4}{-3}=\lambda \text { (say) }
$$
As point $D$ is on line $B C$, so coordinates of $D$ are $(2 \lambda, 8 \lambda-11,-3 \lambda+4)$
Since, $A D \perp B C$
$$
\begin{aligned}
\therefore \quad & A D \cdot B C=0 \\
& (2 \lambda-1,8 \lambda-19,-3 \lambda) \cdot(2,8,-3)=0 \\
\therefore \quad & 2(2 \lambda-1)+8(8 \lambda-19)+(-3)(-3 \lambda)=0 \\
& 4 \lambda-2+64 \lambda-152+9 \lambda=0 \\
& 77 \lambda=154 \\
& \lambda=2
\end{aligned}
$$
Hence, the coordinates of $D$ are $(4,5,-2)$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A bead of mass m can slide without friction along a vertical ring of radius R. One end of a spring of force constant k=3mgR is connected to the bead and the other end is fixed at the centre of the ring. Initially, the bead is at the point A and due to a small push it starts sliding down the ring. If the bead momentarily loses contact with the ring at the instant when the spring makes an angle of 60Â° with the vertical, then the natural length of the spring is

Options:
A 5R9
B 3R4
C 5R6
D 4R7

The correct answer is:
5R6

Solution:
Let's say the bead loses contact with the ring at point B, then N2=0 and

mg cos60Â°+ kx=mv2R

mg2+kx=mv2R

Applying the conservation of mechanical energy principle between A & B

mv22=mgR1-cos60Â°=mgR2

âmv2R=mg

mg2+kx=mg

âx=mg2k=R6

The natural length of the spring l=R-x=R-R6=5R6

Topic: Physics, Laws of Motion, JEE Main
A bead of mass m can slide without friction on a fixed circular horizontal ring of radius 3R having a centre at the point C. The bead is attached to one of the ends of spring of spring constant k. Natural length of spring is R and the other end of the spring is fixed at point O as shown in the figure. If the bead is released from position A, then the kinetic energy of the bead when it reaches point B is

Options:
A 252kR2
B 92kR2
C 8kR2
D 12kR2

The correct answer is:
8kR2

Solution:
Given,
Mass of bead =m
Radius of circular horizontal ring =3R
Natural length of spring =R
According to conservation of energy,
KEi+PEi=KEf+PEf
â 0+12kOA-R2=KEf+12kOB-R2
â 0+12k5R-R2=KEf+12kR-R2
â KEf=8kR2

Topic: Physics, Work Power Energy, JEE Main
A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figure and whose equation is x2 = 4ay. The wire frame is fixed and the bead can slide on it without friction. The bead is released from the point y = 4a on the wire frame from rest. The tangential acceleration of the bead when it reaches the position given by y = a is :

Options:
A g 2
B 3 g 2
C g 2
D g 5

The correct answer is:
g 2

Solution:
x2 = 4ay
Differentiating w.r.t. y, we get
dy dx = x 2 a
It can be considered as an inclined plane
â´ At 2 a, a , dy dx = 1 â hence Î¸ = 4 5 â
the component of weight along tangential direction is mg sin Î¸ .
hence tangential acceleration is g sin Î¸ = g 2

Topic: Physics, Motion In Two Dimensions, JEE Main
A bead of mass m stays at point Pa, b on a wire bent in the shape of a parabola y=4Cx2 and rotating with angular speed ω (see figure). The value of ω is (neglect friction)

Options:
A 22gC
B 2gC
C 2gCab
D 2gC

The correct answer is:
22gC

Solution:
mω2acosθ=mgsinθ

ω=gtanθa

y=4cx2

tanθ=dydx=8xC

(tanθ)a,b=8aC

ω=g×8aca=22gC

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2020 (02 Sep Shift 1)
A bead of weight $w$ can slide on smooth circular wire in a vertical plane. The bead is attached by a light thread to the highest point of the wire and in equilibrium, the thread is taut and make an angle $\theta$ with the vertical then tension of the thread and reaction of the wire on the bead are

Options:
A
$T=w \cos \theta \quad R=w \tan \theta$

B
$\mathrm{T}=2 \mathrm{w} \cos \theta \quad \mathrm{R}=\mathrm{w}$

C
$\mathrm{T}=\mathrm{w} \quad \mathrm{R}=\mathrm{w} \sin \theta$

D
$T=w \sin \theta \quad R=w \cot \theta$

The correct answer is:
$T=w \cos \theta \quad R=w \tan \theta$

Solution:
No Solution Available

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2002
A beaker containing a liquid is kept inside a big closed jar. If the air inside the jar is continuously pumped out, the pressure in the liquid near the bottom of the liquid will

Options:
A Increases
B Decreases
C Remain constant
D First decrease and then increase

The correct answer is:
Decreases

Solution:
Total pressure at (near) bottom of the liquid
$P=P_0+h \rho g$
As air is continuously pumped out from jar (container), $P_0$ decreases and hence $P$ decreases.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A beaker containing mixture of 1 M Y and 1 M Z . A gas X at 1âatm is bubbled through this solution at 25oC . If the reduction potential of Z>Y>X , then

Options:
A Y will oxidize X and not Z
B Y will oxidize Z and not X
C Y will oxidize both X and Z
D Y will reduce both X and Z

The correct answer is:
Y will oxidize X and not Z

Solution:
From the reduction potential, the tendency to gain electron is in the order Z>Y>X
Thus Y+e→Y - ; X→X++e .

Topic: Chemistry, Electrochemistry, JEE Main
A beaker contains $200 \mathrm{gm}$ of water. The heat capacity of the beaker is equal to that of $20 \mathrm{gm}$ of water. The initial temperature of water in the beaker is $20^{\circ} \mathrm{C}$. If $440 \mathrm{gm}$ of hot water at $92^{\circ} \mathrm{C}$ is poured in it, the final temperature (neglecting radiation loss) will be nearest to

Options:
A $58^{\circ} \mathrm{C}$
B $68^{\circ} \mathrm{C}$
C $73^{\circ} \mathrm{C}$
D $78^{\circ} \mathrm{C}$

The correct answer is:
$68^{\circ} \mathrm{C}$

Solution:
Heat lost by hot water $=$ Heat gained by cold water in beaker + Heat absorbed by beaker
$\begin{aligned} & \Rightarrow 440(92-\theta)=200 \times(\theta-20)+20 \times(\theta-20) \\ & \Rightarrow \theta=68^{\circ} \mathrm{C}\end{aligned}$

Topic: Physics, Thermal Properties of Matter, JEE Main
A beaker contains a fluid of density ρkgm3 , specific heat SJkgoC and viscosity η . The beaker is filled up to height h. To estimate the rate of heat transfer per unit area Q˙A by convection when beaker is put on a hot plate, a student proposes that it should depend on η , SΔθh and 1ρg when Δθ ( inoC ) is the difference in the temperature between the bottom and top of the fluid. In that situation the correct option for Q˙A is:

Options:
A SΔθhη
B ηSΔθh1ρ g
C SΔθηh1ρg
D SΔθηh

The correct answer is:
SΔθhη

Solution:
QA=ηxs∆θhy1ρgz . . . (1)
QA=MT-3,η=ML-1T-1
s∆θh=LT-2,1ρg=M-1L2T2
MT-3=MxL-xT-xLyT-2yM-2L2zT2z
MT-3=Mx-zL-x+y+2zT-x-2y+2z
So 1=x-z . . . (2)
-x+y+2z=0. . . (3)
-x-2y+2z=-3. . . (4)
from 2, 3, 4
x=1, y=1, z=0, from (1)
so QA=η s∆θh

Topic: Physics, Units and Dimensions, JEE Main
A beaker contains a fluid of density ρkgm3 , specific heat SJkgoC and viscosity η . The beaker is filled up to height h. To estimate the rate of heat transfer per unit area Q˙A by convection when beaker is put on a hot plate, a student proposes that it should depend on η , SΔθh and 1ρg when Δθ ( inoC ) is the difference in the temperature between the bottom and top of the fluid. In that situation the correct option for Q˙A is:

Options:
A SΔθhη
B ηSΔθh1ρ g
C SΔθηh1ρg
D SΔθηh

The correct answer is:
SΔθhη

Solution:
QA=ηxs∆θhy1ρgz . . . (1)
QA=MT-3,η=ML-1T-1
s∆θh=LT-2,1ρg=M-1L2T2
MT-3=MxL-xT-xLyT-2yM-2L2zT2z
MT-3=Mx-zL-x+y+2zT-x-2y+2z
So 1=x-z . . . (2)
-x+y+2z=0. . . (3)
-x-2y+2z=-3. . . (4)
from 2, 3, 4
x=1, y=1, z=0, from (1)
so QA=η s∆θh

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2015 (11 Apr Online)
A beaker contains water up to a height \( h_{1} \) and kerosene of height \( h_{2} \) above water so that the total height of (water \( + \) kerosene \( ) \) is \( \left(h_{1}+h_{2}\right) . \) Refractive index of water is \( \mu_{1} \) and that of kerosene is \( \mu_{2} . \) The apparent shift in the position of the bottom of the beaker when viewed from above is

Options:
A \( \left(1-\frac{1}{\mu_{1}}\right) h_{2}+\left(1-\frac{1}{\mu_{2}}\right) h_{1} \)
B \( \left(1+\frac{1}{\mu_{1}}\right) h_{1}+\left(1+\frac{1}{\mu_{2}}\right) h_{2} \)
C \( \left(1-\frac{1}{\mu_{1}}\right) h_{1}+\left(1-\frac{1}{\mu_{2}}\right) h_{2} \)
D \( \left(1+\frac{1}{\mu_{1}}\right) h_{2}-\left(1+\frac{1}{\mu_{2}}\right) h_{1} \)

The correct answer is:
\( \left(1-\frac{1}{\mu_{1}}\right) h_{1}+\left(1-\frac{1}{\mu_{2}}\right) h_{2} \)

Solution:
Apparent shift due to a transparent material of thickness h is give by,

Δ h⁡=1-1μh
∴ Apparent shift produced by water
△h1=1-1μ1h1
And apparent shift produced by kerosene
△h2=1-1μ2h2
△h= △h1 +△h2=1-1μ1h1+1-1μ2h2

Topic: Physics, Ray Optics, JEE Main
A beaker of height H is made up of a material whose coefficient of linear thermal expansion is 3 Î± . It is filled up to the brim by a liquid whose coefficient of volume expansion is 3 Î± . If now the beaker along with its contents is uniformly heated through a small temperature T, the level of liquid will reduce by (Given, Î±Tâª1 )

Options:
A αTH
B 3αTH
C 9αTH
D 6αTH

The correct answer is:
3αTH

Solution:
Let A is cross - sectional area of a beaker
Then, final cross - sectional area
A â² = A 1 + 6 Î± T
Final volume of liquid
V â² = AH 1 + 3 Î± T
Final height, H â² = V â² A â²
â H â² = H 1 + 3 Î± T 1 + 6 Î± T â¼ H 1 - 3 Î± T
Reduction in level, H - H â² = 3 Î± TH

Topic: Physics, Thermal Properties of Matter, JEE Main
A beam of clectron passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off, and the same magnetic field is maintained, the electrons move:

Options:
A in a circular orbit
B along a parabolic path
C along a straight line
D in an elliptical orbit.

The correct answer is:
in a circular orbit

Solution:
Since, direction of motion of electron must be perpendicular to the both electric and magnetic fields.When electric field is switch off then the magnetic field is perpendicular to the direction of motion of charge particle and therefore, the electrons starts exhibiting circular motion.

Topic: Physics, Magnetic Effects of Current, JEE Main
A beam of clectron passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off, and the same magnetic field is maintained, the electrons move:

Options:
A in a circular orbit
B along a parabolic path
C along a straight line
D in an elliptical orbit.

The correct answer is:
in a circular orbit

Solution:
Since, direction of motion of electron must be perpendicular to the both electric and magnetic fields.When electric field is switch off then the magnetic field is perpendicular to the direction of motion of charge particle and therefore, the electrons starts exhibiting circular motion.

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2007
A beam of diameter d is incident on a glass hemisphere as shown. If the radius of curvature of the hemisphere is very large in comparison to d, then the diameter of the beam at the base of the hemisphere will be

Options:
A 34d
B d
C d3
D 23d

The correct answer is:
23d

Solution:
μ 2 v - μ 1 u = μ 2 - μ 1 R ⇒ 3 2 v - 1 ∞ = 3 2 - 1 R ⇒ 3 2 v = 1 2 R
v = 3 R ⇒ d 3 R = d ′ 2 R ⇒ d ′ = 2 3 d

Topic: Physics, Ray Optics, JEE Main
A beam of electromagnetic radiation of intensity 6.4× 10 −5 W/ cm 2 is comprised of wavelength, λ=310nm . It falls normally on a metal (work function φ=2eV ) of surface area of 1cm2 . If one in 103 photons ejects an election, total number of electrons ejected in 1s is 10x . hc=1240eVnm,1eV=1.6×10-19J, then x is ___________

Options:

The correct answer is:
11

Solution:
Energy of photon. E=1240310 nm=4eV>2eV (So photoelectric effect will take place)
=4×1.6×10-19=6.4×10-19joule
Number of photons falling per second
=6.4×10-5×16.4×10-19=1014
Number of photoelectron emitted per second
=1014103=1011

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2020 (07 Jan Shift 1)
A beam of electrons of energy E scatters from a target having atomic spacing of 1 Å. The first maximum intensity occurs at θ=60°. Then E(in eV) is........
(Planck's constant h=6.64×10-34 Js, 1 eV=1.6×10-19J, electron mass m=9.1×10-31 kg)

Options:

The correct answer is:
50

Solution:
2dsinθ=nλ

2d32=1λ

d=1 Å

λ=3 Å

λ=3=150V

V=50 volt

E=50 eV

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2020 (05 Sep Shift 1)
A beam of electrons striking a copper target produces X-rays. Its spectrum is as shown. Keeping the voltage same if the copper target is replaced with a different metal, the cut-off wavelength and characteristic lines of the new spectrum will change in comparison with old as

Options:
A Cut-off wavelength will remain unchanged while characteristic lines will be different
B Both cut-off wavelength and characteristic lines will remain unchanged
C Both cut-off wavelength and characteristic lines will be different
D Cut-off wavelength will be different while characteristic lines will remain unchanged

The correct answer is:
Cut-off wavelength will remain unchanged while characteristic lines will be different

Solution:
Cut- off wavelength of emitted X-rays depends on maximum kinetic energy of incident electrons on the target and is independent of nature of target. The characteristic lines depend on nature of material of target.

Topic: Physics, Atomic Physics, JEE Main
A beam of light composed of red and green ray is incident obliquely at a point on the face of rectangular glass slab. When coming out on the opposite parallel face, the red and green ray emerge from:

Options:
A two points preparing in two different non parallel directions
B two points propagating in two different parallel directions
C one point propagating in two different directions
D one point propagating in the same directions

The correct answer is:
two points propagating in two different parallel directions

Solution:
Red and green rays emerge from two points propagating in two different parallel direction.

Topic: Physics, Ray Optics, JEE Main
A beam of light consisting of red, green and blue colours is incident on a right-angled prism on face $A B$. The refractive indices of the material for the above red, green and blue colours are $1.39,1.44$ and $1.47$ respectively. A person looking on surface $A C$ of the prism will see

Options:
A
no light

B
green and blue colours

C
red and green colours

D
red colour only

The correct answer is:
red colour only

Solution:
For light to come out through face 'AC', total internal reflection must not take place.
$$
\begin{aligned}
& \text { i.e., } \theta < c \Rightarrow \sin \theta < \sin c \\
& \Rightarrow \sin \theta < \frac{1}{\mu}
\end{aligned}
$$
or $\quad \mu < \frac{1}{\sin \theta} \Rightarrow \mu < \frac{1}{\sin 45^{\circ}}$
$$
\Rightarrow \mu < \sqrt{2} \Rightarrow \mu < 1.414
$$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2012 (26 May Online)
A beam of light consisting of red, green and blue colours is incident on a right-angled prism. The refractive index of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively.
The prism will:

Options:
A Separate all the three colours from one another
B Not separate the three colours at all
C Separate the red colour part from the green and blue colours
D Separate the blue colour part from the red and green colours

The correct answer is:
Separate the red colour part from the green and blue colours

Solution:
For Total internal reflection, \(i > i_{\mathrm{c}}\)
so, \(\sin i > \sin i_c\)
\(\sin 45^{\circ} > \frac{1}{\mu}\)
⇒μ=2
⇒μ=1.414
Since μ of green and violet is greater than 1.414, so they will undergo total internal reflection. But red colour will be refracted. So the prism will separate red colour from the green and blue colours.

Topic: Physics, Ray Optics, JEE Main
A beam of light consisting of red, green and blue colours is incident on a right-angled prism. The refractive index of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively.

The prism will:

Options:
A Separate all the three colours from one another
B Not separate the three colours at all
C Separate the red colour part from the green and blue colours
D Separate the blue colour part from the red and green colours

The correct answer is:
Separate the red colour part from the green and blue colours

Solution:
â
For TIR I>Ic
so, Sin i>sinâ¡Ic
Sin 45o>1Î¼âÎ¼2âÎ¼=1.414
Since Î¼ of green and violet are greater than 1.414 so they will total internal refrected. But red colour will be vetracted.

Topic: Physics, Ray Optics, JEE Main
A beam of light consisting of two wavelengths, 650 nm and 520 nm is used to obtain interference fringes in Young's double-slit experiment. What is the least distance (in m ) from a central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is 3 mm and the distance between the plane of the slits and the screen is 150 cm.

Options:

The correct answer is:
1.3

Solution:
Let "y' be the linear distance between the centre of screen and the point at which the bright fringes due to both wavelength coincides. Let n1th number of bright fringe with wavelength Î»1 coincides with n2m number of bright fringe with wavelength Î»2 .
â´ân1Î²1=n2Î²2
ân1Î»1Dd=n2Î»2Dd
ân1Î»1=n2Î»2 â¦(i)
Also at first position, the nth bright fringe of one will coincide with (n+1)th bright of other.
If Î»2<Î»1
So then n2>n1
ân2=n1+1 â¦(ii)
Using equation (ii) and equation (i),
n1Î»1=n1+1Î»2
ân1650Ã10-9=n1+1520Ã10-9
â65n1=52n1+52
â13n1=52
ân1=4
Thus, y=n1Î²1
=n1Î»1Dd
=46.5Ã10-71.53Ã10-3
=1.3Ã10-3 m
=1.3 mm

Topic: Physics, Wave Optics, JEE Main
A beam of light consists of two wavelengths, 6300 Ã and 5600 Ã. This beam of light is used to obtain an interference pattern in YDSE. If 4th bright fringe of 6300 Ã coincides with the nth dark fringe of 5600 Ã from the central line, then find the value of n.

Options:

The correct answer is:
5

Solution:
Fringe width Îw1=6300 ÃDd and for 5600 Ã; Îw2=5600 ÃDd
So 4th bright fringe position of 6300 Ã is at 4Ã6300 ÃDd.
So for nth dark fringe of 5600 Ã is at n-125600ÃDd.
â4Ã6300ÃDd=n-12Ã5600Dd
â4Ã6356=n-12
ân-12=92
ân=5

Topic: Physics, Wave Optics, JEE Main
A beam of light has two wavelengths of $4972 Å$ and $6216 Å$ with a total intensity of $3.6 \times 10^{-3}$ $\mathrm{Wm}^{-2}$ equally distributed among the two wavelengths. The beam falls normally on an area of $1 \mathrm{~cm}^2$ of a clean metallic surface of work function $2.3 \mathrm{eV}$. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photoelectrons liberated in $2 \mathrm{~s}$ is approximately:

Options:
A
$6 \times 10^{11}$

B
$9 \times 10^{11}$

C
$11 \times 10^{11}$

D
$15 \times 10^{11}$

The correct answer is:
$9 \times 10^{11}$

Solution:
Given, $\lambda_1=4972 Å$
and $\lambda_2=6216 Å$
and $\mathrm{I}=3.6 \times 10^{-3} \mathrm{Wm}^{-2}$
Intensity associated with each wavelength
$$
=\frac{3.6 \times 10^{-3}}{2}=1.8 \times 10^{-3} \mathrm{Wm}^{-2}
$$
Work function $\phi=\mathrm{hv}=\frac{\mathrm{hc}}{\lambda}$
$$
\begin{aligned}
&=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^8\right)}{\lambda} \\
&=\frac{12.4 \times 10^3}{\lambda} \mathrm{ev}
\end{aligned}
$$
for different wavelengths
$$
\begin{array}{r}
\phi_1=\frac{12.4 \times 10^3}{\lambda_1}=\frac{12.4 \times 10^3}{4972}=2.493 \mathrm{eV} \\
=3.984 \times 10^{-19} \mathrm{~J} \\
\phi_2=\frac{12.4 \times 10^3}{\lambda_2}=\frac{12.4 \times 10^3}{6216}=1.994 \mathrm{eV} \\
=3.184 \times 10^{-19} \mathrm{~J}
\end{array}
$$
Work function for metallic surface $\phi=2.3$ $\mathrm{eV}$ (given)
$\phi_2 < \phi$
Therefore, $\phi_2$ will not contribute in this process.
Now, no. of electrons per $\mathrm{m}^2-\mathrm{s}=$ no. of photons per $\mathrm{m}^2-\mathrm{s}$
no. of electrons per $\mathrm{m}^2-\mathrm{s}$
$$
\begin{aligned}
&=\frac{1.8 \times 10^{-3}}{3.984 \times 10^{-19}} \times 10^{-4} \\
&\left(\because 1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2\right)=0.45 \times 10^{12}
\end{aligned}
$$
So, the number of photo electrons liberated
in 2 sec.
$=0.45 \times 10^{12} \times 2$
$=9 \times 10^{11}$

Topic: Physics, Dual Nature of Matter, JEE Main