datavorous/entrance-exam-dataset huggingface dataset:
Let z represent the principal argument of the complex number z. The, z=3 and argz-1-argz+1=π4 intersect:

Options:
A Exactly at one point
B Exactly at two points
C Nowhere
D At infinitely many points.

The correct answer is:
Nowhere

Solution:
Let z represent the principal argument of the complex number z.

The, z=3 will represent a circle with centre 0,0 and radius 3 units

And argz-1-argz+1=π4 will represent arc of the circle with centre 0,1 and radius 2, now on plotting the diagram,

We can see they are intersecting nowhere.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Let $z$ satisfy $|z|=1$ and $z=1-\bar{z}$.
Statement $1: z$ is a real number.
Statement 2 : Principal argument of z is $\frac{\pi}{3}$

Options:
A
Statement 1 is true Statement 2 is true; Statement 2 is a correct explanation for Statement 1 .

B
Statement 1 is false; Statement 2 is true

C
Statement 1 is true, Statement 2 is false.

D
Statement 1 is true; Statement 2 is true; Statement 2 is not a correct explanation for Statement 1 .

The correct answer is:
Statement 1 is false; Statement 2 is true

Solution:
Let $z=x+i y, \bar{z}=x-i y$
$$
\begin{aligned}
& \text { Now, } z=1-\bar{z} \\
\Rightarrow & x+i y=1-(x-i y) \\
\Rightarrow & 2 x=1 \Rightarrow x=\frac{1}{2} \\
& \text { Now, }|z|=1 \Rightarrow x^2+y^2=1 \Rightarrow y^2=1-x^2 \\
\Rightarrow & y=\pm \frac{\sqrt{3}}{2}
\end{aligned}
$$
Now, $\tan \theta=\frac{y}{x} \quad(\theta$ is the argument $)$
$$
=\frac{\sqrt{3}}{2} \div \frac{1}{2}
$$
(+ve since only principal argument)
$$
\begin{gathered}
\quad=\sqrt{3} \\
\Rightarrow \quad \theta=\tan ^{-1} \sqrt{3}=\frac{\pi}{3}
\end{gathered}
$$
Hence, $z$ is not a real number
So, statement-1 is false and 2 is true.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2013 (25 Apr Online)
Let $z$, $w$ be complex numbers such that $\bar{z}+i \bar{w}=0$ and $\arg z w=\pi$. Then arg $z$ equals

Options:
A
$\frac{\pi}{4}$

B
$\frac{5 \pi}{4}$

C
$\frac{3 \pi}{4}$

D
$\frac{\pi}{2}$

The correct answer is:
$\frac{3 \pi}{4}$

Solution:
Here $\omega=\frac{z}{i} \Rightarrow \arg \left(z \cdot \frac{z}{i}\right)=\pi \Rightarrow 2 \arg (z)-\arg (i)=\pi \Rightarrow \arg (z)=\frac{3 \pi}{4}$.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2004
Let z=x+iy and w=u+iv be complex numbers on the unit circle such that z2+w2=1. Then the number of ordered pairs z,w is

Options:
A 0
B 4
C 8
D infinite

The correct answer is:
8

Solution:
It is given that complex numbers z=x+iy and w=u+iv are on the unit circle such that z2+w2=1 ...(i)

So, z2¯+w2¯=1

⇒ z¯2+w¯2=1

⇒ z¯zz2+w¯ww2=1

⇒1z2+1w2=1⇒z2+w2=z2w2

⇒ z2w2=1 ...(ii)

∵ The number of solution of equations x2+y2=1 and x2y2=1 is eight

∴ Number of ordered pair z,w is also eight.

Topic: Mathematics, Complex Number, JEE Main
Let z=x+iy and w=u+iv be two complex numbers, such that z=w=1 and z2+w2=1. Then, the number of ordered pairs z,w is equal to (where, x,y,u,vâR and i2=-1)

Options:

The correct answer is:
8

Solution:
Let, z=eiα and w=eiβ α,β∈-π,π
z2+w2=1⇒ei2α+ei2β=1
⇒cos⁡2α+cos⁡2β=1 and sin⁡2α+sin⁡2β=0
⇒2cosα+βcosα-β=1 and 2sinα+βcosα-β=0
⇒sinα+β=0⇒α+β=nπ
For α+β=-π, we have cos⁡2α=12⇒α=-5π6,-π6⇒2 pairs of α,β
For α+β=0, we have cos⁡2α=12⇒4 pairs of α,β
For α+β=π, we have cos⁡2α=12⇒2 pairs of α,β

Topic: Mathematics, Complex Number, JEE Main
Let Z=x+iy is a complex number, such that x2+y2=1. In which of the following cases Z1-Z for xâ 1 lies in the ÎÎnd quadrant? âx,yâR, i2=-1

Options:
A x>0
B x<0
C y>0
D y<0

The correct answer is:
y>0

Solution:
Let, Z=cos⁡θ+isin⁡θ
Now, Z1-Z=cos⁡θ+isin⁡θ1-cos⁡θ-isin⁡θ
=12sinθ2cos⁡θ+isin⁡θ-i2sinθ2-icosθ2=i2sinθ2cosθ2+isinθ2
=cosθ+π2+isinθ+π22sinθ2
Which lies in second quadrant,
if θ∈0,π⇒sinθ>0⇒y>0

Topic: Mathematics, Complex Number, JEE Main
Let z=x+iy be a non-zero complex number such that z2=iz2, where i=-1, then z lies on the :

Options:
A line, y=-x
B imaginary axis
C line, y=x
D real axis

The correct answer is:
line, y=x

Solution:
x+iy2=ix2+y2

⇒x2-y2+2ixy=i(x2+y2)

compare real and imaginary parts

⇒x=y

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2020 (06 Sep Shift 2)
Let \( z=\frac{(2 \sqrt{3}+2 i)^{8}}{(1-i)^{6}}+\frac{(1+i)^{6}}{(2 \sqrt{3}-2 i)^{8}} . \) Let \( \theta \) be the argument of \( z \) such that \( \theta \in(-\pi, \pi] \) then \( 4 \sin \theta \) is equal to

Options:

The correct answer is:
2

Solution:
z= z18z̄26+z26z̄18=|z1|16+|z2|12|z̄2|6(z̄1)8
arg(z)=– 8 arg (z̄1)– 6arg (z̄2)+2kπ, k∈I
= 8 arg (z1)+6 arg (z2)+2kπ
= 8. π6+6. π4+2kπ= 4π3+3π2–2π= 5π6
∴ 4 sinθ=4 sin 5π6= 2

Topic: Mathematics, Complex Number, JEE Main
Level of estrogen and progesterone are minimum at the time of

Options:
A follicular phase
B ovulation
C secretory phase
D onset of menstrual phase

The correct answer is:
onset of menstrual phase

Solution:
The menstrual cycle consists of changes in the ovary and the uterus simultaneously. It begins with the menstrual or bleeding phase. This represents the shedding of the endometrium of the uterus and unfertilized secondary oocyte. This occurs because the corpus luteum of the previous cycle is degenerating. Due to this the hormones secreted by corpus luteum also start decreasing. These hormones include estrogen and progesterone. These hormones are responsible for maintaining the uterine lining and since these are decreasing the uterine lining too starts shedding. Thus levels of these hormones are minimum at the beginning of the menstrual cycle. During the follicular phase, FSH is increasing and estrogen and progesterone have started the proliferation of the endometrium. During ovulation LH is at its peak and estrogen is also increasing. The secretory phase in uterus coincides with luteal phase in the ovary and thus because of the formation of corpus luteum, estrogen and progesterone are also high.

Topic: Biology, Human Reproduction, NEET
Leydig cells produce a group of hormones called

Options:
A
androgens

B
estrogens

C
aldosterone

D
gonadotropins

The correct answer is:
androgens

Solution:
The Leydig cells or interstitial cells, which are present in the interstitial spaces of testis produce a group of hormones called androgens mainly testosterone.
Estrogen is secreted by the growing follicles in ovaries and stimulates growth and development of female secondary sex organs.
Aldosterone is secreted by the adrenal gland helps in maintaining electrolytes in the body fluid, volume, osmotic pressure and blood pressure.
Gonadotropins are secreted by anterior pituitary gland and stimulate the gonadal activity. These include LH and FSH.

Topic: Biology, Chemical Coordination and Integration
lf f x = sin lim tâ0 2x Ï cot â1 x t 2 , then â«-Ï2Ï2fx dx is equal to (where, xâ 0)

Options:
A -2
B -1
C 0
D 2

The correct answer is:
-1

Solution:
Let y=limt→0⁡2xπcot-1⁡xt2
Case-I : when x>0 then y=2xπlimt→0⁡cot-1⁡xt2=2xπ×0=0
Case-II : when x<0 then y=2xπ limt→0⁡cot-1⁡xt2=2xπ×π=2x
f(x)=sin0x>0sin2xx<0
Now, ∫ −π 2 π 2 f( x )dx= ∫ − π 2 0 sin2x dx+ ∫ 0 π 2 0   dx=− ( cos2x 2 ) − π 2 0 =− 1 2 ( 1−( −1 ) )=−1

Topic: Mathematics, Definite Integration, JEE Main
LH and FSH are collectively called

Options:
A oxytocin
B somatotropins
C luteotropic
D gonadotropins

The correct answer is:
gonadotropins

Solution:
Trophe $=$ nourishment. Most of the tropic hormones are secreted by anterior lobe of pituitary.
Gonadotropins orgonadotropic hormones are those which stimulate the gonads (testes and ovaries),FSH (Follicle Stimulating Hormone) stimulates growth ofovarian follicles and the secretion ofoestrogens in thefemale andspermatogenesis (formation of sperms) in themale . Luteinizing Hormone(LH) stimulatescorpus luteum of the ovary to secreteprogesterone in the females. In male, it activates theLeydig's (interstitial)cells of the testis to secreteandrogens.
Somatotropins or somatotropic hormone orgrowth hormone stimulates growth directly and in conjugation with other hormones the target cells of these hormones are various body cells which undergo growth.Luteotropic hormones stimulates the corpus luteum of the ovary to secrete progesterone hormone. It is also called Prolactin Hormone (PRL) orMammotropin Hormone (MTH) because its main physiological effect is to activate growth of breast during pregnancy and secretion of mammary glands after child birth.Oxytocin (OT; pitocin) is secreted by posterior lobe of pituitary. It promotes contraction of the uterine muscles and myoepithelial cells of the lactating breast, ejecting the milk after child birth. Oxytocin is calledbirth hormone andmilk ejecting hormone.

Topic: Biology, Chemical Coordination and Integration, JIPMER, JIPMER 2009
Li 3 N +  H 2 O→A  gas
Li 2 CO 3 → Δ B  gas
Hybridisation of the central atom in A and B respectively.

Options:
A sp3, sp
B sp2, sp3
C sp, sp
D sp3, sp3

The correct answer is:
sp3, sp

Solution:
Li3N+3H2O→3LiOH+NH3

Li2CO3→ Δ Li2O+CO2

Topic: Chemistry, s Block Elements, JEE Main
Li7 fuses with a proton according to the nuclear reaction given below:

p + 7Li → 4He + 4He

Given that the atomic masses of 1H, 4He and 7Li are 1.007825 u, 4.002603 u and 7.016004 u respectively, where u=931.5 MeV/c2, then the Q-value of the reaction is

Options:
A 17.35 MeV
B

18.06 MeV

C

177.35 MeV

D

170.35 MeV

The correct answer is:
17.35 MeV

Solution:
The total mass of the initial particles
mi = 1.007825 +7.016004

mi= 8.023829 u
and the total mass of final particles
mf = 2×4.002603 = 8.005206 u
Difference between the initial and final mass of particles
= mi - mf= 8.023829 - 8.005206
= 0.018623 u
The Q value is given by
Q = ∆mc2
= 0.018623×931.5 = 17.35 MeV

Topic: Physics, Nuclear Physics, JEE Main
$\mathrm{LiAlH}_4$ converts acetic acid into

Options:
A Acetaldehyde
B Methane
C Ethyl alcohol
D Methyl alcohol

The correct answer is:
Ethyl alcohol

Solution:
$\mathrm{CH}_3 \mathrm{COOH}+4 \mathrm{H} \xrightarrow{\mathrm{LiAlH}_4} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}+\mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Lichens are symbiotic associations between a fungus and algae. Most of these lichens consist of:

Options:
A

Red algae and ascomycetes

B

Brown algae and phycomycetes

C

Blue-green algae and basidomycetes

D

Blue-green algae and ascomycetes

The correct answer is:

Blue-green algae and ascomycetes

Solution:
The lichen fungus is typically a member of the Ascomycota rarely a member of the basidiomycota. The algal or cyanobacterial cells are photosynthetic, and as in higher plants they reduce atmospheric carbon dioxide into organic carbon sugars to feed both symbionts. Both partners gain water and mineral nutrients mainly from the atmosphere through rain and dust. The fungal partner protects the alga by retaining water, serving as a larger capture area for mineral nutrients and, in some cases, provides minerals obtained from the substratum. If a cyanobacterium is present, as a primary partner or another symbiont in addition to green alga as in certain tripartite lichens, they can fix atmospheric nitrogen, complementing the activities of the green alga.

Topic: Biology, Biological Classification, NEET
Liebig's method is used for the estimation of

Options:
A Nitrogen
B Sulphur
C Carbon and hydrogen
D Halogens

The correct answer is:
Carbon and hydrogen

Solution:
No Solution available

Topic: Chemistry, General Organic Chemistry, JEE Main
Ligament is mainly made up of

Options:
A reticulin
B elastin
C myosin
D collagen

The correct answer is:
collagen

Solution:
Ligament is tough cord or fibrous band of dense regular connective tissue that contains numerous parallel arrangements of collagen fibres. It connects bones or cartilages and serves to strengthen joints.

Topic: Biology, Sexual Reproduction in Flowering Plants, JIPMER, JIPMER 2008
Light consisting of a plane waves of wavelength, $\lambda_1=8 \times 10^{-5} \mathrm{~cm}$ and $\lambda_2=6 \times 10^{-5} \mathrm{~cm}$ generates an interference pattern in Young's double slit experiment. If $n_1$ denotes the $n_1$ th dark fringe due to light of wavelength $\lambda_1$ which coincides with $n_2$ th bright fringe due to light of wavelength $\lambda_2$, then

Options:
A $n_1=3, n_2=1$
B $n_1=4, n_2=5$
C $n_1=1, n_2=2$
D $n_1=3, n_2=2$

The correct answer is:
$n_1=1, n_2=2$

Solution:
Given, $\lambda_1=8 \times 10^{-5} \mathrm{~cm}$ and $\lambda_2=6 \times 10^{-5} \mathrm{~cm}$
Position $n^{\text {h }}$, dark fringe (due to light of wavelength $\lambda_1$ ) is given by,
$x_{n_1}=\left(2 n_1 \times 1\right) \frac{D \lambda_1}{2 d}$
Position of $n_2{ }^{\text {th }}$ bright fringe due to light of wavelength $\lambda_2$ is given by
$x_{n_2}=\frac{n_2 D \lambda_2}{d}$
Since both the fringes are coincide to each other. Hence,
$\begin{aligned}\left(2 n_1+1\right) \frac{D \lambda_1}{2 d} & =\frac{n_2 D \lambda_2}{d} \\ \frac{2 n_1+1}{n_2} & =\frac{2 \lambda_2}{\lambda_1}=\frac{2 \times 6 \times 10^{-5}}{8 \times 10^{-5}} \\ \frac{2 n_1+1}{n_2} & =\frac{3}{2} \\ 4 n_1+2 & =3 n_2 \\ 4 n_1-3 n_2 & =-2 \\ 4 n_1-3 n_2 & =-2\end{aligned}$
when $n_1=1$ and $n_2=2$, then Eq. (i) is satisfied. Hence, for $n_1=1$ and $n_2=2$, given fringes coincide to each other.

Topic: Physics, Wave Optics, JEE Main
Light emerges out of a convex lens when a source of light kept at its focus. The shape of wavefront of the light is :

Options:
A both spherical and cylindrical
B plane
C spherical
D cylindrical

The correct answer is:
plane

Solution:
Light emerges parallel
$\therefore$ planor wavefront

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (05 Apr Shift 1)
Light enters from air into a given medium at an angle of 45° with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of 15° from its original direction. The refractive index of the medium is :

Options:
A 1.732
B 1.333
C 1.414
D 2.732

The correct answer is:
1.414

Solution:
Given here, i=45°

Deviation in light ray, D=i-r, here, r is refracted ray.

⇒15°=45-r⇒r=30°

Using Snell's law,

n1sini=n2sinr

⇒1sin45°=μsin30°

⇒12=μ12

Thus, the refractive index of the medium is μ=2=1.414

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (29 Jul Shift 2)
Light from a point source in air falls on a convex curved surface of radius 20 cm and refractive index 1.5. If the source is located at 100 cm from the convex surface, the image will be formed at ______ cm from the object.

Options:

The correct answer is:
200

Solution:
μ2v-μ1u=μ2−μ1R

⇒1.5v−1−100=1.5−120

⇒v=100 cm

Distance from object

=100+100

=200 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Light green crystals of ferrous sulphate lose water molecule and turn brown on exposure to air. This is due to its oxidation to Extra]

Options:
A $\mathrm{Fe}_2 \mathrm{O}_3$
B $\mathrm{Fe}_2 \mathrm{O}_3 \cdot \mathrm{H}_2 \mathrm{O}$
C $\mathrm{Fe}(\mathrm{OH}) \mathrm{SO}_4$
D $\mathrm{Fe}_2 \mathrm{O}_3+\mathrm{FeO}$

The correct answer is:
$\mathrm{Fe}(\mathrm{OH}) \mathrm{SO}_4$

Solution:
Light green crystals of ferrous sulphate lose water molecule and turn brown on exposure to air. This is due to its oxidation to \(\mathrm{Fe}(\mathrm{OH}) \mathrm{SO}_4\)
\(\stackrel{+2}{\mathrm{Fe}} \mathrm{SO} \xrightarrow{\mathrm{H}_2 \mathrm{O}} ~\stackrel{+3}{\mathrm{Fe}}(\mathrm{OH}) \mathrm{SO}_4\)

Topic: Chemistry, d and f Block Elements, JEE Main
Light green crystals of ferrous sulphate lose water molecule and turn brown on exposure to air. This is due to its oxidation to

Options:
A $\mathrm{Fe}_2 \mathrm{O}_3$
B $\mathrm{Fe}_2 \mathrm{O}_3 \cdot \mathrm{H}_2 \mathrm{O}$
C $\mathrm{Fe}(\mathrm{OH}) \mathrm{SO}_4$
D $\mathrm{Fe}_2 \mathrm{O}_3+\mathrm{FeO}$

The correct answer is:
$\mathrm{Fe}(\mathrm{OH}) \mathrm{SO}_4$

Solution:
$\begin{aligned} & \mathrm{Fe}(\mathrm{OH}) \mathrm{SO}_4 \\ & \mathrm{FeSO}_4 \xrightarrow{\mathrm{H}_2 \mathrm{O}} \mathrm{Fe}(\mathrm{OH}) \mathrm{SO}_4\end{aligned}$

Topic: Chemistry, d and f Block Elements, JEE Main
Light is incident at an angle $i$ on a glass slab. The reflected ray is completely polarised. The angle of refraction is

Options:
A $90^{\circ}-i$
B $180^{\circ}-i$
C $90^{\circ}+i$
D $\bar{i}$

The correct answer is:
$90^{\circ}-i$

Solution:
Brewster discovered that when a beam of unpolarised light is reflected from a transparent medium (refractive index $=\mu)$, the reflected light is completely plane polarised at a certain angle of incidence (called the angle of polarisation $i$ )
From figure it is clear that $i+r=90^{\circ}$
$$
r=(90-i)
$$

Topic: Physics, Wave Optics, JEE Main
Light is incident from a medium into air at two possible angles of incidence (A) $20^{\circ}$ and (B) $40^{\circ}$. In the medium light travels $3.0 \mathrm{~cm}$ in $0.2 \mathrm{~ns}$. The ray will :

Options:
A
suffer total internal reflection in both cases (A) and (B)

B
suffer total internal reflection in case (B) only

C
have partial reflection and partial transmission in case (B)

D
have $100 \%$ transmission in case (A)

The correct answer is:
suffer total internal reflection in case (B) only

Solution:
Velocity of light in medium
$$
\mathrm{V}_{\text {med }}=\frac{3 \mathrm{~cm}}{0.2 \mathrm{~ns}}=\frac{3 \times 10^{-2} \mathrm{~m}}{0.2 \times 10^{-9} \mathrm{~s}}=1.5 \mathrm{~m} / \mathrm{s}
$$
Refractive index of the medium
$$
\mu=\frac{V_{\text {air }}}{V_{\text {med }}}=\frac{3 \times 10^8}{1.5}=2 \mathrm{~m} / \mathrm{s}
$$
As $\mu=\frac{1}{\sin C}$
$$
\therefore \sin \mathrm{C}=\frac{1}{\mu}=\frac{1}{2}=30^{\circ}
$$
Condition of TIR is angle of incidence i must be greater than critical angle. Hence ray will suffer TIR in case of $(B)\left(i=40^{\circ}>30^{\circ}\right)$ only.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2013 (09 Apr Online)
Light is incident normally on the face AB of a prism as shown in the figure. A layer of liquid of refractive index Î¼ is applied on the face AC of the prism. If the prism is made of glass of refractive index 32, then the limit of Î¼ for which total internal reflection takes place on face AC is

Options:
A μ<334
B μ>52
C μ>2
D μ<3

The correct answer is:
μ<334

Solution:
For critical angle at the glass-liquid interface we have

Î¼1sinC=Î¼2sin90Â°

â32sinC=Î¼

âsinC=2Î¼3

For total internal reflection

60Â°>C

âsin60Â°>sinC

â32>2Î¼3

Î¼<334

Topic: Physics, Ray Optics, JEE Main
Light of different colours propagates through air

Options:
A With the velocity of air
B With different velocities
C With the velocity of sound
D Having the equal velocities

The correct answer is:
Having the equal velocities

Solution:
In vacuum, the speed of light is independent of wave length. Thus vacuum (or air) is a non-dispersive medium in which all colours travel with the same speed.

Topic: Physics, Ray Optics, JEE Main
Light of frequency 7.21Ã 10 14 â Hz is incident on a metal surface. Electrons with a maximum speed of 6.0Ã105 m s-1 are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Options:
A 4.74× 10 14   Hz
B 5.74× 10 15   Hz
C 6.84× 10 13   Hz
D 4.84× 10 14   Hz

The correct answer is:
4.74× 10 14   Hz

Solution:
Given, frequency of light, Î½=7.21Ã 10 14 âHz
Mass of electron, m=9.1Ã 10 â31 âkg
Maximum speed of electrons, v max =6Ã 10 5 â m/s
Let Î½ 0 be the threshold frequency.
Use the formula for kinetic energy
KE= 1 2 m v max 2 =hÎ½âh Î½ 0
i.e., 1 2 Ã9.1Ã 10 â31 Ã6Ã 10 5 Ã6Ã 10 5 =6.63Ã 10 â34 ( Î½â Î½ 0 )
or Î½ - Î½ 0 = 3 6 Ã 9 . 1 Ã 1 0 - 2 1 2 Ã 6 . 6 3 Ã 1 0 - 3 4 = 2 . 4 7 Ã 1 0 1 4
or Î½ 0 =7.21Ã 10 14 â2.47Ã 10 14 ( âµâÎ½=7.21Ã 10 14 â Hz )
=4.74Ã 10 14 â Hz.

Topic: Physics, Dual Nature of Matter, JEE Main
Light of two different frequencies whose photons have energies $1 \mathrm{eV}$ and $2.5 \mathrm{eV}$ respectively illuminate a metallic surface whose work function is $0.5 \mathrm{eV}$ successively. Ratio of maximum speeds of emitted electrons will be

Options:
A $1: 2$
B $1: 1$
C $1: 5$
D $1: 4$

The correct answer is:
$1: 2$

Solution:
Kinetic energy
$$
\mathrm{KE}=\phi-\phi_0
$$
Here, $\mathrm{KE}_1=1-0.5=0.5 \mathrm{eV}$
$$
\begin{aligned}
& \mathrm{KE}_2=2.5-0.5=2 \mathrm{eV} \\
& \therefore \frac{\mathrm{KE}_1}{\mathrm{KE}_2}=\frac{0.5}{2}=\frac{1}{4} \\
& \text { or } \quad \frac{y_1^2}{y_2^2}=\frac{1}{4} \\
& \text { or } \quad \frac{y_1}{y_2}=\sqrt{\frac{1}{4}}=\frac{1}{2}
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, JEE Main
Light of two different frequencies whose photons have energies 1 eV and 2.5 eV, respectively, successively illuminate a metallic surface whose work function is 0.5 eV. The ratio of maximum speeds of the emitted electrons will be

Options:
A 1:4
B 1:1
C 1:5
D 1:2

The correct answer is:
1:2

Solution:
According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is Kmax=hν-ϕ0
Where hν is the energy of the incident photon and ϕ0 is the work function.
But Kmax=12mvmax2
∴12mvmax2=hν-ϕ0
As per the question,
12mv2max1⁡=1 eV-0.5 eV=0.5 eV ...(i)
and 12mv2max2⁡=2.5 eV-0.5 eV=2 eV ...(ii)
Dividing equation (i) by equation (ii), we get
v2max1⁡v2max2⁡=0.5 eV2 eV=14
vmax1⁡vmax2⁡=14=12

Topic: Physics, Dual Nature of Matter, JEE Main
Light of two different frequencies whose photons have energies 1 eV and 2.5 eV, respectively, successively illuminate a metallic surface whose work function is 0.5 eV. The ratio of maximum speeds of the emitted electrons will be

Options:
A 1:4
B 1:1
C 1:5
D 1:2

The correct answer is:
1:2

Solution:
According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is Kmax=hυ-ϕ0
Where hυ is the energy of incident photon and ϕ0 is the work function.
But Kmax=12mvmax2
∴12mvmax2=hυ-ϕ0
As per question,
12mv2max1⁡=1 eV-0.5 eV=0.5 eV ...(i)
and 12mv2max2⁡=2.5 eV-0.5 eV=2 eV ...(ii)
Dividing equation (i) by equation (ii), we get
v2max1⁡v2max2⁡=0.5 eV2 eV=14
vmax1⁡vmax2⁡=14=12

Topic: Physics, Dual Nature of Matter, JEE Main
Light of wavelength 0.6 Î¼m falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0.5 V. With the light of wavelength 0.4 Î¼m, stopping potentials is 1.5 V. With this data, the value of he is nÃ10â15 V s. Find the value of n.

Options:

The correct answer is:
4

Solution:
eV=hcλ−W⇒V=(he)cλ−We

V1=(he)cλ1−We ……....……(1)

V2=(he)cλ2−We ……………(2)

Solving these two equations, we get,

he=λ1λ2(V1−V2)c(λ2−λ1)=(0.6×0.4×10−12)(1.0)(3×108)(0.2×10−6)

=4×10−15 V s

n=4

Topic: Physics, Dual Nature of Matter, JEE Main
Light of wavelength $180 \mathrm{nm}$ ejects photoelectron from a plate of a metal whose work function is $2 \mathrm{eV}$. If a uniform magnetic field of $5 \times 10^{-5} \mathrm{~T}$ is applied parallel to plate, what would be theradius of the path followed by electrons ejected normally from the plate with maximum energy?

Options:
A $1.239 \mathrm{~m}$
B $0.149 \mathrm{~m}$
C $3.182 \mathrm{~m}$
D $2.33 \mathrm{~m}$

The correct answer is:
$0.149 \mathrm{~m}$

Solution:
If $\mathrm{v}_{\max }$ is the speed of the fastest electron emitted from the metal surface, then
$$
\begin{aligned}
& \frac{h c}{\lambda}=W_{0}+\frac{1}{2} m v_{\max }^{2} \frac{\left(6.63 \times 10^{-34}\right) \times\left(3 \times 10^{8}\right)}{\left(180 \times 10^{-9}\right)} \\
&=2 \times\left(1.6 \times 10^{-19}\right)+\frac{1}{2}\left(9.1 \times 10^{-31}\right) v_{\max }^{2} \\
\therefore \quad & v=1.31 \times 10^{6} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
The radius of the electron is given by
$$
r=\frac{m v}{q B}=\frac{\left(9.1 \times 10^{-31}\right) \times\left(1.31 \times 10^{6}\right)}{\left(1.6 \times 10^{-19}\right) \times\left(5 \times 10^{-9}\right)}=0.149 \mathrm{~m}
$$

Topic: Physics, Dual Nature of Matter, JEE Main