datavorous/entrance-exam-dataset huggingface dataset:
The terminal velocity \( v_{r} \) of a small steel ball of radius \( r \) falling under gravity through a column of a viscous liquid of coefficient of viscosity \( \eta \) depends on the mass of the ball \( m \), acceleration due to gravity \( g \), coefficient of viscosity \( \eta \) and radius \( r \). Which of the following relations is dimensionally correct?

Options:
A \( v_{r} \propto \frac{m g r}{\eta} \)
B \( v_{r} \propto m g \eta r \)
C \( v_{r} \propto \frac{m g}{r \eta} \)
D \( v_{r} \propto \frac{\eta m g}{r} \)

The correct answer is:
\( v_{r} \propto \frac{m g}{r \eta} \)

Solution:
According to question, terminal velocity vr of ball depends on mass of ball m, acceleration due to gravity g, coefficient of viscosity η and radius of ball r.

Using Dimensional analysis, we can write

vr∝magbηcrd⇒vr=magbηcrd

Substituting the dimensions of each term in the above expression we get

M0L1T-1=M1L0T0aM0L1T-2bM1L-1T-1cM0L1T0d

Now equating the exponent of each of the fundamental dimensions, we get

a+c=0 ...1b+d-c=1 ...2-2b-c=-1 ...3

Solving above system of equations, we get a=b=1 and c=d=-1.

Hence, the dimensionally correct relation is vr∝mgηr.

Topic: Physics, Units and Dimensions, JEE Main
The terms tan⁡80o, tan⁡70o+tan⁡10o and tan⁡10o are in

Options:
A arithmetic progression
B geometric progression
C harmonic progression
D None of these

The correct answer is:
arithmetic progression

Solution:
tan⁡80o=tan⁡70o+10o=tan⁡70o+tan⁡10o1-tan⁡70otan⁡10o
⇒ tan⁡80o-tan⁡80otan⁡70otan⁡10o
=tan⁡70o+tan⁡10o
⇒ tan⁡80o-tan⁡80otan⁡70ocot⁡80o=tan⁡70o+tan⁡10o
⇒ tan⁡80o=2tan⁡70o+tan⁡10o
⇒ tan⁡80o+tan⁡10o=2tan⁡70o+tan⁡10o
⇒ they are in A.P.

Topic: Mathematics, Sequences and Series, JEE Main
The test performed on compound x and their inferences are:

Test	Inference

a 2, 4 - DNP test	Coloured precipitate yellow

b Iodoform test	Yellow precipitate

c Azo-dye test	No dye formation

Compound ′x′ is:

Options:
A
B

C
D

The correct answer is:

Solution:
group gives 2,4-DNP test and iodoform test.

not in the plane of benzene so no dye test.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The test to distinguish primary, secondary and tertiary amines is:

Options:
A Sandmeyer's reaction
B Carbylamine reaction
C Idoform test
D C6H5SO2Cl

The correct answer is:
C6H5SO2Cl

Solution:
The Hinsberg test, which can distinguish primary, secondary, and tertiary amines, is based upon sulfonamide formation. In the Hinsberg test, an amine is reacted with benzene sulfonyl chloride. If a product forms, the amine is either a primary or secondary amine, because tertiary amines do not form stable sulfonamides. If the sulfonamide that forms dissolves in aqueous sodium hydroxide solution, it is a primary amine. If the sulfonamide is insoluble in aqueous sodium hydroxide, it is a secondary amine

Benzene sulphonyl chloride is Hinsberg reagent.

Topic: Chemistry, Amines, JEE Main, JEE Main 2016 (09 Apr Online)
The test used for identifying peptide linkage in proteins is

Options:
A Borsche's test
B Molisch's test
C Ninhydrin test
D Biuret test

The correct answer is:
Biuret test

Solution:
Biuret test identifies peptide linkage in protein

Topic: Chemistry, Biomolecules, JEE Main
The testes in humans are situated outside the abdominal cavity inside a pouch called scrotum. The purpose served is for

Options:
A escaping any possible compression by the visceral organs
B providing more space for the growth of epididymis
C providing a secondary sexual feature for exhibiting the male sex
D maintaining the scrotal temperature lower than the internal body temperature

The correct answer is:
maintaining the scrotal temperature lower than the internal body temperature

Solution:
The testes in humans are situated outside the abdominal cavity in scrotal sacs. This is because the temperature of scrotal sacs is $25^{\circ} \mathrm{C}$ less than internal body temperature.

Topic: Biology, Human Reproduction, NEET, NEET 2011 (Screening)
The testes in humans are situated outside the abdominal cavity, inside a pouch called scrotum. The purpose served is for

Options:
A Maintaining the scrotal temperature lower than the internal body temperature
B Escaping any possible compression by the visceral organs
C Providing more space for the growth of epididymis
D Providing a secondary sexual feature for exhibiting the male sex

The correct answer is:
Maintaining the scrotal temperature lower than the internal body temperature

Solution:
The testes are present in the scrotum which maintains the scrotal temperature below 2oC of the internal body temperature and is required for the process of spermatogenesis.

Topic: Biology, Human Reproduction, NEET
The theory that can completely/properly explain the nature of bonding in NiCO4 is:

Options:
A Werner's theory
B Molecular orbital theory
C Crystal field theory
D Valence bond theory

The correct answer is:
Molecular orbital theory

Solution:
There is synergic bonding present in NiCO4.

As CO is neutral ligand, so oxidation state of nickel is zero i.e.Ni0.

So, electronic configuration of Ni0 =Ar3d8 4s2

As CO is strong field ligand, it will back paired the 4s electrons to 3d orbitals.

So in complex electronic configuration of Ni0 =Ar 3d10 4s0 here inner d orbital is filled.

In ligand CO (carbonyl) the anti-bonding orbital is vacant.

CO: σ1s2, σ*1s2, σ2s2, σ*2s2, σ2pz2, πpx2=πpy2, π*px0=π*py0

The fully filled inner d-orbital of central metal atom gives it's electrons to vacant π anti-bonding orbital, this is called back π bonding, due to this back bonding bond between metal and ligand is very strong.

This back π bonding is also called synergic bonding and this synergic bonding is explained by molecular orbital theory.

Carbon monoxide CO made synergic bonding with Haemoglobin in are blood, as synergic bonding is very strong, so oxygen taking capacity of haemoglobin decrease and it cause death.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (07 Jan Shift 1)
The thermal decomposition of a compound is of first order. If a sample of the compound decomposes $50 \%$ in 120 minutes, in what time will it undergo $90 \%$ decomposition

Options:
A Nearly 240 minutes
B Nearly 480 minutes
C Nearly 450 minutes
D Nearly 400 minutes

The correct answer is:
Nearly 400 minutes

Solution:
Nearly 400 minutes

Topic: Chemistry, Chemical Kinetics, JEE Main
The thermal stability of alkaline earth metal carbonates $\mathrm{MgCO}_3, \mathrm{CaCO}_3, \mathrm{BaCO}_3$ and $\mathrm{SrCO}_3$ decreases as

Options:
A $\mathrm{CaCO}_3\gt\mathrm{SrCO}_3\gt\mathrm{MgCO}_3\gt\mathrm{BaCO}_3$
B $\mathrm{BaCO}_3\gt\mathrm{SrCO}_3\gt\mathrm{MgCO}_3\gt\mathrm{CaCO}_3$
C $\mathrm{BaCO}_3\gt\mathrm{SrCO}_3\gt\mathrm{CaCO}_3\gt\mathrm{MgCO}_3$
D $\mathrm{MgCO}_3\gt\mathrm{CaCO}_3\gt\mathrm{SrCO}_3\gt\mathrm{BaCO}_3$

The correct answer is:
$\mathrm{BaCO}_3\gt\mathrm{SrCO}_3\gt\mathrm{CaCO}_3\gt\mathrm{MgCO}_3$

Solution:
Thermal stability increasing from top to bottom.

Topic: Chemistry, s Block Elements, JEE Main
The thermistors are usually made of

Options:
A
metals with low temperature coefficient of resistivity

B
metals with high temperature coefficient of resistivity

C
metal oxides with high temperature coefficient of resistivity "

D
semiconducting materials having low temperature coefficient of resistivity.

The correct answer is:
metal oxides with high temperature coefficient of resistivity "

Solution:
These are devices whose resistance varies quite markedly with temperature mean having high-temperature coefficient of resistivity. [Their name are derived from thermal resistors]. Depending on their composition they can have either negative temperature coefficient or positive temperature coefficient or positive temperature coefficient or positive temperature coefficient characteristics.
The negative temperature coefficient type consists of a mixture of oxides of iron, nickel, and cobalt with small amounts of other substances. The positive temperature coefficient types are based on barium titanate.

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2004
The thermo emf $E$ in volt of a certain thermo-couple is found to vary with temperature difference $\theta$ in ${ }^{\circ} \mathrm{C}$ between the two junctions according to the relation $E=30 \theta-\frac{\theta^2}{15}$
The neutral temperature for the thermo-couple will be

Options:
A $450^{\circ} \mathrm{C}$
B $400^{\circ} \mathrm{C}$
C $225^{\circ} \mathrm{C}$
D $30^{\circ} \mathrm{C}$

The correct answer is:
$225^{\circ} \mathrm{C}$

Solution:
Given, $\mathrm{E}=30 \theta-\frac{\theta^2}{15}$
At neutral temperature
$\begin{aligned}
\frac{\mathrm{dE}}{\mathrm{d} \theta} & =0 \\
\therefore \quad 30-\frac{2 \theta}{15} & =0 \\
\theta & =225^{\circ} \mathrm{C}
\end{aligned}$

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2010 (Mains)
The thermo e.m.f. of a thermo-couple is $25 \mu \mathrm{V} /{ }^{\circ} \mathrm{C}$ at room temperature. A galvanometer of 40 ohm resistance, capable of detecting current as low as $10^{-5} \mathrm{~A}$, is connected with the thermo couple. The smallest temperature difference that can be detected by this sytem is

Options:
A
$16^{\circ} \mathrm{C}$

B
$12^{\circ} \mathrm{C}$

C
$8^0 \mathrm{C}$

D
$20^{\circ} \mathrm{C}$

The correct answer is:
$16^{\circ} \mathrm{C}$

Solution:
$\Delta \mathrm{T}=\frac{40}{25 \times 10^{-6} \times 10^{-5}} \Rightarrow \Delta \mathrm{T}=16^0 \mathrm{C}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2003
The thermo emf of a thermocouple varies with the temperature $\theta$ of the hot junction as $\mathrm{E}=\mathrm{a} \theta+\mathrm{b} \theta^2$ in volts where the ratio $\mathrm{a} / \mathrm{b}$ is $700^{\circ} \mathrm{C}$. If the cold junction is kept at $0^{\circ} \mathrm{C}$, then the neutral temperature is

Options:
A
$700^{\circ} \mathrm{C}$

B
$350^{\circ} \mathrm{C}$

C
$1400^{\circ} \mathrm{C}$

D
no neutral temperature is possible for this thermocouple.

The correct answer is:
no neutral temperature is possible for this thermocouple.

Solution:
$E=a \theta+b \theta^2$
At neutral temperature $\mathrm{dE} / \mathrm{d} \theta=0$
$\therefore \frac{\mathrm{dE}}{\mathrm{d} \theta}=\mathrm{a}+2 \mathrm{~b} \theta_{\mathrm{n}}=0 ; \theta_{\mathrm{n}}=-\frac{\mathrm{a}}{2 \mathrm{~b}}$
Now $\frac{a}{b}=700^{\circ} \mathrm{C}$ (given)
$\theta_n=-700 / 2=-350^{\circ} \mathrm{C}$
Now $\theta_c=0^{\circ} \mathrm{C}$.
So, $\theta_n>0^{\circ} \mathrm{C}$
But mathematically $\theta_{\mathrm{n}} < 0^{\circ} \mathrm{C}$.

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2004
The thermodynamic process, in which internal energy of the system remains constant is

Options:
A Isochoric
B Adiabatic
C Isothermal
D Isobaric

The correct answer is:
Isothermal

Solution:
An isothermal process is a thermodynamic process in which the temperature of a system remains constant. The transfer of heat into or out of the system happens so slowly that thermal equilibrium is maintained.

ΔT=0, ΔU=nCV∆T=0.

According to first law of thermodynamics,

∆Q=∆U+∆W⇒∆Q=∆W

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2023 (11 Apr Shift 2)
The thermostable enzyme, Taq polymerase used in PCR is obtained from

Options:
A Bacillus subtilis
B Pseudomonas putida
C Thermus aquaticus
D Thiobacillus ferroxidans

The correct answer is:
Thermus aquaticus

Solution:
Taq polymerase, generally used in PCR is isolated from thermophilic bacterium Thermus aquaticus. This enzyme remains active at very high temperature and is not destroyed during high temperature in PCR.

Topic: Biology, Biotechnology - Principles and Processes, NEET
The thickness at the centre of a plano convex lens is 3 mm and the diameter is 6 cm. If the speed of light in the material of the lens is 2×108 m s-1. The focal length of the lens is ___________.

Options:
A 0.30 cm
B 15 cm
C 1.5 cm
D 30 cm

The correct answer is:
30 cm

Solution:
R2=r2+(R-t)2

R2=r2+R2+t2-2Rt

Neglecting t2, we get

R=r22t

∴1f=μ-11R-1∞=μ-1R

f=Rμ-1=r22t(μ-1)=3×10-222×3×10-3×32-1

=9×10-46×10-3×1×2

f=0.3 m=30 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The third ionization enthalpy is minimum for:

Options:
A Co
B Fe
C Ni
D Mn

The correct answer is:
Fe

Solution:
COâ Ar 3 d 7 4 s 2
Feâ Ar 3 d 6 4 s 2

Niâ Ar 3 d 8 4 s 2
Mnâ Ar 3 d 5 4 s 1
In Fe third electron will remove from 3d and extra stability will be attained by this so ionisation energy is minimum for Fe.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The third law of thermodynamics states that:

Options:
A the entropy of a perfectly crystalline pure substance at zero \( \mathrm{K} \) is zero.
B absolute entropy of hydrogen ion is zero at zero \( \mathrm{K} \).
C Net change in entropy in conversionH \( _{2(\mathrm{~g})}(130 \mathrm{~K}) \rightarrow \mathrm{H}_{2(\mathrm{~g})}(200 \mathrm{~K}) \) is zero.
D entropy generally decrease in combustion reactions.

The correct answer is:
the entropy of a perfectly crystalline pure substance at zero \( \mathrm{K} \) is zero.

Solution:
The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero. Specifically, the entropy of a pure crystalline substance (perfect order) at absolute zero temperature is zero.

Topic: Chemistry, Thermodynamics (C), JEE Main
The third line of the Balmer series in the emission spectrum of the hydrogen atom is due to the transition from the

Options:
A fourth Bohr orbit to the first Bohr orbit
B fifth Bohr orbit to the second Bohr orbit
C sixth Bohr orbit to the third Bohr orbit
D seventh Bohr orbit to the third Bohr orbit

The correct answer is:
fifth Bohr orbit to the second Bohr orbit

Solution:
Rydberg's formula for calculation of wave numbers of lines in hydrogen spectrum is
v̄=R1n12-1n22
For Balmer series, n1=2 and n2=3,4,5 , so third line of the Balmer series is due to transition from fifth Bohr orbit to second Bohr orbit.

Topic: Chemistry, Structure of Atom, JEE Main
The three charges q2,q and q2 are placed at the corners A,D and C of a square of side a as shown in figure. The magnitude of electric field E at the corner B of the square, is

Options:
A q4πϵ0a212+12
B q4πϵ0a21+12
C q4πϵ0a21-12
D q4πϵ0a212-12

The correct answer is:
q4πϵ0a212+12

Solution:
EA=kq2a2=kq2a2, EC=kq2a2=kq2a2 and ED=kq2a2=kq2a2

As we can see from diagram, EC & EA are perpendicular.

Therefore, net electric field will be,

=kq2a2+kq2a22+kq2a22=kq2a2+kq2a2=q4πϵ0a212+12

Topic: Physics, Electrostatics, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The three processes in a thermodynamic cycle shown in the figure are : Process $1 \rightarrow 2$ is isothermal; Process $2 \rightarrow 3$ is isochoric (volume remains constant); Process $3 \rightarrow 1$ is adiabatic. The total work done by the ideal gas in this cycle is $10 \mathrm{~J}$. The internal energy decreases by $20 \mathrm{~J}$ in the isochoric process. The work done by the gas in the adiabatic process is $-20 \mathrm{~J}$. The heat added to the system in the isothermal process is

Options:
A $0 \mathrm{~J}$
B $10 \mathrm{~J}$
C $20 \mathrm{~J}$
D $30 \mathrm{~J}$

The correct answer is:
$30 \mathrm{~J}$

Solution:
$\Delta \mathrm{Q}_{1 \rightarrow 2}=\Delta \mathrm{W}_{12}$
$\mathrm{w}_{\text {Total }}=\Delta \mathrm{W}_{12}+\Delta \mathrm{W}_{31}$
$10=\Delta \mathrm{w}_{12}-20$
$\Delta \mathrm{Q}_{12}=\Delta \mathrm{w}_{12}=30 \mathrm{~J}$

Topic: Physics, Thermodynamics, KVPY, KVPY 2013 (SB/SX)
The three resistances $A, B$ and $C$ have values $3 R, 6 R$ and $R$ respectively. When some potential difference is applied across the network, the thermal powers dissipated by $A, B$ and $C$ are in the ratio

Options:
A $2: 3: 4$
B $2: 4: 3$
C $4: 2: 3$
D $3: 2: 4$

The correct answer is:
$4: 2: 3$

Solution:
Thermal power in $A=P_A=\left(\frac{2 i}{3}\right)^2 3 R=\frac{4}{3} i^2 R$
Thermal power in $B=P_B=\left(\frac{i}{3}\right)^2 6 R=\frac{2}{3} i^2 R$
Thermal power in
$\begin{aligned} & C=P_C=i^2 R \\ & \Rightarrow P_A: P_B: P_C \\ & =\frac{4}{3}: \frac{2}{3}: 1=4: 2: 3\end{aligned}$

Topic: Physics, Current Electricity, JEE Main
The three sides of a right angled triangle are in GP (geometric progression). If the two acute angles be $\alpha$ and $\beta$, then tan $\alpha$ and tan $\beta$ are

Options:
A $\frac{\sqrt{5}+1}{2}$ and $\frac{\sqrt{5}-1}{2}$
B $\sqrt{\frac{\sqrt{5}+1}{2}}$ and $\sqrt{\frac{\sqrt{5}-1}{2}}$
C $\sqrt{5}$ and $\frac{1}{\sqrt{5}}$
D $\frac{\sqrt{5}}{2}$ and $\frac{2}{\sqrt{5}}$

The correct answer is:
$\sqrt{\frac{\sqrt{5}+1}{2}}$ and $\sqrt{\frac{\sqrt{5}-1}{2}}$

Solution:
Let $\Delta A B C$ be a right angled triangle at $B$. Let $\angle A$ and $\angle C$ be $\alpha$ and $\beta$

Since, sides are in GP so sides are $a,$ ar, ar $^{2}$
$\begin{array}{lr}\text { Now, } & \begin{array}{r}A C^{2}=A B^{2}+B C^{2} \\ \left(a r^{2}\right)^{2}=a^{2}+(a \cdot r)^{2}\end{array} \\ \Rightarrow \quad a^{2} r^{4}=a^{2}+a^{2} r^{2} \\ \Rightarrow \quad r^{4}-r^{2}-1=0 \\ \Rightarrow \quad r^{2}=\frac{1+\sqrt{5}}{2} \\ \Rightarrow \quad r=\sqrt{\frac{\sqrt{5}+1}{2}}\end{array}$
$\therefore \quad \tan \alpha=\frac{B C}{A B}=\frac{a r}{a}$
$\Rightarrow \quad r=\sqrt{\frac{\sqrt{5}+1}{2}}$
$\begin{aligned} \text { and } \tan \beta &=\frac{A B}{B C}=\frac{a}{a r}=\frac{1}{r} \\ &=\frac{1}{\sqrt{\frac{\sqrt{5}+1}{2}}}=\sqrt{\frac{\sqrt{5}-1}{2}} \end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main
The three straight lines $a x+b y=c, b x+c y=a$ and $c x+a y=b$ are collinear, if

Options:
A $a+b+c=0$
B $b+c=a$
C $c+a=b$
D $a+b=c$

The correct answer is:
$a+b+c=0$

Solution:
Given three straight lines $a x+b y-c=0, b x+c y-a=0, c x+a y-b=0$ are collinear,
$\left|\begin{array}{lll}
a & b & -c \\
b & c & -a \\
c & a & -b
\end{array}\right|=0 \Rightarrow-(a+b+c)\left|\begin{array}{lll}
1 & b & c \\
1 & c & a \\
1 & a & b
\end{array}\right|=0$
Clearly, $(a+b+c)=0$

Topic: Mathematics, Straight Lines, JEE Main
The threshold frequency for a metallic surface corresponds to an energy of $6.2 \mathrm{eV}$, and the stopping potential for a radiation incident on this surface $5 \mathrm{~V}$. The incident radiation lies in

Options:
A
X-ray region

B
ultra-violet region

C
infra-red region

D
visible region

The correct answer is:
ultra-violet region

Solution:
$\lambda=\frac{1242 \mathrm{e} V \mathrm{~nm}}{11.2} \approx 1100 Å$
Ultraviolet region

Topic: Physics, Dual Nature of Matter, JEE Main
The threshold frequency for a metallic surface corresponds to an energy of $6.2 \mathrm{eV}$, and the stopping potential for a radiation incident on this surface $5 \mathrm{~V}$. The incident radiation lies in

Options:
A
X-ray region

B
ultra-violet region

C
infra-red region

D
visible region

The correct answer is:
ultra-violet region

Solution:
$\lambda=\frac{1242 \mathrm{e} V \mathrm{~nm}}{11.2} \approx 1100 Å$
Ultraviolet region

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2006
The threshold frequency for a photo-sensitive metal is $3.3 \times 10^{14} \mathrm{~Hz}$. If light of frequency $8.2 \times 10^{14} \mathrm{~Hz}$ is incident on this metal, the cut-off voltage for the photo-electric emission is nearly

Options:
A $2 \mathrm{~V}$
B $3 \mathrm{~V}$
C $5 \mathrm{~V}$
D $1 \mathrm{~V}$

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

Solution:
$$
\text { } \begin{aligned}
V_0 & =\frac{E-V}{e} \\
& =\frac{h\left(v-v_0\right)}{e} \\
& =\frac{6.62 \times 10^{-34}\left(8.2 \times 10^{14}-3.3 \times 10^{14}\right)}{1.6 \times 10^{-19}} \\
& =\frac{6.62 \times 10^{-34}}{1.6} \times 4.9 \times 10^{33} \\
& =\frac{6.62 \times 4.9 \times 10^{-1}}{1.6}
\end{aligned}
$$
$$
V_0=2 \text { volt }
$$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2011 (Mains)
The threshold frequency of a metal with work function 6.63 eV is :

Options:
A 16×1015 Hz
B 16×1012 Hz
C 1.6×1012 Hz
D 1.6×1015 Hz

The correct answer is:
1.6×1015 Hz

Solution:
The formula to calculate the work function of the metal is given by

ϕ0=hv0 ...1

From equation (1), it follows that

6.63×1.6×10-19=6.63×10-34v0⇒v0=1.6×10-1910-34⇒v0=1.6×1015 Hz

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The threshold frequency of metal is f0. When the light of frequency 2f0 is incident on the metal plate, the maximum velocity of photoelectron is v1. When the frequency of incident radiation is increased to 5f0. the maximum velocity of photoelectrons emitted is v2. The ratio of v1 to v2 is:

Options:
A v1v2=12
B v1v2=18
C v1v2=116
D v1v2=14

The correct answer is:
v1v2=12

Solution:
Maximum kinetic energy is related to the incident frequency f and the threshold frequency f0 as:

Kmax=hf-hf0

When, f=2f0

12mv12=2hf0-hf0=hf0 ...1

When, f=5f0

12mv22=5hf0-hf0=4hf0 ...2

Hence, the ratio, v1v2=12

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (01 Feb Shift 2)
The threshold wavelength for photoelectric emission from a material is 5500 A∘. Photoelectrons will be emitted, when this material is illuminated with monochromatic radiation from a

A. 75 W infra-red lamp

B. 10 W infra-red lamp

C. 75 W ultra-violet lamp

D. 10 W ultra-violet lamp

Choose the correct answer from the options given below :

Options:
A B and C only
B A and D only
C C only
D C and D only

The correct answer is:
C and D only

Solution:
Given here, threshold wavelength λth=5500 Å=550 nm

For photoelectric emission, the wavelength of the light should be less than 550 nm.

Wavelength of infrared radiation extends from 700 nm-1 mm and that of ultraviolet radiation is 100-400 nm.

Since, UV light has less wavelength than 550 nm, so only UV would be able to emit photoelectrons.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (29 Jan Shift 1)
The tidal waves in the sea are primarily due to

Options:
A The gravitational effect of the moon on the earth
B The gravitational effect of the sun on the earth
C The gravitational effect of venus on the earth
D The atmospheric effect of the earth itself

The correct answer is:
The gravitational effect of the moon on the earth

Solution:
The correct answer is Gravitational effect of the Moon.
Key Points:
- The strongest force responsible for the phenomenon of tides is the Gravitational force of the moon.
- The gravitational pulls between the earth, moon, and sun dictate the tides. The moon, however, influences tides the most.
Important Points:
- When two masses are placed at some distance to each other then it applies force on each other. This force is called gravitational force.
- Newton's law of gravitation: According to this law- When two objects having some masses are placed at a certain distance then the gravitational force is directly proportional to the product of masses and it is inversely proportional to the square of the distance between them.
- The moon and sun are the main reason for the rising and falling of ocean tides.
- As the moon is closer to earth so the gravitational pull of the moon is more than that of the gravitational pull of the sun. So the tidal effect of the moon is more than that of the sun.
- Hence the strongest reason for Tidal waves or ocean waves in the sea is the Gravitational effect of the Moon.

Topic: Physics, Gravitation, JEE Main
The time by a photoelectron to come out after the photon strikes is approximately

Options:
A
$10^{-1} \mathrm{~s}$

B
$10^{-4} \mathrm{~s}$

C
$10^{-10} \mathrm{~s}$

D
$10^{-16} \mathrm{~s}$

The correct answer is:
$10^{-10} \mathrm{~s}$

Solution:
$10^{-10}$ sec.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2006
The time dependence of the position of a particle of mass m=2 is given by r →t=2t i^-3t2j^ . Its angular momentum, with respect to the origin, at time t=2 is:

Options:
A 36 k^
B 48 i^+j^
C -48 k^
D -34 k^-i^

The correct answer is:
-48 k^

Solution:
r→=2t i^-3t2 j^
v→=2 i^-6t j^
Momentum p→=mv→
=4 i^-12t j^
∴ angular momentum L→=r→×p→
=2t i^-3t2 j^×4 i^-12t j^
=-24t2 k^+12t2 k^
=-12t2 k^
At time t=2
L→=-48 k^

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (10 Apr Shift 2)