datavorous/entrance-exam-dataset huggingface dataset:
Pure Si at $500 \mathrm{~K}$ has equal number of electron $\left(n_e\right)$ and hole $\left(n_h\right)$ concentrations of $1.5 \times 10^{16} \mathrm{~m}^{-3}$. Doping by indium increases $n_h$ to $4.5 \times 10^{22} \mathrm{~m}^{-3}$. The doped semiconductor is of

Options:
A $n$-type with electron concentration $n_e=5 \times 10^{22} \mathrm{~m}^{-3}$
B $p$-type with electron concentration $n_e=2.5 \times 10^{10} \mathrm{~m}^{-3}$
C n-type with electron concentration $n_e=2.5 \times 10^{23} \mathrm{~m}^{-3}$
D p-type having electron concentration $n_e=5 \times 10^9 \mathrm{~m}^{-3}$

The correct answer is:
p-type having electron concentration $n_e=5 \times 10^9 \mathrm{~m}^{-3}$

Solution:
$$
\begin{aligned}
& n_i^2=n_e n_h \\
& n_e=\frac{\left(n_i\right)^2}{n_h} \\
& n_e=\frac{\left(1.5 \times 10^{16}\right)^2}{\left(4.5 \times 10^{22}\right)} \\
& n_e=5 \times 10^9 \mathrm{~m}^{-3}
\end{aligned}
$$
So, $n_{f_1} \gg n_e$ semiconductor is $p$-type.

Topic: Physics, Semiconductors, NEET, NEET 2011 (Mains)
Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol-1 . The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46 g mol-1 ]

Among the following, the option representing change in the freezing point is

Options:
A
B
C
D

The correct answer is:

Solution:
As when temperature increase, vapour pressure increases so C and D options get rejected.
ΔTf=Kf×m
273-Tf′=2×34.5460.5
So Tf′=270K

Topic: Chemistry, Solutions, JEE Main
P-V plots for two gases during adiabatic expansion are shown in figure. Plots 1 and plots 2 should correspond respectively to

Options:
A He and O2
B O2 and He
C He and Ar
D O2 and N2

The correct answer is:
O2 and He

Solution:
PV 5/3  = constant for He γ =  5 3
PV 7/5  = constant for  O 2 γ =  7 5
∴ For the same decrease in V
PO2<PHe

Topic: Chemistry, Thermodynamics (C), JEE Main
Pyroxenes are class of silicate minerals, which exhibit a polymeric chain structure, as shown below

Its simplest repeating unit is

Options:
A [SiO4]4-
B [SiO3]2-
C [Si2O7]6-
D [Si4O11]6-

The correct answer is:
[SiO3]2-

Solution:
In pyroxenes, there is a total of 3 oxygen atoms per silicon atom, so the simplest repeating unit is SiO 3 2− or SiO 3 n 2n−

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Quartz is a crystalline variety of

Options:
A Silicon carbide
B Sodium silicate
C Silica
D Silicon

The correct answer is:
Silica

Solution:
Quartz is a crystalline variety of silica

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Quercus species are the dominant component in

Options:
A Tropical rain forests
B Temperate deciduous forests
C Alpine forests
D Scrub forests

The correct answer is:
Temperate deciduous forests

Solution:
Temperate deciduous forest
Tropical rain forest - Dipterocarpus, Hopea
Scrub forest - Oak, Eucalyptus
Alpine forest - Rhododendron, Juniperus

Topic: Biology, Organisms and Populations, NEET, NEET 2008 (Mains)
Ques:
$\sigma$ is the uniform surface charge density of a thin spherical shell of radius $R$. The electric field at any point on the surface of the spherical shell is :

Options:
A $\sigma / \epsilon_o R$
B $\sigma / \epsilon_o$
C $\sigma / 4 \epsilon_o$
D $\sigma / 2 \epsilon_o$

The correct answer is:
$\sigma / \epsilon_o$

Solution:
By Gauss law $\int \overrightarrow{\mathrm{E}} \cdot \mathrm{d} \overrightarrow{\mathrm{A}}=\frac{\mathrm{q}_{\mathrm{in}}}{\varepsilon_0}$
$\begin{aligned}
& \mathrm{EdA}=\frac{\sigma \times \mathrm{dA}}{\varepsilon_0} \\
& \mathrm{E}=\frac{\sigma}{\varepsilon_0}
\end{aligned}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (06 Apr Shift 1)
Question 1: In how many ways can you put 7 letters into their respective envelopes such that exactly 3 go into the right envelope?

Options:
A $24$
B $102$
C $315$
D $530$

The correct answer is:
$315$

Solution:
No of ways in which the 3 correct envelopes can be selected $={ }^7 C_3=35$ Derangement of 4 envelopes \& letters = 9 (derangement value for 4 is 9 ) Total No of ways of arrangement $=$ $9 \times 35=315$

Topic: Mathematics, Permutation Combination, JEE Main
Question: How is the observation that charges appear on both a glass rod and a silk cloth when rubbed consistent with the law of conservation of charge? Options:

Options:
A Charges are created during rubbing, violating the law of conservation of charge.
B Charges on the glass rod and silk cloth cancel each other out, maintaining charge neutrality.
C Charges are transferred from one body to another during rubbing, preserving the total charge in the system.
D The law of conservation of charge is not applicable in this scenario.

The correct answer is:
Charges are transferred from one body to another during rubbing, preserving the total charge in the system.

Solution:
Correct Option is : (C)
Charges are transferred from one body to another during rubbing, preserving the total charge in the system.

Topic: Chemistry, Electrostatics, NEET
Question: Let $f(x)$ be a continuous function on the interval $[a, b]$, and let $c$ be a constant such that $a \lt c \lt b$. Which of the following statements is true regarding the definite integrals $\int_a^b f(x) d x$ and $\int_a^c f(x) d x$ ?

Options:
A $\int_a^b f(x) d x=\int_a^c f(x) d x$
B $\int_a^b f(x) d x \lt \int_a^c f(x) d x$
C $\int_a^b f(x) d x\gt\int_a^c f(x) d x$
D The relationship between the two integrals cannot be determined without additional information.

The correct answer is:
$\int_a^b f(x) d x\gt\int_a^c f(x) d x$

Solution:
The correct answer is $\int_a^b f(x) d x\gt\int_a^c f(x) d x$.
Since $c$ is between $a$ and $b$, the area under the curve of $f(x)$ from $a$ to $c$ is a part of the area from $a$ to $b$. Therefore, $\int_a^b f(x) d x$ must be greater than $\int_a^c f(x) d x$.

Topic: Mathematics, Definite Integration, JEE Main
Question: When viewing through a magnifying glass, does the angular magnification change if the eye is moved back?

Options:
A Yes, it increases significantly.
B No, it remains the same.
C Yes, it decreases a little.
D Yes, it decreases significantly.

The correct answer is:
Yes, it decreases a little.

Solution:
Correct Option is : (C)
Yes, it decreases a little.

Topic: Chemistry, Ray Optics, NEET
Question: Why does a bird perching on a bare high-power line remain unharmed while a man standing on the ground touching the same line receives a fatal shock?

Options:
A Birds have insulating feathers.
B The man has a lower resistance than the bird.
C Current passes only when there is a difference in potential.
D Birds are immune to electric shocks.

The correct answer is:
Current passes only when there is a difference in potential.

Solution:
Correct Option is : (C)
Current passes only when there is a difference in potential.

Topic: Chemistry, Capacitance, NEET
∑r=0nr2r+1nCr is equal to

Options:
A 2n-1n2+n+2-1n+1
B 2n-1n2-n-2+1n+1
C 2n-1n2-n+2-1n+1
D 2n-1n2+n-2+1n+1

The correct answer is:
2n-1n2-n+2-1n+1

Solution:
∑r=0nr2r+1nCr=∑r=0nr2-1+1r+1nCr
=∑r=0nr-1+1r+1nCr=∑r=0nr⋅nCr-∑r=0nnCr+∑r=0n1r+1nCr
1r+1nCr=1n+1n+1Cr+1 and r⋅nCr=n·n-1Cr-1
=∑r=1nn·n-1Cr-1-2n+∑r=0n1n+1·n+1Cr+1
=nn-1C0+n-1C1+...+n-1Cn-1-2n+1n+1n+1C1+n+1C2+...+n+1Cn+1
=n⋅2n-1-2n+1n+12n+1-1
=1n+12n+1-1+2n-1n-2n+1=2n-1n2-n+2-1n+1

Topic: Mathematics, Binomial Theorem, JEE Main
∑r=0nrn-rnCr2 is equal to

Options:
A n22n-1Cn
B n22nCn-1
C n22n-1Cn-1
D n22n-2Cn

The correct answer is:
n22n-2Cn

Solution:
∑r=0nr⋅nCrn-rnCn-r
=∑r=1n-1n⋅n-1Cr-1n·n-1Cn-r-1
=n2∑r=1n-1 n-1Cr-1·n-1Cn-r-1
=n2∑r=1n-1n-1Cr-1n-1Cr
=n2n-1C0n-1C1+n-1C1n-1C2+.....+n-1Cn-2n-1Cn-1
=n2× coefficient of xn-2 in 1+xn-11+xn-1
=n22n-2Cn-2=n22n-2Cn

Topic: Mathematics, Binomial Theorem, JEE Main
∑r=1nrr4+r2+1 is equal to

Options:
A n2+n2n2+n+1
B n2+2n2n2+n+1
C 2n2+n2n2+n+1
D n2+nn2+n+1

The correct answer is:
n2+n2n2+n+1

Solution:
Let, Ιr=rr4+r2+1=rr2+r+1r2-r+1
=12r2+r+1-r2-r+1r2+r+1r2-r+1
=-121r2+r+1-1r2-r+1
=-12Vr-Vr-1
⇒∑r=1nΙr=-12Vn-V0
=-121n2+n+1-1
=121-1n2+n+1=12n2+nn2+n+1

Topic: Mathematics, Sequences and Series, JEE Main
$r$ and $n$ are positive integers $r>1, n>2$ and coefficient of $(r+2)^{\text {th }}$ term and $3 r^{\text {th }}$ term in the expansion of $(1+x)^{2 n}$ are equal, then $n$ equals

Options:
A
3r

B
3r + 1

C
2r

D
2r + 1

The correct answer is:
2r

Solution:

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2002
RC H 2 OH → KMn O 4 /O H − P. What is the compound P?

Options:
A

RCOOH

B

RCHO

C

RCH2OH

D

None of these

The correct answer is:

RCOOH

Solution:
RC H 2 OH â KMn O 4 /O H â RâCH=O â KMn O 4 RâCOOH

Primary alcohols are oxidized to carboxylic acids with KMnO4 in neutral, acidic or basic medium.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$\mathrm{R}-\mathrm{CH}_2-\mathrm{CCl}_2-\mathrm{R} \xrightarrow{\text { Reagent }} \mathrm{R}-\mathrm{C} \equiv \mathrm{C}-\mathrm{R}$ The reagent is

Options:
A $\mathrm{Na}$
B $\mathrm{HCl}$ and $\mathrm{H}_2 \mathrm{O}$
C $\mathrm{KOH}$ in $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
D $\mathrm{Zn}$

The correct answer is:
$\mathrm{KOH}$ in $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$

Solution:
$\mathrm{R}-\mathrm{CH}_2-\mathrm{CCl}_2-\mathrm{R} \xrightarrow{\mathrm{KOH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}$ $R-C \equiv C-R+2 \mathrm{HCl}$
This reaction is an example of dehydrohalogenation Hence, alcoholic $\mathrm{KOH}$ is used as a reagent.

Topic: Chemistry, Hydrocarbons, JEE Main
$\mathrm{R}-\mathrm{CH}_2-\mathrm{CCl}_2-\mathrm{R} \xrightarrow{\text { Reagent }} \mathrm{R}-\mathrm{C} \equiv \mathrm{C}-\mathrm{R}$
The reagent is

Options:
A $\mathrm{Na}$
B $\mathrm{HCl}$ and $\mathrm{H}_2 \mathrm{O}$
C $\mathrm{KOH}$ in $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
D $\mathrm{Zn}$

The correct answer is:
$\mathrm{KOH}$ in $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$

Solution:
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3+\mathrm{HBr} \xrightarrow{\text { Peroxide }}$

Topic: Chemistry, Hydrocarbons, JEE Main
$\mathrm{R}-\mathrm{CH}=\mathrm{CH}_2+\mathrm{CO}+\mathrm{H}_2 \xrightarrow[\text { High Pressure }]{\text { High Temp }} \mathrm{RCH}_2 \mathrm{CH}_2 \mathrm{CHO}$.
The above reaction is

Options:
A Mendius reaction
B Oxo process
C Sandorn's reaction
D Stephen's reaction

The correct answer is:
Oxo process

Solution:
when reaction between any alkene and water gas $(\mathrm{CO}+\mathrm{H} 2)$ is occur at high temperature and high pressure the product will be an aldehyde and this process is called oxo process.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
R-CN (i) DIBAL-H(ii)H2O⟶R-Y Consider the above reaction and identify "Y"

Options:
A -CH2NH2
B -CONH2
C - CHO
D -COOH

The correct answer is:
- CHO

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (27 Jul Shift 2)
R. M. S velocity of oxygen molecules at N.T.P is \( 0.5 \mathrm{~km} \mathrm{~s}^{-1} \). The R.M.S velocity for the hydrogen molecule at N.T.P is

Options:
A \( 4 \mathrm{~km} \mathrm{~s}^{-1} \)
B \( 2 \mathrm{~km} \mathrm{~s}^{-1} \)
C \( 3 \mathrm{~km} \mathrm{~s}^{-1} \)
D \( 1 \mathrm{~km} \mathrm{~s}^{-1} \)

The correct answer is:
\( 2 \mathrm{~km} \mathrm{~s}^{-1} \)

Solution:
The root mean square velocity is given by :
Vrms=3RTM
∴ Vrms∝ 1M
∴ Vo2VH2=232=14
∴ VH2=4 Vo2= 2 km s-1

Topic: Physics, Kinetic Theory of Gases, JEE Main
rNa+=195 pm and rCl-=281 pm in NaCl (rock salt) structure. What is the shortest distance between Na+ ions?

Options:
A 778.3 pm
B 673.06 pm
C 195.7 pm
D 390.3 pm

The correct answer is:
673.06 pm

Solution:
The shortest distance between two Na+ion=12 face-diagonal
=12×22rNa++rcr
=2(195+281) pm=673.06 pm

Topic: Chemistry, Solid State, JEE Main
$\mathrm{R}-\mathrm{OH}+\mathrm{HX} \rightarrow \mathrm{R}-\mathrm{X}+\mathrm{H}_2 \mathrm{O}$
In the above reaction, the reactivity of different alcohols is

Options:
A Tertiary > Secondary > Primary
B Tertiary < Secondary < Primary
C Tertiary < Secondary > Primary
D Secondary < Primary < Tertiary

The correct answer is:
Tertiary > Secondary > Primary

Solution:
$R-\mathrm{OX}+\mathrm{HX} \rightarrow \mathrm{R}-\mathrm{X}+\mathrm{H}_2 \mathrm{O}$
Reactivity order of alcohols for this reaction
$3^{\circ}\gt2^{\circ}\gt1^{\circ}$
Reactivity order of halogen acids
$\mathrm{R}-\mathrm{I}\gt\mathrm{R}-\mathrm{Br}\gt\mathrm{R}-\mathrm{Cl}$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
RAAS is a system involving many organs. Which of these statements about RAAS is false?

Options:
A It operates in response to decrease in blood pressure or blood volume
B The JG cells release an enzyme renin which converts angiotensinogen into a peptide angiotensin I
C Angiotensin II inhibits the release of aldosterone from adrenal gland
D Angiotensin II increases blood pressure by causing arterioles to constrict

The correct answer is:
Angiotensin II inhibits the release of aldosterone from adrenal gland

Solution:
RAAS - Renin Angiotensin Aldosterone System is a specialized response of the body in order to maintain the optimal blood pressure of the body. When the blood pressure falls it is sensed by the JG cells in the kidney. They release a hormone called Renin in response to this. Renin converts Angiotensinogen in to Angiotensin I. Angiotensin I is converted in Angiotensin II by enzyme ACE. Angiotensin II causes vascular constriction and stimulates adrenal cortex to release Aldosterone which causes water and salt retention and thus increase the blood pressure.

Topic: Biology, Excretory Products and their Elimination, NEET
Racemic mixture is formed by mixing two

Options:
A
isomeric compounds

B
chiral compounds

C
meso compounds

D
optical isomers

The correct answer is:
optical isomers

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2002
Racemose inflorescence is observed in the family:

Options:
A Oleaceae
B Liliaceae
C Solanaceae
D Fabaceae

The correct answer is:
Fabaceae

Solution:
In racemose type of inflorescence, the main axis continues to grow, the flowers are borne laterally in an acropetal succession. This type of inflorescence is observed in the members of the Fabaceae family, e.g., soybeans, groundnuts, Sesbania, Indigofera, etc.

Topic: Biology, Morphology of Flowering Plants, NEET
Radar waves are sent towards a moving airplane and the reflected waves are received. When the airplane is moving towards the radar, the wavelength of the wave

Options:
A decrease
B increase
C remains the same
D sometimes increase or decrease

The correct answer is:
decrease

Solution:
By Dopplers effect, if source moves towards the observer, frequency received will increase.
As $v \propto \frac{1}{\lambda}$
$\therefore$ wavelength will decrease.

Topic: Physics, Magnetic Properties of Matter, VITEEE, VITEEE 2007
Radial nodes present in 3s and 2p-orbitals are respectively

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

The correct answer is:
2, 0

Solution:
Number of radial nodes =(n-l-1)
For 3s, n=3,l=0 (Number of radial node =3-0-1=2 )
For 2p,
n=2,l=1 Number of radial node =2-1-1=0

Topic: Chemistry, Structure of Atom, JEE Main
Radiation coming from transitions n=2 to n=1 of hydrogen atoms fall on He+ ions in n=1 and n=2 states. The possible transition of helium ions as they absorb energy from the radiation is:

Options:
A n=2⟶n=3
B n=2⟶n=4
C n=2⟶n=5
D n=1⟶n=4

The correct answer is:
n=2⟶n=4

Solution:
E=13.6111-122 eV ...(1)
For Helium:
E=13.6221n12-1n22 eV ...(2)
From equation (1) and (2)
13.6 34=13.6 4 1n12-1n22
1n12-1n22=316
From option, n1=2 and n2=4 are satisfying the equation.

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2019 (08 Apr Shift 1)
Radiation corresponding to the transition n = 4 to n = 2 in hydrogen atoms falls on a certain metal (work function = 2.0 eV ). The maximum kinetic energy of the photoelectrons will be:

Options:
A 0.55 eV
B 2.55 eV
C 4.45 eV
D None

The correct answer is:
0.55 eV

Solution:
En=-13.6n2eV ; E2=-13.622 ; E4=-13.642eV/atom
ΔE=E4-E2=2.55 eV
Absorbed energy = work function of metal + KE
2.55 = 2 + K.E.
K.E. = 0.55 eV

Topic: Chemistry, Structure of Atom, JEE Main
Radiation of wavelength 300 nm and intensity 100 W m-2 falls on the surface of a photosensitive material. If 2% of the incident photons produce photo electron, the number of photoelectrons emitted from an area of 2 cm2 of the surface is nearly

Options:
A 15×1011
B 6.04×1014
C 1.5×1012
D 60.4×1015

The correct answer is:
6.04×1014

Solution:
Power of radiation is expressed as, Power=Nhcλ

Here, N is the number of photons per unit time per unit area.

⇒P=Nhcλ⇒N=Pλhc=100 W m-2×300×10-9 m6.6×10-34 J s×3×108 m s-1=1.51×1020 m-2 s-1

Given that 2% of the incident, photons produce photoelectron therefore, number of photoelectrons produced by photons is =2100×1.51×1020 m-2 s-1

Now, the number of photoelectrons emitted from an area of 2 cm2 of the surface in 1 s is nearly

n=1 s2×10-4 m2×3.02×1018 m-2 s-1n=6.04×1014

Topic: Physics, Dual Nature of Matter, JEE Main
Radiation of wavelength Î» is incident on a photocell. The fastest emitted photoelectron has a speed v. If the wavelength is changed to 3Î»4, the speed of the fastest emitted photoelectron will be

Options:
A =v4312
B =v3412
C >v4312
D <v4312

The correct answer is:
>v4312

Solution:
12mv2=hcλ-ϕ ...(i)
12mv'2=43hcλ-ϕ ...(ii)
From Eqn. (i)

hcλ=12mv2+ϕ
On putting this Equation (ii)
12mv′2=4312mv2+ϕ-ϕ
v'2=43v2+2ϕ3m
⇒v'>43v

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2016 (03 Apr)
Radiation of wavelength Î» is incident on a photocell. The fastest emitted photoelectron has a speed v. If the wavelength is changed to 3Î»4, the speed of the fastest emitted photoelectron will be

Options:
A =v4312
B =v3412
C >v4312
D <v4312

The correct answer is:
>v4312

Solution:
12mv2=hcλ-ϕ ...(i)
12mv'2=43hcλ-ϕ ...(ii)
From Eqn. (i) hcλ=12mv2+ϕ
On putting this Eqn. (ii)
12mv′2=4312mv2+ϕ-ϕ
v'2=43v2+2ϕ3m
⇒v'>43v

Topic: Physics, Dual Nature of Matter, JEE Main
Radiation, with wavelength 6561 Å falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of 3×10-4 T . If the radius of the largest circular path followed by the electrons is 10 mm , the work function of the metal is close to:

Options:
A 1.6 eV
B 0.8eV
C 1.1 eV
D 1.8eV

The correct answer is:
1.1 eV

Solution:
KEmax=E-ϕ
=12400λinÅ-ϕ (in eV)
∴r=2mKEeB
KEmax=r2e2B22m (in J)
=r2eB22m (in eV)
∴ϕ=124006556-r2eB22m
=1.1 eV

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (09 Jan Shift 1)
Radio carbon dating is done by estimating in specimen the

Options:
A Amount of ordinary carbon still present
B Amount of radio carbon still present
C Ratio of amount of 14C6 to 12C6 still present
D Ratio of amount of 12C6 to 14C6 still present

The correct answer is:
Ratio of amount of 14C6 to 12C6 still present

Solution:
Radiocarbon dating relies on a simple natural phenomenon. As the earth's upper is bombarded by cosmic radiation, atmospheric nitrogen is broken down into an unstable isotope of carbon-carbon (C-14).
The unstable isotope is brought to earth by atmospheric activity, such as storms, and becomes fixed in the biosphere. Because it reacts identically to C-12 and C-13, C-14 attached to complex organic molecules through photosynthesis in plants and becomes their molecular makeup. Animals eating those plants in turn absorb carbon-14 as well as stable isotopes. This process of ingesting C-14 continues as long as the plant or animal remains alive.
The C-14 within an organism is continually decaying into stable carbon isotopes, but organism is absorbing more C-14 during its life, the ratio of C-14 to C-12 remains about same as the ratio in the atmosphere. Where the organism dies, the ratio of C-14 within its carcass begins to gradually decrease.

Topic: Physics, Nuclear Physics, NEET
Radio waves coming at ∠α to vertical are received by radar after reflection from a nearby water surface and directly. The height of the antenna from the water surface so that it records a maximum intensity, should be (wavelength =λ)

Options:
A Î» 4cos Î±
B 2λ 4cos α
C 3λ 6cos α
D 5λ 7cos α

The correct answer is:
λ 4cos α

Solution:
Path difference = BN + BA + Î» 2
(because reflection occurs from denser medium)
In ÎACB
hAB=cos Î±
AB=hcos Î±....(1)
from right angle ÎANB
BN AB = cos 2 Î±
BN=ABâcos2Î± ....(2)
Path difference = AB + AB cos 2 Î± + Î» 2
Î» = AB 1 + cos 2 Î± + Î» 2
Î»2=hcos Î±Ã2cos2Î±
Î»2=2h cos Î±
h=Î»4 cos Î±

Topic: Physics, Wave Optics, JEE Main
Radioactivity of a sample ( $\mathrm{z}=22$ ) decreases $90 \%$ after 10 years. What will be the half life of the sample?

Options:
A 5 years
B 2 years
C 3 years
D 10 years

The correct answer is:
3 years

Solution:
Hints: $t=10$ yrs $\quad t_{\frac{1}{2}}=?$
$$
\lambda=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{N}_0}{\mathrm{~N}_{\mathrm{t}}}
$$
Since radioactivity decreases $90 \%$ in 10 yrs. $\Rightarrow N_0=100 \& N_t=10$
Thus $\lambda=\frac{2.303}{10} \log \frac{100}{10} \Rightarrow \lambda=\frac{2.303}{10}$
since $\mathrm{t}_{\frac{1}{2}}=\frac{0.693}{\lambda}=\frac{2.303 \times \log 2}{\lambda} \Rightarrow \mathrm{t}_{\frac{1}{2}}=\frac{2.303 \times \log 2}{2.303 / 10}$
$\Rightarrow \mathrm{t}_{\frac{1}{2}}=(\log 2) \times 10 \simeq 3$ years

Topic: Chemistry, Chemical Kinetics, JEE Main
Radius of a certain orbit of hydrogen atom is $8.48 Ã…$. If energy of electron in this orbit is $E / x$. then $x=$ _____
(Given $\mathrm{a}_0=0.529 Ã…, E=$ energy of electron in ground state).

Options:

The correct answer is:
16

Solution:
We know
$\begin{aligned}
& \mathrm{r}=0.529 \frac{\mathrm{n}^2}{\mathrm{Z}} \Rightarrow 8.48=0.529 \frac{\mathrm{n}^2}{1} \\
& \mathrm{n}^2=16 \Rightarrow \mathrm{n}=4
\end{aligned}$
We know
$\mathrm{E} \propto \frac{1}{\mathrm{n}^2}$
$\begin{aligned}
& E_{n^{\text {th }}}=\frac{E}{16} \\
& x=16
\end{aligned}$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (06 Apr Shift 1)
Radius of a conductor increases uniformly from left end to right end as shown in figure.

Material of the conductor is isotropic and its curved surface is thermally isolated from surrounding. Its ends are maintained at temperatures T1 and T2(T1>T2). If, in steady state, heat flow rate is equal to H, then which of the following graphs is correct?

Options:
A
B
C
D

The correct answer is:

Solution:
Since the curved surface of the conductor is thermally insulated, in steady state, the rate of flow of heat at every section will be the same. Hence, the curve between H and x will be a straight
line parallel to x-axis.

Topic: Physics, Thermal Properties of Matter, NEET
Radius of curvature of concave mirror is $40 \mathrm{~cm}$ and the size of image is twice as that of object, then the object distance is

Options:
A $60 \mathrm{~cm}$
B $20 \mathrm{~cm}$
C $40 \mathrm{~cm}$
D $30 \mathrm{~cm}$

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

Solution:
Focal length, \(\mathrm{f}=-20 \mathrm{~cm}\)
Magnification, \(\mathrm{m}=-2\)
\(\begin{aligned}
& \Rightarrow \frac{-\mathrm{v}}{\mathrm{u}}=-2 \\
& \mathrm{v}=2 \mathrm{u} \\
& \text {Now, } \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\
& \frac{1}{2 \mathrm{u}}+\frac{1}{\mathrm{u}}=\frac{1}{-2 0} \\
& \frac{3}{24}=\frac{-1}{2 0} \\
& \therefore \mathrm{u}=-3 \mathrm{~cm}
\end{aligned}\)

Topic: Physics, Ray Optics, JEE Main
Radius of gyration of a body depends upon

Options:
A axis of rotation
B translational motion
C shape of the body
D area of the body

The correct answer is:
axis of rotation

Solution:
Radius of gyration of a body depends on the axis of rotation.

Topic: Physics, Rotational Motion, JEE Main
Radius of the first Bohr's orbit of hydrogen atom is

Options:
A $1.06 Å$
B $0.22 Å$
C $0.28 Å$
D $0.53 Å$

The correct answer is:
$0.53 Å$

Solution:
Bohr's radius of the hydrogen atom
$r=\frac{n^2 \times 0.529 A}{z} ;$ where $z=$ Atomic number,
$n=$ Number of orbitals

Topic: Chemistry, Structure of Atom, JEE Main