datavorous/entrance-exam-dataset huggingface dataset:
Assume that each born child is equally likely to be a boy or girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is:

Options:
A 112
B 110
C 111
D 117

The correct answer is:
111

Solution:
Family has four children in total.

∴ Total number of ways in which at least two girls are there

is 4C2+4C3+4C4=11

Favourable case in which all are girls is =1

∴ P=111

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (10 Apr Shift 1)
Assume that each diode as shown in the tịgure has a forvard bias resistance of $50 \Omega$ and an infinite reverse bias resistance. The current through the resistance $150 \Omega$ is

Options:
A $0.66 \mathrm{A}$
B $0.05 \mathrm{A}$
C zero
D $0.04 \mathrm{A}$

The correct answer is:
$0.04 \mathrm{A}$

Solution:
The circuit is

As the lower diode attached to $100 \Omega$ resistance is in reversed biased so, $1 t$ is non-conducting
Now, the circuit can be redrawn as,

$\therefore$ Current through the circuit, $I=\frac{V}{R}$
$$
=\frac{10}{250}=0.04 \mathrm{A}
$$

Topic: Physics, Semiconductors, JEE Main
Assume that protons and neutrons have equal masses. Mass of a nucleon is 1.6×10-27 kg and radius of nucleus is 1.5×10-15 A13 m. The approximate ratio of the nuclear density and water density is n×1013. The value of n is _____.

Options:

The correct answer is:
11

Solution:
As we know, density of nuclei= mass of nuclei volume of nuclei

⇒ρ=1.6×10-27A43π1.5×10-153A

=1.6×10-2714.14×10-45=0.113×1018 kg m-3

We also know that, density of water is ρw=103 kg m-3.

Hence,
ρρw=11.31×1013.

Therefore, n≈11

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Assume that the displacement (s) of air is proportional to the pressure difference Δp created by a sound wave. Displacement (s) further depends on the speed of sound (v), density of airρ and the frequency (f). If Δp~10 Pa,n~300 m/s,p~1 kg/m3 f~1000Hz, then s will be of the order of (take the multiplicative constant to be 1 )

Options:
A 3100mm
B 10 mm
C 110mm
D 1 mm

The correct answer is:
3100mm

Solution:
S=P0BK

=P0PV2ωV

=P0PVω

=101×1000×300m

≃3100mm

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2020 (05 Sep Shift 1)
Assume that, the drag force on a football depends only on the density of the air, velocity of the ball and the cross-sectional area of the ball. Balls of different sizes but the same density are dropped in an air column. The terminal velocity reached by balls of masses 250 g and 125 g are in the ratio

Options:
A 21/6
B 21/3
C 21/2
D 22/3

The correct answer is:
21/6

Solution:
Given, drag force Fd depends on air density σ, velocity of ball v and area of cross-section of ball A.
∴Fd∝σavbAc

Substituting dimensions of different physical quantities, we have
MLT-2=ML-3aLT-1b L2c

Equating dimensions of fundamental quantities, we have

a=1....i

-3a+b+2c=1....ii

-b=-2⇒b=2....iii

Putting the values of a and b in Eq (ii), we get

⇒-3×1+2+2c=1⇒c=1

∵a=1,b=2,c=1

So, drag force on football is

Fd=k·σ·v2·A

When football reaches terminal speed, its weight is balanced by drag force.

⇒mg=Fd

⇒mg=kσvT2A

where, uT=terminal speed of football.

⇒mg=kσvT2πR2

⇒mg=kσuT2πm43πρ23

[ from m=43πR3ρ where ρ= density of football]

⇒vT∝m1/6

⇒v1v2=m1m216=25012516=21/6

Topic: Physics, Units and Dimensions, JEE Main
Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is 100 g. The time period of the motion of the particle will be (approximately)

(take g=10 ms-2, radius of earth =6400 km)

Options:
A 24 hours
B 1 hour 24 minutes
C 1 hour 40 minutes
D 12 hours

The correct answer is:
1 hour 24 minutes

Solution:
The gravitational force acting on a particle of mass m at a distance r (<R) from the centre of Earth is

F=-GM'mr2 ....(1)

where M' is the mass of portion of Earth enclosed by a sphere of radius r concentric with Earth. Here - sign indicates radially inward direction of force.

Taking the density of Earth to be uniform

M'4πr3=M4πR3

⇒M'=Mr3R3 ....(2)

Substituting M' from equation 2 into equation 1:

F=-GMmR3 r

Which is similar to the equation of motion F=-kx with k=GMmR3=mgR (∵g=GMR2)

For such motion, the time period is given by:

T=2πmk

Substituting k from above,

T=2πRg

Putting the values, we get

T=2×3.146400×10310=2×3.14×800≈1 hour 24 minutes

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Assume that the earth moves around the sun in a circular orbit of radius $R$ and there exists a planet which also move around the sun in a circular orbit with an angular speed twice as large as that of the earth. The radius of the orbit of the planet is

Options:
A $2^{-2 / 3} R$
B $2^{2 / 3} R$
C $ 2^{-1 / 3} R$
D $\frac{R}{\sqrt{2}}$

The correct answer is:
$2^{-2 / 3} R$

Solution:
According to the Kepler's third law
$$
T^{2} \propto r^{3}
$$
where, $T=$ time period of revolution
$r=$ radius
Now, $\left(\frac{T_{E}}{T_{P}}\right)^{2}=\left(\frac{r_{E}}{r_{P}}\right)^{3}=\frac{r_{E}}{r_{P}}=\left(\frac{T_{E}}{T_{P}}\right)^{2 / 3}$
$$
\frac{r_{E}}{r_{P}}=\left(\frac{2 \pi / \omega_{E}}{2 \pi / \omega_{P}}\right)^{2 / 3} \quad\left[\because T=\frac{2 \pi}{\omega}\right]
$$
$$
\frac{r_{E}}{r_{P}}=\left(\frac{\omega_{P}}{\omega_{E}}\right)^{2 / 3}
$$
According to the question, $\omega_{P}=2 \omega_{E}$ and $r_{E}=R$
$\Rightarrow \quad \frac{R}{r_{P}}=\left(\frac{2 \omega_{E}}{\omega_{E}}\right)^{2 / 3}$
$$
\begin{array}{l}
\frac{R}{r_{P}}=(2)^{2 / 3} \\
r_{P}=\frac{R}{(2)^{2 / 3}}=R(2)^{-2 / 3}
\end{array}
$$

Topic: Physics, Gravitation, JEE Main
Assume that the nuclear binding energy per nucleon (BA) versus mass number (A) is as shown in the figure. Use this plot to choose the correct choice given below.

Options:
A Fusion of two nuclei with mass numbers lying in the range of 1 < A < 50 will release energy.
B Fusion of two nuclei with mass numbers lying in the range of 51 < A < 100 will consume energy.
C Fission of a nucleus lying in the mass range of 100 < A < 200 will release energy when broken into two equal fragments.
D Fission of a nucleus lying in the mass range of 200 < A < 260 will release energy when broken into two equal fragments.

The correct answer is:
Fission of a nucleus lying in the mass range of 200 < A < 260 will release energy when broken into two equal fragments.

Solution:
In fusion, two or more lighter nuclei combine to make a comparatively heavier nucleus.
In fission, a heavy nucleus breaks into more than one comparatively lighter nucleus.
Further, energy will be released in a nuclear process if total binding energy increases.
Therefore Correct option is that the fission of a nucleus lying in the mass range of 200 < A < 260 will release energy when broken into two equal fragments.

Topic: Physics, Nuclear Physics, JEE Main
Assume that the radius of the first Bohr orbit of hydrogen atom is 0.6Å. The radius of the third Bohr orbit of He+ is _____ picometer. (Nearest Integer)

Options:

The correct answer is:
270

Solution:
For hydrogen like species, from Bohr's formulas we get,
r∝n2Z
where n is orbit number, Z is atomic number

Thus,

⇒rHe*=rH×n2Z

⇒rHe+=0.6×322

=2.7Å=270 × 10-12 m

∴rHe+=270pm

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Assume there are two identical simple pendulum Clocks-1 is placed on the earth and Clock-2 is placed on a space station located at a height h above the earth surface. Clock-1 and Clock-2 operate at time periods 4 s and 6 s respectively. Then the value of h is -
(consider radius of earth RE=6400 km and g on earth 10 m s-2)

Options:
A 1200 km
B 1600 km
C 3200 km
D 4800 km

The correct answer is:
3200 km

Solution:
The time period of a simple pendulum is given by, T=2πlg.

Therefore, T∝1g.

The acceleration due to gravity at height h from the surface of the earth is given by, g'=gR+h2

Therefore,

T1T2=g2g1⇒46=RR+h⇒h=R2=3200 km

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Assuming 1μg of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is_____×10-1μg. [Given : ln10=2.303; log2=0.30]

Options:

The correct answer is:
1

Solution:
λt=lnaa-x

we know that half life time t1/2 = ln2λ

So, λ = ln2t1/2

100=30ln2ln1w100=30log2log1w100=300.30log1wlog1w = 1

1w=10

W=0.1μg

=1×10-1μg

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2022 (29 Jul Shift 2)
Assuming 2s â 2p mixing is NOT operative, the paramagnetic species among the following is

Options:
A Be2
B B2
C C2
D N2

The correct answer is:
C2

Solution:
Assuming that no 2s-2p mixing takes place
(i) Be2→σ1s2, σ*1s2, σ2s2, σ*2s2 (diamagnetic)
(ii) B2→σ1s2, σ*1s2, σ2s2, σ*2s2,σ2pz2, π2px0π2py0 (diamagnetic)
(iii) C2→σ1s2, σ*1s2, σ2s2, σ*2s2, σ2pz2,π2px1π2py1, π*2px0π*2py0 , σ*2pz0 (paramagnetic)
(iv) N2→σ1s2, σ*1s2, σ2s2, σ*2s2, σ2pz2, π2px2π2py2 , π*2px0π*2py0 , σ*2pz0 (diamagnetic)

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Assuming complete ionization, same moles of which of the following compounds will require the least amount of acidified KMnO4 for complete oxidation?

Options:
A FeSO4
B FeSO3
C FeC2O4
D FeNO22

The correct answer is:
FeSO4

Solution:
MnO4-⟶Mn2+ ; Change in oxidation no. = 5
In option,
i. Fe2+ ⟶Fe3+
Change in oxidation no. = 1
ii. Fe2+ ⟶Fe3+
Change in oxidation no. = 1
SO32-⟶SO42-
Change in oxidation no. = 2
=1+2=3
iii. Fe2+ ⟶Fe3+
Change in oxidation no. = 1
C2O42-⟶2CO2
Change in oxidation no. = 2
=1+2=3
iv. Fe2+ ⟶Fe3+
Change in oxidation no. = 1
2NO2-⟶2NO3-
Change in oxidation no. = 4
=1+4=5

Topic: Chemistry, Redox Reactions, JEE Main
Assuming f to be the frequency of the electromagnetic wave corresponding to the first line in Balmer series, the frequency of the immediate next line is

Options:
A 0.5f
B 1.35f
C 2.05f
D 2.70f

The correct answer is:
1.35f

Solution:
For Balmer series, nf =2 and ni=3,4,5,....
Frequency, of 1st spectral line of Balmer series
f = RZ2 c122-132
or f=RZ2 cÃ536 ...(i)
Frequency, of 2nd spectral line of Balmer series
fâ = RZ2 c122-142
or fâ²=RZ2 cÃ316 ....(ii)
Form eqs. (i) and (ii), we have
ffâ² =2027
â´ fâ²=2720 f=1.35 f

Topic: Physics, Atomic Physics, JEE Main
Assuming fully decomposed, the volume of $\mathrm{CO}_2$ released at STP on heating $9.85 \mathrm{~g}$ of $\mathrm{BaCO}_3$ (Atomic mass of $B a=137)$ will be

Options:
A $0.84 L$
B $2.24 L$
C $4.06 L$
D $1.12 L$

The correct answer is:
$1.12 L$

Solution:
\(\mathrm{BaCO}_3 \rightarrow \mathrm{BaO}+\mathrm{CO}_2 \uparrow\)
Molecular weight of \(\mathrm{BaCO}_3=137+12+3 \times 16=197\)
\(\therefore \quad 197 \mathrm{gm}\) produces 22.4 L at S.T.P.
\(\therefore \quad 9.85 \mathrm{gm}\) produces \(\frac{22.4}{197} \times 9.85=1.12 \mathrm{~L}\) at S.T.P.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Assuming ideal gas behavior, the ratio of density of ammonia to that of hydrogen chloride at same temperature and pressure is: (Atomic weight of Cl is 35.5 u)

Options:
A 1.46
B 1.64
C 0.46
D 0.64

The correct answer is:
0.46

Solution:
d=P(M.W.)RT
dNH3dHCl= (M.W.)NH3(M.W.)HCl= 1736.5=0.46

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2018 (16 Apr Online)
Assuming that about 20 MeV of energy is released per fusion reaction 1H2+â1H3ââ0n1+â2He4, then the mass of 1H2 consumed per day in a fusion reactor of power 1 MW will approximately be

Options:
A 0.001 g
B 0.1 g
C 10.0 g
D 1000 g

The correct answer is:
0.1 g

Solution:
Energy produced U=Pt
=106×24×36×102
=24×36×108 J
The energy released per fusion reaction
=20 MeV
=20×106×1.6×10-19
=32×10-13 J
The energy released per atom of 1H2
=32×10-13 J
Number of 1H2 atoms used =24×36×10832×10-13
=27×1021
Mass of 6×1023 atom =2 g
∴ Mass of 27×1021 atoms
=26×1023×27×1021
≈0.1g

Topic: Physics, Nuclear Physics, JEE Main
Assuming that four protons combine to form a helium nucleus and two positrons each of mass 0.000549 a.m.u., calculate the energy released. Given, the mass of 1H1=1.007825 a.m.u. and mas of 2He4 =4.002603 a.m.u.

Options:
A 25.71 MeV

B 20.71 MeV

C 22.75 MeV

D 23.50 MeV

The correct answer is:
25.71 MeV

Solution:
The nuclear fusion reaction can be written as:
1H1+1H1+1H1+1H1→2He4+2 1e0+Q
If mN 1H1 and mN2He4 represent the masses of 1H1 and 2He4 nuclei respectively, then
Q=14 mN1H1-mN2He4-2me×931.5 MeV .....(i)
Where me represents the mass of the positron 1e0 .
Here, mN 2He4 4.002603 a.m.u.; me=0.0005499 a.m.u.
Substituting for mN1He1, mN2He4 and me in the equation (i), we have
Q=4 ×1.007825-4.002603-2 ×0.0000549 ×931.5
=0.027599 ×931.5=25.71 MeV

Topic: Physics, Nuclear Physics, JEE Main
Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecule $\mathrm{B}_2$ is

Options:
A
1 and diamagnetic

B
0 and diamagnetic

C
1 and paramagnetic

D
0 and paramagnetic

The correct answer is:
1 and diamagnetic

Solution:
$\mathrm{B}_2$ : Total electrons $=10$
Configuration : $\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^1=\pi 2 p_y^1$
If Hunds rule is violated, then
$$
\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \quad \pi 2 p_x^2=\pi 2 p_y^0
$$
So, diamagnetic
$$
\text { Bond order }=\frac{6-4}{2}=1
$$
Chemical bonding
Conceptual
II

Topic: Chemistry, Structure of Atom, JEE Main
Assuming that no two consecutive digits are same, the number of $n$ digit numbers, is

Options:
A $n$ !
B $9$ !
C $9^n$
D $n^9$

The correct answer is:
$9^n$

Solution:
There are $\{0,1,2, \ldots . .10\}, 10$ different digits of which $1^{\text {st }}$ place is filled in 9 ways, $2^{\text {nd }}$ place in 9 ways (since 0 can be included and first digit is not repeated) and so on $n$ times.
Thus number of ways is
$=9 \times 9 \times \ldots \ldots \ldots \ldots=9^n$

Topic: Mathematics, Permutation Combination, JEE Main
Assuming that the degree of hydrolysis is small, the $\mathrm{pH}$ of $0.1 \mathrm{M}$ solution of sodium acetate $\left(\mathrm{K}_{\mathrm{a}}=1.0 \times 10^{-5}\right)$ will be:

Options:
A
$5.0$

B
$6.0$

C
$8.0$

D
$9.0$

The correct answer is:
$9.0$

Solution:
Sodium acetate is a salt of strong base and weak acid.
$$
\begin{aligned}
&\therefore \mathrm{pH}=7+\frac{1}{2} \mathrm{p} K_a+\frac{1}{2} \log c \quad \text { where } \\
&\mathrm{p} k_a=-\log K_a=-\log 10^{-5}=5 \\
&\log \mathrm{c}=\log 10^{-1}=-1 \\
&\mathrm{pH}=7+\frac{5}{2}-\frac{1}{2}=9.0
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2014 (11 Apr Online)
Assuming that the human pupil has a radius of 0.25 cm and a comfortable viewing distance of 25 cm. The minimum separation between two point objects that the human eye can resolve for the light of wavelength 500 nm is

Options:
A 300 μm
B 1 μm
C 30 Î¼m
D 100 μm

The correct answer is:
30 Î¼m

Solution:
By Fraunhofer diffraction through a circular aperture, θ=1.22λd
d= diameter of pupil =2×0.25=0.5 cm
λ=500 nm. The first dark ring is formed by the light diffracted from the hole at an angle θ with the axis. Viewing distance, D=25 cm∴ The minimum separation between 2 objects =Dθ=25×10-2×1.22×500×10-95×10-3 m
Dθ=30×10-6 m⇒Dθ=30 μm.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2015 (04 Apr)
Assuming that the human pupil has a radius of 0.25 cm and a comfortable viewing distance of 25 cm. The minimum separation between two point objects that the human eye can resolve for the light of wavelength 500 nm is

Options:
A 300 μm
B 1 μm
C 30 Î¼m
D 100 μm

The correct answer is:
30 Î¼m

Solution:
By fraunhofer diffraction through a circular aperture, Î¸=1.22Î»d
d = diameter of pupil =2Ã0.25=0.5 cm
Î»=500 nm
First dark ring is formed by the light diffracted from the hole at an angle Î¸ with the axis.
Viewing distance, D=25 cm
â´ Minimum separation between 2 objects =DÎ¸
=25Ã10-2Ã1.22Ã500Ã10-95Ã10-1
=30Ã10-6 m
=30 Î¼m

Topic: Physics, Wave Optics, JEE Main
Assuming that water vapour is an ideal gas, the internal energy $(\Delta \mathrm{U})$ when $1\mathrm{~mol}$ of water is vapourised at $1 \mathrm{~bar}$ pressure and $100^{\circ} \mathrm{C}$, (Given: Molar enthalpy of vapourization of water at $1 \mathrm{~bar}$ and $373 \mathrm{~K}=41 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ ) will be

Options:
A
$4.100 \mathrm{~kJ} \mathrm{~mol}^{-1}$

B
$3.7904 \mathrm{~kJ} \mathrm{~mol}^{-1}$

C
$37.904 \mathrm{~kJ} \mathrm{~mol}^{-1}$

D
$41.00 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$37.904 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
$\begin{aligned}
& \mathrm{H}_2 \mathrm{O}(\ell) \stackrel{\text { vaporisation }}{\longrightarrow} \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \\
& \Delta \mathrm{n}_{\mathrm{g}}=1-0=1 \\
& \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\
& \Delta \mathrm{U}=\Delta \mathrm{H}-\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\
& =41-8.3 \times 10^{-3} \times 373 \\
& =37.9 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}$
Hence, (C) is correct.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2007
Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from $10$ white, $9$ green and $7$ black balls is

Options:
A
$880$

B
$629$

C
$630$

D
$879$

The correct answer is:
$879$

Solution:
Number of ways of selecting one or more balls from $10$ white, $9$ green, and $7$ black balls $=(10+1)(9+1)(7+1)-1=11 \times 10 \times 8-1=879.$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2012 (Offline)
Assuming the earth to be a sphere of uniform density, the acceleration due to gravity inside the earth at a distance of $r$ from the centre is proportional to

Options:
A
$r$

B
$r^{-1}$

C
$r^2$

D
$r^{-2}$

The correct answer is:
$r$

Solution:
Acceleration due to gravity at depth d from the surface of the earth or at a distance $r$ from the centre ' $O$ ' of the earth $g^{\prime}=\frac{4}{3} \pi \rho G r$ Hence $g^{\prime} \propto r$

Topic: Physics, Gravitation, JEE Main, JEE Main 2012 (12 May Online)
Assuming the earth to be a sphere of uniform mass density, a body weighed $300 \mathrm{~N}$ on the surface of earth. How much it would weigh at R/4 depth under surface of earth ?

Options:
A $75 \mathrm{~N}$
B $300 \mathrm{~N}$
C $375 \mathrm{~N}$
D $225 \mathrm{~N}$

The correct answer is:
$225 \mathrm{~N}$

Solution:
$\begin{aligned} & \text { At surface: } \mathrm{mg}=300 \mathrm{~N} \\ & \mathrm{~m}=\frac{300}{\mathrm{~g}_{\mathrm{s}}} \\ & \text { At Depth } \frac{\mathrm{R}}{4}: \mathrm{g}_{\mathrm{d}}=\mathrm{g}_{\mathrm{s}}\left[1-\frac{\mathrm{d}}{\mathrm{R}}\right] \\ & \mathrm{g}_{\mathrm{d}}=\mathrm{g}_{\mathrm{s}}\left[1-\frac{\mathrm{R}}{4 \mathrm{R}}\right] \\ & \mathrm{g}_{\mathrm{d}}=\frac{3 \mathrm{~g}_{\mathrm{s}}}{4}\end{aligned}$
weight at depth $\quad=\mathrm{m} \times \mathrm{g}_{\mathrm{d}}$
$\begin{aligned} & =\mathrm{m} \times \frac{3 \mathrm{~g}_{\mathrm{s}}}{4} \\ & =\frac{3}{4} \times 300 \\ & =225 \mathrm{~N}\end{aligned}$

Topic: Physics, Gravitation, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Assuming the earth to be a sphere of uniform mass density, the weight of a body at a depth d=R2 from the surface of earth, if its weight on the surface of earth is 200 N, will be : (Given R= radius of earth)

Options:
A 300 N
B 100 N
C 400 N
D 500 N

The correct answer is:
100 N

Solution:
The acceleration due to gravity with depth is

g'=g1-dR

The given data is

W=200 Nd=R2

Let the weight be at the given depth, W'. Hence, the value is

mg'=mg1-R2R=mg2

The Weight W' =W2=2002=100 N

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Assuming the earth to have a constant density, point out which of the following curves show the variation of acceleration due to gravity from the centre of earth to the points far away from the surface of earth

Options:
A

B

C

D None of these

The correct answer is:

Solution:
$F_{in}=\frac{G M r}{R^3}$
$F_{in}$ ${\alpha r}$
$F_{out}$ $=\frac{G M}{r^2}$
$F_{out}$ $\alpha \frac{1}{r^2}$

Topic: Physics, Gravitation, JEE Main
Assuming the gas to be ideal the work done on the gas in taking it from $A$ to $B$ is

Options:
A
$200 \mathrm{R}$

B
$300 \mathrm{R}$

C
$400 \mathrm{R}$

D
$500 \mathrm{R}$

The correct answer is:
$400 \mathrm{R}$

Solution:
$$
\begin{aligned}
& W_{A B}=\Delta Q-\Delta U={ }^{\prime} C_p d T-n C_v d T \text { (at } \\
& \text { constant pressure }) \\
& =n\left(C_p-C_v\right) d t \\
& =n R d T=2 \times R \times(500-300)=400 \mathrm{R}
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2009
Assuming the nitrogen molecule is moving with r.m.s. velocity at 400 K, the de-Broglie wave length of nitrogen molecule is close to : (Given : nitrogen molecule weight : 4.64×10−26 kg, Boltzman constant : 1.38×10−23 J K-1, Planck constant : 6.63×10−34 J s)

Options:
A 0.24 A∘
B 0.20 A∘
C 0.34 A∘
D 0.44 A∘

The correct answer is:
0.24 A∘

Solution:
λ=h2mk=hm×32kT=h3mkT

=6.63×10-343×4.64×10-26×1.38×10-13×400=0.24 A°

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2020 (06 Sep Shift 2)
Assuming the sun to be a spherical body of radius R at a temperature of T K, evaluate the total radiant power, incident on earth, at a distance r from the sun. (r0 is the radius of the earth and Ï is Stefan's constant)

Options:
A 4πr02R2σT4/r2
B πr02R2σT4/r2
C r02R2σT4/4πr2
D R2σT4/r2

The correct answer is:
πr02R2σT4/r2

Solution:
From Stefan's law, the rate at which energy is radiated by sun at its surface is P=ÏÃ4ÏR2T4
[Sun is a perfect black body as it emits radiations of all wavelengths and so for it e=1]
The intensity of this power at earth's surface is
I= P 4Ï r 2 = ÏÃ4Ï R 2 T 4 4Ï r 2 = Ï R 2 T 4 r 2
The area of earth which receives this energy is only one-half of total surface area of earth, whose projection would be Ïr02.
â´ Total radiant power as received by earth
= Ïr02ÃI
= Ïr02ÃÏR2T4r2=Ïr02R2ÏT4r2

Topic: Physics, Thermal Properties of Matter, JEE Main
Assuming the sun to be a spherical body of radius $\mathrm{R}$ at a temperature of $\mathrm{T} \mathrm{K}$, evaluate the total radiant power, incident on Earth, at a distance $r$ from the Sun.

Options:
A
$\frac{\mathrm{R}^2 \sigma \mathrm{T}^4}{\mathrm{r}^2}$

B
$\frac{4 \pi r_0^2 R^2 \sigma T^4}{r^2}$

C
$\frac{\pi r_0^2 R^2 \sigma T^4}{r^2}$

D
$\frac{r_0^2 R^2 \sigma T^4}{4 \pi r^2}$

The correct answer is:
$\frac{\pi r_0^2 R^2 \sigma T^4}{r^2}$

Solution:
$\frac{\pi r_0^2}{4 \pi r^2}\left(\sigma \mathrm{T}^4 \cdot 4 \pi \mathrm{R}^2\right)=\frac{\sigma \pi \mathrm{T}^4 \mathrm{R}^2 \mathrm{r}_0^2}{\mathrm{r}^2}$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2006
Astigmatism (for a human eye) can be removed by using

Options:
A concave lens
B convex lens
C cylindrical lens
D prismatic lens

The correct answer is:
cylindrical lens

Solution:
When the lens is perfectly spherical rays refracts equally from each portion and converge at one point. But when lens is not spherical rays from one side of the lens gets refracted more than the other. This causes blurry image. Person will not be able to see the whole image of the object but can see only some portion of it at a time. This type of optical defect of eye is called Astigmatism. This can be corrected by cylindrical lenses in the spectacles.

Topic: Physics, Ray Optics, JEE Main
Astronomers have observed that light coming from far away stars do not show the same spectrum as observed on Earth. Each wavelength is shifted slightly towards the red end of the spectrum. It is observed that the 1st line of Lyman series shows 121 nm on Earth, but from the star it shows 122 nm. The speed of the star with respect to Earth is close to

Options:
A 2.5Ã106âm s-1
B 2Ã106âm s-1
C 1.5Ã106âm s-1
D 0.5Ã106âm s-1

The correct answer is:
2.5Ã106âm s-1

Solution:
1Î»=câvc+v 1Î»0
â Î»0=Î»câvc+v
121=122câvc+v
â vâ2.5Ã106 m sâ1

Topic: Physics, Wave Optics, JEE Main