datavorous/entrance-exam-dataset huggingface dataset:
The pyramid of biomass shows a sharp _________________ at higher trophic levels.

Options:
A decrease in biomass
B increase in biomass
C decrease in number
D increase in number

The correct answer is:
decrease in biomass

Solution:
In most ecosystems, all the pyramids, of number, of energy and biomass are upright, i.e., producers are more in number and biomass than the herbivores, and herbivores are more in number and biomass than the carnivores. Also energy at a lower trophic level is always more than at a higher level.
In an upright pyramid of biomass in a grassland ecosystem, the pyramid shows a sharp decrease in biomass at higher trophic levels.

Topic: Biology, Ecosystem, NEET
The pyramid of energy in a forest ecosystem is

Options:
A always upright.
B always inverted.
C both upright and inverted.
D None of the above

The correct answer is:
always upright.

Solution:
Pyramid of energy is always upright in all types of ecosystem as energy goes on decreasing with each and every trophic level.

Topic: Biology, Ecosystem, NEET
The pyramid of numbers in a pond ecosystem is

Options:
A inverted
B upright
C spindle-shaped
D irregular

The correct answer is:
upright

Solution:
In a pond ecosystem, phytoplanktons are producers which are maximum in number, though their biomass is less as compared to consumers. Thus, the pyramid of number is always upright for any aquatic ecosystem.

Topic: Biology, Ecosystem, NEET
The QRS complex in an ECG is not related to:

Options:
A Ventricular contraction
B A measure of determining the heartbeat
C Stroke volume of the cardiac cycle
D Ventricular repolarisation

The correct answer is:
Ventricular repolarisation

Solution:
The QRS complex in ECG represents the depolarisation of the ventricles, which initiates the ventricular contraction. By counting the number of QRS complex in a given time period, the heart beat of a person can be determined.

Topic: Biology, Body Fluids and Circulation, NEET
The quantities A and B are related by the relation, m=AB, where m is the linear density and A is the force. The dimensions of B are of

Options:
A Pressure
B Work
C Latent heat
D None of the above

The correct answer is:
Latent heat

Solution:
m=linear density=mass per unit length=ML
A=force=[MLT-2]
[B]=[A][m]=[MLT-2][ML-1]=[L2T-2]
This is same dimension as that of latent heat.

Topic: Physics, Units and Dimensions, NEET
The quantities of heat required to raise the temperature of two solid copper spheres of radii r1 and r2r1=1.5 r2 through 1 K are in the ratio:

Options:
A 94
B 32
C 53
D 278

The correct answer is:
278

Solution:
Heat supplied ΔQ=MsΔT

For same material 's' same.

ΔQ∝M and M=43πr3ρ

ΔQ∝r3

ΔQ1ΔQ2=r1r23=1.513=278

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2020 (Phase 1)
The quantities which have the same dimensions as those of solid angle are:

Options:
A stress and angle
B strain and arc
C angular speed and stress
D strain and angle

The correct answer is:
strain and angle

Solution:
Solid angle $d \Omega=\frac{d A}{r^2}$ has dimensions $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]$
Strain $=\frac{\Delta l}{l}$ has dimensions $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]$
Angle measured in radians is also dimensionless $\left[\mathrm{M}^0 \mathrm{~L}^{\circ} \mathrm{T}^0\right]$
$\theta=\frac{I}{r}$

Topic: Physics, Units and Dimensions, NEET, NEET 2024
The quantity of a charge that will be transferred by a current flow of 20 A over 1 h 30 min period is

Options:
A 10.8Ã103 C
B 10.8Ã104 C
C 5.4Ã103 C
D 1.8Ã104 C

The correct answer is:
10.8Ã104 C

Solution:
Electric current, I=dQdt
dQ=Idt
t1=0,t2=1 h 30 min
=90minâ¡ =90Ã60=5400 s
Electric charge, Q=I.t
=20Ã5400 C
=10.8Ã 10 4 C

Topic: Physics, Magnetic Effects of Current, NEET
The quantity PVkT represents

Options:
A mass of the gas
B K.E. of the gas
C number of moles of the gas
D number of molecules of the gas

The correct answer is:
number of molecules of the gas

Solution:
PV=nNARTNA
nkT .N A k =  R N A
PV kT  =  nN A  = number of molecules

Topic: Chemistry, States of Matter, NEET
The quantum numbers of four electrons are given below :
I. $n=4 ; I=2 ; m_1=-2 ; s=-\frac{1}{2}$
II. $n=3 ; I=2 ; m_1=1 ; s=+\frac{1}{2}$
III. $n=4 ; I=1 ; m_1=0 ; s=+\frac{1}{2}$
IV. $n=3 ; I=1 ; m_1=-1 ; s=+\frac{1}{2}$
The correct decreasing order of energy of these electrons is

Options:
A IV $\gt$ II $\gt$ III $\gt$ I
B I $\gt$ III $\gt$ II $\gt$ IV
C III $\gt$ I $\gt$ II $\gt$ IV
D I $\gt$ II $\gt$ III $\gt$ IV

The correct answer is:
I $\gt$ III $\gt$ II $\gt$ IV

Solution:
(I) $n=4, I=2, m_I=-2, s=-\frac{1}{2}$; represents $4 d(n+I=6)$
(II) $n=3, I=2, m_l=1, s=+\frac{1}{2}$; represents $3 d(n+I=5)$
(III) $n=4, I=1, m_I=0, s=+\frac{1}{2}$; represents $4 p(n+I=5)$
(IV) $n=3, I=1, m_l=-1, s=+\frac{1}{2}$; represents $3 p(n+I=4)$
Order of energy depends on the $(n+I)$, greater is the $(n+I)$ value greater is the energy, if $(n+I)$ is same, then it depends on $n$; if ' $n$ ' is more, energy is more.
Step-1: According to $(\mathrm{n}+\mathrm{I}$ )
Energy $=(\mathrm{I})\gt(\mathrm{II})=($ III $)\gt(\mathrm{IV})$
Step-2 : If $\mathrm{n} \uparrow$, then energy increases
Energy $=(\mathrm{I})\gt($ III $)\gt($ II $)\gt($ IV)

Topic: Chemistry, Structure of Atom, NEET, NEET 2024 (Re-NEET)
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists: 1. n = 4, l = 2, ml= –2 , ms= –1/2 2. n = 3, l = 2, ml= 1 , ms= +1/2 3. n = 4, l = 1, ml= 0 , ms= +1/2 4. n = 3, l = 2, ml= –2 , ms= –1/2 5. n = 3, l = 1, ml= –1 , ms= +1/2 6. n = 4, l = 1, ml= 0 , ms= +1/2

Options:
A (v) > (ii) = (iv) > (vi) = (iii) > (i)
B (v) < (ii) < (iv) < (vi) < (iii) < (i)
C (v) < (ii) = (iv) < (vi) = (iii) < (i)
D (v) >(ii) > (iv) > (vi) > (iii) > (i)

The correct answer is:
(v) < (ii) = (iv) < (vi) = (iii) < (i)

Solution:
Correct Option is : (C)
(v) < (ii) = (iv) < (vi) = (iii) < (i)

Topic: Chemistry, Structure of Atom, NEET
The quietest sound that the ear of a man can detect is known as

Options:
A Acoustic zoning
B Threshold of hearing
C Threshold of pain
D Green muffler

The correct answer is:
Threshold of hearing

Solution:
Wrong sound at the wrong place leads to noise pollution and is responsible for many physiological and psychological problems. The intensity of sound is measured in decibels. Based on the decibel of sound the intensity of sound is classified ranging from Silent or quiet to uncomfortable sounds. The range of Silence/Quiet is 30 dB - 50 dB and is seen in hospitals, schools, colleges, libraries, etc. The threshold of hearing is the quietest sound that can be detected by the human ear.

Topic: Biology, Environmental Issues, NEET
The radiation corresponding to 3 â 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 Ã 1 0 - 4 T . If the radius of the largest circular path followed by these electrons is 10 mm, the work function of the metal is close to :

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

The correct answer is:
1.1 eV

Solution:
We know that
r = m Ï Bq
â´ Ï = Bqr m
â´ KE = 1 2 m v 2
= B 2 q 2 R 2 2m
= 0.79 eV
Now from photoelectric equation
E = Ï + KE
â´ Ï = E - KE
= 1.89 - 0.79
= 1.1 eV

Topic: Physics, Dual Nature of Matter, NEET
The radiation emitted by a star A is 10000 times that of the sun. If the surface temperatures of the sun and star A are 6000 K and 2000 K respectively, the ratio of the radii of the star A and the sun is

Options:
A 300Â :1
B 600Â :1
C 900 :1
D 1200Â :1

The correct answer is:
900 :1

Solution:
ε∝AT4 ; ε∝R2T4 Or R∝εT4
R2R1=ε2ε1.T14T24=100001×600020004
=102×312=900 :1

Topic: Physics, Thermal Properties of Matter, NEET
The radiation emitted by a star A is 10000 times that of the sun. If the surface temperatures of the sun and star A are 6000 K and 2000 K, respectively, the ratio of the radii of the star A and the sun is

Options:
A 300 : 1
B 600 : 1
C 900 : 1
D 1200 : 1

The correct answer is:
900 : 1

Solution:
Radiation emitted, ε∝AT4;ε∝R2T4
Therefore radius, R∝εT4
R2R1=ε2ε1.T14T24=100001×600020004
R2R1=102×312=900:1

Topic: Physics, Thermal Properties of Matter, NEET
The radiation emitted by a star A is 10,000 times that of the sun. If the surface temperatures of the sun and the star A are 8000 K and 2000 K respectively. The ratio of the radii of the star A and the sun is

Options:
A 1600:1
B 1800:1
C 16:1
D 25:1

The correct answer is:
1600:1

Solution:
As E=Q=AT4⇒Q=Ïr2T4
â´QstarQsun=rstar2.Tstar4rsun2.Tsun4=100001
=rstarrsun2200080004
10000Ã256=rstarrsun2
rstarrsun=16001

Topic: Physics, Thermal Properties of Matter, NEET
The radiation emitted, when an electron jumps from n=3 to n=2 orbit is a hydrogen atom, falls on a metal to produce photoelectron. The electrons from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of 1320 T in a radius of 10-3 m . Find the work function of metal.

Options:
A 1.03 eV
B 1.89 eV
C 0.86 eV
D 2.0 3 eV

The correct answer is:
1.03 eV

Solution:
1.03 eV
E3-E2=13.6122-132=13.6Ã536=1.89 eV
Photoelectrons with kEmax are moving on circular path.
r=mvqB
âmv=qBr
âP=qBr
âP=1.6Ã10-19Ã13200Ã10-3
â12Ã10-24=5Ã10-25kgâ¡m/s
Energy of photoelectron âKEmax=P22 m
âKEmax=25Ã10-502Ã9.1Ã10-31Ã1.6Ã10-19eV
âKEmax=0.86 eV
Now use Einstein equation
hv=Ï+KEmax
â1.89=0.56+Ï;
âÏ=1.03 eV

Topic: Physics, Dual Nature of Matter, NEET
The radiation having the least wavelength out of the following options is

Options:
A γ -rays
B β -rays
C α -rays
D X -rays

The correct answer is:
γ -rays

Solution:
Gamma rays have the least wavelength.

Topic: Physics, Atomic Physics, NEET
The radiation with maximum frequency is

Options:
A X - rays
B Radio waves
C UV rays
D IR rays

The correct answer is:
X - rays

Solution:
Increasing orde of wavelength is consimc rays <Î³ rays <X rays < u.v rays < visible <I.R. rays < micro waves < Television waves < Radio waves
â´ Ï Î±1Î»
â´ X - Rays have maximum frequency

Topic: Chemistry, Structure of Atom, NEET
$$
\text { The radical }
because it has
$$

Options:
A 6 p-orbitals and 6 unpaired electrons
B 7 p-orbitals and 6 unpaired electrons
C 7 p-orbitals and 7 unpaired electrons
D 6 p-orbitals and 7 unpaired electrons

The correct answer is:
6 p-orbitals and 6 unpaired electrons

Solution:
Because of the presence of 6 p-orbitals and 6 unpaired electrons, it is aromatic in nature as these unpaired electrons delocalise in p-orbitals.

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2013 (All India)
The radii and Young's moduli of two uniform wires A and B are in the ratio 2:1 and 1:2 respectively. Both wires are subjected to the same longitudinal force. If the increase in the length of the wire A is one per cent, the percentage increase in the length of the wire B is

Options:
A 1.0
B 1.5
C 2.0
D 3.0

The correct answer is:
2.0

Solution:
Increase in length l=FLAY
or l=FLπ r2Y
Percent increase in length
∆x=lL×100=Fπr2Y
Here, same longitudinal force is applied.
So, ∆x1∆x2=r2r12⋅Y2Y1
1∆x2=122⋅21=14×21
1∆x2=12
∆x2=1×2=2%

Topic: Physics, Mechanical Properties of Solids, NEET
The radii of circular orbits of two satellites A and $B$ of the earth are $4 R$ and $R$, respectively. If the speed of satellite $A$ is $3 v$, then the speed of satellite $B$ will be

Options:
A $3 y / 4$
B $6 v$
C $12 y$
D $3 v / 2$

The correct answer is:
$6 v$

Solution:
$$
\begin{aligned}
& \text { Orbital velocity of satellite } \mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}} \\
& \begin{array}{ll}
\Rightarrow & \frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\sqrt{\frac{\mathrm{r}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{A}}}}=\sqrt{\frac{\mathrm{R}}{4 \mathrm{R}}}=\frac{1}{2} \\
\therefore & \frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}}=\frac{3 \mathrm{v}}{\mathrm{v}_{\mathrm{B}}}=\frac{1}{2} \\
\therefore & \mathrm{v}_{\mathrm{B}}=6 \mathrm{v}
\end{array}
\end{aligned}
$$

Topic: Physics, Gravitation, NEET, NEET 2010 (Screening)
The radii of two planets are R1 and R2 and their densities are Ï1 and Ï2 respectively. If g1 and g2 represent the acceleration due to gravity at their respective surfaces, then g1g2 is

Options:
A ρ1R22ρ2R12
B ρ1R12ρ2R22
C ρ2R1ρ1R2
D ρ1R1ρ2R2

The correct answer is:
ρ1R1ρ2R2

Solution:
The value of g at surface
g= Gm e R e 2 = G R e 2 . 4Ï 3 â R e 3 Ï
g=43ÏGRÏ
So, g1g2=43ÏGR1Ï143ÏGR2Ï2â g1g2=R1Ï1R2Ï2

Topic: Physics, Gravitation, NEET
The radii of two planets are respectively R1 & R2 and their densities are respectively ρ1 & ρ2. The ratio of the acceleration due to gravity at their surface is -

Options:
A g1: g2=Ï1R12 : Ï2R22
B g1:g2=R1R2:Ï1Ï2
C g1:g2=R1Ï2:R2Ï1
D g1:g2=R1Ï1:R2Ï2

The correct answer is:
g1:g2=R1ρ1:R2ρ2

Solution:
The value of g at surface
g=GMρR2=GR2.4π3 R3ρ
g=43πGRρ
So, g1g2=43πGR1ρ143πGR2ρ2
⇒    g 1 g 2 = R 1 ρ 1 R 2 ρ 2

Topic: Physics, Gravitation, NEET
The radii of two soap bubbles are r1 and r2. In isothermal conditions, two meet together in a vacuum. Then the radius of the resultant bubble is given by

Options:
A R⁡ = r 1 + r 2 / 2
B R⁡ = r⁡ 1 ( r⁡ 1 r⁡ 2 + r⁡ 2 )
C R⁡ 2 = r⁡ 1 2 + r⁡ 2 2
D R&ApplyFunction; = r&ApplyFunction; 1 + r&ApplyFunction; 2

The correct answer is:
R⁡ 2 = r⁡ 1 2 + r⁡ 2 2

Solution:
Apply mole conservation
n 1 + n 2 =n P 1 V 1 + P 2 V 2 =PV
outside pressure is zero
(4Sr1)43πr13+4Sr243πr23=4SR43πR3r12+r22=R2

Topic: Physics, Mechanical Properties of Fluids, NEET
The radioactive decay of uranium into thorium is represented by the equation U92238âTh90234+X. The unknown species X is

Options:
A an electron
B a proton
C an alpha particle
D a neutron

The correct answer is:
an alpha particle

Solution:
The mass number of X must be 238-234=4 and atomic number of X must be 92-90=2.
Hence, X must be an alpha particle.

Topic: Physics, Nuclear Physics, NEET
The radioactive isotope ${ }_{27}^{60} \mathrm{Co}$ which is used in the treatment of cancer can be made by $(n, p)$ reaction. For this reaction the target nucleus is:

Options:
A ${ }_{28}^{59} \mathrm{Ni}$
B ${ }_{27}^{59} \mathrm{Co}$
C ${ }_{28}^{60} \mathrm{Ni}$
D ${ }_{27}^{60} \mathrm{Co}$

The correct answer is:
${ }_{28}^{60} \mathrm{Ni}$

Solution:
${ }_{28}^{60} \mathrm{Ni}+{ }_0 n^1 \longrightarrow{ }_{27}^{60} \mathrm{Co}+{ }_1^1 \mathrm{P}$
Related Theory
Cobalt therapy is the medical use of gamma rays from the radioisotope cobalt-60 to treat cancer.
Caution
$(n, p)$ means that neutron attacks and proton liberated respectively.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2004
The radioactive isotope X with a half-life of 109 years decays to Y which is stable. A sample of rocks was found to contain both the elements X and Y in the ratio 1:7. The age of the rocks is

Options:
A 2×109 years
B 3×109 years
C 6×109 years
D 7×109 years

The correct answer is:
3×109 years

Solution:
NN0=11+7=18
NN0=12n=18
∴ n=3
t=nT=3×109 years

Topic: Physics, Nuclear Physics, NEET
The radius of a wheel is R and its radius of gyration about its axis passing through its centre and perpendicular to its plane is K. If the wheel is rolling without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is

Options:
A K 2 R 2
B R 2 K 2
C R 2 R 2 + K 2
D K 2 R 2 + K 2

The correct answer is:
K 2 R 2

Solution:
Rotational KE Translational KE = 1 2 I Ï 2 1 2 M v 2 = K 2 R 2
( âµ I = MK 2 , v = R Ï )

Topic: Physics, Rotational Motion, NEET
The radius of an atom is 100 pm. If this element crystallizes in FCC lattice, the edge length of unit cell is

Options:
A 280 pm
B 150 pm
C 141.4 pm
D none of the above

The correct answer is:
141.4 pm

Solution:
For FCC structure
a2=4r
So a=4r2=4×100pm2
=2×100pm=1.414×100 pm
=141.4 pm

Topic: Chemistry, Solid State, NEET
The radius of an electron in second stationary orbit in Bohr's atom is R. The radius of the third orbit will be

Options:
A 3R
B 2.25R
C 9R
D R3

The correct answer is:
2.25R

Solution:
r=0.529n2Z
R=0.529×4Z
R′=0.529×9Z
R′R=94

Topic: Physics, Atomic Physics, NEET
The radius of circle, the period of revolution, initial position and sense of revolution are indicated in the figure.
y-projection of the radius vector of rotating particle P is

Options:
A yt=-3cos2⁡πt, where y in m
B yt=4sin⁡πt2, where y in m
C yt=3cos⁡3πt2, where y in m
D yt=3cos⁡πt2, where y in m

The correct answer is:
yt=3cos⁡πt2, where y in m

Solution:
T=4
ω=2π4=π2
A=3 m
At t=0, y=3 m
∴ y=3cosπ2t

Topic: Physics, Oscillations, NEET, NEET 2019 (All India)
The radius of germanium (Ge) nuclide is measured to be twice the radius of ${ }_4^9 \mathrm{Be}$.
The number of nucleons in Ge are:

Options:
A 72
B 73
C 74
D 75

The correct answer is:
72

Solution:
Nuclear radii $R=R_0 A^{1 / 3}$
Where $\quad R_0=1.2 \mathrm{fm}$
or $\quad R \propto(A)^{1 / 3}$
$\therefore \quad \frac{R_{H_e}}{R_{G_e}}=\frac{(a)^{\frac{1}{3}}}{(A)^{\frac{1}{3}}}$
or $\quad \frac{R_{H_e}}{2 R_{B^{\varepsilon}}}=\frac{(q)^{\frac{1}{3}}}{(\lambda)^{\frac{1}{3}}}$
Given $\quad R_{G e}=2 R_{B C}$
$(A)^{1 / 3}=2 \times(9)^{1 / 3}$
or $A=23 \times 9=8 \times 9=72$
$\therefore$ Number of nucleus is $\mathrm{Ge}$ is 72

Topic: Physics, Nuclear Physics, NEET, NEET 2006
The radius of gyration of a solid sphere of mass 5 kg about $X Y$ is 5 m as shown in figure. The radius of the sphere is $\frac{5 x}{\sqrt{7}} \mathrm{~m}$, then the value of $x$ is:

Options:
A 5
B $\sqrt{2}$
C $\sqrt{2}$
D $\sqrt{5}$

The correct answer is:
$\sqrt{5}$

Solution:
$I_{X Y}=I_{C M}+M R^2=\frac{2}{5} M R^2+M R^2=\frac{7}{5} M R^2=\frac{7}{5} \times 5 R^2=7 R^2$...(1)
$I_{X Y}=M K^2=5 \times 5^2$...(ii)
$\therefore 5 \times 5^2=7 \times R^2 \quad[$ From (1) and (2)]
$\Rightarrow R=\sqrt{\frac{5}{7}} \times 5=\frac{5 x}{\sqrt{7}} \quad$ (Given)
$\therefore x=\sqrt{5}$

Topic: Physics, Rotational Motion, NEET, NEET 2024 (Re-NEET)
The radius of hydrogen atom in its ground state is 5.3Ã10-11 m. After collision with an electron it is found to have a radius of 21.2 Ã 10-11 m. What is the principal quantum number n of the final state of the atom?

Options:
A n=4
B n=2
C n=16
D n=3

The correct answer is:
n=2

Solution:
r∝n2
rfri=nfni2
21.2 ×10-115.3×10-11 = n12
n2=4
n=2

Topic: Physics, Atomic Physics, NEET
The radius of hydrogen atom in its ground state is 5.3×10-11 m. After collision with an electron it is found to have a radius of 21.2 × 10-11 m. What is the principal quantum number n of the final state of the atom?

Options:
A n=4
B n=2
C n=16
D n=3

The correct answer is:
n=2

Solution:
r∝n2
rfri=nfni2
21.2 ×10-115.3×10-11 = n12
n2=4
n=2

Topic: Physics, Atomic Physics, NEET
The radius of inner most orbit of hydrogen atom is 5.3×10-11 m. What is the radius of third allowed orbit of hydrogen atom?

Options:
A 4.77 A°
B 0.53 A°
C 1.06 A°
D 1.59 A°

The correct answer is:
4.77 A°

Solution:
The formula to calculate the radius of the nth orbit is given by

rn=n2Z ...1

Substitute proper values of the orbit numbers in order to calculate the formula for the radius of the innermost and third allowed orbital radii.

r1=121= 1 ...2r3= 32Z= 9 ...3

Divide equation (2) by equation (3) and solve to calculate the radius of the third allowed orbit.

r1r3=19r3=9r1=9×5.3×10-11 m=47.7×10-11 m= 4.77 A∘

Topic: Physics, Atomic Physics, NEET, NEET 2023 (All India)
The radius of La 3+ (Atomic number of La=57 ) is 1.06 Å. Which one of the following given values will be closest to the radius of Lu 3+(Atomic number of Lu=71) ?

Options:
A 1.60 A â
B 1.40 A â
C 1.06 A â
D 0.85 A â

The correct answer is:
0.85 A â

Solution:
Due to lanthanide contraction , the ionic radii of Ln 3+ ( lanthanide ions ) decreases from La 3+ to Lu 3+ . Thus the lowest value ( here 0.85 A â is the ionic radius of Lu 3+ .

Topic: Chemistry, d and f Block Elements, NEET
The radius of Na+cation is less than that of Na atom. Give a reason.

Options:
A The removal of an electron reduces electron-electron repulsions, leading to a smaller cation.
B The addition of an electron increases electron-electron repulsions, causing a larger cation.
C The cation loses an electron, reducing the nuclear charge, resulting in a smaller size.
D The cation gains an electron, increasing the nuclear charge, causing a larger size.

The correct answer is:
The removal of an electron reduces electron-electron repulsions, leading to a smaller cation.

Solution:
Correct Option is : (A)
The removal of an electron reduces electron-electron repulsions, leading to a smaller cation.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
The radius of planet is twice the radius of earth. Both have almost equal averge mass densities $V_P$ and $V_E$ are escape velocities of the planet and the earth respectively, then:

Options:
A $V_P=1.5 V_E$
B $V_P=2 \mathrm{~V}_{\mathrm{E}}$
C $V_E=3 \mathrm{~V}_{\mathrm{P}}$
D $V_E=1.5 \mathrm{~V}_{\mathrm{P}}$

The correct answer is:
$V_P=2 \mathrm{~V}_{\mathrm{E}}$

Solution:
According to the question
$$
R_P=2 R_E, \rho_E=\rho_P
$$
Escape velocity of earth
$$
\begin{aligned}
V_E & =\sqrt{\frac{2 G M_E}{R_E}}=\sqrt{\frac{2 G}{R_E}\left(\frac{4}{3} \pi R_E^3 \rho_E\right)} \\
& =R_E \sqrt{\frac{8}{3} \pi G \rho_E}
\end{aligned}
$$
Similarly
Escape velocity of planet
$$
\begin{aligned}
V_P & =\sqrt{\frac{2 G M_P}{R_P}}=\sqrt{\frac{2 G}{R_P}\left(\frac{4}{3} \pi R_p{ }^3 \rho_p\right)} \\
& =R_P \sqrt{\frac{8}{3} \pi G \rho_p}
\end{aligned}
$$
from eq. (i) and eq. (ii)
$$
\frac{V_E}{V_P}=\frac{R_E}{R_P} \sqrt{\frac{\rho_E}{\rho_P}}=\frac{R_E}{2 R_E} \sqrt{\frac{\rho_Q}{\rho_E}}=\frac{1}{2}
$$
(given)
$$
\Rightarrow V_p=2 V_E
$$

Topic: Physics, Gravitation, NEET, NEET 2013 (Karnataka)
The radius of the 2He4 nucleus is 3 fermi. The radius of 82Pb206 nucleus will be

Options:
A 5 fermi
B 6 fermi
C 11.16 fermi
D 8 fermi

The correct answer is:
11.16 fermi

Solution:
We have, r ∝ A1/3
⇒ r2r1=A2A11/3=20641/3
∴ r2=320641/3=11.16 fermi

Topic: Physics, Atomic Physics, NEET
The radius of the earth is about 6400 km and that of the mars is 3200 km. The mass of the earth is about 10 times the mass of the mars. An object weighs 200 N on the surface of the earth, its weight on the surface of mars will be

Options:
A 8 N

B 20 N
C 40 N
D 80 N

The correct answer is:
80 N

Solution:
Weight on mars
=mÃ410mg=410Ã200=80 N
g e = GM e R e 2 g m = GM m R m 2
g e g m M e Â· R m 2 M m R e 2 = 1 0 1 2 2 = 1 0 4
weight on mars = 2 0 0 Â· 4 1 0 = 8 0 N

Topic: Physics, Gravitation, NEET
The radius of the earth is R. The height of a point vertically above the earth's surface at which acceleration due to gravity becomes 1% of its value at the surface is

Options:
A 8 R
B 9 R
C 10 R
D 20 R

The correct answer is:
9 R

Solution:
Acceleration due to gravity at a height h from earth's surface
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â g′=GMR+h2
Since,Â Â Â Â Â Â Â Â Â Â g′=g100
orÂ Â Â Â Â Â Â Â Â Â Â Â Â g100=Â GMR+h2
orÂ Â Â Â R+h2100=GMg
orÂ Â Â Â Â R+h2100=R2Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â â´g=GMR2
orÂ Â Â Â Â Â Â Â Â Â Â R+h=10R
â¹Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â h=9R

Topic: Physics, Gravitation, NEET
The radius of the earth is R. The height of a point vertically above the earth's surface at which acceleration due to gravity becomes 1% of its value at the surface is

Options:
A 8R
B 9R
C 10R
D 20R

The correct answer is:
9R

Solution:
Acceleration due to gravity at a height h from earth's surface
g′=GMR+h2
Since g′=g100
or g100= GMR+h2
or R+h2100=GMg
R+h2100=R2 ( since g=GMR2 )
R+h=10R
h=9R

Topic: Physics, Gravitation, NEET
The radius of the first orbit of H2 is a 0.53 Aâ, the radius of 25th orbit will be

Options:
A 121.5 A∘
B 111 Aâ
C 331.25 Aâ
D 333 Aâ

The correct answer is:
331.25 Aâ

Solution:
r n ∝ n 2
∴ r 2 5 ∝ 2 5 2
∴ r 1 r 2 5 = 1 2 2 5 2 = 1 6 2 5
∴ r25=625×0.53=331.25 A∘

Topic: Physics, Atomic Physics, NEET
The radius of which of the following orbit is same as that of the first bohr’s orbit of hydrogen atom?

Options:
A Be3+(n=2)
B Li2+(n=2)
C He+(n=2)
D Li2+(n=3)

The correct answer is:
Be3+(n=2)

Solution:
Here Be3+ in second orbit will have radius = radius of first bohr’s orbit of hydrogen atom
rH=0.529n2ZÅ=0.529×11Å=0.529Å
rBe3+(n=2)=0.529×224=0.529Å

Topic: Chemistry, Structure of Atom, NEET