datavorous/entrance-exam-dataset huggingface dataset:
A metal rod of length \(10 \mathrm{~cm}\) and area of cross-section \(2.8 \times 10^{-4} \mathrm{~m}^2\) is covered with a non-conducting substance. One end of it is maintained at \(80^{\circ} \mathrm{C}\), while the other end is put in ice at \(0^{\circ} \mathrm{C}\). It is found that \(20 \mathrm{gm}\) of ice melts in \(5 \mathrm{~min}\). The thermal conductivity of the metal in Js" \(\mathrm{m}^{-1} \mathrm{~K}^{-1}\) is (Latent heat of ice is \(80 \mathrm{cal} \mathrm{g}^{-1}\).)

Options:
A 70
B 80
C 90
D 100

The correct answer is:
100

Solution:
Given,
length of rod, \(l=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
area of cross-section of rod, \(A=2.8 \times 10^{-4} \mathrm{~m}^2\)
temperature at one end, \(T_1=80^{\circ} \mathrm{C}\)
temperature at other end, \(T_2=0^{\circ} \mathrm{C}\)
quantity of melted ice, \(\mathrm{m}=20 \mathrm{gm}\)
time taken to melt ice, \(t=5 \mathrm{~min}=300 \mathrm{sec}\)
and latent heat of ice, \(s=80 \mathrm{cal} \mathrm{g}^{-1}\)
Now, rate of the heat flow \(=\frac{m \times s \times 4.184}{t}\)
\(=\frac{20 \times 80 \times 4.184}{300}=22.314 \mathrm{~J} / \mathrm{s}\)
Rate of heat flow in the rod, \(\Delta Q=\frac{k A \cdot \Delta T}{l}\)
\(\begin{aligned}
& 22.314=\frac{k\left(2.8 \times 10^{-4}\right) \times 80}{0.1} \\
\therefore \quad & k=99.61 \approx 100 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}
\end{aligned}\)

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2019 (20 Apr Shift 1)
A metal rod of length L and cross-sectional area A is heated through T Â°C. What is the force required to prevent the expansion of the rod lengthwise?
[Y = Young's modulus of the material of rod, Î±= coefficient of linear expansion]

Options:
A YAαT(1-αT)
B YAαT(1+αT)
C (1-αT)YAαT
D (1+αT)YAαT

The correct answer is:
YAαT(1+αT)

Solution:
∆L=αLT ...... (i)
Also Y=FA.L(1+αt)∆L
Or ∆L=F.L(1+αt)A.Y ........ (ii)
Equating (i) and (ii)
F.L1+αtA.Y=αLT
∴ F=AYαT(1+αt)

Topic: Physics, Mechanical Properties of Solids, JEE Main
A metal rod of length $\mathrm{L}$ and cross-sectional area $\mathrm{A}$ is heated through $\mathrm{T}^{\circ} \mathrm{C}$. What is the force required to prevent the expansion of the rod lengthwise?
(Y= Young's modulus of material of the rod, $\alpha=$ coefficient of linear expansion of
the rod.)

Options:
A $\mathrm{YA} \alpha / \mathrm{T}(1+\alpha \mathrm{T})$
B $\mathrm{YA} \alpha \mathrm{T} /(1-\alpha \mathrm{T})$
C $\mathrm{YA} \alpha \mathrm{T} /(1+\alpha \mathrm{T})$
D $\mathrm{YA} \alpha /(1-\alpha \mathrm{T})$

The correct answer is:
$\mathrm{YA} \alpha \mathrm{T} /(1+\alpha \mathrm{T})$

Solution:
The increase in length due to heating is given by
$\mathrm{L}=\mathrm{L}_{0}(1+\alpha \mathrm{T}) \quad$ or $\quad \mathrm{L}-\mathrm{L}_{0}=\Delta \mathrm{L}=\mathrm{L}_{0} \alpha \mathrm{T}$
To compress the rod by $\Delta \mathrm{L}$, the force required is given by
$\mathrm{F}=\frac{\mathrm{YA} \Delta \mathrm{L}}{\mathrm{L}}$
Substituting the value of $\mathrm{L}$ and $\Delta \mathrm{L}$ we get
$\mathrm{F}=\frac{\mathrm{YA} \times \mathrm{L}_{0} \alpha \mathrm{T}}{\mathrm{L}_{\mathrm{o}}(1+\alpha \mathrm{T})}=\frac{\mathrm{YA} \alpha \mathrm{T}}{1+\alpha \mathrm{T}}$

Topic: Physics, Thermal Properties of Matter, JEE Main
A metal rod of Young's modulus Y and coefficient of linear expansion Î± is held at its two ends such that its length remains constant on changing the temperature. If its temperature is raised by t Â°C, the thermal stress developed in it is

Options:
A Î±tY
B YÎ±E
C YÎ±t
D 1Yαt

The correct answer is:
Yαt

Solution:
Linear expansion
∆L= αL∆T=FLAY

⇒ αt=FAY
Stress = FA = Yα t

Topic: Physics, Thermal Properties of Matter, NEET
A metal sample carrying a current along $\mathrm{X-}$ axis with density $\mathrm{J}_{\mathrm{x}}$ is subjected to a magnetic field $\mathrm{B}_z$ (along $\mathrm{z-}$axis). The electric field $\mathrm{E}_{\mathrm{y}}$ developed along $\mathrm{Y}$-axis is directly proportional to $\mathrm{J}_{\mathrm{x}}$ as well as $\mathrm{B}_{\mathrm{z}}$. The constant of proportionality has SI unit.

Options:
A
$\frac{m^2}{A}$

B
$\frac{m^3}{A s}$

C
$\frac{m^2}{A s}$

D
$\frac{A s}{m^3}$

The correct answer is:
$\frac{m^3}{A s}$

Solution:
According to question
$\mathrm{E}_{\mathrm{y}} \propto \mathrm{J}_{\mathrm{x}} \mathrm{B}_{\mathrm{Z}}$
$\therefore$ Constant of proportionality
$\mathrm{K}=\frac{\mathrm{E}_{\mathrm{y}}}{\mathrm{B}_{\mathrm{Z}} \mathrm{J}_{\mathrm{x}}}=\frac{\mathrm{C}}{\mathrm{J}_{\mathrm{x}}}=\frac{\mathrm{m}^3}{\mathrm{As}}$
$\left[\right.$ As $\frac{E}{B}=C$ (speed of light) and $\left.J=\frac{I}{\text { Area }}\right]$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2013 (25 Apr Online)
A metal sphere is hung by a string fixed to a wall. The sphere is pushed away from the wall by a stick. The forces acting on the sphere are shown in the second diagram. Which of the following statements is wrong

Options:
A $P=W \tan \theta$
B $\dot{T}+\dot{P}+\dot{W}=0$
C $T^2=P^2+W^2$
D $T=P+W$

The correct answer is:
$T=P+W$

Solution:
As the metal sphere is in equilibrium under the effect of three forces therefore $\vec{T}+\vec{P}+\vec{W}=0$
From the figure $T \cos \theta=W$ ...(i)
$T \sin \theta=P$ ...(ii)
From equation (i) and (ii) we get $P=W \tan \theta$ and $T^2=P^2+W^2$

Topic: Physics, Mathematics in Physics, JEE Main
A metal sphere of radius $1 \mathrm{~m}$ is charged with $10^{-2} \mathrm{C}$ in air. Its bulk modulus is
$10^{11} / 4 \pi^{2}$. The volume strain in the sphere is $\left(\epsilon_{0}=\right.$ pemittivity of free space)

Options:
A $\frac{10^{-1}}{6 \in_{0}}$
B $\frac{10^{-14}}{8 \epsilon_{0}}$
C $\frac{10^{-15}}{8 \epsilon_{0}}$
D $\frac{10^{-12}}{4 \in_{0}}$

The correct answer is:
$\frac{10^{-15}}{8 \epsilon_{0}}$

Solution:
surface charge density
$\sigma=\frac{q}{4 \pi r^{2}}=\frac{10^{-2}}{4 \pi \times 1}=\frac{10^{-2}}{4 \pi} \mathrm{C} / \mathrm{m}^{2}$
strain $=$
$\frac{F}{A}=\frac{1}{2} \frac{\sigma^{2}}{\varepsilon_{2}}=\frac{1}{2 \varepsilon_{0}} \times\left(\frac{10^{-2}}{4 \pi}\right)^{2}=\frac{10^{-4}}{32 \pi^{2} \varepsilon_{0}}$
strain $=\frac{\text { stress }}{k}=\frac{1}{2} \frac{\sigma^{2}}{\varepsilon_{0} k}$
where $k=\frac{10^{11}}{4 \pi^{2}}$
substituting and calculating we get strain =
$\frac{10^{-15}}{8 \varepsilon_{0}}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A metal sphere of radius r and specific heat c is rotated about an axis passing through its centre at a speed of n rotations per second. It is suddenly stopped and 50% of its energy is used in increasing its temperature. Then the rise in temperature of the sphere is

Options:
A 25 Ï2n2r2c
B 110Â Ï2n2r2c
C 78Ïr2n2c
D 18 Ï2n2r2c

The correct answer is:
25 Ï2n2r2c

Solution:
Specific heat of metal =C
Temperature change =Î¸
Here, 50100 (KE of rotation) = c m Î¸
1 2 ( 1 2 I Ï 2 )=CâmâÎ¸,
â 1 4 ( 2 5 m r 2 ) ( 2Ïn ) 2 =CâmâÎ¸
âÎ¸= 2 5 Ï 2 n 2 r 2 c

Topic: Physics, Thermodynamics, JEE Main
A metal sphere of radius r and specific heat S is rotated about an axis passing through its centre, at a speed of n rotations per second. It is suddenly stopped and 50 % of its energy is used in increasing its temperature. Then, the rise in temperature of the sphere is

Options:
A 2Ï2n2r25S
B π2n210r2S
C 78 πr2n2S
D 5πrn214S

The correct answer is:
2π2n2r25S

Solution:
I=25mr2, Ï=2Ïn
k=12IÏ2=12Ã25mr2Ã4Ï2n2
=45Ï2mr2n2
ÎÎ¸=ÎQmS=k2mS
=25Ï2mr2n2mS
=2Ï2r2n25S

Topic: Physics, Thermodynamics, JEE Main
A metal sphere of radius 'R' cm is charged with $4 \pi \mu \mathrm{C}$ situated in air. If ' $\sigma$ ' is surface density of charge , 'E' is electric intensity at a distance 'r' from the centre of sphere then 'r' is $\left(\epsilon_{0}=\right.$ permittivity of free space)

Options:
A $\mathrm{R} \sqrt{\frac{\epsilon_{0} \mathrm{E}}{\sigma}}$
B $\mathrm{R} \sqrt{\frac{\sigma}{\epsilon_{0} \mathrm{E}}}$
C $\sqrt{\frac{\epsilon_{0} \mathrm{E}}{\mathrm{R} \sigma}}$
D $\sqrt{\frac{\mathrm{R} \sigma}{\epsilon_{0} \mathrm{E}}}$

The correct answer is:
$\mathrm{R} \sqrt{\frac{\sigma}{\epsilon_{0} \mathrm{E}}}$

Solution:
$\mathrm{E}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{\mathrm{q}}{\mathrm{r}^{2}} ; \quad \sigma=\frac{\mathrm{q}}{4 \pi \mathrm{R}^{2}}$
$\therefore \mathrm{q}=4 \pi \mathrm{R}^{2} \sigma$
$\therefore \mathrm{E}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{4 \pi \mathrm{R}^{2} \sigma}{\mathrm{r}^{2}}$
$\therefore r^{2}=\frac{R^{2} \sigma}{\epsilon_{0} E} \quad \therefore r=R \sqrt{\frac{\sigma}{\epsilon_{0} E}}$

Topic: Physics, Electrostatics, JEE Main
A metal surface having a work function Ï=2.2Ã10-19 J, is illuminated by the light of wavelength 1320 Ao. What is the maximum kinetic energy (in eV) of the emitted photoelectron? [Take h=6.6Ã10-34 J s]

Options:

The correct answer is:
8

Solution:
K.E·=hcλ-ϕ0

=6.6×10-34×3×1081320×10-10-2.2×1019

=15×10-19-2.2×10-19

=12.8×10-19J

K.E·=12.8×10-191.6×10-19eV=8eV

Topic: Physics, Dual Nature of Matter, JEE Main
A metal surface, having work function Ï=2.2Ã10-19 J, is illuminated by a light of wavelength 1320 Ao. The maximum kinetic energy (in eV) of the emitted photoelectron is [Take h=6.6Ã10-34 J s]

Options:

The correct answer is:
8

Solution:
K.EÂ·=hcÎ»-Ï0

=6.6Ã10-34Ã3Ã1081320Ã10-10-2.2Ã10-19

=15Ã10-19-2.2Ã10-19

=12.8Ã10-19 J

K.EÂ·=12.8Ã10-191.6Ã10-19 eV=8 eV

Topic: Physics, Dual Nature of Matter, JEE Main
A metal surface is exposed to 500 nm radiation. The threshold frequency of the metal for photoelectric current is 4.3×1014 Hz. The velocity of ejected electron is ___×105 ms-1 (Nearest integer)
Use:h=6.63×10-34Js,me=9.0×10-31 kg

Options:

The correct answer is:
5

Solution:
λ=500 nm→v=Cλ=3×108500×10-9=6×1014 Hz

v0=4.3×1014 Hz

For photo electric effect, hv=hv0+KE

KE=hv-hv0=hv-v0=6.6×10-346×1014-4.3×1014

=6.6×1.7×10-20 J

K.E. =12mv2;v=2×K·Em

V=2×6.6×1.7×10-209.1×10-31=24.65×1010=5×105 m/s

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2021 (26 Aug Shift 2)
A metal surface is illuminated by a light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum kinetic energy of the emitted photoelectrons would become

Options:
A unchanged
B half of the original value
C twice of the original value
D four times of the original value

The correct answer is:
unchanged

Solution:
In photoelectric emission, maximum kinetic energy of emitted photoelectrons does not depend upon the intensity of incident radiation. It depends upon the frequency of incident radiation. Hence, when intensity of incident radiation is reduced to one-fourth, then maximum kinetic energy of emitted photoelectrons remains unchanged.

Topic: Physics, Dual Nature of Matter, AP EAMCET, AP EAMCET 2020 (22 Sep Shift 1)
A metal surface is illuminated by a radiation of wavelength 4500 A∘. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90° with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately

Options:
A 1.36 eV
B 1.69 eV
C 2.78 eV
D 2.23 eV

The correct answer is:
1.36 eV

Solution:
From the Einstein's photoelectric equation K=hcλ-ϕ⇒ϕ=hcλ-K

Now R=mvBq⇒v=RBqm⇒K=R2B2q22m

Therefore, ϕ=hcλ-R2B2q22m=6.63×10-34×3×1084500×10-10-2×10-3×2×10-3×1.6×10-1922×9.1×10-31⇒ϕ=4.42-2.25×10-19 J=2.271.6 eV=1.36 eV

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (26 Jun Shift 2)
A metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value then the maximum KE of the emitted photoelectrons would be

Options:
A Twice the original value
B Four times the original value
C One fourth of the original value
D Unchanged

The correct answer is:
Unchanged

Solution:
The maximum kinetic energy of photoelectrons is given by KEmax=hv-v0 …. (i)
Where h = Planck’s constant,
v= frequency of radiation
and v0 = threshold frequency.
It can be seen from Eq. (i), that the maximum KE of emitted photoelectrons is proportional to the frequency of radiation and is independent of the intensity of radiation, so it remains unchanged.

Topic: Physics, Dual Nature of Matter, MHT CET, MHT CET 2019 (Shift 2)
A metal surface is illuminated by light of two different wavelengths 248ânm and 310ânm . The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2 respectively. If the ratio u1u2=21 and hcâ=â1240âeVnm , the work function of the metal is nearly

Options:
A 3.7 eV
B 3.2 eV
C 2.8 eV
D 2.5 eV

The correct answer is:
3.7 eV

Solution:
As K.Eâv2
ratio of kinetic energy =41, as speed are in the ratio =21
K.E = Energy of incident photon â Work function (W)
4 KE=1240248-W .....(i)
and KE=1240310-W .....(ii)
By solving (i) and (ii)
W=3.7 eV

Topic: Chemistry, Structure of Atom, JEE Main
A metal surface of 100 cm2 area has to be coated with nickel layer of thickness 0.001 mm. A current of 2A was passed through a solution of NiNO32 for ‘x’ seconds to coat the desired layer. The value of x is _________. (Nearest integer)

(ρNi (density of Nickel) is 10 g mL–1, Molar mass of Nickel is 60 g mol-1 F=96500 Cmol-1)

Options:

The correct answer is:
16

Solution:
One equivalent of substance deposited for 96500C(one Faraday).

i×t Columbus of charge can be utilised to get w g of substance.

Millimoles of Ni coated

=(2x)10096500×2

=10x965

Mass of Ni=x965600

=600x965mg

Volume=60x965×10-3 cm3

=100×10-4

⇒x=0.1608×103

≈161

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (13 Apr Shift 1)
A metal wire has mass (0.4±0.002) g, radius (0.3±0.001) mm and length (5±0.02) cm. The maximum possible percentage error in the measurement of density will nearly be:

Options:
A 1.4%
B 1.2%
C 1.3%
D 1.6%

The correct answer is:
1.6%

Solution:
Density=massvolume

⇒ρ=mπr2l

Therefore, percentage error can be written as

⇒dρρ×100=dmm+2drr+dll×100=0.0020.4+20.0010.3+0.025×100≈1.6%

Topic: Physics, Mathematics in Physics, JEE Main
$A$ metal wire of circular cross-section has a n'sistance $R_{1}$. The wire is now stretched without breaking, so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now beromes $R_{2},$ then $R_{2}: R_{1}$ is

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

The correct answer is:
$4: 1$

Solution:
As we know that, $R_{1}=\rho \frac{1}{a}=\rho \frac{l^{2}}{V}$
where, $1=$ length of wire $a=$ area of cross-section of the wire and $V=$ volume of the wire $R_{1} \propto l^{2}$
$\Rightarrow \quad \frac{R_{1}}{R_{2}}=\left(\frac{l_{1}}{I_{2}}\right)^{2}=\left(\frac{1}{2}\right)^{2} \Rightarrow R_{2}: R_{1}=4: 1$

Topic: Physics, Current Electricity, JEE Main
$A$ metal wire of circular cross-section has a n'sistance $R_{1}$. The wire is now stretched without breaking, so that its length is doubled and the density is assumed to remain the same. If the resistance of the wire now beromes $R_{2},$ then $R_{2}: R_{1}$ is

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

The correct answer is:
$4: 1$

Solution:
As we know that, $R_{1}=\rho \frac{1}{a}=\rho \frac{l^{2}}{V}$
where, $1=$ length of wire $a=$ area of cross-section of the wire and $V=$ volume of the wire $R_{1} \propto l^{2}$
$\Rightarrow \quad \frac{R_{1}}{R_{2}}=\left(\frac{l_{1}}{I_{2}}\right)^{2}=\left(\frac{1}{2}\right)^{2} \Rightarrow R_{2}: R_{1}=4: 1$

Topic: Physics, Current Electricity, WBJEE, WBJEE 2015
A metal wire of length 0.5 m and cross-sectional area 10-4 m2 has breaking stress 5×108 N m-2. A block of 10 kg is attached at one end of the string and is rotating in a horizontal circle. The maximum linear velocity of block will be _____ m s-1.

Options:

The correct answer is:
50

Solution:
Here in horizontal circular motion, the only centripetal force is tension. So,T=mω2r=mv2l=10×v20.5=20v2

Now, Tmax= Breaking stress × Area

⇒20v2=5×108×10-4

⇒20v2=5×104

Thus, the maximum linear velocity of block is

v=14104=50 m s-1

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2022 (29 Jul Shift 2)
A metal wire of length L and radius r has a resistance R. If a wire of the same metal of length 2L and radius 3r is taken, then what will be its resistance?

Options:
A 29R
B 23R
C 29πR
D 23πR

The correct answer is:
29R

Solution:
As metal used in both the cases is same, therefore, ρ will remain same,

Now, for first case,

R1=ρL1A1⇒R=ρLπr2 ...Eq1

Similarly, for second case we can write,

R2=ρL2A2⇒R2=ρ2Lπ3r2 ...Eq2

Dividing equation (2) with equation (1), we get

R2R=29⇒R2=29R

Topic: Physics, Current Electricity, TS EAMCET, TS EAMCET 2022 (18 Jul Shift 1)
A metal wire of linear mass density of $9.8 \mathrm{~g} / \mathrm{m}$ is stretched with a tension of $10 \mathrm{~kg}$-wt between two rigid supports 1 metre apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency $n$. The frequency $n$ of the alternating source is

Options:
A
$50 \mathrm{~Hz}$

B
$100 \mathrm{~Hz}$

C
$200 \mathrm{~Hz}$

D
$25 \mathrm{~Hz}$

The correct answer is:
$50 \mathrm{~Hz}$

Solution:
$n=\frac{1}{2 l} \sqrt{\frac{T}{\mu}} ; \quad I=1 m$
$\mathrm{T}=10 \mathrm{Kg}$ wt. $=10 \times 10=100 \mathrm{~N}$
$\mu=9.8 \mathrm{~g} / \mathrm{m}=9.8 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \quad \mathrm{n}=50 \mathrm{hz}$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2003
A metal wire of resistance 3 Ω is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60o at the center, the equivalent resistance between these two points will be:

Options:
A 53 Ω
B 125 Ω
C 72 Ω
D 52 Ω

The correct answer is:
53 Ω

Solution:
We know, R=ρlA=ρl2Al=ρl2Volume
R∝l2 , if the volume is constant.
∴ New resistance =4R
R1=4R360×300=10R3
R2=4R360×60=2R3
∴ Equivalent resistance Req=R1R2R1+R2
=10R3×2R310R3+2R3=5R9
=53 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (09 Apr Shift 2)
A metal wire of resistance 3 Ω is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60o at the center, the equivalent resistance between these two points will be:

Options:
A 53 Ω
B 125 Ω
C 72 Ω
D 52 Ω

The correct answer is:
53 Ω

Solution:
We know, R=8lA=8l2Al=8l2Volume
R∝l2 if the volume is constant.
∴ New resistance =4R.
R1=4R360×300=10R3
R2=4R360×60=2R3
∴ Equivalent resistance Req=R1R2R1+R2
=10R3×2R310R3+2R3=5R9
=53 Ω

Topic: Physics, Current Electricity, JEE Main
A metal wire of uniform mass density having length $L$ and mass $M$ is bent to form a semicircular arc and a particle of mass $\mathrm{m}$ is placed at the centre of the arc. The gravitational force on the particle by the wire is :

Options:
A $\frac{\mathrm{GmM} \pi^2}{\mathrm{~L}^2}$
B $\frac{\mathrm{GMm} \pi}{2 \mathrm{~L}^2}$
C $0$
D $\frac{2 \mathrm{GmM} \pi}{\mathrm{L}^2}$

The correct answer is:
$\frac{2 \mathrm{GmM} \pi}{\mathrm{L}^2}$

Solution:
We have $R=\frac{L}{\pi}$
$\begin{aligned}
& \mathrm{g}_0=\frac{2 \mathrm{G} \frac{\mathrm{M}}{\mathrm{L}}}{\mathrm{R}}=\frac{2 \mathrm{GM} \pi}{\mathrm{L}^2} \\
& \therefore \mathrm{F}_{\mathrm{m}}=\mathrm{mg}_0=\frac{2 \mathrm{GM} \pi \mathrm{m}}{\mathrm{L}^2}
\end{aligned}$
Hence option (4)

Topic: Physics, Gravitation, JEE Main, JEE Main 2024 (04 Apr Shift 1)
A metallic cube of side 15 cm moving along y-axis at a uniform velocity of 2 m s-1. In a region of uniform magnetic field of magnitude 0.5 T directed along z- axis. In equilibrium the potential difference between the faces of higher and lower potential developed because of the motion through the field will be mV.

Options:

The correct answer is:
150

Solution:
Using the equation

qE=qvB ⇒E=vB

Using the relation ∆V=Ed

⇒∆V=vBd

=2×12×0.15 V
=0.15 V=150 mV

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (11 Apr Shift 2)
A metallic element crystallizes into a lattice containing a sequence layers of ABABAB ..... . Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space?

Options:

The correct answer is:
26

Solution:
Packing fraction in hcp =6×43πr3Volume
=6×4πr33×Area×Height
Area (of base) of unit cell in hcp =63r2 (follow text)
height of unit cell in hcp =4r×23 (follow text)
So packing fraction =6×4πr33×63r2×4r×23
=π32=0.74
or = 74%
Thus, empty space in hcp = 1 - 0.74 = 0.26
or = 26%.

Topic: Chemistry, Solid State, JEE Main
A metallic element crystallizes into a lattice containing a sequence of hexagonal layers of ABABAB....... What percentage by volume of this lattice is empty space?

Options:
A 26%
B 74%
C 50%
D 85%

The correct answer is:
26%

Solution:
Number of atoms in hcp is 6 per unit cell.
Packing fraction = 6× 4 3 πr 3 Base area×C
Base area =6Ã344r2
C=4r23
Packing fraction =Ï32=0.74
Void % =1-0.74Ã100=26%

Topic: Chemistry, Solid State, JEE Main
A metallic element exists in cubic lattice. Each edge of unit cell is 4 A o The density of metal is 6.25 g  /m 3 . How many unit cells will be present in 100 g of metal?

Options:
A 1×1022
B 2.5×1029
C 5×1023
D 2×1023

The correct answer is:
2.5×1029

Solution:
dâ=â weight Totalâvolume
6.25= 100 nÃ 4Ã 10 â10 3
100 6.25Ã64Ã 10 â30
n = 2.5 ×  10 29

Topic: Chemistry, Solid State, JEE Main
A metallic loop is placed in a uniform magnetic field $\mathbf{B}$ with the planc of the loop perpendicular to $\mathbf{B}$. Under which condition given an emf will be induced in the loop "If the loop is ....."

Options:
A moved along the direction of $\mathbf{B}$
B squeezed to a smaller area
C rotated about its axis
D rotated about one of its diameters

The correct answers are:
squeezed to a smaller area, rotated about one of its diameters

Solution:
According to the question.
Given, A metallic loop is placed in uniform magnetic field B. Magnetic field $\mathbf{B}$ and metallic loop are perpendicular to each other. Then, if metallic loop is moved along $\mathbf{B}$ or rotated about its own axis. the net flux associated with it remains constant. Thus, no emf will be induced in these cases. However, when the loop is squeezed to a smaller area or rotated about one of its diameters. Then its flux changes. Thus, emf is induced. So, option (b) and (d) both are correct.

Topic: Physics, Electromagnetic Induction, WBJEE, WBJEE 2019
A metallic oxide reacts with water to form its hydroxide, hydrogen peroxide and also liberates
oxygen. The metallic oxide could be

Options:
A \( \mathrm{O} \mathrm{K} \mathrm{O}_{2} \)
B \( \mathrm{Na}_{2} \mathrm{O}_{2} \)
C \( \mathrm{CaO} \)
D \( \mathrm{Li}_{2} \mathrm{O} \)

The correct answer is:
\( \mathrm{O} \mathrm{K} \mathrm{O}_{2} \)

Solution:
Potassium Superoxide react with water to produce potassium hydroxide, hydrogen peroxide and oxygen. This reaction
takes place at a temperature near \( 0^{\circ} \mathrm{C} \). Whereas, other metal oxide reacts with water but does not produce peroxide.
\( 2 \mathrm{KO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{KOH}+\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{O}_{2} \)
Reaction of other metal
\( \mathrm{Na}_{2} \mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 4 \mathrm{NaOH}(a q)+\mathrm{O}_{2}(g) \) \( \mathrm{CaO}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_{2} \) \( \mathrm{Li}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{LiOH} \)

Topic: Chemistry, Solutions, JEE Main