datavorous/entrance-exam-dataset huggingface dataset:
A metal block of base area 0.20 m2 is placed on a table, as shown in figure. A liquid film of thickness 0.25 mm is inserted between the block and the table. The block is pushed by a horizontal force of 0.1 N and moves with a constant speed. If the viscosity of the liquid is 5.0×10–3 Pl, the speed of block is ______ ×10-3 m s-1.

Options:

The correct answer is:
25

Solution:
Viscous force is given by,

|F|=ηAΔvΔh

⇒0.1=5×10-3×0.2×v0.25×10-3

⇒v=0.025 m s-1

⇒v=25×10-3 m s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2023 (29 Jan Shift 2)
A metal block of mass m is suspended from a rigid support through a metal wire of diameter 14 mm. The tensile stress developed in the wire under equilibrium state is 7×105 N m–2. The value of mass m is ______ kg.

(Take g=9.8 m s-2 and π=227)

Options:

The correct answer is:
11

Solution:
The data given is

T=7×105 N m-2r=7×10-3 m

Using the equation

mgA=T⇒mg=TA⇒mg=Tπr2

⇒mg=7×105×227×72×10-6

⇒mg=49×2210

⇒m=49×2298=11 kg

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2023 (06 Apr Shift 2)
A metal chloride contains 55.0% of chlorine by weight. 100 mL vapours of the metal chloride at STP weigh 0.57 g. The molecular formula of the metal chloride is

(Given: Atomic mass of chlorine is 35.5 u)

Options:
A MCl4
B MCl3
C MCl2
D MCl

The correct answer is:
MCl2

Solution:
Let the formula of metal chloride be MClx

Mass of 100 mL of vapours of MClx at STP =0.57 g

The mass of substance which occupies 22400 ml of volume at STP is molar mass(M)

∴ 100×M22400=0.57

∴ Molar mass of MClx=127.68 g mol–1

The mass percentage of chlorine=mass of chlorinemass of the compound×100

% of Cl=35.5x×100127.68=55

⇒ x=2

∴ Formula of metal chloride is MCl2

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2023 (12 Apr Shift 1)
A metal conductor of length $1 \mathrm{~m}$ rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth's magnetic field is $0.3 \times 10^{-4} \mathrm{~T}$, then the e.m.f. developed between the two ends of the conductor is

Options:
A
depends on the nature of the metal used

B
depends on the intensity of the radiation

C
depends both on the intensity of the radiation and the metal used

D
is the same for all metals and independent of the intensity of the radiation.

The correct answer is:
depends on the intensity of the radiation

Solution:
emf. developed is given by
$$
\varepsilon_{\text {ind }}=\frac{1}{2} \mathrm{~B} \omega \mathrm{R}^2=50 \mu \mathrm{V} \text {. }
$$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2004
A metal crystallizes in a face centred cubic structure. If the edge length of its unit cell is 'a', the closest approach between two atoms in the metallic crystal will be

Options:
A 22a
B 2a
C a2
D 2a

The correct answer is:
a2

Solution:
FCC unit cell: eight corners with six faces

r + 2r +r =2a4r =2aClosest distance in fcc2r =2a2 = 2a2×2=a2

Topic: Chemistry, Solid State, JEE Main, JEE Main 2017 (02 Apr)
A metal crystallizes into two cubic systems-face centred cubic (fcc) and simple cubic (SC), whose unit cell lengths are 4.5 and 1.5  A o respectively. Calculate the ratio of densities of face centred cubic and Simple cubic.

Options:
A 0.15
B 1.44
C 2.25
D 3.25

The correct answer is:
0.15

Solution:
face centred cubic unit cell length =â4.5â A o
Simple cubic unit cell length =â1.5â A o
Density in face centred cubic =n1Ãatomic weightV1ÃAvogadro number
Density simple cubic (SC) =n2Ãatomic massV2ÃAvogardro number
DfccDbcc=n1n2ÃV2V1
n1 for face centred cubic = 4; Also V1=a3=4.5Ã10-83
n2 for simple cubic = 1; Also V2=a3=1.5Ã10-83
DfccDsc=4Ã(1.5Ã10-8)31Ã(4.5Ã10-8)3=0.15

Topic: Chemistry, Solid State, JEE Main
A metal disc of radius : 'R' rorates with an angular velocity ' $\omega$ ' 'about an axis perpendicular to its plane passing through its centre in a magnetic' field of induction ' $\mathrm{B}$ ' acting perpendicular to the plane of the disc. The magnitude of induced. e.m.f. between the rim and axis of the disc is

Options:
A $\pi \mathrm{BR}^2$
B $\frac{2 \pi^2 \mathrm{BR}^2}{\omega}$
C $\pi \mathrm{BR}^2 \omega$
D $\frac{\mathrm{BR}^2 \omega}{2}$

The correct answer is:
$\frac{\mathrm{BR}^2 \omega}{2}$

Solution:
Induced e.m.f.,
$\mathrm{e}=\frac{-\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}}$
$=-\mathrm{B} \frac{\mathrm{dA}}{\mathrm{dt}}$
$\ldots .(\because \mathrm{B}=$ constant $)$
Area swept between axis and the rim, $\mathrm{dA}=\pi \mathrm{R}^2$ Time during which the change in flux taking place, $\mathrm{dt}=\frac{2 \pi}{\omega}$
$\therefore \quad \mathrm{e}=\frac{-\mathrm{B} \pi \mathrm{R}^2}{2 \pi / \omega}=\frac{-\mathrm{B} \omega \mathrm{R}^2}{2}$
$\therefore \quad|\mathrm{e}|=\frac{\mathrm{B} \omega \mathrm{R}^2}{2}$

Topic: Physics, Electromagnetic Induction, JEE Main
A metal has $9 \times 10^{28}$ conduction electrons per $\mathrm{m}^3$ and its resistivity is $1 \times 10^{-8} \Omega$. m. If the drift speed of an electron in the metal is $1.6 \times 10^6 \mathrm{~m} / \mathrm{s}$ then its mean free path is (mass of electron $=9 \times 10^{-31} \mathrm{~kg}$ and charge of electron $=1.6 \times 10^{-19} \mathrm{C}$ )

Options:
A $55.5 \mathrm{~nm}$
B $78.0 \mathrm{~nm}$
C $40.0 \mathrm{~nm}$
D $62.5 \mathrm{~nm}$

The correct answer is:
$62.5 \mathrm{~nm}$

Solution:
Mean free path is given by
$$
\begin{aligned}
& \lambda=\frac{\mathrm{mv}}{1 \mathrm{Ve}^2 \mathrm{p}} \\
& \lambda=\frac{9 \times 10^{-31} \times 1.6 \times 10^6}{9 \times 10^{28} \times\left(1.6 \times 10^{-19}\right)^2 \times 10^{-8}} \\
& \lambda=6.25 \times 10^{-8} \mathrm{~m} \\
& \lambda=62.5 \mathrm{~nm}
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
A metal has a work function ϕ0 and its corresponding threshold wavelength is λ0. If the threshold wavelength for a metal whose work function is ϕ03 is nλ0, then what is the the value of n?

Options:

The correct answer is:
3

Solution:
We know, Ï0=âcÎ»n

â´ Ï0Ï0'=Î»0'Î»0

Here, Ï0'=Ï03

â´ Î»0'Î»0=3

â´ Î»0'=3Î»0

â´ n=3

Topic: Physics, Dual Nature of Matter, JEE Main
A metal has half life period of 10 days, $A$ sample originally has a mass of $1000 \mathrm{mg}$,
then the mass remaining after 50 days is

Options:
A $\frac{225}{8} m g$
B $\frac{125}{8} \mathrm{mg}$
C $\frac{125}{4} \mathrm{mg}$
D $\frac{225}{4} m g$

The correct answer is:
$\frac{125}{4} \mathrm{mg}$

Solution:
A mass has half life period of 10 days. It means every ten days, mass remaining is half of the mass before 10 days.
Initial mass $=1000 \mathrm{mg}$.
$\therefore$ Mass after, 10 days $=\frac{1}{2} \times 1000=500 \mathrm{mg}$
Mass after 20 days $=\frac{1}{2} \times 500=250 \mathrm{mg}$
Mass after 30 days $=\frac{1}{2} \times 250=125 \mathrm{mg}$
Mass after 40 days $=\frac{1}{2} \times 125=\frac{125}{2} \mathrm{mg}$
Mass after 50 days $=\frac{1}{2} \times \frac{125}{2}=\frac{125}{4} \mathrm{mg}$

Topic: Mathematics, Differential Equations, MHT CET, MHT CET 2020 (13 Oct Shift 2)
A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated (P) by the metal. The sensor has a scale that displays log2â¡PP0 , where P0 is a constant. When the metal surface is at a temperature of 487 Â°C, the sensor shows a value 1. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to 2767 Â°C?

Options:

The correct answer is:
9

Solution:
P=eAσT4 where T is in kelvin
log2⁡eAσ487+2734P0=1 .....(i)
log2⁡eAσ2767+2734P0=x ......(ii)
(ii) - (i)
log2⁡30407604=x-1
∴ x=9

Topic: Physics, Thermal Properties of Matter, JEE Main
A metal loop of area $10 \mathrm{~cm}^2$ is placed in a region such that its area vector points along $\hat{\mathrm{k}}$. The region contains a uniform magnetic field of magnitude $1.73 \mathrm{~T}$ that points in the direction $\hat{i}+\hat{j}+\hat{k}$. When the magnetic field is switched off, the field decreases to zero at a steady rate in $10 \mathrm{~s}$, then the magnitude of emf induced in the loop is

Options:
A $0.10 \mathrm{mV}$
B $0.17 \mathrm{mV}$
C $1 \mathrm{mV}$
D $1.7 \mathrm{mV}$

The correct answer is:
$0.10 \mathrm{mV}$

Solution:
We have
$\mathrm{e}=-\frac{\Delta \phi}{\Delta \mathrm{t}}$
$\Rightarrow \quad \mathrm{e}=\frac{-\left(\phi_{\mathrm{f}}-\phi_{\mathrm{i}}\right)}{\Delta \mathrm{t}}=\frac{\phi_{\mathrm{i}}-\phi_{\mathrm{f}}}{\Delta \mathrm{t}}$
$=\frac{1.73\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\right) \cdot\left(10 \times 10^{-4}\right) \hat{\mathrm{k}}-0}{10}$
$=\frac{\frac{1.73}{\sqrt{3}} \times 10^{-3}}{10}=10^{-4}$ volt $=0.1 \mathrm{mV}$

Topic: Physics, Electromagnetic Induction, JEE Main
A metal M and its compound can give the following observable changes in a consequence of reactions
M→diluteHNO3Colourlesssolutions→aqueousNaOHWhiteppt.→excessNaOHaq.Colourlesssolutions
→H2SWhiteppt.

Options:
A Mg
B Pb
C Zn
D Sn

The correct answer is:
Zn

Solution:
Here the metal is Zn and the reactions are as follows

Topic: Chemistry, d and f Block Elements, JEE Main
A metal M crystallizes into two lattices :- face centred cubic (fcc) and body centred cubic (bcc) with unit cell edge length of 2.0 and 2.5A° respectively. The ratio of densities of lattices fcc to bec for the metal M is _______(Nearest integer)

Options:

The correct answer is:
4

Solution:
The density of the crystal lattice can be calculated as follows,

d=Z×MNAa3

Z=effective number of atoms in unit cell(For BCC=2 and FCC=4)

M=Molar mass

NA=Avogadro's number

a=Edge length

dFCCdBCC=4×MwNA×(2)32×MwNA×(2.5)3=3.90

Topic: Chemistry, Solid State, JEE Main, JEE Main 2023 (01 Feb Shift 2)
A metal, $M$ forms chlorides in its $+2$ and $+4$ oxidation states. Which of the following statements about these chlorides is correct?

Options:
A
$\mathrm{MCl}_2$ is more volatile than $\mathrm{MCl}_4$

B
$\mathrm{MCl}_2$ is more soluble in anhydrous ethanol than $\mathrm{MCl}_4$

C
$\mathrm{MCl}_2$ is more ionic than $\mathrm{MCl}_4$

D
$\mathrm{MCl}_2$ is more easily hydrolysed than $\mathrm{MCl}_4$

The correct answer is:
$\mathrm{MCl}_2$ is more ionic than $\mathrm{MCl}_4$

Solution:
No Solution Available

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2006
A metal M forms hexagonal close-packed structure. The total number of voids in 0.02 mol of it is _____ ×1021 (Nearest integer)

(Given NA=6.02×1023)

Options:

The correct answer is:
36

Solution:
From the diagram of voids in HCP:
There are 6 octahedral voids and 12 tetrahedral voids in a HCP unit cell.
One unit cell of HCP contains =18 voids

Thus, No. of voids in 0.02 mol of hcp

=186×6.02×1023×0.02

≈3.6×1022

≈36×1021

Topic: Chemistry, Solid State, JEE Main, JEE Main 2023 (29 Jan Shift 2)
A metal M forms the sulphate M2SO43. A 0.596 gram sample of the sulphate reacts with excess BaCl2 to give 1.220 g BaSO4. What is the atomic weight of M (in g/mol)?

Options:
A 26.9
B 69.7
C 55.8
D 23

The correct answer is:
26.9

Solution:
Mole of BaSO4=1.22233.3
Moles of M2SO43=1.22233.3Ã13
=1.743Ã10-3
Wt. of M2SO43=0.596
â´ 1.743Ã10-3 2M+96Ã3=0.596
M=26.9

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A metal $M$ on heating in nitrogen gas gives $Y$. $Y$ on treatment with $\mathrm{H}_2 \mathrm{O}$ gives a colourless gas which when passed through $\mathrm{CuSO}_4$ solution gives a blue colour. $Y$ is

Options:
A
$\mathrm{NH}_3$

B
$\mathrm{Mg}\left(\mathrm{NO}_3\right)_2$

C
$\mathrm{Mg}_3 \mathrm{~N}_2$

D
$\mathrm{MgO}$

The correct answer is:
$\mathrm{Mg}_3 \mathrm{~N}_2$

Solution:
$\underset{M}{3 \mathrm{Mg}}+\mathrm{N}_2 \longrightarrow \underset{Y}{\mathrm{Mg}_3} \mathrm{~N}_2$
$$
\begin{aligned}
& \underset{Y}{\mathrm{Mg}_3 \mathrm{~N}_2}+6 \mathrm{H}_2 \mathrm{O} \longrightarrow 3 \mathrm{Mg}(\mathrm{OH})_2+\underset{\text { colourless }}{2 \mathrm{NH}_3 \uparrow} \\
& \mathrm{CuSO}_4+4 \mathrm{NH}_3 \longrightarrow\left[\mathrm{Cu}\left(\mathrm{NH}_3\right)_4\right] \mathrm{SO}_4 \\
&
\end{aligned}
$$
Blue complex

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2012 (12 May Online)
A metal (M) produces a gas (N) on reaction with alkalies like NaOH and KOH. Same gas is produced when the metal reacts with dilute sulphuric acid. Gas (N) reacts with another toxic gas (P) to form methanol at high temperature and pressure. (N) also reacts with reacts metals like (Q) to form electrovalent hydrides. M, N, P and Q respectively are

Options:
A Zn, H2, CO, Na
B Na, H2, Cl2, Ca
C Al, H2,H2S, B
D Mg, H2,NO2, Al

The correct answer is:
Zn, H2, CO, Na

Solution:
ZnM+2NaOH→Na2ZnO2+H2N
Zn+H2SO4→ZnSO4+H2
CO(P)+2H2→700K, 200 atmCo catalystCH3OH,
2Na(Q)+H2→2NaH

Topic: Chemistry, d and f Block Elements, JEE Main
A metal M reacts with nitrogen gas to afford M 3 N , M 3 N on heating at high temperature gives back M and on reaction with water produces a gas B. Gas B reacts with an aqueous solution of CuSO4 to form a deep blue compound. M and B respectively are

Options:
A Li and NH3.
B Ba and NH3.
C Na and N2.
D Al and N2.

The correct answer is:
Li and NH3.

Solution:
The given sequence of unbalanced reactions represents the above process.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2017 (08 Apr Online)
A metal $\mathrm{M}$ readily forms its sulphate $\mathrm{MSO}_4$ which is water - soluble. It forms its oxide MO which becomes inert on heating. It forms an insoluble hydroxide $\mathrm{M}(\mathrm{OH})_2$ which is soluble in $\mathrm{NaOH}$ solution. Then $M$ is

Options:
A
Mg

B
Ba

C
Ca

D
Be

The correct answer is:
Be

Solution:
Beryllium shows anomalous properties due to its small size

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2002
A metal $M$ readily forms its sulphate $\mathrm{MSO}_4$ which is water-soluble. It forms its oxide $M O$ which becomes inert on heating. It forms its insoluble hydroxide $\mathrm{M}(\mathrm{OH})_2$ which is soluble in $\mathrm{NaOH}$ solution. Then $M$ is

Options:
A $\mathrm{Mg}$
B Ba
C Ca
D $\mathrm{Be}$

The correct answer is:
$\mathrm{Be}$

Solution:
The solubility of hydroxides of alkaline earth metals in water increases on moving down the group.

Topic: Chemistry, s Block Elements, JEE Main
A metal of density 7.5×103 kg m-3 has an fcc crystal structure with lattice parameter a = 400 pm. Calculate the number of unit cells present in 0.015 kg of the metal.

Options:
A 6.250×1022
B 3.125×1023
C 3.125×1022
D 1.563×1022

The correct answer is:
3.125×1022

Solution:
V1=a3=(400×10-12m)3=64×10-30m3
V2=massdensity=0.015kg7.5×10-3m-3
=2×10-6m3
where V1 is the volume of the unit cell and V2 that of the metal sample.
So number of unit cell =TotalvolumeVolumeofunitcell=2×10-6m364×10-30m3
=3.125×1022

Topic: Chemistry, Solid State, JEE Main
A metal on combustion in excess air forms X . X upon hydrolysis with water yields H2O2 and O2 along with another product. The metal is:

Options:
A Na
B Mg
C Li
D Rb

The correct answer is:
Rb

Solution:
K, Rb and Cs form super oxides on reaction with excess air
Rb+O2âRbO2 (excess)
2RbO2+2H2Oâ2RbOH+H2O2+O2â

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2019 (12 Jan Shift 1)
A metal on combustion in excess air forms X. X upon hydrolysis with water yields H2O2 and O2 along with another product the metal is

Options:
A Li
B Na
C Rb
D Mg

The correct answer is:
Rb

Solution:
It is Rb and the reactions are as follows
Rb+O2→RbO2'X'
2RbO2+2H2O→2RbOH+H2O2+O2

Topic: Chemistry, s Block Elements, JEE Main
A metal oxide has the formula Z2O3 . It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is

Options:
A 27.9
B 159.6
C 79.8
D 55.8

The correct answer is:
55.8

Solution:
Z2O3+3H2→2Z+3H2O
Valency of metal in Z2O3=3
0.1596 g of Z 2 O 3 react with 6 mg of H 2
So 1 g of H2 react with =0.15960.006=26.6 g of Z2O3
So Eq. wt. of Z2O3=26.6
Now, Eq. wt. of Z+ Eq. wt. of O= Eq. wt. of Z+8=26.6
So Eq. wt. of Z=26.6-8=18.6
So At. wt. of Z=18.6×3=55.8

Topic: Chemistry, Redox Reactions, JEE Main
A metal (P) on heating in nitrogen gas gives compound Q. Q on treatment with H2O gives a colourless gas which when passed through CuSO4 solution gives a dark blue-violet coloured solution. P and Q respectively, are

Options:
A Na and Na3N
B Na and NaNO3
C Mg and Mg3N2
D Mg and Mg(NO3)2

The correct answer is:
Mg and Mg3N2

Solution:
P+ N 2 ânitride(Q) â H 2 O NH 3 + CuSO 4 â Blue-violet coloured sol.
3 Mg P  +  N 2 → Mg 3 N 2 Q → H 2 O Mg(OH) 2  +  NH 3
Hence P and Q are Mg and Mg3N2 respectively.

Topic: Chemistry, s Block Elements, JEE Main
A metal plate gets heated, when cathode rays strike against, it due to

Options:
A Kinetic energy of cathode rays
B Potential energy of cathode rays
C Linear velocity of cathode rays
D Angular velocity of cathode rays

The correct answer is:
Kinetic energy of cathode rays

Solution:
Kinetic energy of cathode rays
When cathode rays strike the metal plate, they transfer their energy to plate hence, metal gets heated.

Topic: Physics, Atomic Physics, JEE Main
A metal plate of area 1×10-4 m2 is illuminated by a radiation of intensity 16 milli Wm2 . The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electron per second and their maximum energy, respectively, will be: 1 eV=1.6×10-19J

Options:
A 1014 and 10 eV
B 1012 and 5 eV
C 1011 and 5 eV
D 1010 and 5 eV

The correct answer is:
1011 and 5 eV

Solution:
Einstein's photoelectric equation
Maximum KE=hCλ-ϕ
=10-5
=5 eV
Intensity I=NphCλtA⇒Npt=IAhCλ
and number of electrons per second. =Npt×10%
=IAhCλ×110=16×10-3×10-410×1.6×10-19×10
=1011 per sec

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (10 Jan Shift 2)
A metal plate of thickness $d / 2$ is introduced in between the plates of a parallel plate air capacitor with plate separation of $d$, then capacity

Options:
A decreases 2 times
B increases 2 times
C remains same
D becomes zero

The correct answer is:
increases 2 times

Solution:
Given, $t=\frac{d}{2}$
The capacitance is given by
$C=\frac{\varepsilon_{0} A}{d}$
For dielectric medium between plates,
$C^{\prime}=\frac{\varepsilon_{0} A}{d-t+\frac{t}{K}}$
For metal, $K=$ infinity
$\Rightarrow \quad C^{\prime}=\frac{\varepsilon_{0} A}{d-\frac{d}{2}}=\frac{2 \varepsilon_{0} A}{d}=2 C$
Hence, capacitance increases 2 times.

Topic: Physics, Capacitance, COMEDK, COMEDK 2016
A metal rod $2 \mathrm{~m}$ long increases in length by $1.6 \mathrm{~mm}$, when heated from $0^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$. The coefficient of linear expansion of metal rod is

Options:
A $1.33 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
B $1.66 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
C $1.33 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
D $1.66 \times 10^{-3} /{ }^{\circ} \mathrm{C}$

The correct answer is:
$1.33 \times 10^{-5} /{ }^{\circ} \mathrm{C}$

Solution:
We know,
Coefficient of Linear expansion $\alpha=\frac{\mathrm{L}_2-\mathrm{L}_1}{\mathrm{~L}_1 \Delta \mathrm{T}}... (i)$
Given: $\Delta \mathrm{T}=60-0=60^{\circ} \mathrm{C}$
$\mathrm{L}_1=2 \mathrm{~m}$ and $\mathrm{L}_2=2.0016$
Substituting the given values into (i),
$\alpha=\frac{.0016}{120}=1.33 \times 10^{-5} /{ }^{\circ} \mathrm{C}$

Topic: Physics, Thermal Properties of Matter, JEE Main
A metal rod having linear expansion coefficient 2Ã10-5 Â°C-1 has a length of 1 m at 20 Â°C. The temperature at which it is shortened by 1 mm is

Options:
A -20 °C
B -15 °C
C -30 °C
D -25 °C

The correct answer is:
-30 °C

Solution:
From, l2=l11+αt2-t1
t2=t1+l2-l1l1α
= 20+-10-31.0×2×10-5=-30 °C

Topic: Physics, Thermal Properties of Matter, JEE Main
A metal rod moves at a constant velocity in a direction perpendicular to its length. A constant uniform magnetic field exists in space in a direction perpendicular to the rod as well as its velocity. Select the correct statement(s) from the following

Options:
A The entire rod is at the same electric potential
B There is an electric field in the rod
C The electric potential is highest at the centre of the rod and decreases towards its ends
D The electric potential is lowest at the centre of the rod and increases towards its ends

The correct answer is:
There is an electric field in the rod

Solution:
A motional emf $e=B v l$ is induced in the rod, or we can say, a potential difference is induced between the two ends of the rod $A B$, with $P$ at higher potential and $Q$ at lower potential. Due to this potential difference, there is an electric field in the rod.

Topic: Physics, Electromagnetic Induction, JEE Main
A metal rod of cross-sectional area $3 \times 10^{-6} \mathrm{~m}^{2}$ is suspended vertically from one
end has a length $0 \cdot 4 \mathrm{~m}$ at $100^{\circ} \mathrm{C}$. Now the rod is cooled upto $0^{\circ} \mathrm{C}$, but prevented
from contracting by attaching a mass ' $\mathrm{m}$ ' at the lower end. The value of ' $\mathrm{m}$ ' is
$\left(\mathrm{Y}=10^{11} \mathrm{~N} / \mathrm{m}^{2}\right.$, coefficient of linear expansion $\left.=10^{-5} / \mathrm{K}, \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$

Options:
A $40 \mathrm{~kg}$
B $20 \mathrm{~kg}$
C $30 \mathrm{~kg}$
D $10 \mathrm{~kg}$

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

Solution:
$\Delta \mathrm{L}=\mathrm{L} \propto \Delta \mathrm{T}=\frac{\mathrm{FL}}{\mathrm{AY}}$
$\therefore \mathrm{F}=\mathrm{AY} \propto \Delta \mathrm{T}$
$\therefore \mathrm{Mg}=\mathrm{AY} \propto \Delta \mathrm{T}$
$\mathrm{M}=\frac{\mathrm{AY} \propto \Delta \mathrm{T}}{\mathrm{g}}=\frac{3 \times 10^{-6} \times 10^{11} \times 10^{-5} \times 100}{10}=30 \mathrm{~kg}$

Topic: Physics, Thermal Properties of Matter, JEE Main
A metal rod of cross-sectional area 10-4 m2 is hanging in a chamber kept at 20°C with a weight attached to its free end. The coefficient of thermal expansion of the rod is 2.5×10-6 K-1 and its Young's modulus is 4×1012 N/m2. When the temperature of the chamber is lowered to T, then a weight of 5000 N needs to be attached to the rod, so that its length is unchanged. Then, T is

Options:
A 15°C
B 12°C
C 5°C
D 0°C

The correct answer is:
15°C

Solution:
Length of rod given remains unchanged. This means contraction due to cooling is equals to elongation due to hanging of weight.

As, thermal strain = strain caused by weight

⇒αΔθ=Δll

⇒Δθ=Δl/lα=FYAα ∵Y=F/AΔl/l

Note Δθ in K is same as Δθ in C∘.

∴ 20-T=5°C or T=15°C

Topic: Physics, Mechanical Properties of Solids, KVPY, KVPY 2019 (SB/SX)
A metal rod of length 2 m has cross-sectional areas 2A and A as shown in the figure. The ends are maintained at temperatures 100 Â°C and 70 Â°C. The temperature at middle point C (in Â°C) is

Options:

The correct answer is:
90

Solution:
Let Î¸ be temperature of the middle point C and in the series, rate of heat flow is the same.
â K2A100â-Î¸=KAÎ¸-70â
â 200-2Î¸=Î¸-70â3Î¸=270âÎ¸=90 Â°C

Topic: Physics, Thermal Properties of Matter, JEE Main
A metal rod of length $2 m$ is rotating with an angular velocity of 100 $\mathrm{rad} / \mathrm{sec}$ in a plane perpendicular to a uniform magnetic field of $0.3 \mathrm{~T}$. The potential difference between the ends of the rod is

Options:
A $30 \mathrm{~V}$
B $40 \mathrm{~V}$
C $60 \mathrm{~V}$
D $600 \mathrm{~V}$

The correct answer is:
$60 \mathrm{~V}$

Solution:
$e=\frac{1}{2} B I^2 \omega=\frac{1}{2} \times 0.3 \times(2)^2 \times 100=60 \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, JEE Main