datavorous/entrance-exam-dataset huggingface dataset:
A conducting sphere A having a radius a is charged to a potential V1. Now it is surrounded by another conducting spherical shell of the radius b. The potential V2 acquired by the sphere A after it is connected to shell B by a thin conducting wire is -

Options:
A V1ab
B V1ba
C V1(a+b)b
D V1aa+b

The correct answer is:
V1ab

Solution:
V1 = Kqa
V2 = Kq′b = Kqb
= KV1aKb = V1ab

Topic: Physics, Electrostatics, JEE Main
A conducting sphere $S_1$ of radius $r_1$ is connected by a conducting wire to another conducting sphere $S_2$ of radius $r_2$, where $r_1=3 \mathrm{~cm}$ and $r_2=2$ $\mathrm{cm}$. Before they are connected, $s_1$ carries charge of 10 units. The electric potential at the point which is at a distance $4 \mathrm{~cm}$ from the centre of $S_1$ and a distance $3 \mathrm{~cm}$ from the centre of $S_2$ is

Options:
A $\frac{1}{4 \pi \varepsilon_0} \frac{17}{6}$
B $\frac{1}{4 \pi \varepsilon_0} \frac{3}{2}$
C $\frac{1}{4 \pi \varepsilon_0} \frac{1}{6}$
D $\frac{1}{4 \pi \varepsilon_0} \frac{17}{12}$

The correct answer is:
$\frac{1}{4 \pi \varepsilon_0} \frac{17}{6}$

Solution:
When spheres are connected, charges are redistributed such that potential on their surfaces are same.
So, we have
$$
\begin{aligned}
q_1+q_2 & =10 \\
\frac{k q_1}{r_1} & =\frac{k q_2}{r_2} \Rightarrow \frac{q_1}{3}=\frac{q_2}{2}
\end{aligned}
$$
and
$$
\frac{k q_1}{r_1}=\frac{k q_2}{r_2} \Rightarrow \frac{q_1}{3}=\frac{q_2}{2}
$$
Hence, $q_1=6$ units and $q_2=4$ units.
Now, potential at some point $P$ distant $4 \mathrm{~cm}$ from $s_1$ and $3 \mathrm{~cm}$ from $s_2$ is
$$
\begin{aligned}
V & =\frac{k q_1}{r_1}+\frac{k q_2}{r_2}=k \frac{6}{4}+k \frac{4}{3} \\
& =k\left(\frac{3}{2}+\frac{4}{3}\right)=k \frac{17}{6} V=\frac{1}{4 \pi \varepsilon_0} \frac{17}{6} V
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main
A conducting square frame of side a and a long straight wire carrying current I are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity V. The e.m.f induced in the frame (when the centre of the frame is at a distance x from the wire) will be proportional to :

Options:
A 1x2
B 12x-a2
C 12x+a2
D 12x-a2x+a

The correct answer is:
12x-a2x+a

Solution:
The potential difference across AB is
VA-VB=B1 a.V.
⇒ μi2π x-a2 aV
The potential difference across CD is
VC-VD=B2 a.V
B2=μ0i2πx+a2
VC-VD= μ0i2πx+a2 aV
Net Potential difference =μi aV2 π 1x - a2- 1x + a2
( V A − V B )−( V C − V D )= μia 2π ( 2a x 2 − a 2 4 ) ∝ 14x2-a2 ∝ 12x+a(2x-a)

Topic: Physics, Electromagnetic Induction, JEE Main
A conducting square frame of side a and a long straight wire carrying current I are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity V. The e.m.f induced in the frame (when the centre of the frame is at a distance x from the wire) will be proportional to :

Options:
A 1x2
B 12x-a2
C 12x+a2
D 12x-a2x+a

The correct answer is:
12x-a2x+a

Solution:
The potential difference across AB is
VA-VB=B1 a.V.
⇒ μi2π x-a2 aV
The potential difference across CD is
VC-VD=B2 a.V
B2=μ0i2πx+a2
VC-VD= μ0i2πx+a2 aV
Net Potential difference =μi aV2 π 1x - a2- 1x + a2
( V A − V B )−( V C − V D )= μia 2π ( 2a x 2 − a 2 4 )
∝ 14x2-a2 ∝ 12x+a(2x-a)

Topic: Physics, Electromagnetic Induction, JEE Main
A conducting square loop of side $L$ and resistance $R$ moves in its plane with a uniform velocity $v$ perpendicular to one of its sides. A magnetic induction B constant in time and space, pointing perpendicular and into the plane at the loop exists everywhere with half the loop outside the field, as shown in figure. The induced emf is

Options:
A
zero

B
RvB

C
VBL/R

D
VBL

The correct answer is:
VBL

Solution:
No Solution Available

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2002
A conducting wire ab of length l, resistance r and mass m starts sliding down at t=0 down a smooth vertical, thick pair of perfectly conducting rails as shown in the figure. A uniform magnetic field B exist in the space in a direction perpendicular to the plane of the rails. What will be the final constant speed of wire?

Options:
A mgrB2l2
B mgr2B2l2
C mgr2Bl2
D mgrBl2

The correct answer is:
mgrB2l2

Solution:
If there are no dissipating forces then the system will oscillate but if there is some small resistance then the acceleration of the conductor is related to the net force as
mdvdt=mg-ilB=mg-B2l2vr
When the rod attains the steady constant speed, dvdt is zero. Hence

v0=mgrB2l2

Topic: Physics, Electromagnetic Induction, JEE Main
A conducting wire of length l, area of crosssection A and electric resistivity ρ is connected between the terminals of a battery. A potential difference V is developed between its ends, causing an electric current. If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be___

Options:
A 14VAρl
B 34VAρl
C 14ρlVA
D 4VAρl

The correct answer is:
14VAρl

Solution:
As per the question

Resistance =ρ2lA/2=4ρl A

⇒ Current =VR=VA4ρl

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (16 Mar Shift 1)
A conductivity cell with two electrodes (dark side) are half filled with infinitely dilute aqueous solution of a weak electrolyte. If volume is doubled by adding more water at constant temperature, the molar conductivity of the cell will -

Options:
A decrease sharply
B increase sharply
C remain same or can not be measured accurately
D depend upon type of electrolyte

The correct answer is:
remain same or can not be measured accurately

Solution:
Solution is already infinitely dilute, hence no change in molar conductivity upon addition of water

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2024 (06 Apr Shift 1)
A conductor $10 \mathrm{~cm}$ long is moved with a speed $1 \mathrm{~m} / \mathrm{s}$ perpendicular to a field of strength 1000 $\mathrm{A} / \mathrm{m}$. The e.m.f. induced in the conductor is [Given : $\mu_0=4 \pi \times 10^{-7} \mathrm{~Wb} / \mathrm{Am}$ ]

Options:
A $\pi \mathrm{mV}$
B $2 \pi \mathrm{mV}$
C $40 \pi \mu \mathrm{V}$
D $4 \pi \mu \mathrm{V}$

The correct answer is:
$40 \pi \mu \mathrm{V}$

Solution:
Motional emf
$\mathrm{e}=\mathrm{Bv} \mathrm{l}$
Magnetic field $B=\mu_0 H$
$\therefore \quad \mathrm{e}=\mu_0 \mathrm{H} \mathrm{v} l$
$\begin{aligned} & \mathrm{e}=4 \pi \times 10^{-7} \times 1000 \times 0.01 \times 1 \\ & \mathrm{e}=40 \pi \mu \mathrm{V}\end{aligned}$

Topic: Physics, Electromagnetic Induction, JEE Main
A conductor AB of length l carrying a current i is placed perpendicular to a long straight conductor XY carrying a current I as shown. The force on AB will be

Options:
A μ0Ii2πln2
B μ0Ii2πln3
C 3μ0Ii2π
D 2μ0Ii3π

The correct answer is:
μ0Ii2πln3

Solution:
dF=idxB
dF=idxμ0I2πx
FAB=∫l23l2μ0iIdx2πx
FAB=μ0Ii2πln3

Topic: Physics, Magnetic Effects of Current, JEE Main
A conductor is carrying a current i. The magnetic field intensity at the point O which is the common centre for three arcs is

Options:
A 5Âµ0iÎ¸24ÏR
B µ0iθ24πR
C 11µ0iθ24πR
D zero

The correct answer is:
5µ0iθ24πR

Solution:
Find B due to each arc at O and add them using vector addition.
Since magnetic field at the centre of an arc is equal to B=Âµ0I4ÏrÎ¸
hence, net B=Âµ0i4Ï 1R-12R+13RÎ¸=5Âµ0iÎ¸24ÏR

Topic: Physics, Magnetic Effects of Current, JEE Main
A conductor lies along the z-axis at - 1.5 ≤ z <1.5 m and carries a fixed current of 10.0 A in - a ^ z direction (see figure). For a field B → = 3.0 × 1 0 - 4 e - 0.2x a ^ y T, find the power required to move the conductor at constant speed to x = 2.0 m, y = 0 m in 5 × 1 0 - 3 s . Assume parallel motion along the x-axis.

Options:
A 1.57 W
B 2.97 W
C 14.85 W
D 29.7 W

The correct answer is:
2.97 W

Solution:
Power P = W Δ t
Force on conductor
F→=I(l→×b→)
F → =-10( 3 a ^ z )×( 3× 10 −4 e −0.2x a ^ y )
⇒F→=90×10-4(e-0.2x) along with x-axis
work done on the conductor in moving along x-axis.
W=∫x=02F→.dx→
W = ∫ 0 2 Bi ℓ dx = ∫ 0 2 3 × 10 - 4 e - 0.2x × 3 × 10dx =90×10-4e-0.2x-0.202
= 450 × 10 - 4 [ 1 - e - 0.4 ] = 180 × 10 - 4 ( Using e - x = 1 - x )
P = W Δ t = 180 × 10 - 4 5 × 10 - 3 = 3.6 W.
Hence, closest option is 2.97 W.
(if we used exact value of exponential term will get exact answer).

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2014 (06 Apr)
A conductor moving in the magnetic field $B$ and the induced current is $I$. If the magnetic field is doubled the induced current will

Options:
A remain the same
B be half
C be doubled
D be four times

The correct answer is:
be doubled

Solution:
Given, initial magnetic field $=B$
Initial induced current $=I$
Final magnetic field, $B^{\prime}=2 B$
Final current $=I^{\prime}$
Since, emf, $\varepsilon=B l v \sin \theta=I R$
where, $B$ is magnetic field,
$l$ is length of conductor,
$v$ is velocity,
$I$ is current,
and $R$ is resistance.
$\therefore \quad I \propto B$
Hence, if $B$ doubled current $I$ also doubled.

Topic: Physics, Electromagnetic Induction, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 1)
A conductor of $3 \mathrm{~m}$ in length is moving perpendicularly to magnetic field of $10^{-3}$ tesla with the speed of $10^2 \mathrm{~m} / \mathrm{s}$, then the e.m.f. produced across the ends of conductor will be

Options:
A $0.03$ volt
B $0.3$ volt
C $3 \times 10^{-3}$ volt
D $3$ volt

The correct answer is:
$0.3$ volt

Solution:
$e=B v l=3 \times 10^{-3} \times 10^2=0.3 \mathrm{~volt}$

Topic: Physics, Electromagnetic Induction, JEE Main
A conductor wire carrying current i is placed parallel to a long conducting sheet having a current per unit width j0 and width d, as shown in the figure. the force per unit length on the conductor wire will be

Options:
A Î¼0j0iÏtan-1d2h-k^
B μ0j0iπtan-12hd-k^
C j0iμ0πtan-12hd-k^
D j0iμ0πtan-1d2h-k^

The correct answer is:
μ0j0iπtan-1d2h-k^

Solution:
â
F = 2 â« dF sin Î¸
= 2 â« Î¼ 0 4 Ï 2 j 0 dx . i h 2 + x 2 . sin Î¸
= Î¼ 0 Ï j 0 i â« â dx h 2 + x 2 Â· h h 2 + x 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@51FD@
= Î¼ 0 Ï j 0 i Â· h â« 0 d / 2 dx h 2 + x 2
= Î¼ 0 Ï j 0 ih 1 h tan -1 x h 0 d / 2
F â = Î¼ 0 j 0 i Ï tan -1 d 2h - k ^

Topic: Physics, Magnetic Effects of Current, JEE Main
A conductor wire carrying current i is placed symmetrically and parallel to a long conducting sheet having a current per unit width j0 and width d, as shown in the figure. the force per unit length on the conductor wire will be

Options:
A Î¼0j0iÏtan-1d2h-k^
B μ0j0iπtan-12hd-k^
C j0iμ0πtan-12hd-k^
D j0iμ0πtan-1d2h-k^

The correct answer is:
μ0j0iπtan-1d2h-k^

Solution:
â
F = 2 ∫ dF sin θ
= 2 ∫ μ 0 4 π 2 j 0 dx . i h 2 + x 2 . sin θ
= μ 0 π j 0 i ∫ ​ dx h 2 + x 2 · h h 2 + x 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@51FD@
= μ 0 π j 0 i · h ∫ 0 d / 2 dx h 2 + x 2
= μ 0 π j 0 ih 1 h tan -1 x h 0 d / 2
F→=μ0j0iπtan-1d2h-k^

Topic: Physics, Magnetic Effects of Current, JEE Main
A cone having fixed volume has semi-vertical angle of Ï4. At an instant when its height is decreasing at the rate of 2m/s, its radius increases at a rate equal to

Options:
A 2m/s
B 4m/s
C 1m/s
D 8m/s

The correct answer is:
1m/s

Solution:
As, V=13πr2h
⇒dVdt=π3r2dhdt+h⋅2rdrdt=0
⇒rdhdt=-2hdrdt

Also, rh=tanπ4⇒r=h
∴dhdt=-2drdt⇒drdt=-12dhdt
⇒drdt=+1
⇒ radius increases at 1m/s.

Topic: Mathematics, Application of Derivatives, JEE Main
A cone made of insulating material has a total charge Q spread uniformly over its sloping surface having a slant length L. The energy required to take a test charge q from infinity to apex A of the cone will be

Options:
A Qq 2 π ε 0 L
B Qq 4 π ε 0 L
C - Qq 3 π ε 0 L
D Qq 8 π ε 0 L

The correct answer is:
Qq 2 π ε 0 L

Solution:
Ï=QÏRL
L= Slant height
dQ=ÏÃ2Ïrdx
= Charge on the elementary ring
dQ=QÏRLÃ2Ïrdx=QLR2rdx ... (1)
By similarity of triangles :
RL=rx âr=RLx ... (2)
By (1) and (2) we get
dQ=QLRÃ2RLxdx =2QL2xdx
â´ the electric potential at A due to ring is
dV=kdQx=kx2QL2xdx
â´ Total potential at A is
V=â«dv=2kQL2â«0Ldx=2kQL2ÃL
V=2kQL
â´ Work done to shift a charge q from infinity to point A is = qV
W=2kqQL =2Ã14ÏÏµ0qQL=qQ2ÏÏµ0L

Topic: Physics, Magnetic Effects of Current, JEE Main
A cone of base radius $R$ and height $h$ is located in a uniform electric field $\overrightarrow{\mathrm{E}}$ parallel to its base. The electric flux entering the cone is:

Options:
A
$\frac{1}{2} \mathrm{EhR}$

B
EhR

C
$2 \mathrm{EhR}$

D
$4 \mathrm{EhR}$

The correct answer is:
EhR

Solution:
Electric flux, $\phi=E(d s) \cos \theta$ $=E(h R) \cos 0^{\circ}=E h R$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2014 (11 Apr Online)
A cone of radius R and height H, is hanging inside a liquid of density Ï by means of a string as shown in the figure. The force, due to the liquid acting on the slant surface of the cone is

Options:
A ρ π g H R 2
B π ρ H R 2
C 4 3 π ρ g H R 2
D 2 3 π ρ g H R 2

The correct answer is:
2 3 π ρ g H R 2

Solution:
Buoyoncy = force on base-force on slant surface (fây)
Force on slant surface = force on base-buoyancy
Force on slant surface = π R 2 H ρ g - 1 3 π R 2 H ρ g
Force on slant surface = 2 3 π R 2 H ρ g

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A conical pendulum of length l makes an angle θ=45° with respect to Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.4 m and its center is vertically below O. The speed of the pendulum, in its circular path, will be - Take g=10 m s-2

Options:
A 0.2 m s-1
B 0.4 m s-1
C 2 m s-1
D 4 m s-1

The correct answer is:
2 m s-1

Solution:
Tsin⁡θ=mv2r
Tcos⁡θ=mg
tan⁡θ=v2rg
θ=45o
v2=rg
v=rg=0.4×10
v=2 m s-1

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2017 (09 Apr Online)
A conical portion of radius R and height H is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and its mass is M. It is suspended by a string in a liquid of density Ï where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is

Options:
A Mg
B Mg-Vρg
C Mg+πR2hρg
D ρgV+πR2h

The correct answer is:
ρgV+πR2h

Solution:
Net pressure force = weight of liquid displaced
⇒ F-ρghπR2=ρgV
⇒ F=ρgV+πR2h

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A constant 60 V supply is connected across the two resistors as shown in diagram. Calculate the reading of the voltmeter which has a resistance of 200 kÎ©

Options:
A 12 V
B 15Â V
C 20Â V
D 30Â V

The correct answer is:
12 V

Solution:
Voltmeter of resistance 200 kΩ and resistance 200 kΩ are in parallel combination
V â²=60Ã100500=12V

Topic: Physics, Current Electricity, JEE Main
A constant current was passed through a solution of AuCl4- ion between gold electrodes. After a period of 10.0 minutes, the increase in mass of cathode was 1.314 g. The total charge passed through the solution is ___×10-2 F. (Given atomic mass of Au=197)

Options:

The correct answer is:
2

Solution:
The half-cell reaction is,

Au3+ + 3e− → Au

Given W = 1.314 g

Atomic mass of Au=197

When one Faraday of charge is passed through a circuit, then one equivalent of charge is deposited at the respective electrode.

WE= charge1 F

1.3141973=Q1F

Q=2×10-2 F

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2024 (29 Jan Shift 2)
A constant force acting on a body of mass $3.0 \mathrm{~kg}$ changes its speed from $2.0 \mathrm{~m} / \mathrm{s}^{-1}$ to $3.5 \mathrm{~m} / \mathrm{s}^{-1}$ in $25 \mathrm{~s}$. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

Options:

None

Solution:
Here, $m=3.0 \mathrm{~kg}, u=2.0 \mathrm{~ms}^{-1}$ and $v=3.5 \mathrm{~ms}^{-1}$, $t=25 \mathrm{~s}$
Now, $F=m a=m(v-u) / t$
$$
=3.0(3.5-2.0) / 25=0.18 \mathrm{~N}
$$
$\therefore$ the force is along the direction of motion.

Topic: Physics, Laws of Motion
A constant magnetic field of 1 T is applied in the x>0 region. A metallic circular ring of radius 1 m is moving with a constant velocity of 1 m s-1 along the x-axis. At t=0 s, the centre O of the ring is at x=-1 m. What will be the value of the induced emf in the ring at t=1 s ? (Assume the velocity of the ring does not change.)

Options:
A 0 V
B 2 V
C 2π V
D 1 V

The correct answer is:
2 V

Solution:
Recall the formula of induced E.M.F. in terms of the length of the rod, and uniform magnetic field, E=Bvl, here in the case of the ring, l=2R,⇒E=Bv2R, when half ring enters, E=1×1×2×1=2 V.

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2021 (27 Aug Shift 2)
A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time t is proportional to :-

Options:
A t23
B t32
C t
D t12

The correct answer is:
t32

Solution:
P=C⇒FV=C⇒MdVdtV=C
⇒V22∝t
⇒V∝t1/2
⇒dxdt∝t1/2
⇒x∝t3/2

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (18 Mar Shift 1)
A constant power is supplied to a disc that can rotate freely about a fixed vertical axis. Angular velocity Ï of the disc varies with the number of rotations (n) made by the disc after starting from rest as Ï=knx. Hence x is: (K is a constant)

Options:
A 13
B 32
C 23
D 2

The correct answer is:
13

Solution:
τ=Iα
P=τω
P=I α×ω
P=Iω2dωdθ
ω2dω∝dθ⇒ω33∝θ
ω×θ1/3 .

Topic: Physics, Rotational Motion, JEE Main
A constant power \( \mathrm{p} \) is applied to a car starting from rest. If \( \mathrm{v} \) is the velocity of the car at time \( \mathrm{t} \), then

Options:
A \( v \propto t \)
B \( v \propto \frac{1}{t} \)
C \( v \propto \sqrt{t} \)
D \( v \propto \frac{1}{\sqrt{t}} \)

The correct answer is:
\( v \propto \sqrt{t} \)

Solution:
Power,
p=m×a×v
p=m×v2t
If p is constant, then for a given body v2∝t
Or v∝t

Topic: Physics, Work Power Energy, JEE Main
A constant torque acting on a uniform circular wheel changes its angular momentum from \( A_{0} \) to \( 4 \mathrm{~A}_{0} \) in \( 4 \mathrm{~s} \). The magnitude
of this torque is

Options:
A \( \frac{3 \mathrm{~A}_{0}}{4} \)
B \( 4 \mathrm{~A}_{0} \)
C \( A_{0} \)
D \( 12 \mathrm{~A}_{0} \)

The correct answer is:
\( \frac{3 \mathrm{~A}_{0}}{4} \)

Solution:
The angular momentum is changing, means an angular impulse will arise in the system, which is given by,

angular impulse=change in angular momentum

τ×t=Lf-Li,

where, τ is the torque and

t is the time taken.

4 τ=4 A0-A0

⇒ τ=3 A04.

Topic: Physics, Rotational Motion, JEE Main
A constant torque of $31.4 \mathrm{~N}-\mathrm{m}$ is exerted on a pivoted wheel. If angular acceleration of wheel is $4 \pi \mathrm{rad} / \mathrm{s}^{2},$ then the moment of inertia of the wheel is

Options:
A $2.5 \mathrm{~kg} \mathrm{~m}^{2}$
B $3.5 \mathrm{~kg} \mathrm{~m}^{2}$
C $4.5 \mathrm{~kg} \mathrm{~m}^{2}$
D $5.5 \mathrm{~kg} \mathrm{~m}^{2}$

The correct answer is:
$2.5 \mathrm{~kg} \mathrm{~m}^{2}$

Solution:
$I=\frac{\tau}{\alpha}=\frac{31.4}{4 \pi}=\frac{31.4}{4 \times 3.14}=2.5 \mathrm{~kg} \mathrm{~m}^{2}$.

Topic: Physics, Rotational Motion, JEE Main
A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if

Options:
A both the length and the radius of the wire are halved.
B both the length and the radius of the wire are doubled.
C the radius of the wire is doubled.
D the length of the wire is doubled.

The correct answer is:
both the length and the radius of the wire are doubled.

Solution:
The heat produced is given by

$$

H=\frac{V^{2}}{R} \text { and } R=\frac{\rho \ell}{\pi r^{2}}

$$

$\therefore H=V^{2}\left(\frac{\pi r^{2}}{\rho \ell}\right)$

or $\quad H=\left(\frac{\pi V^{2}}{\rho}\right) \frac{r^{2}}{\ell}$

Thus heat $(H)$ is doubled if both length $(\ell)$ and radius $(r)$ are doubled.

Topic: Physics, Current Electricity, JEE Main
A constant voltage is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be:

Options:
A
Increased 8 times

B
Doubled

C
Halved

D
Unchanged

The correct answer is:
Increased 8 times

Solution:
Rate of heat i.e., Power developed in the
$$
\text { wire }=P=\frac{V^2}{R}
$$
Resistance of the wire of length, $L$
$$
R_1=\frac{\rho L}{A}=\frac{\rho L}{\pi r^2}
$$
$\therefore$ Power, $P_1=\frac{V^2}{R_1}$
Resistance of the wire when length is halved i.e., $L / 2$
$$
R_2=\frac{\rho \frac{L}{2}}{\pi(2 r)^2}=\frac{\rho L}{\pi 8 r^2}=\frac{R_1}{8}
$$
$\therefore$ Power, $P_2=\frac{V}{\frac{R_1}{8}}=\frac{8 V}{R_1}$
or, $\mathrm{P}_2=8 \mathrm{P}_1$ i.e., power increased 8 times of previous or original wire.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A constant voltage of $25 \mathrm{~V}$ is applied to a series $L-R$ circuit at $t=0$, by closing a switch. What is the potential difference across the resistor and the inductor at time $t=0$ ?

Options:
A $0 \mathrm{~V}, 25 \mathrm{~V}$
B $12.5 \mathrm{~V}, 1.25 \mathrm{~V}$
C $10 \mathrm{~V}, 15 \mathrm{~V}$
D $25 \mathrm{~V}, 0 \mathrm{~V}$

The correct answer is:
$0 \mathrm{~V}, 25 \mathrm{~V}$

Solution:
At, $t=0$, switch closes
Current $i=i_0\left[1-e^{-t R / L}\right]$
So, at $t=0, i=0$
So, potential across inductor $=25 \mathrm{~V}$ and across resistor $=0 \mathrm{~V}$

Topic: Physics, Alternating Current, JEE Main
A Container having 1 mole of a gas at a temperature $27^{\circ} \mathrm{C}$ has a movable piston which maintains at constant pressure in container of $1 \mathrm{~atm}$. The gas is compressed until temperature becomes $127^{\circ} \mathrm{C}$. The work done is ( $C_\rho$ for gas is $\left.7.03 \mathrm{cal} / \mathrm{mol}^{-} \mathrm{K}\right)$

Options:
A $703 \mathrm{~J}$
B $814 \mathrm{~J}$
C $121 \mathrm{~J}$
D $2035 \mathrm{~J}$

The correct answer is:
$814 \mathrm{~J}$

Solution:
At constant pressure
$W=P \Delta V=\mu R \Delta T=1 \times 8.31 \times 100=831 \approx 814 \mathrm{~J}$

Topic: Physics, Thermodynamics, JEE Main
A container is divided into two chambers by a partition. The volume of first chamber is 4.5 litre and second chamber is 5.5 litre. The first chamber contain 3.0 moles of gas at pressure 2.0 atm and second chamber contain 4.0 moles of gas at pressure 3.0 atm. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is x×10-1 atm. Value of x (nearest integer) is

Options:

The correct answer is:
26

Solution:
Let common equilibrium pressure of mixture is P atmp. then

f2P1 V1+f2P2 V2=f2PV1+V2

f224.5+f235.5=f2P4.5+5.5

⇒P=2.55=x×10-1 atmp

So x=25.5≈26 (Nearest integer)

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2021 (26 Feb Shift 1)