datavorous/entrance-exam-dataset huggingface dataset:
Two identical bar magnets are fixed with their centres at a distance $d$ apart. A stationary charge $Q$ is placed at $P$ in between the gap of the two magnets at a distance $D$ from the centre $O$ as shown in the figure
The force on the charge $Q$ is

Options:
A zero
B directed along $O P$
C directed along $P O$
D directed perpendicular to the plane of paper

The correct answer is:
zero

Solution:
The force of attraction on $Q$ by the north pole of first magnet is $F$. The force of attraction on the same charge by $n$ the north pole of another magnet must be opposite to that of the first magnet. Thus, they concel the effect of one other and hence force on the charge $Q$ is zero.

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2010 (Mains)
Two identical beakers A and B contain equal volumes of two different liquids at 60Â°C each and left to cool down. Liquid in A has density of 8Ã102 kg m-3 and specific heat of 2000 J kg-1K-1 while the liquid in B has density 103 kg m-3 and specific heat of 4000 J kg-1K-1. Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)

Options:
A
B
C
D

The correct answer is:

Solution:
From Newton's law of cooling,
Rate of cooling: -dTdt=4σeAT03T-T0ms
∴ρA<ρB
mA<mB
And sA<sB
mAsA<mBsB
At t=0,-dTdtA>-dTdtB
Initially, the rate of cooling for A will be higher than B and temperature (T) vs time (t) graph will be a curve.

Topic: Physics, Thermal Properties of Matter, JEE Main
Two identical blocks A and B each of mass m resting on the smooth horizontal floor are connected by a light spring of natural length L and spring constant K. A third block C of mass m moving with a speed v along the line joining A and B collides with A.The maximum compression in the spring is

Options:
A vm2K
B mv2K
C mvK
D m2K

The correct answer is:
vm2K

Solution:
C comes to rest

Vcm of A & B=v2

⇒12 is vret2=12kx2

x=μ×v2k=m2k×v.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (17 Mar Shift 2)
Two identical blocks each of mass 1 kg are joined together with a compressed spring. Sometime later after the release of the system, the two blocks are seen to be moving with unequal speeds in the opposite directions as shown in the figure. Choose the correct statement

Options:
A it is not possible
B whatever may be the speed of the blocks the centre of mass will remain stationary
C the centre of mass of the system is moving with a velocity of 2 m s-1â¡
D the centre of mass of the system is moving with a velocity of 1 m s-1â¡

The correct answer is:
the centre of mass of the system is moving with a velocity of 1 m s-1â¡

Solution:
vcm=15+1-31+1
vcm=1 m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two identical blocks of ice move in opposite directions with equal speed and collide with cach other. What will be the minimum speed required to make both the blocks melt completely, if the initial temperatures of the blocks were $-8^{\circ} \mathrm{C}$ each? (Specific heat of ice is $2100 \mathrm{Jkg}^{-1} \mathrm{K}^{-1}$ and latent heat of fusion of ice is $3.36 \times 10^{5} \mathrm{Jkg}^{-1}$ )

Options:
A $840 \mathrm{ms}^{-1}$
B $420 \mathrm{ms}^{-1}$
C $84 m s^{-1}$
D $84 \mathrm{ms}^{-1}$

The correct answer is:
$840 \mathrm{ms}^{-1}$

Solution:
Maximum loss in $K . E=$ Total $K . E$ energy before collision of two ice blocks
$$
K . E_{\max } \text { loss }=\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}=m v^{2}
$$
This maximum loss in kinetic energy will be equal to loss in heat to melt two ice block.
$$
\Rightarrow \quad m v^{2}=(m s \Delta \theta+m L) 2
$$
According to question,
$$
\begin{aligned}
L &=336 \times 10^{5} \mathrm{J} \mathrm{kg}^{-1} \\
S=& 2100 \mathrm{J} \mathrm{kg}^{-1} \mathrm{K}^{-1} \\
\Delta \theta=8^{\circ} \mathrm{C}
\end{aligned}
$$
From Eq.
(i) $v=\sqrt{2(s-\Delta \theta+L)}$
$$
\begin{array}{l}
=\sqrt{2\left(2100 \times 8+336 \times 10^{5}\right)} \\
v=840 \mathrm{ms}^{-1}
\end{array}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two identical blocks of mass m are suspended from a beam balance whose scale pans differ in vertical height by h≪R, If R and ρ are the radius and density of the earth, then the error in weighing is

Options:
A 23πρR3Gm
B 83πρGmh
C 83πρR3Gm
D 43πρGm2h

The correct answer is:
83πρGmh

Solution:
As with height g varies as
and in accord with the figure h1>h2 so W1 will be lesser than W2 and
W2-W1=mg2-mg1=2mgh1R-h2R
or W2-W1=2mGMR2hR
as g=GMR2 and h1-h2=h
or W2-W1=2mhGR343ÏR3Ï
=83 ÏÏGmh as M=43ÏR3Ï

Topic: Physics, Gravitation, JEE Main
Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100 oC, while the other one is at 0 oC. If the two bodies are brought into contact, then assuming no heat loss, the final common temperature is

Options:
A 50 oC
B >50 oC
C <50 oC but >0 oC
D 0 oC

The correct answer is:
>50 oC

Solution:
Loss of heat by the hot body = gain of heat by the cold body.

Since, the heat capacity increases with temperature, the hotter body will lose more energy when its temperature falls down from 100 oC to 50 oC compared to the energy required by the cold body to raise its temperature from 0 oC to 50 oC. Thus, the temperature will be more than 50 oC.

Topic: Physics, Thermal Properties of Matter, JEE Main
Two identical calorimeters A and B contain an equal quantity of water at 20 Â°C. A 5 g piece of metal X of specific heat 0.2 cal g-1 Â°C-1 is dropped into A and 5 g piece of metal Y is dropped into B. The equilibrium temperature in A is 22 Â°C and that in B is 23 Â°C. The initial temperature of both the metals was 40 Â°C. The specific heat of metal Y (in cal g-1 Â°C-1) is

Options:
A 2785
B 5485
C 8185
D 0.4

The correct answer is:
2785

Solution:
For X:5Ã0.2Ã(40-22)
=mÃCÃ22-20
For Y:5ÃCÃ(40-23)
=mÃCÃ(23-20)
âC=2785

Topic: Physics, Thermodynamics, JEE Main
Two identical capacitors A and B are connected as shown in the circuit. Initially the switch 'S' is closed. Now the switch is opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. The ratio of total electrostatic energy stored in the capacitors before and after introduction of the dielectric is

Options:
A $3: 1$
B $5: 1$
C $3: 5$
D $5: 3$

The correct answer is:
$3: 5$

Solution:
Let the capacitance of each capacitor be $\mathrm{C}$.
Total electrostatic energy stared in capacitor
$$
\begin{aligned}
\mathrm{U}_1 & =\frac{1}{2} \mathrm{CE}^2+\frac{1}{2} \mathrm{CE}^2 \\
\mathrm{U}_1 & =\mathrm{CE}^2
\end{aligned}
$$
Now dielectric intorduces after pening the switch s.
Energy stored in $\mathrm{A}=\frac{1}{2} \mathrm{KCE}^2$
Let the charge stared in capacitor B initially be Q. In an open circuit, the charge will remain same.
$$
\mathrm{Q}=\mathrm{EC}
$$
Energy stored in $B=\frac{1}{2} \frac{Q^2}{K C}$
$$
=\frac{1}{2} \frac{(\mathrm{EC})^2}{\mathrm{KC}}=\frac{1}{2} \frac{\mathrm{E}^2 \mathrm{C}^2}{\mathrm{KC}}=\frac{1}{2} \frac{\mathrm{CE}^2}{\mathrm{~K}}
$$
Total energy, $\mathrm{U}_2=\frac{1}{2} \mathrm{KCE}^2+\frac{1}{2} \frac{\mathrm{CE}^2}{\mathrm{~K}}$
$$
=\frac{1}{2} \mathrm{CE}^2\left(\mathrm{~K}+\frac{1}{\mathrm{~K}}\right)
$$
Ratio, $\frac{\mathrm{U}_1}{\mathrm{U}_2}=\frac{\mathrm{CE}^2}{\frac{1}{2} \mathrm{CE}^2\left(\mathrm{~K}+\frac{1}{\mathrm{~K}}\right)}$
Here, $\mathrm{K}=3$
$$
=\frac{2}{3+\frac{1}{3}}=\frac{6}{10}=\frac{3}{5}
$$

Topic: Physics, Capacitance, AP EAMCET, AP EAMCET 2023 (19 May Shift 1)
Two identical capacitors A and B, charged to the same potential 5V are connected in two different circuits as shown below at time t=0. If the charge on capacitors A and B at time t=CR is QA and QB respectively, then (Here e is the base of natural logarithm)

Options:
A QA=VCe,QB=CV2
B QA=VC,QB=CV
C QA=VC,QB=VCe
D QA=CV2,QB=VCe

The correct answer is:
QA=VC,QB=VCe

Solution:
Charge on capacitor after charging with a battery of potential V, Q=CV

Now, capacitor A is connected as shown in the figure.

As diode is in reverse bias, so capacitor charge remains same after time t=CR.

∴ QA=CV

In case B, capacitor is connected as shown in the figure.

Here, diode is in forward bias, so capacitor discharges and its charge decays as

QB=Q·e-tRC

where, t=RC

Topic: Physics, Alternating Current, JEE Main, JEE Main 2020 (09 Jan Shift 2)
Two identical capacitors A and B of capacitance C each, are charged to the same potential V and are connected in two circuits at t=0, as shown in the figure i and ii respectively. The charge on the capacitors at time t=CR are respectively

Options:
A VC, VC
B VCe, VC
C VC,VCe
D VCe,VCe

The correct answer is:
VCe, VC

Solution:
Time t=CR is known as time constant. It is time in which charge on the capacitor decreases to 1e times of its initial charge (steady-state charge).
In figure (i) PN junction diode is in forward bias, so current will flow in the circuit i.e., charge on the capacitor decrease and in time t it becomes Q=1eQo ; where Q0=CV⇒Q=CVe
In figure (ii) P-N junction diode is in reverse bias, so no current will flow through the circuit hence change on the capacitor will not decay and it remains same i.e. CV after time t.

Topic: Physics, Semiconductors, NEET
Two identical capacitors A and B of capacitance C each, are charged to the same potential V and are connected in two circuits at t=0, as shown in the figure i and ii respectively. The charge on the capacitors at time t=CR are respectively

Options:
A VC, VC
B VCe, VC
C VC,VCe
D VCe,VCe

The correct answer is:
VCe, VC

Solution:
Time t=CR is known as time constant. It is time in which charge on the capacitor decreases to 1e times of its initial charge (steady-state charge).
In figure (i) PN junction diode is in forward bias, so current will flow in the circuit i.e., charge on the capacitor decrease and in time t it becomes Q=1eQo ; where Q0=CV⇒Q=CVe
In figure (ii) P-N junction diode is in reverse bias, so no current will flow through the circuit hence change on the capacitor will not decay and it remains same i.e. CV after time t.

Topic: Physics, Semiconductors, JEE Main
Two identical capacitors have same capacitance C. One of them is charged to the potential V and other to the potential 2V. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is :

Options:
A 14CV2
B 2CV2
C 12CV2
D 34CV2

The correct answer is:
14CV2

Solution:
The common potential after connection would be, VC=qnetCnet=CV+C2V2C

⇒VC=3V2

Therefore, loss of energy

=12CV2+12C2V2−122C3V22

=CV24

Topic: Physics, Capacitance, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Two identical capacitors, have the same capacitance $\mathrm{C}$. One of them is charged to potential $\mathrm{V}_{1}$ and the other to $\mathrm{V}_{2}$. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is -

Options:
A $\frac{1}{4} \mathrm{C}\left(\mathrm{V}_{1}^{2}-\mathrm{V}_{2}^{2}\right)$
B $\frac{1}{4} C\left(V_{1}^{2}+V_{2}^{2}\right)$
C $\frac{1}{4} \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}$
D $\frac{1}{4} C\left(V_{1}+V_{2}\right)^{2}$

The correct answer is:
$\frac{1}{4} \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}$

Solution:
Initial energy of combined system

$$

\mathrm{U}_{1}=\frac{1}{2} \mathrm{CV}_{1}^{2}+\frac{1}{2} \mathrm{CV}_{2}^{2}

$$

Final common potential, $\mathrm{V}=\frac{\mathrm{V}_{1}+\mathrm{V}_{2}}{2},$

Final energy of system,

$$

\mathrm{U}_{2}=2 \times \frac{1}{2} \mathrm{C}\left(\frac{\mathrm{V}_{1}+\mathrm{V}_{2}}{2}\right)^{2}

$$

Hence loss of energy $=\mathrm{U}_{1}-\mathrm{U}_{2}$

$$

=\frac{1}{4} \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}

$$

Topic: Physics, Capacitance, JEE Main
Two identical cells each of emf 1.5 V are connected in parallel across a parallel combination of two resistors each of resistance 20 Ω. A voltmeter connected in the circuit measures 1.2 V. The internal resistance of each cell is :

Options:
A 2.5 Ω
B 4 Ω
C 5 Ω
D 10 Ω

The correct answer is:
5 Ω

Solution:
Equivalent resistance of the parallel 20 Ω branches will be 10 Ω.

Equivalent EMF of the two branches which are in parallel will be, =1.5r+1.5r1r+1r=1.5 V and the equivalent internal resistance for same branches will be, r2.

For loop containing voltmeter and 10 Ω resistor, 1.2-10i=0 ⇒i=0.12 A.

Reading of voltmeter should be,

1.2=1.5-ir2 ⇒ir2=0.3 ⇒r=0.6i=0.60.12= 5 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (24 Jun Shift 1)
Two identical cells each of emf 1.5 V are connected in series across a 10 Ω resistance. An ideal voltmeter connected across 10 Ω resistance reads 1.5 V. The internal resistance of each cell is _____ Ω.

Options:

The correct answer is:
5

Solution:
Let R1=10 Ω.

Using V=iR

i=310+2r

The voltmeter reading is V'=1.5 V.

Thus,
iR1=V'⇒10×310+2r=1.5
⇒20=10+2 r
⇒2r=10
⇒r=5 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (11 Apr Shift 2)
Two identical cells, when connected either in parallel or in series gives same current in an external resistance 5 Ω. The internal resistance of each cell will be ______ Ω.

Options:

The correct answer is:
5

Solution:
Let r be the internal resistance of each cell.

In above shown figure, first represents the series connection of cells. Second and third figures represents the parallel connection of cells.

Now, in case of series connection, εnet=2ε and rnet=2r. So, current i=2ε5+2r .....(1)

In case of parallel connection, εnet=εr+εrr+r=ε and rnet=r2. So, current i=εr2+5 .....(2)

Equating (1) and (2), we get

2ε5+2r=εr2+5

⇒r+10=5+2r

⇒r=5 Ω.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Two identical charged particles each having a mass 10 g and charge 2.0×10-7C are placed on a horizontal table with a separation of L between them such that they stay in limited equilibrium. If the coefficient of friction between each particle and the table is 0.25, find the value of L. [Use g=10 ms-2 ]

Options:
A 12 cm
B 10 cm
C 8 cm
D 5 cm

The correct answer is:
12 cm

Solution:
Normal reaction applied by table on object will be, N=mg.

Both charges are same, hence a repulsive force will act between them.

Value of force acting will be, kQ2L2.

Now the maximum value of friction force which table can provide is,

=μN=μmg

For equilibrium, kQ2L2=μmg.

L=kQ2μmg=k0.25mgQ=2kmgQ

=29×10910×10-3×102×10-7=0.12 m=12 cm

Topic: Physics, Electrostatics, JEE Main, JEE Main 2022 (24 Jun Shift 2)
Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of $30^{\circ}$ with each other. When suspended in a liquid of density $0.8 \mathrm{~g} \mathrm{~cm}^{-3}$, the angle remains the same. If density of the material of the sphere is $16 \mathrm{~g} \mathrm{~cm}^{-3}$, the dielectric constant of the liquid is

Options:
A
4

B
3

C
2

D
1

The correct answer is:
2

Solution:
From F.B.D of sphere, using Lami's theorem
$$
\frac{\mathrm{F}}{\mathrm{mg}}=\tan \theta
$$
when suspended in liquid, as $\theta$ remains same,
$\therefore \quad \frac{\mathrm{F}^{\prime}}{\mathrm{mg}\left(1-\frac{\rho}{\mathrm{d}}\right)}=\tan \theta$
using (i) and (ii)
$$
\frac{F}{m g}=\frac{F^{\prime}}{m g\left(1-\frac{\rho}{d}\right)} \text { where, } F^{\prime}=\frac{F}{K}
$$
$\therefore \quad \frac{\mathrm{F}}{\mathrm{mg}}=\frac{\mathrm{F}^{\prime}}{\mathrm{mg} \mathrm{K}\left(1-\frac{\rho}{\mathrm{d}}\right)}$
or $\quad \mathrm{K}=\frac{1}{1-\frac{\rho}{d}}=2$

Topic: Physics, Electrostatics, JEE Main
Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of $30^{\circ}$ with each other. When suspended in a liquid of density $0.8 \mathrm{~g} \mathrm{~cm}^{-3}$, the angle remains the same. If density of the material of the sphere is $16 \mathrm{~g} \mathrm{~cm}^{-3}$, the dielectric constant of the liquid is

Options:
A
4

B
3

C
2

D
1

The correct answer is:
2

Solution:
From F.B.D of sphere, using Lami's theorem
$$
\frac{\mathrm{F}}{\mathrm{mg}}=\tan \theta
$$
when suspended in liquid, as $\theta$ remains same,
$\therefore \quad \frac{\mathrm{F}^{\prime}}{\mathrm{mg}\left(1-\frac{\rho}{\mathrm{d}}\right)}=\tan \theta$
using (i) and (ii)
$$
\frac{F}{m g}=\frac{F^{\prime}}{m g\left(1-\frac{\rho}{d}\right)} \text { where, } F^{\prime}=\frac{F}{K}
$$
$\therefore \quad \frac{\mathrm{F}}{\mathrm{mg}}=\frac{\mathrm{F}^{\prime}}{\mathrm{mg} \mathrm{K}\left(1-\frac{\rho}{\mathrm{d}}\right)}$
or $\quad \mathrm{K}=\frac{1}{1-\frac{\rho}{d}}=2$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2010
Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle θ with each other. When suspended in water the angle remains the same. If density of the material of the sphere is 1.5 g/cc, the dielectric constant of water will be ______.

(Take density of water =1 g/cc)

Options:

The correct answer is:
3

Solution:
In air:

tanθ2=Fmg=q24πε0r2mg

In water:

tanθ2=F'mg'=q24πε0εrr2mgeff

Equate both equations, we get

ε0 g=ε0εrg1-11.5

⇒εr=3

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Two identical charged spheres of material density $\rho$, suspended from the same point by inextensible strings of equal length make an angle $\theta$ between the strings. When suspended in a liquid of density $\sigma$ the angle $\theta$ remains the same. The dielectric constant $K$ of the liquid is

Options:
A $\frac{\rho}{\rho-\sigma}$
B $\frac{\rho-\sigma}{\rho}$
C $\frac{\rho}{\rho+\sigma}$
D $\frac{\rho+\sigma}{\rho}$

The correct answer is:
$\frac{\rho}{\rho-\sigma}$

Solution:
$\begin{aligned} E_{r} &=\frac{F_{a}}{F_{m}} \\ &=\frac{m g}{\left(m-m_{1}\right) g}=\frac{\rho V}{(\rho-\sigma) V}=\frac{\rho}{(\rho-\sigma)} \end{aligned}$

Topic: Physics, Electrostatics, KCET, KCET 2011
Two identical charged spheres suspended from a common point by two massless strings of length I are initially a distance $d(d< < 1)$ apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity $v$. Then as a function of distance $x$ between them,

Options:
A
$v \propto x^{-1}$

B
$v \propto x^{1 / 2}$

C
$v \propto x$

D
$v \propto x^{-1 / 2}$

The correct answer is:
$v \propto x^{-1 / 2}$

Solution:
At any instant of separation between charges is $x$.
$$
\begin{aligned}
& \text { equilibrium condition }=\mathrm{K} \frac{\mathrm{Q}^2}{\mathrm{x}^2}=\omega \frac{\mathrm{x}}{2 \ell} \\
& \Rightarrow \mathrm{Q}^2=\mathrm{Cx}^3 \\
& \Rightarrow 2 \mathrm{Q} \frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{C} 3 \mathrm{x}^2 \frac{\mathrm{dx}}{\mathrm{dt}} \\
& \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}} \propto \frac{\mathrm{x}^{3 / 2}}{\mathrm{x}^2} \propto \mathrm{x}^{-1 / 2}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2011
Two identical circular coils $A$ and $B$ are kept on a horizontal tube side by side without touching each other. If the current in the coil $A$ increases with time, in response, the coil $B$

Options:
A is attracted by $A$
B remains stationary
C is repelled
D rotates

The correct answer is:
is repelled

Solution:
No Solution Available

Topic: Physics, Electromagnetic Induction, AIIMS, AIIMS 2018 (27 May)
Two identical circular coils $A$ and $B$ are kept on a horizontal tube side by side without touching each other. If the current in the coil $A$ increases with time, in response, the coil $B$

Options:
A is attracted by A
B remains stationary
C is repelled
D rotates

The correct answer is:
is repelled

Solution:
As the current in coil $A$ changes, the magnetic flux linked with coil $B$ changes, hence as a result current induces in coil $B$. The direction of induced current in coil $B$ is opposite to that of the direction of current in coil $A$ as shown in figure
Due to currents in opposite directions in the near by sides of the coil, the coil $B$ is repelled.

Topic: Physics, Electromagnetic Induction, KCET, KCET 2013
Two identical circular wires of radius 20 cm and carrying current 2A are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wires is________×10-8T.

(Take π=3.14)

Options:

The correct answer is:
628

Solution:
The formula to calculate the magnetic field B at the centre of a current carrying circular loop is given by

B= μ0i2R ...(1)

where, μ0 is the permeability of free space, i is the current and R is the radius of the loop.

BV, BH are the magnetic field due to the vertical and the horizontal loop, which by magnitude are equal.

The resultant magnetic field Bnet at the centre is, then, given by

Bnet =B2+B2=2B=2×4π×10-7×22×0.2

=628×10-8 T

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Two identical condensers $M$ and $N$ are connected in series with a battery. The space between the plates of $M$ is completely filled with a dielectric medium of dielectric constant 8 and a copper plate of thickness $\frac{d}{2}$ is introduced between the plates of $N$. ( $d$ is the distance between the plates). Then potential differences across $M$ and $N$ are, respectively, in the ratio

Options:
A 1 : 4
B 4 : 1
C 3 : 8
D 1 : 6

The correct answer is:
1 : 4

Solution:
Let the capacitance of condensor, $M=C$
The new capacitance of condensor, $C_M=8 C$
The capacitance of condensor, $N=C$
The new capacitance of condensor, $C_N=\frac{C^{\prime}}{2}$
where $\quad C^{\prime}=\frac{A \varepsilon_0}{d / 4}=\frac{4 A \varepsilon_0}{d}=4 C$
$\therefore \quad C_N=\frac{C^{\prime}}{2}=\frac{4 C}{2}=2 C$
Since the two condensers are connected in series
$\begin{aligned}
\therefore \quad \quad V_M: V_N & =\frac{1}{C_M}: \frac{1}{C_N} \\
& =\frac{1}{8 C}: \frac{1}{2 C}=1: 4 \\
\Rightarrow \quad V_M: V_N & =1: 4
\end{aligned}$

Topic: Physics, Capacitance, JEE Main
Two identical conducting rods \( A B \) and \( C D \) are connected to a circular conducting ring at two diametrically opposite points \( B \) and \( C \). The radius of the ring is equal to the length of rods \( A B \) and \( C D \). The area of cross-section and thermal conductivity of the rod and ring are equal. Points \( \mathrm{A} \) and \( D \) are maintained at temperatures of \( 100^{\circ} \mathrm{C} \) and \( 0{ }^{\circ} \mathrm{C} \). Temperature of point \( C \) will be,

Options:
A \( 62^{\circ} \mathrm{C} \)
B \( { }^{37}{ }^{\circ} \mathrm{C} \)
C \( 28^{\circ} \mathrm{C} \)
D \( 45^{\circ} \mathrm{C} \)

The correct answer is:
\( 28^{\circ} \mathrm{C} \)

Solution:
Given that the radius of ring is equal to length of rods AB and CD. Suppose area of cross -section and thermal conductivity is same .

The temperature at A is 100 °C and at point D is 0 °C.

Resistance R∝l (length of conductor)

We can write,

RAB=RCD=R

Circumference of semicircle =πl (given radius is equal to length of rod)

∴ The resistance of semicircle =πR, because R∝l.

There are two resistors which are in parallel combination along the circle.

So, the equivalent resistance of parallel combination is

RBC=πR2.

Heat current along wire AD,

H=∆TReqand Req=R+π2R+R

Flow rate of heat along wire AD, HAD=100-0R+π2·R+R=1004+πR2=28R units

In steady state, flow rate of heat current will be same throughout the length of wire. So we can write,

HAD=HCD

28R=θC-0R

∴ θC=28 °C

Topic: Physics, Thermal Properties of Matter, JEE Main
Two identical conducting spheres A and B carry an equal charges. They are separated by a distance much larger than their diameters, and the force between them is F. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and then removed. As a result, the force between A and B would be equal to:

Options:
A 3F4
B F2
C 3F8
D F

The correct answer is:
3F8

Solution:
F=kq2r2 when A and C are touched charge on both will be q2. Then when B and C are touched, qB= q2+q2=3q4
F′=kqAqBr2= k ×q2 ×3q4r2=38kq2r2=38F.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2018 (16 Apr Online)