datavorous/entrance-exam-dataset huggingface dataset:
Two identical conducting wires $\mathrm{A O B}$ and $\mathrm{C O D}$ are placed at right angles to each other. The wire $\mathrm{AOB}$ carries an electric current $\mathrm{I_1}$ and $\mathrm{COD}$ carries a current $\mathrm{I_2}$. The magnetic field on a point lying at a distance ' $\mathrm{d}$ ' from $\mathrm{O}$, in a direction perpendicular to the plane of the wires $\mathrm{A O B}$ and $\mathrm{C O D}$, will be given by

Options:
A
$\frac{\mu_0}{2 \pi}\left(\frac{\mathrm{I}_1+\mathrm{I}_2}{d}\right)^{1 / 2}$

B
$\frac{\mu_0}{2 \pi \mathrm{d}}\left(\mathrm{I}_1^2+\mathrm{I}_2^2\right)^{1 / 2}$

C
$\frac{\mu_0}{2 \pi d}\left(\mathrm{I}_1+\mathrm{I}_2\right)$

D
$\frac{\mu_0}{2 \pi d}\left(\mathrm{I}_1^2+\mathrm{I}_2^2\right)$

The correct answer is:
$\frac{\mu_0}{2 \pi}\left(\frac{\mathrm{I}_1+\mathrm{I}_2}{d}\right)^{1 / 2}$

Solution:
$\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{d}} \sqrt{\left(\mathrm{I}_1^2+\mathrm{I}_2^2\right)}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2007
Two identical containers A and B having the same volume of an ideal gas at the same temperature have the mass of the gas as m1 and m2 respectively and 2m1=3m2. The gas in each cylinder expands isothermally to double of its volume. If the change in pressure in A is 300 Pa, then the change in pressure in B is

Options:
A 200 Pa

B 300 Pa

C 400 Pa

D 500 Pa

The correct answer is:
200 Pa

Solution:
2m1=3m2
P1V=m1MRT and P2V=m2MRT=23m1MRT
P1P2=32 â¦â¦â¦(i)
The process is isothermal P1V1=P2V2
Hence, âP1=P1-P12âP1=2âP1
âP2=P2-P22=P22=1223P1=P13
âP2=23âP1=23.300
âP2=200 Pa

Topic: Physics, Thermodynamics, JEE Main
Two identical containers $A$ and $B$ with frictionless pistons contain the same ideal gas at the same temperature and the same volume $V$. The mass of the gas in $A$ is $m_A$ and that in $B$ is $m_B$. The gas in each cylinder is now allowed to expand isothermally to the same final volume $2 \mathrm{~V}$. The changes in the pressure in $A$ and $B$ are found to be $\Delta P$ and $1.5 \Delta P$ respectively. Then

Options:
A $4 m_A=9 m_B$
B $2 m_A=3 m_B$
C $3 m_A=2 m_B$
D $9 m_A=3 m_B$

The correct answer is:
$3 m_A=2 m_B$

Solution:
Process is isothermal. There fore $T=$ constant,
$\left(P \propto \frac{1}{V}\right)$ volume is increasing, therefore pressure will decreases.
In chamber $A$ :
$\Delta P=P_i-P_f=\frac{\mu_A R T}{V}-\frac{\mu_A R T}{2 V}=\frac{\mu_A R T}{2 V}$ ...(i)
In chamber $B$ :
$1.5 \Delta P=P_i-P_f=\frac{\mu_B R T}{V}-\frac{\mu_B R T}{2 V}=\frac{\mu_B R T}{2 V}$ ...(ii)
from equations (i) and (ii) $\frac{\mu_{\mathrm{A}}}{\mu_B}=\frac{1}{1.5}=\frac{2}{3}$
$\Rightarrow \frac{m_A / M}{m_B / M}=\frac{2}{3} \Rightarrow 3 m_A=2 m_B$.

Topic: Physics, Thermodynamics, JEE Main
Two identical containers of the same emissivity contain liquid A and B at same initial temperature 60 â o C and densities ÏA & ÏBÏA<ÏB. Which of the following plot best represents the temperature variation of both with time? (Given SA=1000 J kg-1 K-1 & SB=2000 J kg-1 K-1)

Options:
A
B
C
D

The correct answer is:

Solution:
From Newton's law of cooling, rate of cooling: -dTdt=4ÏeAT03T-T0ms
âµÏA<ÏB
mA<mB
And sA<sB
â´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 cylindrical vessels are kept on the ground and each contain the same liquid of density d. The area of the base of both vessels is S but the height of liquid in one vessel is x1 and in the other x2. When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is :

Options:
A gdSx22+x12
B gdSx2+x12
C 34gdSx2-x12
D 14gdSx2-x12

The correct answer is:
14gdSx2-x12

Solution:
Initial height of liquid in container's of same cross section are x1 and x2 respectively. Now value is opened find loss in potential energy when water level be become same

loss in PE=Ui-Uf

=ρ(A)x1x12+ρAx2x22g-ρAx1+x22×x1+x24×2g

=ρAgx122+x222-x1+x124=ρAgx1-x224

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2020 (04 Sep Shift 2)
Two identical electric lamps marked $500 \mathrm{~W}, 220 \mathrm{~V}$ are connected in series and then joined to a $110 \mathrm{~V}$ line. The power consumed by each lamp is

Options:
A $\frac{125}{4} W$
B $\frac{25}{4} w$
C $\frac{225}{4} W$
D $125 \mathrm{~W}$

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

Solution:
Voltage across each bulb $V=\frac{110}{2}=55 \mathrm{~W}$ so,power consumed by each bulb will be
$\begin{aligned} P^{\prime} & =\left(\frac{55}{220}\right)^2 \times 500 \\ & =\frac{125}{4} W\end{aligned}$

Topic: Physics, Current Electricity, JEE Main
Two identical electric point dipoles have dipole have dipole moments p→1=pi^ and p→2=−pi^ and are held on the x-axis at distance 'a' from each other. When released, they move along the x-axis with the direction of their dipole moments remaining unchanged. If the mass of each dipole is 'm', their speed when they are infinitely far apart is :

Options:
A Pa1πε0ma
B Pa12πε0ma
C Pa2πε0ma
D Pa32πε0ma

The correct answer is:
Pa12πε0ma

Solution:
ki+ui=kf+uf

0-2kpr3p2·cos180°=12mv2+12mv2+0

mv2=2kp1p2r3

⇒θ=2kp1p2mr3

p1=p2=p&r=a

v=pa12πϵ0ma

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (06 Sep Shift 2)
Two identical glass bulbs are interconnected by a thin glass tube. A gas is filled in these bulbs at NTP. If one bulb is placed in ice and another bulb is placed inside a hot bath, then the pressure of the gas becomes 1.5 times. The temperature of the hot bath will be

Options:
A 100 °C

B 182 °C

C 256 °C

D 546 Â°C

The correct answer is:
546 Â°C

Solution:
Quantity of gas in these bulbs is constant, i.e.,
initial no. of moles in both the bulbs = final number of moles
μ 1 + μ 2 = μ 1 ′ + μ 2 ′
PVR273+PVR273=1.5PVR273+1.5PVRT
⇒ 2273=1.5273+1.5T
⇒ T=819K=546 °C

Topic: Physics, Thermodynamics, JEE Main
Two identical glass spheres filled with air are connected by a horizontal glass tube. The glass tube contains a pellet of mercury at its mid-point. Air in one sphere is at $0^{\circ} \mathrm{C}$ and the other is at $20^{\circ} \mathrm{C}$. If both the vessels are heated through $10^{\circ} \mathrm{C}$, then neglecting the expansions of the bulbs and the tube

Options:
A the mercury pellet gets displaced towards the sphere at lower temperature.
B the mercury pellet gets displaced towards the sphere at higher temperature.
C the mercury pellet does not get displaced at all.
D the temperature rise causes the pellet to expand without any displacement.

The correct answer is:
the mercury pellet does not get displaced at all.

Solution:
Let $n_1$ and $n_2$ be the number of moles in the bulbs at $0^{\circ} \mathrm{C}$ and $20^{\circ} \mathrm{C}$, respectively. Then,
$p V=273 n_1 R=293 n_2 R$
$\therefore \frac{n_1}{n_2}=\frac{293}{273}...(i)$
When the vessels are heated, let the volume of low temperature be $V_1$ and that of the other be $V_2$. Since, pressure are same, hence
$V_1 p=283 n_1 R$ and $V_2 p=303 n_2 R$
$\therefore \frac{V_1}{V_2}=\frac{283 n_1}{303 n_2}...(ii)$
From Eqs. (i) and (ii), we get
$\frac{V_1}{V_2}=\frac{283}{303} \times \frac{293}{273}=1$
Thus, the mercury pellet remains at the same position.

Topic: Physics, Kinetic Theory of Gases, JIPMER, JIPMER 2017
Two identical heater filaments are connected first in parallel and then in series. At the same applied voltage, the ratio of heat produced in same time for parallel to series will be:

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

The correct answer is:
4:1

Solution:
Let the resistance of the filaments be R and the voltage be V.

For the series combination the power is

P1=V22R

For parallel combination,

P2=V2R2=2V2R

Hence, the ratio is

⇒P2P1=41

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Two identical ladders are arranged as shown in the figure. Mass of the block is m and the mass of each ladder is M. The length of each of the ladder is L. The system is in equilibrium. What is the magnitude of frictional force acting at A or B?

Options:
A mg2cosθ
B Mg2cosθ
C M-m2g cos θ
D M+m2gcotθ

The correct answer is:
M+m2gcotθ

Solution:
Drawing force diagrams of the rod, we have
This is the equilibrium of coplanar forces, hence using the equations
∑ F x = 0 , ∑ F y = 0 and

Net moment about point O = 0

we have the equations,

N+N2=Mg+mg2... (i)

N1 = f ...(ii)
N = N2+Mg+mg2 ... (iii)

and MgL2 cos θ + fLsin θ = NLcos θ ... (iv)
Solving these four equations, we have

f⁡=M+m2gcotθ

Topic: Physics, Laws of Motion, JEE Main
Two identical light sources $S_1$ and $S_2$ emit light of same wavelength $\lambda$. These light rays will exhibit interference if

Options:
A Their phase differences remain constant
B Their phases are distributed randomly
C Their light intensities remain constant
D Their light intensities change randomly

The correct answer is:
Their phase differences remain constant

Solution:
For interference phase difference must be constant.

Topic: Physics, Wave Optics, JEE Main
Two identical light sources $S_{1}$ and $S_{2}$ emit light of same wavelength $\lambda .$ These light rays will exhibit interference if

Options:
A their phase differences remain constant
B their phases are distributed randomly
C their light intensities remain constant
D their light intensities change randomly

The correct answer is:
their phase differences remain constant

Solution:
For inter ference phase difference must be constant.

Topic: Physics, Electrostatics, VITEEE, VITEEE 2010
Two identical masses are as shown in the diagram. One is thrown upwards with velocity 20 m s-1 and another is just dropped simultaneously. If the collision between them is elastic, find the time interval between their striking with ground

Options:
A Zero
B 2 s

C 1 s

D 3 s

The correct answer is:
2 s

Solution:
In elastic collision between two identical masses, the masses just exchange their velocities. So, time interval would remain the same as would have been it they had not collided.
t1=2hg=2×2010=2 s
t2=2ug=2×2010=4 s
∴ Δt=t2-t1=2 s
The only difference will be that now the lower ball will collide at 2 s and the upper ball at 4 s.

Topic: Physics, Center of Mass Momentum and Collision, NEET
Two identical metal plates are given positive charges Q1 and Q2(<Q1) respectively. If they are now brought close together to form a parallel plates capacitor with capacitance C, the potential difference between them is

Options:
A Q1+Q22C
B Q1+Q2C
C Q1-Q2C
D Q1-Q22C

The correct answer is:
Q1-Q22C

Solution:
For charge Q1, electric field is
E1=Q12ε0A
where A is area.
For charge Q2 electric field is
E2=Q22ε0A
Resultant electric field
E=E1-E2=(Q1-Q2)2ε0A
Potential difference between plates when they are brought close together to form a parallel plate capacitor is
V=Ed=(Q1-Q2)2ε0Ad
Since, C=ε0Ad for parallel plate capacitor.
V=(Q1-Q2)2C

Topic: Physics, Electrostatics, JEE Main
Two identical metal wires of thermal conductivities K1 and K2 respectively are connected in series. The effective thermal conductivity of the combination is:

Options:
A 2 K1 K2 K1+K2
B K1+K22 K1 K2
C K1+K2 K1 K2
D K1 K2 K1+K2

The correct answer is:
2 K1 K2 K1+K2

Solution:
Reff=l K1 A+l K2 A=2l KeqA

Keq=2 K1 K2 K1+K2

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2021 (17 Mar Shift 1)
Two identical metallic square loops $L_{1}$ and $L_{2}$ are placed next to each other with their sides parallel on a smooth horizontal table. Loop $L_{1}$ is fixed and a current which increases as a function of time is passed through it. Then loop $\mathrm{L}_{2}$ :

Options:
A rotates about its center of mass
B moves towards $\mathrm{L}_{1}$
C remains stationary
D moves away from $L_{1}$

The correct answer is:
moves away from $L_{1}$

Solution:

Topic: Physics, Electromagnetic Induction, KVPY, KVPY 2014 (SB/SX)
Two identical objects A and B are at temperatures TA and TB respectively. Both objects are placed in a room with perfectly absorbing walls maintained at a temperature T(TA>T>TB). The objects A and B attain the temperature T eventually. Select the correct statements from the following

Options:
A A only emits radiations, while B only absorbs it until both attain the temperature T
B A loses more heat by radiation than it absorbs, while B absorbs more radiation than it emits until they attain the temperature T
C Both A and B only absorb radiation but do not emit it, until they attain the temperature T.
D B only emits radiations, while A only absorbs it until both attain the temperature T

The correct answer is:
A loses more heat by radiation than it absorbs, while B absorbs more radiation than it emits until they attain the temperature T

Solution:
Every object emits and absorbs the radiations simultaneously. If the energy emitted is more than the energy absorbed temperature falls and vice versa.

Topic: Physics, Thermal Properties of Matter, NEET
Two identical parallel plate air capacitors are connected in series to a battery of emf V. If one of the capacitor is completely filled with dielectric material of constant K, then potential difference of the other capacitor will become

Options:
A KVK+1
B VK+1
C K-1VK
D VKK+1

The correct answer is:
KVK+1

Solution:
Both the capacitance will have same charge V 1 = q C and V 2 = q KC
But V 1 + V 2 =V
⇒ q C + q KC =V
⇒ q C . K+1 K =V
Potential difference across capacitor without dielectric is
V'= q C = K K+1  V

Topic: Physics, Electrostatics, JEE Main
two identical parallel plate capacitors of capacitance $C$ each are connected in series with abattery of emf, E as shown, If one of the capacitors is now filled with a dielectric of dielectric constant $\mathrm{k}$, the amount of charge which will flow through the battery is (neglect internal resistance of the battery)

Options:
A $\frac{k+1}{2(k-1)} C E$
B $\frac{k-1}{2(k+1)} C E$
C $\frac{\mathrm{k}-2}{\mathrm{k}+2} \mathrm{CE}$
D $\frac{\mathrm{k}+2}{\mathrm{k}-2} \mathrm{CE}$

The correct answer is:
$\frac{k-1}{2(k+1)} C E$

Solution:
$\begin{array}{l}
Q=\frac{C}{2} E \\
C e q=\frac{C \cdot K C}{C+K C}=\frac{C^{2} K}{C(1+k)}=\frac{C K}{1+K} \\
Q=\frac{C K}{1+k} E \\
\Delta Q=\left(\frac{C K}{1+k}-\frac{C}{2}\right) E
\end{array}$
$\begin{aligned} \Delta Q &=\left(\frac{C K}{1+k}-\frac{C}{2}\right) E \\ &=\frac{2 c k-c(k-1)}{2(1+k)} \\ &=\frac{2 c k-c k-c}{2(1+k)}=\frac{c k-c}{2(1+k)} \end{aligned}$

Topic: Physics, Capacitance, JEE Main
Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d . The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K1, K2 and K3 . The first capaciitor is filled as shown in figure I, and the second one is filled as shown in figure II.
If these two modified capacitors are charged by the same potential V , the ratio of the energy stored in the two, would be (E1 refers to capacitor I and E2 to capacitor (II) ):

Options:
A E1E2=9K1K2K3(K1K2K3)(K2K3 + K3K1K1+K1K2)
B E1E2=(K1K2K3)(K2K3 + K3K1K1+K1K2)K1K2K3
C E1E2=(K1K2K3)(K2K3 + K3K1K1+K1K2)9K1K2K3
D E1E2=K1K2K3(K1K2K3)(K2K3 + K3K1K1+K1K2)

The correct answer is:
E1E2=9K1K2K3(K1K2K3)(K2K3 + K3K1K1+K1K2)

Solution:
Energy stored in a capacitor E=12CV2
⇒E1E2=C1C2
C1=d3 ε0k1A+d3ε0Ak2+d3ε0Ak3-1
=d3Aε0-11k1+1k2+1k3-1
=d3Aε0-1k1k2+k2k3+k3k1k1k2k3-1
C1=3Aε0dk1k2k3k1k2+k2k3+k3k1
C2=ε0A3k1d+ε0A3k2d+ε0A3k3d
∴C1C2=E1E2=9 k1k2k3k1+k2+k3k1k2+k2k3+k3k1

Topic: Physics, Capacitance, JEE Main, JEE Main 2019 (12 Apr Shift 1)
Two identical particles each of mass m go round a circle of radius a under the action of their mutual gravitational attraction. The angular speed of each particle will be :

Options:
A Gma3
B Gm8a3
C Gm4a3
D Gm2a3

The correct answer is:
Gm4a3

Solution:
The gravitational force FG between the particles is given by

FG=Gm22a2= Gm24a2 ...1

The centripetal force FC of each particle can be written as

FC=mω2a ...2

Under balanced condition, equate equation (1) and equation (2) and simplify to obtain the angular speed of each particle.

mω2a=Gm24a2⇒ω2=Gm4a3⇒ω=Gm4a3

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (15 Apr Shift 1)
Two identical particles move towards each other with velocity $2 v$ and $v$ respectively. The velocity of centre of mass is

Options:
A
v

B
v/3

C
v/2

D
zero

The correct answer is:
v/2

Solution:
Conserving Linear Momentum
$$
2 \mathrm{Mv}_{\mathrm{c}}=2 \mathrm{Mv}-\mathrm{Mv} \Rightarrow \mathrm{v}_{\mathrm{c}}=\mathrm{v} / 2
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2002
Two identical particles of mass 1 kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is:

Options:
A G2R3
B 12GR3
C 12R1G
D 2GR3

The correct answer is:
12GR3

Solution:
F=Gm2(2R)2=mRω2

ω=12GR3

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (27 Jul Shift 2)
Two identical photocathodes receive light of frequencies $\mathrm{f}_1$ and $\mathrm{f}_2$. If the velocities of the photo electrons (of mass $\mathrm{m}$ ) coming out are respectively $\mathrm{v}_1$ and $\mathrm{v}_2$, then

Options:
A
$v_1^2-v_2^2=\frac{2 h}{m}\left(f_1-f_2\right)$

B
$v_1+v_2=\left[\frac{2 h}{m}\left(f_1+f_2\right)\right]^{1 / 2}$

C
$v_1^2+v_2^2=\frac{2 h}{m}\left(f_1+f_2\right)$

D
$\mathrm{v}_1-\mathrm{v}_2=\left[\frac{2 h}{\mathrm{~m}}\left(\mathrm{f}_1-\mathrm{f}_2\right)\right]^{1 / 2}$

The correct answer is:
$v_1^2-v_2^2=\frac{2 h}{m}\left(f_1-f_2\right)$

Solution:
No Solution Available

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2003
Two identical piano strings of length 0.750 m are each, tuned exactly to 500 Hz . The tension in one of the strings is then increased by 21% . If they are now struck, what is the beat frequency between the fundamental mode of the two strings?

Options:
A 45
B 50
C 60
D 40

The correct answer is:
50

Solution:
f1=12lTμ=500
f2=12l1.21Tμ=1.1500
f2-f1=50

Topic: Physics, Waves and Sound, JEE Main
Two identical piano wires have a fundamental frequency of 600 cycle per second when kept under the same tension. What fractional increase in the tension of one wires will lead to the occurrence of 6 beats per second when both wires vibrate simultaneously?

Options:
A $0.01$
B $0.02$
C $0.03$
D $0.04$

The correct answer is:
$0.02$

Solution:
When both the wires vibrate simultaneously, beats per second,
$$
\begin{array}{c}
n_{1} \pm n_{2}=6 \\
\text { or } \frac{1}{2 l} \sqrt{\frac{T}{m}} \pm \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}=6 \\
\Rightarrow \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-\frac{1}{2 l} \sqrt{\frac{T}{m}}=6 \\
\Rightarrow \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}-600=6 \Rightarrow \frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}=606...(i)
\end{array}
$$
Given that fundamental frequency
$$
\frac{1}{2 l} \sqrt{\frac{T}{m}}=600...(ii)
$$
Dividing Eq. (i) by(ii),
$$
\begin{array}{l}
\frac{\frac{1}{2 l} \sqrt{\frac{T^{\prime}}{m}}}{\frac{1}{2 l} \sqrt{\frac{T}{m}}}=\frac{606}{600} \\
\Rightarrow \sqrt{\frac{T^{\prime}}{T}}=(1.01) \Rightarrow \frac{T^{\prime}}{T}=(1.02) \% \\
\Rightarrow T^{\prime}=T(1.02)
\end{array}
$$
Increase in tension
$$
\begin{array}{l}
\Delta T^{\prime}=T \times 1.02-T=(0.02 T) \\
\therefore \Delta T=0.02
\end{array}
$$

Topic: Physics, Waves and Sound, JEE Main
Two identical piano wires have a fundamental frequency of 600 cycle per second when kept under the same tension. What fractional increase in the tension of wires will lead to the occurrence of 6 beats per second when both wires vibrate simultaneously?

Options:
A 0.01
B 0.02
C 0.03
D 0.04

The correct answer is:
0.02

Solution:
Beats per second when both the wires vibrate simultaneously

n1 ± n2 = 6

or 12l⁡Tm±12l⁡T'm=6

or 12l⁡T'm-12l⁡Tm=6

or 12l⁡T'm-600=6 ...(i)
Given that fundamental frequency
12lTm=600

Dividing Eq.(i) by Eq. (ii), we get

12l⁡T'm12l⁡Tm=606600

Or T'T=1.01

Or T⁡ ′ T⁡ = 1 . 0 2 %

Or T' =T (1.02)

Increase in tension
ΔT' = (T × 1.02) – T
= 0.02T
Hence, ΔT' = 0.02

Topic: Physics, Waves and Sound, JEE Main
Two identical piano wires kept under the same tension $T$ have a fundamental frequency of $600 \mathrm{~Hz}$. The fractional increase in the tension of one of the wires which will lead to occurrence of 6 beat/s when both the wires oscillate together would be

Options:
A 0.02
B 0.03
C 0.04
D 0.01

The correct answer is:
0.02

Solution:
According to law of tension the frequency of the string varies directly as the square root of its tension
$$
\text { or } \quad \begin{aligned}
n & \propto \sqrt{T} \\
\frac{\Delta n}{n} & =\frac{1}{2} \cdot \frac{\Delta T}{T} \\
\frac{\Delta T}{T} & =2 \times \frac{\Delta n}{n} \\
\frac{\Delta T}{T} & =2 \times \frac{6}{600} \\
& =0.02
\end{aligned}
$$

Topic: Physics, Waves and Sound, NEET, NEET 2011 (Mains)
Two identical plates of different metals are joined to form a single plate whose thickness is double the thickness of each plate. If the coefficients of conductivity of each plate are 2 and 3 respectively, then the conductivity of composite plate will be

Options:
A $5$
B $2.4$
C $1.5$
D $1.2$

The correct answer is:
$2.4$

Solution:
Thermal conductivity of composite plate
$K_{\text {eq }}=\frac{2 K_1 K_2}{K_1+K_2}=\frac{2 \times 2 \times 3}{2+3}=\frac{12}{5}=2.4$

Topic: Physics, Thermal Properties of Matter, JEE Main
Two identical point charges of charge \( -q \) are fixed as shown in the figure below. A third charge \( +q \) is placed midway between the two charges at the point P. Suppose this charge \( +q \) is displaced a small distance from the point P in the directions indicated by the arrows, in which direction(s) will \( +q \) be stable with respect to the displacement?

Options:
A \( A_{1} \) and \( A_{2} \)
B \( B_{1} \) and \( B_{2} \)
C Both directions
D Not stable

The correct answer is:
\( B_{1} \) and \( B_{2} \)

Solution:
When the charge is at B1 and B2, the net force acting on the charge will be towards the centre as shown. Therefore, at B1 and B2, there is stable equilibrium. When the charge is at A1, the force of attraction from the charge on the left side will be more, so it moves towards left. When the charge is at A2, the force of attraction from the charge on the right will be more, so it moves towards right. Hence, A1 and A2 are unstable.

Topic: Physics, Electrostatics, JEE Main
Two identical satellites are moving around the earth in circular orbits at heights 3R and R respectively where R is the radius of the Earth. The ratio of their kinetic energies is

Options:
A 2Â :Â 1
B 1 : 2
C 3Â :Â 1
D 2Â :Â 3

The correct answer is:
1 : 2

Solution:
The kinetic energy of satellites in a circular orbit is given as
Ek=GMm2r⇒ Ek1Ek2=r2r1=R+RR+3R=2R4R=12

Topic: Physics, Gravitation, JEE Main
Two identical short bar magnets each having magnetic moment M are placed at a distance of 2d with their axes perpendicular to each other in a horizontal plane. The magnetic induction at a point midway between them is

Options:
A 2μ0M4πd3
B 3μ0M4πd3
C μ0M2πd3
D 5μ0M4πd3

The correct answer is:
5μ0M4πd3

Solution:
At
magnetic field is a vector quantity. at point p two magnetic field are acting at 90o
Bnet=B12+B22
=2μ0M4πd32+μ0M4πd32=μ0M4πd3 5

Topic: Physics, Magnetic Properties of Matter, JEE Main
Two identical solid spheres each of mass 2 kg and radii 10 cm are fixed at the ends of a light rod. The separation between the centres of the spheres is 40 cm. The moment of inertia of the system about an axis perpendicular to the rod passing through its middle point is______×10–3 kg m2.

Options:

The correct answer is:
176

Solution:
The moment of inertia of a sphere is given by the formula I=25MR2.

Since there are two masses, the total moment of inertia about the axis perpendicular to the rod passing through the centre is

I=2×25MR2+Ml2+r2 ...(i)

The given data is

M=2 kgl=0.4-0.2 m=0.2 mR=0.1 m

Substituting the values in equation (i)

I=45(2×0.12)+2×2(0.22+0.1)2 kg m2⇒I=45×0.02+4×0.04 kg m2⇒I=0.016+0.16 kg m2=0.176 kg m2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2023 (06 Apr Shift 1)