datavorous/entrance-exam-dataset huggingface dataset:
Two long conductors, separated by a distance $\mathrm{d}$ carry current $\mathrm{I}_1$ and $\mathrm{I}_2$ in the same direction. They exert a force $F$ on each other. Now the current in one of them increased to two times and its direction reversed. The distance is also increased to $3 \mathrm{~d}$. The new value of the force between them is

Options:
A
$-2 \mathrm{~F}$

B
$\mathrm{F} / 3$

C
$-2 F / 3$

D
$-\mathrm{F} / 3$

The correct answer is:
$-2 F / 3$

Solution:
Force between two long conductor carrying current
$$
F=\frac{\mu_0}{2 \pi} \frac{l_1 l_2}{d} \ell
$$
According to question
$$
F^{\prime}=\frac{\mu_0}{2 \pi} \frac{\left(-2 I_1\right)\left(I_2\right)}{d} \ell
$$
From equation (i) and (ii), $\mathrm{F}^{\prime}=-\frac{3}{2} \mathrm{~F}$.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2004
Two long current carrying conductors are placed parallel to each other at a distance of 8 cm between them. The magnitude of magnetic field produced at mid-point between the two conductors due to current flowing in them is 300 μT. The equal current flowing in the two conductors is :

Options:
A 30 A in the same direction.

B 60 A in the opposite direction.
C 30 A in the opposite direction.
D 300 A in the opposite direction.

The correct answer is:
30 A in the opposite direction.

Solution:
Magnetic field in parallel wires is given by B=μ0I2πd

At midway, distance=d2

∵ μ0I2πd2×2=B

⇒μ0π×2Id=B

⇒4×10-7×2I0.08=300×10-6

⇒I=3×10-4×0.082×4×10-7=30 A

As Bnet≠0, so the wires are carrying current in opposite direction.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Two long currents carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle ' θ ' with the vertical. If wires have a mass λ per unit length then the value of I is:
(g= gravitational acceleration)

Options:
A πλgLμ0tanθ
B sinθπλgLμ0 cosθ
C 2sinθπλgLμ0cosθ
D 2πgLμ0tanθ

The correct answer is:
2sinθπλgLμ0cosθ

Solution:
Two wires will repel each other due to magnetic force, then the magnetic force per unit length is, dfdl=μ0I22π2Lsinθ=μ0I24πLsinθ. And mass per unit length of each wire=dmdl=λ. So, the magnetic force on the total length L of the wire is fm=μ0I2L4πLsinθ, and weight =λLg. By equilibrium of wire, Tsinθ=fm  & Tcosθ = W = λl'g⇒TsinθTcosθ=fmmg⇒fm=λl'gtanθ
⇒μ0I24πLsinθL=λLgsinθcosθ
⇒I2=λgπLμ0cosθ4sin2θ⇒I=2sinθλπgLμ0cosθ

Topic: Physics, Electrostatics, JEE Main, JEE Main 2015 (04 Apr)
Two long parallel wires are at a distance $2 \mathrm{~d}$ apart. They carry steady equal current flowing out of the plane of the paper as shown. The variation of the magnetic field along the line $\mathrm{XX}$ ' is given by

Options:
A

B

C

D

The correct answer is:

Solution:
The magnetic field in between because of each will be in opposite direction
$$
\begin{aligned}
B_{\text {in between }} & =\frac{\mu_0 i}{2 \pi x} \hat{j}-\frac{\mu_0 i}{2 \pi(2 d-x)}(-\hat{j}) \\
& =\frac{\mu_0 i}{2 \pi}\left[\frac{1}{x}-\frac{1}{2 d-x}\right](\hat{j})
\end{aligned}
$$
at $\quad x=d, \quad B_{\text {in between }}=0$
for $\quad x < d, \quad B_{\text {in between }}=(\hat{j})$
for $\quad x>d, \quad B_{\text {in between }}=(-\hat{j})$
towards $\mathrm{x}$ net magnetic field will add up and direction will be $(-\hat{\mathrm{j}})$
towards $\mathrm{x}^{\prime}$ net magnetic field will add up and direction will be $(\hat{\mathrm{j}})$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2010
Two long parallel wires carry currents i1 and i2 such that i1>i2. When the currents are in the same direction, the magnetic field at a point midway between the wires is 6Ã10-6 T. If the direction of i2 is reversed, the field becomes 3Ã10-5 T. The ratio i1i2 is

Options:
A 12
B 23
C 32
D 15

The correct answer is:
32

Solution:
When the currents are in the same direction then the magnetic field
B=μ04π×2d i1-i2
6×10-6= μ04π×2d i1-i2 …(i)
When the currents are in the reversed direction then the magnetic field
B2=μ04π×2d i1-(-i2)
or B2=μ04π×2d i1+i2
or 3×10-5=μ04π×2di1+i2 …(ii)
Dividing Eq. (i) by Eq. (ii)
i1-i2i1+i2=6×10-63×10-5
or i1-i2i1+i2=210
or 5i1-5i2=i1+i2
or 4i1=6i2
i1i2=64
i1i2=32

Topic: Physics, Magnetic Effects of Current, JEE Main
Two long parallel wires carrying currents 8 A and 15 A in opposite directions are placed at a distance of 7 cm from each other. A point P is at equidistant from both the wires such that the lines joining the point P to the wires are perpendicular to each other. The magnitude of magnetic field at P is ______ ×10-6 T. (Given: 2=1.4)

Options:

The correct answer is:
68

Solution:
Here, the magnetic fields produced by the two wires will be perpendicular to one another.

Magnetic field due to wire carrying current 8 A is B1=μ0i12πd

and magnetic field due to wire carrying current 15 A is B2=μ0i22πd

The net magnetic field is

Bnet=B12+B22=μ02πdi12+i22

=4π×10-72π×72×10-2×82+152 d=72 cm

=68×10-6 T.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Two long parallel wires separated by $0.1 \mathrm{m}$ carry currents of 1 A and 2 A, respectively in opposite directions. A third current-carrying wire parallel to both of them is placed in the same plane such that it feels no net magnetic force. It is placed at a distance of

Options:
A $0.5 \mathrm{m}$ from the 1 st wire, towards the 2nd wire
B $0.2 \mathrm{m}$ from the 1 st wire, towards the 2nd wire
C $0.1 \mathrm{m}$ from the 1 st wire, away from the 2nd wire
D $0.2 \mathrm{m}$ from the 1 st wire, away from the 2nd wire

The correct answers are:
$0.1 \mathrm{m}$ from the 1 st wire, away from the 2nd wire

Solution:
According to question, the third wire should not feel magnetic force due to wire (1) and
(2). It can be only possible in the condition
that fields due to wire (1) and (2) must act opposite to each other and must be equal. The magnetic field due to long straight wire,
$$
B=\frac{\mu_{0} \cdot I}{2 \pi x}
$$
$\therefore \quad B_{1}=B_{2}$
$$
\frac{\mu_{j_{1}}}{2 \pi x}=\frac{\mu_{d_{2}}}{2 \pi(0 \cdot 1+x)}
$$
Here
$$
\begin{aligned}
i_{1} &=1 \mathrm{A} \\
i_{2} &=2 \mathrm{A} \\
\frac{\mu_{0} \times 1}{2 \pi x} &=\frac{\mu_{0} \times 2}{2 \pi(0 \cdot 1+x)} \\
2 x &=(0 \cdot 1+x) \\
x &=0 \cdot 1 \mathrm{m}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
Two long straight parallel wires, carrying (adjustable) currents I1 and I2 , are kept at a distance d apart. If the force F between the two wires is taken as 'positive' when the wires repel each other and 'negative' when the wires attract each other, the graph showing the dependence of F, on the product I1I2 , would be:

Options:
A
B
C
D

The correct answer is:

Solution:
Force between two parallel current carrying wires is given as
f=μ0I1I22πdl
F∝I1I2
When I1, I2 are in same direction the I1I2 is+ve ⇒ attraction ⇒ -ve
When I1I2 are in opposite direction then
I1I2=-ve
∴ Force = repulsive = +ve

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2015 (11 Apr Online)
Two long straight wires P and Q carrying equal current 10 A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F1. If distance between wires is halved and currents on them are doubled, force F2 on 10 cm length of wire P will be:

Options:
A 8F1
B 10 F1
C F18
D F110

The correct answer is:
8F1

Solution:
As we know, force per unit length between two parallel straight wires is given by, F=μ0i1i22πd.

Therefore, required ratio

F1 F2=μ0(10)22π(5×10-2)μ0(20)22π52×10-2=18

⇒F2=8F1

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (24 Jan Shift 1)
Two loops P and Q are made from a uniform wire. The radii of P and Q are R1 and R2, respectively, and their moments of inertia about their axis of rotation are I1 and I2, respectively. If I2I1=4 , then R2R1 is

Options:
A 423
B 413
C 4-23
D 4-13

The correct answer is:
413

Solution:
I of loop =MR2
=λ 2πR. R2 ( λ= linear mass density)
=λ2πR3
⇒ I1=λ2πR13I2=λ2πR23 made up of same wire
I2I1=R23R13=4
⇒ R2R1=413

Topic: Physics, Rotational Motion, JEE Main
Two loops P and Q are made from a uniform wire. The radii of P and Q are r1 and r2 respectively, and their moments of inertia are I1 and I2 respectively. If I2=4I1 , then r2r1 equals

Options:
A 423
B 413
C 4-23
D 4-13

The correct answer is:
413

Solution:
Moment of inertia of a loop is,
I=Mr2=2πrAρr2
Therefore I∝r3
or r∝I13
For the two loops
r2r1=I2I113
As I2=4I1
r2r1=4113

Topic: Physics, Rotational Motion, JEE Main
Two loudspeakers are being compared. One is perceived to be 32 times louder than the other. The difference in intensity levels between the two, when measured in decibels, is

Options:
A 60
B 40
C 50
D 30

The correct answer is:
50

Solution:
âÎ²=Î²2-Î²1=10logI2I0-10logI1I0
=10logI2I1
Increasing the intensity by a factor 10 changes the sound level by 10 dB, increasing by 100 changes Î² by 20 dB.

Since, 32 = 25
âÎ²=510=50 dB

Topic: Physics, Waves and Sound, JEE Main
Two loudspeakers L1 and L2, driven by a common oscillator and amplifier, are arranged as shown. The frequency of the oscillator is gradually increased from zero and the detector at D records a series of maxima and minima. If the speed of sound is 330 ms-1, then at what frequency (in Hz) does the first maximum is observed?

Options:

The correct answer is:
330

Solution:
Let Δx = path difference
⇒ Δ x = L 2 D - L 1 D
⇒ Δ x = 40 2 + 9 2 - 40
⇒ Δ x = 41 - 40
⇒ Δ x = 1 m
For first maximum, n = 1
Δx=nλ
Δx=n V f
f= nV x
f=330 Hz

Topic: Physics, Waves and Sound, JEE Main
Two magnetic dipoles X and Y are kept at a distance d apart, with their axes perpendicular to each other. The dipole moment of Y is twice that of X. P is a point along the horizontal line which is at the midpoint of d. What is the magnitude of the force of particle of charge q passing through P at angle Î¸=45Â°? (d is very much larger in comparison to dipole)

Options:
A 2μ04πM(d/2)3×qv
B μ04π2M(d/2)3×qv
C μ0πM(d/2)3×qv
D 0

The correct answer is:
0

Solution:
B1=2μ04π·M(d/2)3; B2=μ04π·2M(d/2)3
⇒B1=B2

∴ Net magnetic field will be at 45°.

∴The direction of Bnet and velocity of charge is same, hence charge will experience no force.

Topic: Physics, Magnetic Effects of Current, JEE Main
Two magnetic dipoles X and Y are placed at a separation d , with their axes perpendicular to each other. The dipole moment of Y is twice that of X . A particle of charge q is passing through their mid-point P , at angle θ=45o with the horizontal line, as shown in figure. What would be the magnitude of force on the particle at that instant? ( d is much larger than the dimension of the dipole)

Options:
A 0
B 2 μ04πMd23×qv
C μ04πMd23×qv
D μ04π2Md23×qv

The correct answer is:
0

Solution:
BX=2μ04πMd/2 along x -axis
BY=μ04π2Md/2 along y -axis
BX=BY

Hence Bnet will be at 45o from x -axis.
Now, velocity and magnetic field are in same direction. Hence no force will act on the particle.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Two magnets held together in earth's magnetic field when the same polarity together causes 12 vib/min and when opposite poles 4 vib/min. What is the ratio of magnetic moments? If your answer is K, mark 4K as answer after rounding off to nearest integer.

Options:

The correct answer is:
5

Solution:
Here we use sum and difference method.
M2M1=T22-T12T22+T12
Where T1 and T2 are time periods
T â1n (âµn is frequency.)
â M1M2=n12 + n22n12 - n22
=122+42122-42=144+16144-16=108=1.25

Topic: Physics, Magnetic Properties of Matter, JEE Main
Two magnets held together in earth's magnetic field when the same polarity together causes 12 vib/min and when opposite poles 4 vib/min. What is the ratio of magnetic moments?

Options:

The correct answer is:
1.25

Solution:
Here we use sum and difference method.
M2M1=T22-T12T22+T12
Where T1 and T2 are time periods
T â1n (âµn is frequency.)
â M1M2=n12 + n22n12 - n22
=122+42122-42=144+16144-16=108=1.25

Topic: Physics, Magnetic Properties of Matter, JEE Main
Two magnets of equal mass are joined at 90Â° each other as shown in figure. Magnet N1 S1 has a magnetic moment 3 times that of N2 S2. The arrangement is pivoted so that it is free to rotate in horizontal plane. When in equilibrium, what angle should N1 S1 make with magnetic meridian?

Options:
A 75Â°
B 60Â°
C 30°
D 45Â°

The correct answer is:
30°

Solution:
In equilibrium, the resultant magnetic moment will be along magnetic meridian. Let N1S1 make ∠θ with resultant
tanθ=M2M1=M3M=13 ∴θ=30°

Topic: Physics, Magnetic Properties of Matter, JEE Main
Two man of mass 50 kg and 60 kg on diametrically opposite point of a turntable of mass 90 kg . The turntable is free to rotate about vertical axis. The two men start moving towards each other along the circumference of table. The angle rotated by table when the two men meet each other -

Options:
A 8.2°
B 4.5°
C 2.8°
D 6.8°

The correct answer is:
4.5°

Solution:
Let âÎ¸â be angle rotated by turntable
Let mass of man 1=m1 =50 kg
Let mass of man 2=m2 = 60 kg
Let mass of turntable=m = 90 kg
â´ m1 90Â°+Î¸+m2 -90Â°- Î¸ +m . Î¸= 0
â Î¸= 4.5Â°

Topic: Physics, Rotational Motion, JEE Main
Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability, that first drawn marble is red and second drawn marble is white, is

Options:
A 225
B 425
C 23
D 475

The correct answer is:
475

Solution:
Given marbles are:

10→R, 30→W, 20→B, 15→O

We know that, PE=Number of favourable outcomesTotal number of outcomes

Also, PA and B=PA×PB, where A and B are independent events.

Probability of drawing first marble red and then white is given by,

PE=1075×3075

⇒PE=475

Topic: Mathematics, Probability, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Two masses A and B connected with an inextensible string of length l lie on a smooth horizontal plane. A is given a velocity of v m s-1 along the ground perpendicular to line AB as shown in the figure. Find the tension in a string during their subsequence motion.

Options:
A 2mv23l

B 3mv22l
C mv22l
D 4mv23l

The correct answer is:
2mv23l

Solution:
From conservation of linear momentum
â v cm = m × 0 + 2mv m + 2m ⇒ v cm = 2v 3
position of the centre of mass
ycm=0×2m+m×l2m⇒ycm=l3
blocks will perform circular motion in the center of mass frame
velocity of A in COM frame is v A ′ = v - 2 v/3 ⇒ v A ′ = v/3
Tension will provide the required centripetal force
T = 2m v A ′ 2 ℓ / 3 ⇒ T = 2m v/3 2 ℓ / 3 ⇒ T = 2mv 2 3 ℓ

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two masses A and B, each of mass M are fixed together by a massless spring, A force acts on the mass B as shown in figure. If the mass A starts moving away from mass B with acceleration a, then the acceleration of mass B will be :

Options:
A

MFF+Ma

B F+MaM
C Ma-FM
D F-MaM

The correct answer is:
F-MaM

Solution:
acm=m1a1+m2a2m1+m2

F2M=Ma+MaB2M

aB=F-MaM

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Two masses m1=5 kg and m2=10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is:

Options:
A 10.3 kg
B 18.3 kg
C 27.3 kg
D 43.3 kg

The correct answer is:
27.3 kg

Solution:
T=5g⇒μ10+m≥5g⇒10+m≥50.15
⇒m≥23.33 kg⇒m=27.3 kg.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2018 (08 Apr)
Two masses $m_1=5 \mathrm{~kg}$ and $m_2=4.8 \mathrm{~kg}$ tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when lift free to move $\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right)$

Options:
A
$0.2 \mathrm{~m} / \mathrm{s}^2$

B
$9.8 \mathrm{~m} / \mathrm{s}^2$

C
$5 \mathrm{~m} / \mathrm{s}^2$

D
$4.8 \mathrm{~m} / \mathrm{s}^2$

The correct answer is:
$0.2 \mathrm{~m} / \mathrm{s}^2$

Solution:
$\mathrm{a}=\left(\frac{\mathrm{m}_1-\mathrm{m}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right) \mathrm{g}=0.2 \mathrm{~m} / \mathrm{s}^2$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2004
Two masses M1 and M2 are arranged as shown in the figure. Let 'a' be the magnitude of the acceleration of the mass M1. If the mass of M1 is doubled and that of M2 is halved, then the acceleration of the system is
(Treat all surfaces as smooth : masses of pulley and rope are negligible )

Options:
A M1+M24M1+M2a
B 2M1+M24M1+M2a
C M1+2M24M1+2M2a
D M1+2M2M1+M2a

The correct answer is:
M1+M24M1+M2a

Solution:
Free body diagrams of M1 and M2 are shown below.

From the free body diagram of M2

M2gsinθ-T=M2a

And from the free body diagram of M1

T=M1a

Therefore,

a=M2gsinθM1+M2

When mass of M1 is doubled and mass of M2 is halved,

a'=M22gsinθ2M1+M22⇒a'=M2gsinθ4M1+M2=M1+M24M1+M2a

Topic: Physics, Laws of Motion, JEE Main
Two masses m1 and m2 are connected by a string of length l. They are held in a horizontal plane at a height H above two heavy plates A and B made of different materials placed on the floor. Initially, the distance between the two masses is a<l. When the masses are released they fall freely and collide with the plates simultaneously. The coefficient of restitution for the collision of the first ball-plate pair is 0.8 and for the second ball-plate pair is 0.4. The time after the collision when the string becomes taut is

Options:
A t=52l2-a22gH
B t=54l2-a22gH
C t=52la2gH
D t=54la2gH

The correct answer is:
t=52l2-a22gH

Solution:
The string will become taut when

h1-h2=l2-a2

h1=0.8vt-12gt2

h2=0.4vt-12gt2

h1-h2=0.4vt

0.4vt=l2-a2, where v=2gH

t=52l2-a22gH

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two masses M1 and M2 are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M2 is twice that of M1. the acceleration of the system is a1. When the mass M2 is thrice that of M1. The acceleration of The system is a2. The ratio a1a2 will be

Options:
A 13
B 23
C 32
D 12

The correct answer is:
23

Solution:
In both cases m1>m2. Therefore, m1 will accelerate downward and m2 will accelerate upward.

For m1,

m1g-T=m1a

For m2,

T-m2g=m2a

Adding the above equations we get,

a=m1-m2m1+m2g

⇒a1=m1-m2m1+m2g=2m2-m22m2+m2g=g3

and

⇒a2=m1-m2m1+m2g=3m2-m23m2+m2g=g2

⇒a1a2=23

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Two masses m and m2 are connected at the two ends of a massless rigid rod of length l. The rod is suspended by a thin wire of torsional constant k at the centre of mass of the rod-mass system (see figure). Because of torsional constant k, the restoring torque is τ=kθ for angular displacement θ. If the rod is rotated by θ0 and released, the tension in it when it passes through its mean position will be:

Options:
A kÎ¸02
B 3kÎ¸0Â Â 2l
C 2kÎ¸0Â Â 2l
D kÎ¸0 2l

The correct answer is:
kÎ¸0 2l

Solution:
When rod is rotated by angle Î±, the twist in the wire is Ï.
angular frequency of wire, Î©=kI
angular frequency of rod, Ï=Î¸0.Î©
Max. Tension in the rod =mÏ2.l3
=m Î¸0 2.kI.l3 , here I=ml23
=Î¸0 2kl

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Two masses m and M are attached with strings as shown in the figure. For the system to be in equilibrium, we must have

Options:
A tanθ=1+2Mm
B tanθ=1-2mM
C tanθ=1-M2m
D tanθ=1+m2M

The correct answer is:
tanθ=1+2Mm

Solution:
mg=2Tsin45°
⇒ mg=2T
T1 cosθ=Tcos45∘
⇒ T1cosθ=T2=mg2 ...(i)
Further, Mg+T cos 45∘=T1 sinθ ∵T=mg2
⇒ T1 sinθ=Mg+mg212 ...(ii)
⇒ tanθ=Mg+mg2mg2=1+2Mm

{dividing Eq. (ii) by Eq. (i)}

Topic: Physics, Laws of Motion, JEE Main
Two masses m and M are attached with strings as shown in the figure. For the system to be in equilibrium, we must have

Options:
A tan Î¸ = 1 + 2 M m
B tan θ = 1 - 2 m M
C tan θ = 1 - M 2 m
D tan θ = 1 + m 2 M

The correct answer is:
tan θ = 1 + 2 M m

Solution:
mg=2Tsin45°
⇒ mg = 2 T
T 1 cos θ = T cos 4 5 ∘
⇒ T 1 cos θ = T 2 = mg 2 ...(i)
Further, Mg + T cos 4 5 ∘ = T 1 sin θ ∵ T = mg 2
⇒ T 1 sin θ = Mg + mg 2 1 2 ...(ii)
⇒ tan θ = Mg + mg 2 mg 2 = 1 + 2 M m {dividing Eq. (ii) by Eq. (i)}

Topic: Physics, Laws of Motion, JEE Main
Two masses m and M (>m) are joined by a light string passing over a smooth light pulley. The centre of mass of the system moves with an acceleration

Options:
A gM-mM+m downward
B gM-mM+m2 downward
C gM-mM+m2 upward if M < m
D zero

The correct answer is:
gM-mM+m2 downward

Solution:
a com = m 1 a 1 ¯ + m 2 a 2 ¯ m 1 + m 2
a com = 1 m+M [ M( M−m ) m+M - m( M−m ) m+M ]g
= 1 m+M [ ( M−m ) 2 m+M ]g= ( M−m M+m ) 2 g

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two masses of $1 \mathrm{gm}$ and $4 \mathrm{gm}$ are moving with equal kinetic energies. The ratio of the magnitudes of their linear momenta is

Options:
A $4: 1$
B $\sqrt{2}: 1$
C $1: 2$
D $1: 16$

The correct answer is:
$1: 2$

Solution:
$P=\sqrt{2 m E}$. If $E$ are same then $P \propto \sqrt{m}$
$\Rightarrow \frac{P_1}{P_2}=\sqrt{\frac{m_1}{m_2}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two masses of $5 \mathrm{~kg}$ and $3 \mathrm{~kg}$ are suspended with the help of massless inextensible strings as shown in figure below.
When whole system is going upwards with acceleration $2 \mathrm{~m} / \mathrm{s}^2$, the value of $T_1$ is (use, $g=9.8 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A $35.4 \mathrm{~N}$
B $23.6 \mathrm{~N}$
C $59 \mathrm{~N}$
D $94.4 \mathrm{~N}$

The correct answer is:
$94.4 \mathrm{~N}$

Solution:
$$
\text { The given situation is shown below }
$$
Equation of motion for body $m_2$,
$$
\begin{aligned}
T_2-m_2 g & =m_2 a \\
T_2-3 g & =3 \times 2 \\
T_2 & =3 g+6=3 \times 9.8+6=35.4 \mathrm{~N}
\end{aligned}
$$
Equation of motion of mass of $5 \mathrm{~kg}$,
$$
\begin{aligned}
& & T_1-T_2-m_1 g & =m_1 a \\
\Rightarrow & & T_1-35.4-5 \times 9.8 & =5 \times 2 \\
\Rightarrow & & T_1-84.4 & =10^{\circ} \\
\Rightarrow & & T_1 & =10+84.4=94.4 \mathrm{~N}
\end{aligned}
$$

Topic: Physics, Laws of Motion, JEE Main
Two massless string of length 5 m hang from the ceiling very near to each other as shown in the figure. Two balls A and B of masses 0.25 kg and 0.5 kg are attached to the string. The ball A is released from rest at a height of 0.45 m as shown in the figure. The collision between two balls is completely elastic. Immediately after the collision, the kinetic energy of the ball B is 1 J. The velocity of ball A just after the collision is

Options:
A 5 m s-1 to the right

B 5 m s-1 to the left
C 1 m s-1 to the right

D 1 m s-1 to the left

The correct answer is:
1 m s-1 to the left

Solution:
KE B = 1 2 m B v B 2
1 = 1 2 × 0 . 5 × v B 2
vB= 2 m s-1
u A = 2 gh
= 2 × 1 0 × 0 . 4 5
= 3 m s-1
From the conservation of linear momentum
0 . 2 5 × 3 + 0 = 0 . 2 5 × v A + 0 . 5 × 2
0 . 7 5 = 0 . 2 5 v A + 1
0 . 2 5 v A = - 0 . 2 5
vA=-1 m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two materials having coefficients of thermal conductivity 3K and K and thickness d and 3d respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are θ2 and θ1 respectively, θ2>θ1. The temperature at the interface is

Options:
A θ2+θ12
B Î¸16+5Î¸26
C Î¸13+2Î¸23
D Î¸110+9Î¸210

The correct answer is:
θ110+9θ210

Solution:
Let the temperature of the junction T Â°C .

Rate of heat flow in Rod 1 = rate of heat flow in Rod 2

3kAdÎ¸2-T=kA3dT-Î¸1

â9Î¸2-T=T-Î¸1

â10T=9Î¸2+Î¸1

âT=9Î¸2+Î¸110=Î¸110+9Î¸210

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2019 (09 Apr Shift 2)
Two medians drawn from acute angles vertices of a right angled triangle intersects at an angle π6. If the length of the hypotenuse of the triangle is 3 units, then area of the triangle (in sq. units) is

Options:
A 3
B 3
C 2
D 9

The correct answer is:
3

Solution:
Slope of AG=â b 2a ,âtan 30 o = 3b 2a 1+ b 2 a 2
a2+b2=9â13=3b.a2a2+b2
â 12 ab=a2+b233=933=3

Topic: Mathematics, Properties of Triangles, JEE Main