datavorous/entrance-exam-dataset huggingface dataset:
Two bar magnets A and B are placed one over the other and are allowed to vibrate in a vibration magnetometer. They make 20 oscillations per minute when the similar poles of A and B are on the same side, while they make 15 oscillations per minute when their opposite poles lie on the same side. If $M_{A}$ and $M_{B}$ are the magnetic moments of $A$ and $B$ and if $M_{A}>M_{B}$, the ratio of $\mathrm{M}_{\mathrm{A}}$ and $\mathrm{M}_{\mathrm{B}}$ is

Options:
A $4: 3$
B $25: 7$
C $7: 5$
D $25: 16$

The correct answer is:
$25: 7$

Solution:
Ratio of magnetic moments
$$
\begin{aligned}
\frac{M_{A}}{M_{B}} &=\frac{T_{d}^{2}+T_{s}^{2}}{T_{d}^{2}-T_{s}^{2}}=\frac{v_{s}^{2}+v_{d}^{2}}{v_{s}^{2}-v_{d}^{2}} \\
&=\frac{\left(\frac{1}{20}\right)^{2}+\left(\frac{1}{15}\right)^{2}}{\left(\frac{1}{15}\right)^{2}-\left(\frac{1}{20}\right)^{2}}=\frac{400+225}{400-225} \\
M_{A}: & M_{B}=25: 7
\end{aligned}
$$

Topic: Physics, Magnetic Properties of Matter, JEE Main
Two bar magnets A and B are placed one over the other and are allowed to Vibrate in a vibration magnetometer. They make 20 oscillations per minute when the similar poles of A and B are on the same side, while they make 15 oscillations per minute when their opposite poles lie on the same side. If MA and MB are the magnetic moments of A and B and if MA>MB, the ratio of MA and MB is

Options:
A 4 : 3
B 25 : 7
C 7 : 5
D 25 : 16

The correct answer is:
25 : 7

Solution:
Ratio of magnetic moments of two magnets of equal size when in same and different position is

M A M B = T d 2 + T s 2 T d 2 - T s 2 = vâ¡ d 2 + vâ¡ s 2 vâ¡ d 2 - vâ¡ s 2

M A M B = 2 0 2 + 1 5 2 2 0 2 - 1 5 2

=400+225400-225

MA:MB=25:7

Topic: Physics, Magnetic Properties of Matter, JEE Main
Two bar magnets having geometry with magnetic moments $M$ and $2 M$ are first placed in such a way that their similar poles are same side then its time period of oscillation is $T_1$. Now the polarity of one of the magnet is reversed then the time period of oscillation is $T_2$, so:

Options:
A $T_1 < T_2$
B $T_1=T_2$
C $T_1>T_2$
D $T_2=\infty$

The correct answer is:
$T_1 < T_2$

Solution:
We can write
$T=2 \pi \sqrt{\frac{1}{M B}} \Rightarrow T \propto \frac{1}{\sqrt{M}}$
Case I : $M_1=2 M+M$
Case II : $M_2=2 M-M$
$\begin{aligned}
& \therefore \frac{T_1}{T_2}=\sqrt{\frac{M}{3 M}}=\frac{1}{\sqrt{3}} \\
& \Rightarrow T_2=\sqrt{3} T_1
\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
Two bar magnets oscillate in a horizontal plane in earth's magnetic field with time periods of 3 s and 4 s respectively. If their moments of inertia are in the ratio of 3:2 then the ratio of their magnetic moments will be

Options:
A 2:1
B 8:3
C 1:3
D 27:16

The correct answer is:
8:3

Solution:
Torque on a bar magnet kept in a magnetic field is given by τ=MBsinθ.

Hence, for small oscillations the time period of bar magnets is given by,

T=2πIMB⇒M∝IT2

⇒M1M2=I1I2×T2T12

⇒M1M2=32×432=83

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Two batteries $A$ and $B$ each of e.m.f. $2 \mathrm{~V}$ are connected in series to an external resistance $R=1 \mathrm{ohm}$. If the internal resistance of battery $A$ is $1.9 \mathrm{ohms}$ and that of $B$ is $0.9 \mathrm{ohm}$, what is the potential difference between the terminals of battery $\mathrm{A}$

Options:
A $2 \mathrm{~V}$
B $3.8 \mathrm{~V}$
C Zero
D None of the above

The correct answer is:
Zero

Solution:
$i=\frac{2+2}{1+1.9+0.9}=\frac{4}{3.8} \mathrm{~A}$
For cell $A \quad E=V+$ ir $\Rightarrow V=2-\frac{4}{3.8} \times 1.9=0$.

Topic: Physics, Current Electricity, JEE Main
Two batteries of different EMFs and different internal resistances are connected as shown. The voltage across AB in volt is.

Options:

The correct answer is:
5

Solution:
i = 3 3 = 1 A

VA-6+1-VB=0

VA-VB=5 V

Topic: Physics, Current Electricity, JEE Main
Two batteries of emf E1 and E2 E2>E1 and internal resistances r1 and r2 respectively are connected in parallel as shown in the figure. Then, which of the following statements is correct?

Options:
A The equivalent emf E is smaller than E1.
B The equivalent emf E=E1+E2.
C The equivalent emf E is greater than E2.
D The equivalent emf E of two cells is between E1 and E2 always.

The correct answer is:
The equivalent emf E of two cells is between E1 and E2 always.

Solution:
The equivalent internal resistance of the two cells between A and B is

R=r1r2r1+r2

If E is the equivalent emf of the two cells in parallel, then

ER=E1r1+E2r2=E1r2+E2r1r1r2
∴E=(E1r2+E2r1r1r2)R=E1r2+E2r1r1+r2
As, E2>E1⇒E1<E<E2 [For all values of r1 and r2

Topic: Physics, Current Electricity, JEE Main
Two batteries, one of emf 18 volts and internal resistance $2 \Omega$ and the other of emf 12 volts and internal resistance $1 \Omega$, are connected as shown. The voltmeter $V$ will record a reading of:

Options:
A 30 volt
B 18 volt
C 15 volt
D 14 volt

The correct answer is:
14 volt

Solution:
Reading of voltmeter
$$
=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}=\frac{E_1 r_2+E_2 r_1}{r_1+r_2}
$$
Putting value we get 14 volt.

Topic: Physics, Current Electricity, JEE Main
Two batteries $\mathrm{V}_{1}$ and $\mathrm{V}_{2}$ are connected to three resistors as shown below.
If $V_{1}=2 V$ and $V_{2}=0 \mathrm{~V}$, the current $I=3 \mathrm{~mA}$. If $V_{1}=0 \mathrm{~V}$ and $V_{2}=4 V$, the current $I=4 \mathrm{~mA}$. Now, if $\mathrm{V}_{1}=10 \mathrm{~V}$ and $\mathrm{V}_{2}=10 \mathrm{~V}$, the current I will be-

Options:
A $7 \mathrm{~mA}$
B $15 \mathrm{~mA}$
C $20 \mathrm{~mA}$
D $25 \mathrm{~mA}$

The correct answer is:
$25 \mathrm{~mA}$

Solution:
$\mathrm{I}=\frac{\mathrm{V}_{\mathrm{eq}}}{\mathrm{R}+\mathrm{R}_{\mathrm{eq}}}$
In each case $R_{e q} \& R$ is same only $V_{1} \& V_{2}$ is changing $\therefore V_{\text {eq }}$ is changing
$\mathrm{V}_{\mathrm{eq}}=\frac{2 \times \mathrm{R}_{2}+0 \times \mathrm{R}_{1}}{\mathrm{R}_{1}+\mathrm{R}_{2}} \quad\left[\mathrm{~V}_{1}=2, \mathrm{~V}_{2}=0\right]$
$\mathrm{V}_{\mathrm{eq}}=\frac{2 \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$
Case $-2 \quad \mathrm{~V}_{\mathrm{eq}}=\frac{4 \mathrm{R}_{1}}{\mathrm{R}_{1}+\mathrm{R}_{2}} \quad\left[\mathrm{~V}_{1}=0, \mathrm{~V}_{2}=4\right]$
$\frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{3}{4}=\frac{2 \mathrm{R}_{2}}{4 \mathrm{R}_{1}} \quad \frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{3}{2}$
Case - $3 \quad \mathrm{~V}_{\mathrm{eq}}=\frac{10 \mathrm{R}_{1}+10 \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$ $\frac{3}{\mathrm{I}^{\prime}}=\frac{2 \mathrm{R}_{2}}{10\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)} \Rightarrow \frac{3}{\mathrm{I}^{\prime}}=\frac{2 \times 1.5 \mathrm{R}_{1}}{10\left(2.5 \mathrm{R}_{1}\right)}$ or $\mathrm{I}^{\prime}=25 \mathrm{~mA}$
$\frac{3}{\mathrm{I}^{\prime}}=\frac{2 \mathrm{R}_{2}}{10\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)} \Rightarrow \frac{3}{\mathrm{I}^{\prime}}=\frac{2 \times 1.5 \mathrm{R}_{1}}{10\left(2.5 \mathrm{R}_{1}\right)}$ or $\mathrm{I}^{\prime}=25 \mathrm{~mA}$

Topic: Physics, Current Electricity, JEE Main
Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of 10 Ω. The internal resistance of the two batteries are 1 Ω and 2 Ω respectively. The voltage across the load lies between,

Options:
A 11.7 V and 11.8 V
B 11.6 V and 11.7 V
C 11.5 V and 11.6 V
D 11.4 V and 11.5 V

The correct answer is:
11.5 V and 11.6 V

Solution:
As shown in figure two cells are connected in parallel with the load resistance of R=10 Ω. Let current i1, i2 drawn from both cells as shown, from K.C.L. i = i1+ i2 ...1

Apply K.V.L. for loop ACDF,

E1-i R-i1 r1=0⇒12-10i-i1=0 ...2

From equations 1 and 2

12-11 i1-10i2=0 ...3

Now, for loop BCDE,

Apply K.V.L.

E2-10i-i2r2=0⇒13-10 i-2i2=0⇒∵i=i1+i2∴ 13-10 i1-12 i2=0 ...4

From equation 3 and 4,

i1=716 A, i2=2332 A

So, net current in external load R=10 Ω is

I=i1+i2=716+2332=3732 A

Then potential difference for a given load,

V=IR=3732×32=11·56 V

So, the potential difference lies between, 11·5 V and 11·6 V.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2018 (08 Apr)
Two beads each of mass m are fixed on a light rigid rod of length 2l which is free to rotate in a horizontal plane. The bead on the far end is given some velocity v as shown in the figure. If Kcm represents the kinetic energy of the centre of mass of the system and KR represents the rotational kinetic energy of the system, then what is the value of KcmKR?

Options:

The correct answer is:
0.9

Solution:
The distance of the centre of mass from the hinge,

rcm=ml+m(2l)m+m=32l

The angular velocity of the system,

Ï=v2l

The velocity of the centre of mass of the system,

vcm=3l2Ï=34v

âKcm=122mv2cm=9mv216

The moment of inertia of the system about the hinge,

I=ml2+m2l2=5ml2

The rotational kinetic energy of the system,

KR=12IÏ2=125ml2v2l2

âKR=5mv28

KcmKR=9mv216Ã85mv2=0.9

Topic: Physics, Rotational Motion, JEE Main
Two beaker A and B present in a closed vessel. Beaker A contains 152.4 g aqueous solution of urea, containing 12 g of urea. Beaker B contains 196.2 g glucose solution, containing 18 g of glucose. Both solutions allowed to attain the equilibrium. Determine wt. % of glucose in it's solution at equilibrium:

Options:
A 6.71
B 14.49
C 16.94
D 20

The correct answer is:
14.49

Solution:
Mole fraction of urea in its solution
=12601260+140.418=0.025
Mole fraction of glucose
=1818018180+178.218=0.01
Mole fraction of glucose is less so vapour pressure above the glucose solution will be higher than the pressure above urea solution, so some H2O molecules will transfer from glucose to urea side in order to make the solutions of equal mole fraction to attain equilibrium, let x moles H2O transferred
â´ 0.20.2+7.8+x=0.10.1+9.9-xâx=4
Now mass of glucose solution
â 196.2-4Ã18=124.2
wt% of glucose =18124.2Ã100=14.49

Topic: Chemistry, Solutions, JEE Main
Two beams, A and B of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30o makes the two beams appear equally bright. If the initial intensities of the two beams are IA and IB respectively, then I A I B equals :

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

The correct answer is:
1 3

Solution:
I A cos 2 θ = I B cos 2 9 0 - θ I = I 0 cos 2 θ by malus equation
∴ I A I B = tan 2 θ = 1 3
where θ = 3 0 o

Topic: Physics, Wave Optics, JEE Main, JEE Main 2014 (06 Apr)
Two beams of light having intensities I and 4I, interfere to produce a fringe pattern on a screen. The phase difference between the beams is Ï2, at a point A and Ï , at a point B. The difference between the resultant intensities at A and B is

Options:
A 2I
B 4I
C 5I
D 7I

The correct answer is:
4I

Solution:
Resultant intensity due to superposition of two waves on the screen is given as
I Ï = I 1 + I 2 + 2 I 1 I 2 cos Ï
Here, I 1 = I and I 2 = 4 I
At point A, Ï = Ï 2 â´ I A = I + 4 I = 5 I
At point B, Ï = Ï â´ I B = I + 4 I - 4 I = I
â´ I A - I B = 4 I

Topic: Physics, Wave Optics, JEE Main
Two beams of light having intensities $I$ and $4 I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{\pi}{2}$ at point $A$ and $\pi$ at point $B$. Then the difference between the resultant intensities at $A$ and $B$ is

Options:
A $2I$
B $4I$
C $5I$
D $7I$

The correct answer is:
$4I$

Solution:
At point $A$, resultant intensity
$\begin{aligned} & I_A=I_1+I_2=5 I \text {; and at point } B \\ & I_B=I_1+I_2+2 \sqrt{I_1 I_2} \cos \pi=5 I+4 I \\ & I_B=9 I \text { so } I_B-I_A=4 I .\end{aligned}$

Topic: Physics, Wave Optics, JEE Main
Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the two beams are π2 and π3 at points A and B respectively. The difference between the resultant intensities at the two points is xI. The value of x will be _____ .

Options:

The correct answer is:
2

Solution:
Given here, phase difference at point A and B are ϕA=π2and ϕB=π3.

Resultant intensity at point A,

IA=I+4I+2I4Icosπ2

⇒IA=5I+4I0=5I

Resultant intensity at point B,

IB=I+4I+2I4Icos60° =5I+4I×12=7I

Difference in intensities is IB-IA=7I-5I=2I.

Thus, value of x=2.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Two billiard balls of equal mass 30 g strike a rigid wall with same speed of 108 kmph (as shown) but at different angles. If the balls get reflected with the same speed, then the ratio of the magnitude of impulses imparted to ball a and ball b by the wall along X direction is:

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

The correct answer is:
2:1

Solution:
Impulse = change in momentum

Ball (a) |Δp→|=2mu=J1

Ball (b) |Δp→|=2mucos45°=J2

J1 J2=1cos45°=2

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Two billiard balls of mass 0.05 kg each moving in opposite directions with 10 ms-1 collide and rebound with the same speed. If the time duration of contact is t=0.005 s, then what is the force exerted on the ball due to each other?

Options:
A 100 N
B 200 N
C 300 N
D 400 N

The correct answer is:
200 N

Solution:
After collision, the balls change their directions of motion without changing the magnitudes of their velocity.

Change in momentum of any one ball

Δp→=2×0.05×10

⇒Δp→=1 kg m s-1

The force exerted on the ball is F→av =Δp→Δt=10.005

Fav =200 N

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Two black bodies $A$ and $B$ have equal surface areas and are maintained at temperatures $27^{\circ} \mathrm{C}$ and $177^{\circ} \mathrm{C}$ respectively. What will be the ratio of the thermal energy radiated per second by $A$ to that by $B ?$

Options:
A $4: 9$
B $2: 3$
C $16: 81$
D $27:177$

The correct answer is:
$16: 81$

Solution:
According to the question, Area of both bodies $A$ and $B=A$
Temperature of body $A=27^{\circ} \mathrm{C}=27+273 \mathrm{K}$
Temperature of body $B=177^{\circ} \mathrm{C}=177+273 \mathrm{K}$
Now, by Stefan-Boltzmann law, thermal energy radiated per second by a body
$$
Q=\sigma A T^{4}
$$
where, $A=$ Area
$T=$ temperature and $\quad \sigma=$ Stefan-Boltzmam's constant
So, the ratio of thermal energy radiated per second by $A$ to that by $B$ is
$$
\frac{Q_{1}}{Q_{2}}=\frac{\sigma A}{\sigma A}\left(\frac{T_{1}}{T_{2}}\right)^{4}
$$
Now, $\frac{Q_{1}}{Q_{2}}=\left(\frac{T_{1}}{T_{2}}\right)^{4}$
$$
\begin{array}{l}
\frac{Q_{1}}{Q_{2}}=\left(\frac{273+27}{273+177}\right)^{4} \\
\frac{Q_{1}}{Q_{2}}=\left(\frac{300}{450}\right)^{4}
\end{array}
$$
$$
\frac{Q_{1}}{Q_{2}}=\left(\frac{2}{3}\right)^{4}=\left(\frac{16}{81}\right)
$$
Ratio of thermal energy radiated per second
$$
Q_{1}: Q_{2}=16: 81
$$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2019
Two blocks $A$ and $B$ each of $20 \mathrm{~kg}$ lying on a frictionless table are connected by a light string. The system is pulled horizontally with an acceleration of $2 \mathrm{~ms}^{-2}$ by a force $F$ on $B$. The tension in the string will be

Options:
A $10 \mathrm{~N}$
B $40 \mathrm{~N}$
C $100 \mathrm{~N}$
D $120 \mathrm{~N}$

The correct answer is:
$40 \mathrm{~N}$

Solution:
The given situation can be shown as
$\text { The tension } T=M_{A} \times a$
here,
$\begin{aligned}
M_{A} &=20 \mathrm{~kg} \text { and } a=2 \mathrm{~ms}^{-2} \\
T &=20 \times 2=40 \mathrm{~N}
\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main
Two blocks A and B of equal mass are released on two sides of a fixed wedge C as shown in the figure. Find the acceleration of the centre of mass of blocks A and B. Neglect friction.

Options:
A g2, downwards
B g, downwards
C g2, upwards
D 3 4 g , downwards

The correct answer is:
g2, downwards

Solution:
Acceleration of both the blocks will be g sin45° or g 2 at right angles to each other. Now,

a→com=mAa→A+mBa→BmA+mB
Here, mA = mB
∴ acom=12aA+aB

=12g (downwards)

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two blocks A and B of equal masses are released from an inclined plane of inclination 45Â° at t=0. Both the blocks are initially at rest. The coefficient of kinetic friction between the block A and the inclined plane is 0.2 while it is 0.3 for block B. Initially the block A is 2 m behind the block B. At what time in seconds will their front faces come in a line, (Take g=10 m s-2)

Options:

The correct answer is:
2

Solution:
Acceleration of A down the plane,

aA = g sin 45Â° -Î¼A g cos 45o

=(10)(12)-(0.2)(10)(12)

= 4â2m/s2
Similarly acceleration of B down the plane, aB =g (sin 45Â°- Î¼B cos 45Â°)

=(10)(12)-(0.3)(10)(12)=3.52ms2

The front face of A and B will come in a line when,

sa = sB + â2
or 12uAt2=12aBt2+2

12Ã42Ãt2=12Ã3.52Ãt2+2
Solving this equation, we get t = 2 s
Additional information, sA=12aAt2=12Ã42Ã(2)2=82m

So, both the blocks will come in a line after A has travelled a distance 8â2 m down the plane.

Topic: Physics, Laws of Motion, JEE Main
Two blocks A and B of masses 1.5 kg and 0.5 kg respectively are connected by a massless inextensible string passing over a frictionless pulley as shown in the figure. Block A is lifted until block B touches the ground and then block A is released. The initial height of block A is 80 cm when block B just touches the ground. The maximum height reached by block B from the ground after the block A falls on the ground is

Options:
A 80 cm
B 120 cm
C 140 cm
D 160 cm

The correct answer is:
120 cm

Solution:
Common acceleration of the system is given by-

asys=1.5-0.51.5+0.5 g=g2=5 m s-2

When block of mass 1.5 kg hits the ground, at that time block of mass 0.5 kg will be at a height of 80 cm from the ground. After this point string become slack and block B will move with deceleration g=10 m s-2.

Using third equation of motion, speed of block B, When block A Reaches the ground-

v2=0+2 g2 80100=8

⇒ v=8 m s-1

Again using third equation of motion, suppose block B will further move upto a height of h m.

0=v2+2 -10 h

⇒ h=v220=0.4 m=40 cm

hence, total height of block B from the ground

=80+40=120 cm

Topic: Physics, Laws of Motion, JEE Main
Two blocks A and B of masses 3 kg and 6 kg are connected by a massless spring of force constant 1800 N m-1 and then they are placed on a smooth horizontal surface. The blocks are pulled apart to stretch the spring by 5 cm and then released. What is the the relative velocity (in m s-1) of the blocks when the spring comes to its natural length?

Options:

The correct answer is:
1.5

Solution:
Let Î¼ is reduced mass of the two blocks, then

Î¼=m1m2m1+m2=3Ã63+6=189=2 kg

Let vr be the relative velocity of the two blocks, then from conservation of mechanical energy we get

12Î¼vr2=12kx2
vr=xkÎ¼=5Ã10-218002

âvr=1.5 m s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two blocks A and B of masses 3m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively

Options:
A g,g3
B g3, g
C g,Â g
D g3,g3

The correct answer is:
g3, g

Solution:
Before the string is cut, tension is acting in the string in downward position.
kx=T+3mg ......(i)
T=mg ......(ii)
⇒ kx=4mg
After the string is cut, T=0
a=kx-3mg3m
a=4mg-3mg3m
Acceleration of Ais a=g3↑ and acceleration of B is a=mgm=g↓.

Topic: Physics, Laws of Motion, JEE Main
Two blocks A and B of masses mA=1 kg and mB=3 kg are kept on the table as shown in figure. The coefficients of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F that can be applied on B horizontally, so that the block A does not slide over the block B is : [Take g=10 m/s2 ]

Options:
A 16 N
B 12 N
C 40 N
D 8 N

The correct answer is:
16 N

Solution:
Maximum possible acceleration of 1 kg block a=µmgm=µg=2 m/s2
For 1 g block not to slide the maximum acceleration of both blocks should be
Fmax-µmg=mamax
Fmax-0.2×4×10=4×2
Fmax- 8=8
Fmax=16 N

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Two blocks A and B of masses m and 2m are connected by a massless spring of force constant k and are placed on a smooth horizontal plane. The spring is stretched by an amount x and then released. The relative velocity of the blocks when the spring comes to its natural length is

Options:
A x3k2m
B x2k3m
C xk3m
D x2km

The correct answer is:
x3k2m

Solution:
Let us assume the velocity of block A and B are v1 and v2 respectively.
Relative velocity will be v1+v2
Using conservation of linear momentum
mv1=2mv2
⇒ v1=2v2
Using conservation of energy
12kx2=12mv12+122mv22
12kx2=12m2v22+mv22
12kx2=2mv22+mv22
3mv22=kx22
v22=kx26m
v2 k6m.x
Relative velocity =3v2
=3k6m.x
= 3k2m.x

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two blocks A and B of masses m and 2m respectively, attached at opposite ends of a spring of constant K, placed on a smooth horizontal surface. Spring is initially at its natural length l. A is given a velocity 2V0 and B given velocity V0 as shown.

Maximum separation between m and centre of mass of the system will be:

Options:
A l3+8mV023K
B l3+2mV023K
C 2l3+2mV023K
D 2l3+8mV023K

The correct answer is:
2l3+8mV023K

Solution:
l1=2ml2m+m=2l3
K1l1=KlâK12l3=KlâK1=3K2
x1 max=m2V023K2=8mV023K
d1 max=l1+x1 max=2l3+8mV023K

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two blocks are connected by a string as shown in the diagram. The upper block is hung by another string. A force $F$ applied on the upper string produces an acceleration of $2 \mathrm{~m} / \mathrm{s}^2$ in the upward direction in both the blocks. If $T$ and $T^{\prime}$ be the tensions in the two parts of the string, then

Options:
A $T=70.8 \mathrm{~N}$ and $T^{\prime}=47.2 \mathrm{~N}$
B $T=58.8 N$ and $T^{\prime}=47.2 N$
C $T=70.8 \mathrm{~N}$ and $T^{\prime}=58.8 \mathrm{~N}$
D $T=70.8 \mathrm{~N}$ and $T^{\prime}=0$

The correct answer is:
$T=70.8 \mathrm{~N}$ and $T^{\prime}=47.2 \mathrm{~N}$

Solution:
FBD of mass $2 \mathrm{~kg} F B D$ of mass $4 \mathrm{~kg}$
$T-T^{\prime}-20=4$
$T^{\prime}-40=8$
By solving (i) and (ii) $T^{\prime}=47.23 \mathrm{~N}$ and $T=70.8 \mathrm{~N}$

Topic: Physics, Laws of Motion, JEE Main
Two blocks of $2 \mathrm{~kg}$ and $1 \mathrm{~kg}$ are in contact on a frictionless table. If a force of $3 \mathrm{~N}$ is applied on $2 \mathrm{~kg}$ block, then the force of contact between the two blocks will be :

Options:
A $0 \mathrm{~N}$
B $1 \mathrm{~N}$
C $2 \mathrm{~N}$
D $3 \mathrm{~N}$

The correct answer is:
$1 \mathrm{~N}$

Solution:
Hints: Common acceleration $=\frac{3}{3}=1 \mathrm{~m} / \mathrm{sec}^2$
$$
\mathrm{N}_1=1 \mathrm{~N}
$$

Topic: Physics, Laws of Motion, JEE Main