datavorous/entrance-exam-dataset huggingface dataset:
Two nuclei have mass numbers in the ratio 8 : 125. The ratio of their nuclear radii is:

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

The correct answer is:
2:5

Solution:
Here ,
R= R 0 A 1/3
Since, R= R 0 A 1/3
â´ R 1 R 2 = A 1 1 3 A 2 1 3 = ( A 1 A 2 ) 1 3 âââââââââââââââââââââââââ= ( 8 125 ) 1 3 âââââââââââââââââââââââââ= 2 5
â´ The ratio of their radii = 2:5

Topic: Physics, Nuclear Physics, JEE Main
Two nuclei have their mass numbers in the ratio of $1: 3$. The ratio of their nuclear densities would be

Options:
A $1: 1$
B $1: 3$
C $3: 1$
D $(3)^{1 / 3}: 1$

The correct answer is:
$1: 1$

Solution:
Density is independent of mass number of nuclei.

Topic: Physics, Nuclear Physics, NEET, NEET 2008 (Mains)
Two nucleotides are joined together by a linkage known as :

Options:
A Phosphodiester linkage
B Glycosidic linkage
C Disulphide linkage
D Peptide linkage

The correct answer is:
Phosphodiester linkage

Solution:
The bond joining two successive nucleotides in the same strand of DNA is called the phosphodiester bond. the phosphodiester bond is the linkage between the 3' carbon atom of one sugar molecule and the 5' carbon atom of another, deoxyribose in DNA and ribose in RNA.

Phosphodiester linkage

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Two numbers a and b are chosen simultaneously from the set of integers 1, 2, 3,…, 39, then the probability that the equation 7a-9b=0 is satisfied is

Options:
A 1247
B 2247
C 4741
D 5741

The correct answer is:
4741

Solution:
Total cases = Selection of two numbers simultaneously
=C392=39Ã19=741
For favorable cases
a=9b7âb must be a multiple of 7 and aâ¤39
âb=7,a=9
b=14,a=18
b=21,a=27
b=28,a=36
Only 4 possible cases
Hence, the required probability =4741

Topic: Mathematics, Probability, JEE Main
Two numbers are selected randomly from set S=1,2,3,4,5,6 without replacement one by one. The probability that a minimum of two number is less than 4, is

Options:
A 115
B 1415
C 15
D 45

The correct answer is:
45

Solution:
Total cases =C2=156
Minimum of 2 numbers <4= Total – (both numbers must be ≥4)
=15-C2=123
The required probability =1215=45

Topic: Mathematics, Probability, JEE Main
Two objects A and B are placed at 15 cm and 25 cm from the pole in front of a concave mirror having radius of curvature 40 cm. The distance between images formed by the mirror is:

Options:
A 40 cm
B 60 cm
C 160 cm
D 100 cm

The correct answer is:
160 cm

Solution:
Let's consider the following diagram-

Let's assume that vA, vB indicates the distances of the images of objects A and B respectively.

Thus, by mirror formula, for object A, it can be written that

1v+1u=1f1vA+1-15=1-201vA=1-20+115=-3+460vA=60 cm

Similarly, for object B, using mirror formula,

1vB+1-25=1-201vB=-120+125=-5+4100=-1100vB=-100 cm

Hence, the distance d between the images can be calculated as follows:

d= vA+vB= 60+100 cm= 160 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Two objects are located at height 10 m above the ground. At some point of time, the objects are thrown with initial velocity 22 m s-1 at an angle 45° and 135° with the positive X-axis respectively. Assuming g=10 m s-2, the velocity vectors will be perpendicular to each other at time equal to

Options:
A 0.2 s
B 0.4 s
C 0.6 s
D 0.8 s

The correct answer is:
0.4 s

Solution:
Given, U1=U2=22 m s-1, g=10 m s-2

The velocity vectors are perpendicular ot each other at time t,

v→1=U1cos45°i^+U1sin45°-gtj^

v→1=2212i^+2212-10tj^

=2i^+2-10tj^ ...1

v2→=U2cos135°i^+U2sin135°-10tj^

=-2i^+2-10tj^ ...2

v→1·v→2=0 (∴ Both are perpendicular)

⇒2i^+2-10tj^·-2i^+2-10tj^=0

⇒-42-10t2=0

On solving,

100t2-40t=0

10t10t-4=0

∴ t=0.4 s

Topic: Physics, Motion In Two Dimensions, JEE Main
Two objects of equal masses placed at certain distance from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other object, then the new force will be

Options:
A 29F
B 169F
C 89F
D F

The correct answer is:
89F

Solution:
The gravitational force between particles is given by, F=Gm1m2r2

∵ F=Gm2r2

When one-third mass is transferred to other object

m1=2m3, m2=4m3

F2=8Gm29r2

⇒F2=89F

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Two objects of masses $200 \mathrm{~g}$ and $500 \mathrm{~g}$ possess velocities $10 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}$ and $3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}} \mathrm{m} / \mathrm{s}$ respectively. The velocity of their centre of mass in $\mathrm{m} / \mathrm{s}$ is :

Options:
A $5 \hat{\mathbf{i}}-25 \hat{\mathbf{j}}$
B $\frac{5}{7} \hat{\mathbf{i}}-25 \hat{\mathbf{j}}$
C $5 \hat{\mathbf{i}}+\frac{25}{7} \hat{\mathbf{j}}$
D $25 \hat{\mathbf{j}}-\frac{5}{7} \hat{\mathbf{j}}$

The correct answer is:
$5 \hat{\mathbf{i}}+\frac{25}{7} \hat{\mathbf{j}}$

Solution:
$m_1=200 \mathrm{~g}, m_2=500 \mathrm{~g}$
$v_1=10 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}, v_2=(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}$
velocity of centre of mass,
$v_{\mathrm{CM}}=\frac{m_1 v_1+m_2 v_2}{m_1+m_2}$
$=\frac{200 \times 10 \hat{\mathbf{i}}+500(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}})}{200+500}$
$=\frac{2000 \hat{\mathbf{i}}+1500 \hat{\mathbf{i}}+2500 \hat{\mathbf{j}}}{700}$
$=\frac{3500 \hat{\mathbf{i}}+2500 \hat{\mathbf{j}}}{700}$
$=5 \hat{\mathbf{i}}+\frac{25}{7} \hat{\mathbf{j}}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two objects P and Q, travelling in the same direction start from rest. While the object P starts at time t=0 and object Q starts later at t=30 min. The object P has an acceleration of 40 km/h2. To catch P at a distance of 20 km, the acceleration of Q should be

Options:
A 40 km/h2
B 80Â km/h2
C 100Â km/h2
D 160 km/h2

The correct answer is:
160 km/h2

Solution:
According to question,
Object P starts at t=0 and acceleration of P=40 km/h2
The time taken by object P to cover 20 km will be,
s=ut+12at2
Where, u=0, s=12at2
t=2sa
Here, a=40 km/h2
and s=20 km
t=2 Ã 2040
=1 hour
Time taken by Q will be 12 hour, because it started after 30 min.
The acceleration in Q,
a=2st2=2 Ã 20122
=2Ã4Ã20
=160 km/h2

Topic: Physics, Motion In One Dimension, JEE Main
Two objects suspended from the arms of a balance are in equilibrium when completely submerged in water. The mass of one body is 36 g and its density is 9 g cm-3â. If the mass of the other object is 48 g, then its density in g cm-3 is

Options:
A 43
B 3 2
C 3
D 5

The correct answer is:
3

Solution:
Apparent weight = V Ï - Ï g = m Ï Ï - Ï g
where m = mass of the body, Ï = density of the body and Ï = density of water If two bodies are in equilibrium then their apparent weight must be equal.
â´ m 1 Ï 1 Ï 1 - Ï g = m 2 Ï 2 Ï 2 - Ï g
â 3 6 9 9 - 1 = 4 8 Ï 2 Ï 2 - 1 g .
By solving we get Ï 2 = 3 .

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non volatile solute are together sealed in a container. Over time:

Options:
A the volume of the solution increases and the volume of the solvent decreases
B the volume of the solution decreases and the volume of the solvent increases
C the volume of the solution and the solvent does not change the volume of the solution does not
D change and the volume of the solvent decreases

The correct answer is:
the volume of the solution increases and the volume of the solvent decreases

Solution:
Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non volatile solute are together sealed in a container, the volume of the solution increases and the volume of the solvent decreases.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Two open organ pipes of lengths $60 \mathrm{~cm}$ and $90 \mathrm{~cm}$ resonate at $6^{\text {th }}$ and $5^{\text {th }}$ harmonics respectively. The difference of frequencies for the given modes is _______ $\mathrm{Hz}$. (Velocity of sound in air $=333 \mathrm{~m} / \mathrm{s}$ )

Options:

The correct answer is:
740

Solution:
The difference in frequency in open organ pipe $=$
$\begin{aligned}
& \mathrm{f}=\frac{\mathrm{nv}}{2 \mathrm{~L}} \\
& \Delta \mathrm{f}=\frac{6 \mathrm{v}}{2 \times 0.6}-\frac{5 \mathrm{v}}{2 \times 0.9} \\
& \mathrm{v}=333 \mathrm{~m} / \mathrm{s} \\
& \Delta \mathrm{f}=740 \mathrm{~Hz}
\end{aligned}$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Two opposite and equal charges $4 \times 10^{-8}$ coulomb when placed $2 \times 10^{-2} \mathrm{~cm}$ away, form a dipole. If this dipole is placed in an external electric field $4 \times 10^8$ newton/coulomb, the value of maximum torque and the work done in rotating it through $180^{\circ}$ will be

Options:
A $64 \times 10^{-4} \mathrm{Nm}$ and $64 \times 10^{-4} \mathrm{~J}$
B $32 \times 10^{-4} \mathrm{Nm}$ and $32 \times 10^{-4} \mathrm{~J}$
C $64 \times 10^{-4} \mathrm{Nm}$ and $32 \times 10^{-4} \mathrm{~J}$
D $32 \times 10^{-4} \mathrm{Nm}$ and $64 \times 10^{-4} \mathrm{~J}$

The correct answer is:
$32 \times 10^{-4} \mathrm{Nm}$ and $64 \times 10^{-4} \mathrm{~J}$

Solution:
Dipole moment $p=4 \times 10^{-8} \times 2 \times 10^{-4}=8$ $\times1.0^{-12} \mathrm{~m}$
Maximum torque $=p E=8 \times 10^{-12} \times 4 \times 1$
$=32 \times 10^{-4} \mathrm{Nm}$
Work done in rotating through $1.80^{\circ}=2 p E$
$=2 \times 32 \times 10^{-4}=64 \times 10^{-4} \mathrm{~J}$

Topic: Physics, Electrostatics, JEE Main
Two oxides of a $X$ contain $50 \%$ and $40 \%$ of non-metal respectively. If the formula of the first oxide is $\mathrm{XO}_2$. Then the formula of second oxide is

Options:
A $X_2 \mathrm{O}_3$
B $X_2 \mathrm{O}_5$
C $X \mathrm{O}_3$
D $X_2 \mathrm{O}$

The correct answer is:
$X \mathrm{O}_3$

Solution:
In $\mathrm{XO}_2, \mathrm{X}: \mathrm{O}:: 50: 50$
Mass of oxygen $=2 \times 16=32 \mathrm{~g}$
Hence, mass of $X=32 \mathrm{~g}$
In $\mathrm{XO}_n, \mathrm{X}: \mathrm{O}:: 40: 60$
Mass of $X$, in this case also remains $=32 \mathrm{~g}$
$40 \% \equiv 32 \mathrm{~g}$
$60 \% \equiv \frac{32}{40} \times 60=48 \mathrm{~g}$
Mass of oxygen $=48 \mathrm{~g}$
$\Rightarrow \quad n=\frac{48}{16}=3$
Formula of oxide $=\mathrm{XO}_3$

Topic: Chemistry, Some Basic Concepts of Chemistry, AP EAMCET, AP EAMCET 2017 (26 Apr Shift 1)
Two oxides of nitrogen, NO and NO2 are allowed to react together at 243 K and form coloured compound of nitrogen (X). When compound (X) reacts with water to yield another compound of nitrogen (Y). The shape of the anion (Y) molecule is

Options:
A angular
B triangular pyramidal
C tetrahedron
D square planar

The correct answer is:
angular

Solution:
Here X is N2O3 (light blue coloured) and Y is Nitrous (HNO2) acid
NO (g) +  NO 2 (g) → 243 K N 2 O 3 X Blue coloured solid → H 2 O HNO 2 Y
Anion of HNO 2 is NO 2 − and it is a angular or bent in shape

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Two parabolas with a common vertex and with axes along the x-axis and y-axis respectively, intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is :

Options:
A 3x+y+4=0
B 82x+y+3=0
C x+2y+3=0
D 4x+y+3=0

The correct answer is:
4x+y+3=0

Solution:
Equation of two parabola are y2=3x and x2=3y.

Let equation of tangent to y2=3x is y=mx+34m is also tangent to x2=3y

⇒ x2=3mx+94m

⇒4mx2-12m2x-9=0 have equal roots

⇒D=0

⇒144m4=44m(-9)

⇒m4+m=0⇒m=-1

Hence, common tangent is y=-x-34

⇒4x+y+3=0

Topic: Mathematics, Parabola, JEE Main, JEE Main 2018 (15 Apr)
Two parabolas with a common vertex and with axes along $x$-axis and $y$-axis, respectively, intersect each other in the first quadrant. if the length of the latus rectum of each parabola is 3 , then the equation of the common tangent to the two parabolas is?

Options:
A
$3(x+y)+4=0$

B
$8(2 x+y)+3=0$

C
$4(x+y)+3=0$

D
$x+2 y+3=0$

The correct answer is:
$4(x+y)+3=0$

Solution:
As origin is the only common point to $x-$ axis and $y$-axis, so, origin is the common vertex Let the equation of two of parabolas be $y^2=4 a x$ and $x^2=4 b y$
Now latus rectum of both parabolas $=3$
$$
\begin{aligned}
&\therefore 4 a=4 b=3 \\
&\Rightarrow a=b=\frac{3}{4}
\end{aligned}
$$
$\therefore$ Two parabolas are $y^2=3 x$ and $x^2=3 y$ Suppose $y=m x+c$ is the common tangent.
$$
\begin{aligned}
&\therefore y^2=3 x \Rightarrow(m x+c)^2=3 x \\
&\Rightarrow m^2 x^2+(2 m c-3) x+c^2=0
\end{aligned}
$$
As, the tangent touches at one point only
So, $b^2-4 a c=0$
$$
\begin{aligned}
&\Rightarrow(2 m c-3)^2-4 m^2 c^2=0 \\
&\Rightarrow 4 m^2 c^2+9-12 m c-4 m^2 c^2=0
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow c=\frac{9}{12 m}=\frac{3}{4 m} \\
&\therefore x^2=3 y \Rightarrow x^2=3(m x+c) \\
&\Rightarrow x^2-3 m x-3 c=0
\end{aligned}
$$
Form (i) and ( ii)
$$
\begin{aligned}
&m^2=\frac{-4 c}{3}=\frac{-4}{3}\left(\frac{3}{4 m}\right) \\
&\Rightarrow m^3=-1 \Rightarrow m=-1 \Rightarrow c=\frac{-3}{4}
\end{aligned}
$$
Hence, $y=m x+c=-x-\frac{3}{4}$
$$
\Rightarrow 4(x+y)+3=0
$$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
Two parallel glass plates are dipped partly in the liquid of density ' $d$ keeping them vertical. If the distance between the plates is ' $x$ ', surface tension for liquids is $T$ and angle of contact is $\theta$, then rise of liquid between the plates due to capillary will be

Options:
A $\frac{T \cos \theta}{x d}$
B $\frac{2 T \cos \theta}{x d g}$
C $\frac{2 T}{x d g \cos \theta}$
D $\frac{T \cos \theta}{x d g}$

The correct answer is:
$\frac{2 T \cos \theta}{x d g}$

Solution:
Let the width of each plate is b and due to surface tension liquid will rise upto height h then upward force due to surface tension
$=2 T b \cos \theta$...(i)
Weight of the liquid rises in between the plates
$=V d g=(b x h) d g$...(ii)
Equating (i) and (ii) we get, $2 T \cos \theta=b x h d g$
$\therefore h=\frac{2 T \cos \theta}{x d g}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of x A in the same direction. If the force of attraction per meter of each wire is 2×10-6 N, then the value of x is approximately

Options:
A 1
B 2.4
C 1.4
D 2

The correct answer is:
1.4

Solution:
Force per unit length between two parallel wires is Fl=μ0i1i22πd

Here, i1=i2=i and Fl=2×10-6 N

Then, 2×10-6=4π2×10-7i22π×0.2

⇒i2=2⇒i=1.4 A

Hence, x=1.4.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Two parallel metal plates having charges $+Q$ and $-Q$ face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will

Options:
A become zero
B increase
C decrease
D remain same

The correct answer is:
decrease

Solution:
The electric field in vacuum
$$
\mathrm{E}_0=\frac{\sigma}{2 \varepsilon_0}
$$
When the plates are dipped in kerosene oil tank, then the electric field between the plate will be
$$
\begin{array}{rlrl}
& & \mathrm{E} & =\frac{\sigma}{2 \varepsilon_0 \mathrm{k}} \\
\because & \mathrm{k} & >1 \\
\therefore & & \mathrm{E} & < \mathrm{E}_0
\end{array}
$$
So, the electric field between the plates will decrease.

Topic: Physics, Electrostatics, NEET, NEET 2010 (Mains)
Two parallel plate air capacitors of the same capacity C are connected in series to a battery of emf E. Then one of the capacitors is completely filled with the dielectric material of constant K. The change in the effective capacitance of the series combination is

Options:
A C2K-1K+1
B C4K-1K+1
C C2K+1K-1
D C2K-1K+12

The correct answer is:
C2K-1K+1

Solution:
1C1=1C+1C=2C⇒C1=C2
1C2=1C+1KC
1C2=1C1+1K
∴ C2=C1+1K=CK(K+1)
∆C= CK(K+1)-C2=CKK+1-12
=C2K-K-12(K+1)=C2K-1K+1

Topic: Physics, Electrostatics, JEE Main
Two parallel plate capacitors C1 and C2 each having capacitance of 10 μF are individually charged by a 100 V D.C. source. Capacitor C1 is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor C2 is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor C1 is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be _____ V.

(Assuming Dielectric constant =10)

Options:

The correct answer is:
55

Solution:
As source is connected in the process therefore, additional charge will be provided. Therefore, charge on C1=KCV.

Since source is disconnected, so its charge will remain the same. Then, charge on C2=CV.

New capacitance in both cases will become, KC.

When they are connected in parallel charge will be equally divided so charge on one capacitor is q=K+12CV.

So common potential,

V'=qKC=K+12KV=55 V.

Topic: Physics, Capacitance, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Two parallel plate capacitors of capacity C and 3C are connected in parallel combination and charged to a potential difference 18 V. The battery is then disconnected and the space between the plates of the capacitor of capacity C is completely filled with a material of dielectric constant 9. The final potential difference across the combination of capacitors will be _____ V.

Options:

The correct answer is:
6

Solution:
The charge on the capacitor is given by q=CV.

Initial charge on C=18CV

And initial charge on 3C=54CV

Let final common potential difference =V'

When a dielectric of dielectric constant k is introduced between the plates of the capacitor the capacitance becomes k times.

Applying conservation of charge,

9CV'+3CV'=18CV+54CV

⇒12CV'=72CV⇒V'=6 V

Topic: Physics, Capacitance, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Two parallel wires are carrying electric currents of equal magnitude and in the same direction. They exert

Options:
A An attractive force on each other
B A repulsive force on each other
C No force on each other
D A rotational torque on each other

The correct answer is:
An attractive force on each other

Solution:
An attractive force on each other
Two straight conductors carry current in same direction, then attractive force acts between them.

Topic: Physics, Magnetic Effects of Current, JEE Main
Two particle are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is

Options:
A zero
B $\frac{2 \pi}{3}$
C $\pi$
D $\frac{\pi}{6}$

The correct answer is:
$\frac{2 \pi}{3}$

Solution:
$$
\begin{aligned}
y_1 & =\frac{A}{2}=A \sin \omega t \\
\Rightarrow \quad \omega t & =30^{\circ} \\
y_2 & =\frac{A}{2}=A \sin \left(\omega t+\frac{\pi}{2}\right) \\
\Rightarrow \quad \omega t+\frac{\pi}{2} & =150^{\circ}
\end{aligned}
$$
$\therefore$ phase difference
$$
\begin{aligned}
\phi & =150^{\circ}-30^{\circ} \\
& =120=\frac{2 \pi}{3} \mathrm{rad}
\end{aligned}
$$

Topic: Physics, Oscillations, NEET, NEET 2011 (Mains)
Two particles A and B are located at points 0, 103 and (0, 0) in x-y plane. They start moving simultaneously at the time t=0 s with constant velocities VâA=5 i^ m s-1 and VâB=-53i^ m s-1, respectively. The time when they are closest to each other is found to be K/2 second. Find K. All distances are given in meter.

Options:

The correct answer is:
3

Solution:
As observed by B motion of A is along AM and BM is the shortest distance between them. Relative displacement of A w.r.t to B is AM=ABcosâ¡30o
VâAB=5i^+53j^
Time taken t=SâABVâAB=AMVAB=103cosâ¡30o10
=1.5 s

Topic: Physics, Motion In Two Dimensions, JEE Main
Two particles $A$ and $B$ (both initially at rest) start moving towards each other under a mutual force of attraction. At the instant, when the speed of $A$ is $v$ and the speed of $B$ is $2 v$, the speed of the centre of mass is

Options:
A zero
B $v$
C $\frac{3 v}{2}$
D $-\frac{3 v}{2}$

The correct answer is:
zero

Solution:
In the absence of external force the speed
$$
u_{\mathrm{co}}=0
$$
$F_{ext}=0$ then
$$
v_{\mathrm{cm} .}=0 \text { always }
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles A and B each of mass m are attached by a light inextensible string of length 2l. The whole system lies on a smooth horizontal table with B initially at a distance l from A. The particle at the end B is projected across the table with speed u perpendicular to AB. The velocity of the ball A just after the string is taut, is

Options:
A u34
B u3
C u32
D u2

The correct answer is:
u34

Solution:
When the string jerk tight both particles begin to move with velocity components v in the direction AB . Using conservation of momentum in the direction AB
mucosâ¡ 30Â°=mv+mv
or v=u34
Hence, the velocity of ball A just after the jerk is u34

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles A and B having charges 20 μC and -5 μC respectively are held fixed with a separation of 5 cm At what position a third charged particle should be placed so that it does not experience a net electric force ?

Options:
A At 5 cm from -5 μC on the right side
B At 5 cm from 20 μC on the left side of system
C At 1.25 cm from -5 μC between two charges
D At midpoint between two charges

The correct answer is:
At 5 cm from -5 μC on the right side

Solution:
Charge should be placed near -5μC at x distance
K(20)Q(5+x)2=K(5)QX2
4x2=(5+x)2
4x2=25+x2+10x
3x2-10x-25=0 ⇒x=5

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (31 Aug Shift 1)
Two particles, $A$ and $B$, having equal charges, after being accelerated through the same potential difference enter into a region of uniform magnetic field and the particles describe circular paths of radii $R_{1}$ and $R_{2}$ resporctively. The ratio of the masses of $A$ and $B$ is

Options:
A $\sqrt{R_{1}} / R_{2}$
B $R_{1} / R_{2}$
C $\left(R_{1} / R_{2}\right)^{2}$
D $\left(R_{2} / R_{1}\right)^{2}$

The correct answer is:
$\left(R_{1} / R_{2}\right)^{2}$

Solution:
Radius of circular path followed by charged particle is given by
$$
R=\frac{m v}{q B}=\frac{\sqrt{2 m K}}{q B} \quad[\because p=m v=\sqrt{2 m K}]
$$
where, $K$ is kinetic energy of particle. Charged particle $q$ is accelerated through some potential difference $V$, such that kinetic energy of particle is
$\therefore$
$$
\begin{aligned}
K &=q V \\
R &=\frac{\sqrt{2 m q V}}{q B}
\end{aligned}
$$
As the two charged particles of same magnitude and being accelerated through same potential, enters into a uniform magnetic field region, then $R \propto \sqrt{m}$
So,
$$
\frac{R_{1}}{R_{2}}=\sqrt{\frac{m_{A}}{m_{B}}} \Rightarrow \frac{m_{A}}{m_{B}}=\left(\frac{R_{1}}{R_{2}}\right)^{2}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
Two particles $A$ and $B$, initially at rest, move to wards each other under mutual force of attraction. At the instant when the speed of $A$ is $v$ and the speed of $B$ is $2 v$, the speed of the centre of mass of the system is

Options:
A 3v
B v
C 1.5v
D Zero

The correct answer is:
Zero

Solution:
The correct option is $\mathbf{4}$
Zero
As net external force on the system is zero and initial velocity of $\mathrm{CM}$ is zero, it will remain at rest.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles \( A \) and \( B \), initially at rest, move towards each other under a mutual force of attraction. At the instant when the speed of \( A \) is \( v \) and the speed of \( B \) is \( 2 v \), the speed of the center of mass of the system is:

Options:
A \( 3 v \)
B \( 2 v \)
C 1. \( 5 v \)
D zero

The correct answer is:
zero

Solution:
The two bodies are moving only because of a mutual force of attraction. Therefore, the net external force of the masses as a system is zero. Therefore, the acceleration of the center of mass is zero, so as the center of mass is initially at rest, it will continue to be at rest.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main