datavorous/entrance-exam-dataset huggingface dataset:
Two particles of masses m and 2m are attached to the massless rod of length 2l as shown in figure. The rod is hinged at its midpoint O and is free to rotate in vertical plane about hinge. The minimum speed v at the given instant so that it can complete the circle.

Options:
A gl
B 4gl
C 5gl
D 4gl3

The correct answer is:
4gl3

Solution:
Wg=Kf-Ki
mg2l-2mg2l=0-12mv2+122mv2
-2mgl=-32mv2
v=4gl3

Topic: Physics, Rotational Motion, JEE Main
Two particles of masses m and 2m are kept at a distance a. Find the speed of particles and their relative velocity of approach when separation becomes a/2.

Options:
A 23a2Gm
B a2Gm
C 22Gm3a
D 6Gma

The correct answer is:
6Gma

Solution:
In arrangements 1 and 2,
Momentum conservation,
0=mv1-2mv2 ...(i)
Energy conservation
-Gmâ2ma=12mv12+122mv22-Gmâ2ma/2 ...(ii)
From Equation (i), we get
v1=2v2
Putting this value v1=2v2 in Equation (ii), we get
-2Gm2a=12m2v22+122mv22-4Gm2a
3mv22=2Gm2a
v2=2Gm3a
Relative velocity v 1,2 =v1+v2
=32Gm3a
=6Gma

Topic: Physics, Gravitation, JEE Main
Two particles of masses M and 2M are moving with speeds of 10 m s-1 and 5 m s-1, as shown in the figure. They collide at the origin and after that they move along the indicated directions with speeds v1 and v2, respectively. The values of v1 and v2 are, nearly

Options:
A 6.5 m s-1 and 3.2 m s-1
B 3.2 m s-1 and 12.6 m s-1
C 13.02 m s-1 and 19.7 m s-1
D 3.2 m s-1 and 6.3 m s-1

The correct answer is:
13.02 m s-1 and 19.7 m s-1

Solution:
From conservation of linear momentum,
Along x-axis:
10âMâ32+â10M12=M32âv1+2Mv212
32âv1+2v2=5(3+2) â¦ (1)
Along y-axis:
-10M12+10M12=Mv112-2Mv212
âv12-2v2=5(2-1) â¦ (2)
From equations (1) and (2)
v132+12=53+2+5(2-1)
v1=13.02âm/s
v2=53+52-13.02Ã322
=19.72âm s-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles of masses m and 3m approach each other with different velocities. After collision, the particle of mass m has velocity vâ in their centre of mass frame. Velocity of particle of mass 3m in the centre of mass frame is

Options:
A -2v→
B -v→2
C -vâ3
D -v→4

The correct answer is:
-v→3

Solution:
In the centre of mass frame, the total linear momentum of the system is zero.

Hence, mvâ+3mvâ'=0

or vâ'=-vâ3

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles of masses \( m_{1} \) and \( m_{2} \) have equal kinetic energies. The ratio of their momentum is:

Options:
A \( m_{1}: m_{2} \)
B \( m_{2}: m_{1} \)
C \( \sqrt{m}_{1}: \sqrt{m}_{2} \)
D \( m_{1}^{2}: m_{2}^{2} \)

The correct answer is:
\( \sqrt{m}_{1}: \sqrt{m}_{2} \)

Solution:
Kinetic energy, K=P22m
∴P2=2mK
P=2mK
P1P2=2m1K2m2K

P1P2=m1m2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles of the same mass m are moving in circular orbits because of force, given by F r=-16r-r3. The first particle is at a distance r=1, and the second, at r=4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to

Options:
A 3 ×10-3
B 6 ×102
C 6 ×10-2
D 10-2

The correct answer is:
6 ×10-2

Solution:
Centripetal force is given as Fr=mv2r.

Given in the problem, mV2r=16r+r3 ...i
Kinetic energy is stated as KE0=12mv2.
KE0=1216+r4 (From equationi)
The ratio of kinetic energies of first and second particle is K1K2=16+1216+2562=17272
K1K2 ≃6 ×10-2.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2018 (16 Apr Online)
Two particles P1 and P2 are performing SHM along the same line about the same mean position. Initially, they are at their positive extreme position. If the time period of each particle is 12 s and the difference of their amplitudes is 12 cm then find the minimum time (in seconds) after which the separation between the particles become 6 cm.

Options:

The correct answer is:
2

Solution:
The x coordinates of the particles are
x1=A1cosâ¡Ït, x2=A2cosâ¡Ït
Separation x1-x2=A1-A2cosâ¡Ït=12cosâ¡Ït
Now x1-x2=6=12cosâ¡Ït
âÏt=Ï3â2Ï12 Ãt=Ï3
ât=2 s

Topic: Physics, Oscillations, JEE Main
Two particles $\mathrm{P}$ and $\mathrm{Q}$ describe S.H.M. of same amplitude a, same frequency $\mathrm{f}$ along the same straight line. The maximum distance between the two particles is $\mathrm{a} \sqrt{2}$. The initial phase difference between the particle is $-$

Options:
A zero
B $\pi / 2$
C $\pi / 6$
D $\pi \beta$

The correct answer is:
$\pi / 2$

Solution:
$\mathrm{x}_{1}=\mathrm{a} \sin \left(\omega \mathrm{t}+\phi_{1}\right), \mathrm{x}_{2}=\mathrm{a} \sin \left(\omega \mathrm{t}+\phi_{2}\right)$

$\Rightarrow \quad\left|\mathrm{x}_{1}-\mathrm{x}_{2}\right|=2 \mathrm{a} \sin \left(\omega \mathrm{t}+\frac{\phi_{1}+\phi_{2}}{2}\right) \cos \left(\frac{\phi_{1}-\phi_{2}}{2}\right)$

Tomaximize $\left|\mathrm{x}_{1}-\mathrm{x}_{2}\right|: \sin \left(\omega \mathrm{t}+\frac{\phi_{1}+\phi_{2}}{2}\right)=1$

$\Rightarrow \mathrm{a} \sqrt{2}=2 \mathrm{a} \times 1 \times \cos \left(\frac{\phi_{1}-\phi_{2}}{2}\right)$

$\Rightarrow \frac{1}{\sqrt{2}}=\cos \left(\frac{\phi_{1}-\phi_{2}}{2}\right) \Rightarrow \frac{\pi}{4}=\frac{\phi_{1}-\phi_{2}}{2}$

$\Rightarrow \phi_{1}-\phi_{2}=\frac{\pi}{2}$

Topic: Physics, Oscillations, JEE Main
Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity $\overrightarrow{\mathrm{u}}$ and the other from rest with uniform acceleration $\overrightarrow{\mathrm{f}}$. Let $\alpha$ be the angle between their directions of motion. The relative velocity of the second particle w.r.t. the first is least after a time.

Options:
A
$\frac{u \cos \alpha}{f}$

B
$\frac{u \sin \alpha}{f}$

C
$\frac{\mathrm{f} \cos \alpha}{\mathrm{u}}$

D
u $\sin \alpha$.

The correct answer is:
$\frac{u \cos \alpha}{f}$

Solution:
Using ${ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r={ }^n C_{r+1}+\underbrace{n} C_{r-1}+{ }^n C_r+{ }^n C_r={ }^n C_{r+1}+{ }^{n+1} C_r+{ }^n C_r$
${ }^{n+1} C_{r+1}+{ }^{n+1} C_r \Rightarrow{ }^{n+2} C_{r+1}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2003
Two particles which are initially at rest, move towards each other under the action of their internal attraction. If their speeds are $\mathrm{v}$ and $2 \mathrm{v}$ at any instant, then the speed of centre of mass of the system will be

Options:
A $2 \mathrm{v}$
B 0
C $1.5 \mathrm{v}$
D $\mathrm{v}$

The correct answer is:
0

Solution:
As initially both the particles were at rest therefore velocity of centre of mass was zero and there is no external force on the system so speed of centre of mass remains constant $i e$, it should be equal to zero.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two particles which are initially at rest, move towards each other under the action of their internal attraction. If their speeds are $\mathrm{v}$ and $2 \mathrm{v}$ at any instant, then the speed of centre of mass of the system will be

Options:
A $2 \mathrm{v}$
B 0
C $1.5 \mathrm{v}$
D $\mathrm{v}$

The correct answer is:
0

Solution:
As initially both the particles were at rest therefore velocity of centre of mass was zero and there is no external force on the system so speed of centre of mass remains constant $i e$, it should be equal to zero.

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2010 (Screening)
Two particles which are initially at rest move towards each other under the action of their
mutual attraction. If their speeds are \( v \) and \( 2 v \) at any instant, then the speed of center of mass
of the system is,

Options:
A \( 1.5 \mathrm{v} \)
B \( 2 \mathrm{v} \)
C \( \mathrm{v} \)
D zero

The correct answer is:
zero

Solution:
(D)
At initial position both \( A \) and \( B \) are at rest, hence, momentum is zero. It is given that no external force isacting on them.
So, momentum remains zero. Since mass cannot be zero, therefore velocity is zero.

Topic: Physics, Center of Mass Momentum and Collision, KCET, KCET 2019
Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter, they enter normally in a region of uniform magnetic field and describes circular paths of radii R1 and R2 respectively. The mass ratio of X and Y is :

Options:
A R2R12
B R1R22
C R1R2
D R2R1

The correct answer is:
R1R22

Solution:
Energy gained due to potential difference by charged particle:

KE=qV

Relation between momentum & kinetic energy:

p=2mKE

Now, radius of the circular path under magnetic field will be given by,

R=mvqB=pqB=2m(KE)qB=2mqVqB

Therefore,

R∝m⇒m∝R2

⇒m1 m2=R1R22

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is

Options:
A 9
B 10
C 8
D 11

The correct answer is:
11

Solution:
Time period of a pendulum,

T∝l

⇒T1T2=l1l2

If n is the number of vibrations after which the pendulums are again in phase, the number of vibration of the longer pendulum will be n-1. Therefore,

T1×n-1T2×n=1 ⇒121n-1100n=1

⇒11n-11=10n ⇒n=11

Topic: Physics, Oscillations, JEE Main
Two periodic waves of intensities I1 and I2 pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

Options:
A I&ApplyFunction; 1 - I&ApplyFunction; 2 2
B 2 I⁡ 1 + I⁡ 2
C I 1 + I 2
D I&ApplyFunction; 1 + I&ApplyFunction; 2 2

The correct answer is:
2 I⁡ 1 + I⁡ 2

Solution:
Other factors such as ω and v remaining the same,
I⁡ = A⁡ 2 × constant K, or A⁡ = I⁡ K⁡
On superposition
Amax = A1 + A2 and Amin = A1 – A2
∴A⁡max2=A⁡12+A⁡22+2A⁡1A⁡2⇒I⁡maxK⁡=I⁡1K⁡+I⁡2K⁡+2I⁡1I⁡2K⁡
A⁡min2=A⁡12+A⁡22-2A⁡1A⁡2⇒I⁡minK⁡=I⁡1K⁡+I⁡2K⁡-2I⁡1I⁡2K⁡
∴ I max + I min = 2 I 1 + 2 I 2

Topic: Physics, Waves and Sound, NEET
Two periodic waves of intensities $I_1$ and $I_2$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

Options:
A $2\left(\mathrm{I}_1+\mathrm{I}_2\right)$
B $\mathrm{I}_1+\mathrm{I}_2$
C $\left(\sqrt{I_1}+\sqrt{I_2}\right)^2$
D $\left(\sqrt{I_1}-\sqrt{I_2}\right)^2$

The correct answer is:
$2\left(\mathrm{I}_1+\mathrm{I}_2\right)$

Solution:
$$
\begin{aligned}
& I_{\text {max }}=\left(\sqrt{l_1}+\sqrt{l_2}\right)^2 \\
& I_{\text {min }}=\left(\sqrt{l_1}-\sqrt{I_2}\right)^2 \\
\therefore \quad & I_{\text {max }}+I_{\text {min }}=2\left(I_1+I_2\right)
\end{aligned}
$$

Topic: Physics, Waves and Sound, NEET, NEET 2008 (Mains)
Two periodic waves of intensities $I_1$ and $I_2$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

Options:
A $I_1+I_2$
B $\left(\sqrt{I_1}+\sqrt{I_2}\right)^2$
C $\left(\sqrt{I_1}-\sqrt{I_2}\right)^2$
D $2\left(I_1+I_2\right)$

The correct answer is:
$2\left(I_1+I_2\right)$

Solution:
Resultant intensity of two periodic waves is given by
$I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \delta$
where $\delta$ is the phase difference between the waves.
For maximum intensity, $\delta=2 n \pi ; n=0,1,2 \ldots$ etc.
Therefore, for zero order maxima, $\cos \delta=1$
$I_{\text {max }}=I_1+I_2+2 \sqrt{I_1 I_2}=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2$
For minimum intensity, $\delta=(2 n-1) \pi$;
$n=1,2 \ldots \text { etc }$
Therefore, for Ist order minima, $\cos \delta=-1$
$\begin{aligned}
I_{\min } & =I_1+I_2-2 \sqrt{I_1 I_2} \\
& =\left(\sqrt{I_1}-\sqrt{I_2}\right)^2
\end{aligned}$
Therefore,
$\begin{aligned}
I_{\max }+I_{\min } & =\left(\sqrt{I_1}+\sqrt{I_2}\right)^2+\left(\sqrt{I_1}-\sqrt{I_2}\right)^2 \\
& =2\left(I_1+I_2\right)
\end{aligned}$

Topic: Physics, Current Electricity, JEE Main
Two perpendicular tangents to $y^2=4 a x$ always intersect on the line, if

Options:
A $x=a$
B $x+a=0$
C $x+2 a=0$
D $x+4 a=0$

The correct answer is:
$x+a=0$

Solution:
We know that tangent to the parabola at points $t_1$ and $t_2$ are $t_1 y=x+a t_1^2$ and $t_2 y=x+a t_2^2$. Since
tangents are perpendicular to the parabola, therefore, $\frac{1}{t_1} \cdot \frac{1}{t_2}=-1$ or $t_1 t_2=-1$
We also know that their point of intersection $=\left(a t_1 t_2, a\left(t_1+t_2\right)\right)=\left(-a, a\left(t_1+t_2\right)\right)$
Thus these points lie on directrix $x=-a$ or $x+a=0$

Topic: Mathematics, Parabola, JEE Main
Two persons $A$ and $B$ are located in $X-Y$ plane at the points $(0,0)$ and $(0,10)$ respectively. (The distances are measured in MKS unit). At a time $t=0$, they start moving simultaneously with velocities $\quad \overrightarrow{\mathbf{v}}_A=2 \mathbf{j} \mathrm{ms}^{-1} \quad$ and $\quad \overrightarrow{\mathbf{v}}_B=2 \hat{\mathbf{i}} \mathrm{ms}^{-1}$ respectively. The time after which $A$ and $B$ are at their closest distance is

Options:
A 2.5 s
B 4 s
C 1 s
D $\frac{10}{\sqrt{2}} \mathrm{~s}$

The correct answer is:
4 s

Solution:
Let after the time $(t)$ the position of $A$ is $\left(0, v_A t\right)$ and position of $B=\left(v_B t, 10\right)$. Distance between them
$\begin{aligned} y & =\sqrt{\left(0-v_B t\right)^2+\left(v_A t-10\right)^2} \\ \text { or } \quad y^2 & =(2 t)^2+(2 t-10)^2\end{aligned}$
or $\quad y^2=l=4 t^2+4 t^2+100-40 t$ $\Rightarrow \quad l=8 t^2+100-40 t$
Now, $\quad \frac{d l}{d t}=(16 t-40)=0$
$t=\frac{40}{16}=2.5 \mathrm{~s}$
As $\frac{d^2 l}{d t^2}=16=(+\mathrm{ve})$
Hence, $l$ will be minimum.

Topic: Physics, Motion In One Dimension, JEE Main
Two persons A and B have equal number of sons. There are three cinema tickets which are to be distributed among sons of A and B. The probability that all the tickets go to sons of A is 120. The number of sons to each of them is

Options:
A 3
B 4
C 5
D 6

The correct answer is:
3

Solution:
Let A,B have n sons each
Total cases =2nC3
Favourable cases =nC3
Probability =nC32nC3=120
ânn-1n-22n2n-12n-2=120
â10n2-3n+2=4n2-6n+2
â10n2-30n+20=4n2-6n+2
â6n2-24n+18=0
â6n2-4n+3=0
ân=1, 3 n=1 is rejected

Topic: Mathematics, Probability, JEE Main
Two persons A and B perform same amount of work in moving a body through a certain distance d with application of forces acting at angles 45° and 60° with the direction of displacement respectively. The ratio of force applied by person A to the force applied by person B is 1x. The value of x is _________.

Options:

The correct answer is:
2

Solution:
Given, W1=W2 and, we know that work done is a scalar quantity it means, F1Scos45°=F2Scos60°⇒F1F2=12=1x⇒x=2.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2021 (27 Aug Shift 1)
Two persons of masses $55 \mathrm{~kg}$ and $65 \mathrm{~kg}$ respectively, are at the opposite ends of a boat. The length of the boat is $3.0 \mathrm{~m}$ and weighs $100 \mathrm{~kg}$. The $55 \mathrm{~kg}$ man walks up to the $65 \mathrm{~kg}$ man and sits with him. If the boat is in still water the centre of mass of the system shifts by

Options:
A $3.0 \mathrm{~m}$
B $2.3 \mathrm{~m}$
C zero
D $0.75 \mathrm{~m}$

The correct answer is:
zero

Solution:
Here on the entire system net external force on the system is zero hence centre of mass remains unchanged.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two persons of masses $55 \mathrm{~kg}$ and $65 \mathrm{~kg}$ respectively, are at the opposite ends of a boat. The length of the boat is $3.0 \mathrm{~m}$ and weighs $100 \mathrm{~kg}$. The $55 \mathrm{~kg}$ man walks up to the $65 \mathrm{~kg}$ man and sits with him. If the boat is in still water the centre of mass of the system shifts by

Options:
A $3.0 \mathrm{~m}$
B $2.3 \mathrm{~m}$
C zero
D $0.75 \mathrm{~m}$

The correct answer is:
zero

Solution:
Here on the entire system net external force on the system is zero hence centre of mass remains unchanged.

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2012 (Screening)
Two persons pull a wire towards themselves. Each person exerts a force of $200 \mathrm{~N}$ on the wire. Young's modulus of the material of wire is $1 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$. Original length of the wire is $2 \mathrm{~m}$ and the area of cross section is $2 \mathrm{~cm}^2$. The wire will extend in length by ______ $\mu \mathrm{m}$.

Options:

The correct answer is:
20

Solution:
$200 \mathrm{~N} \longleftrightarrow O$
$\begin{aligned} & \frac{\mathrm{F}}{\mathrm{A}}=\mathrm{Y} \frac{\Delta \ell}{\ell} \Rightarrow \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}} \\ & \Delta \ell=\frac{200 \times 2}{2 \times 10^{-4} \times 10^{11}}=2 \times 10^{-5}=20 \mu \mathrm{m}\end{aligned}$

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2024 (09 Apr Shift 1)
Two persons start from the junction of two perpendicular roads with velocities 3 m/sec and 4 m/sec. The rate at which the two persons separate is

Options:
A 5m/sec
B 25m/sec
C 4m/sec
D 3m/sec

The correct answer is:
5m/sec

Solution:
Let, at time t, the distance between the two persons be x m
⇒x=3t2+4t2
⇒x=5t
⇒dxdt=5 m/sec

Topic: Mathematics, Application of Derivatives, JEE Main
Two pi and half sigma bonds are present in:

Options:
A O2-
B N2+
C N2
D Â O2

The correct answer is:
N2+

Solution:
O 2 − =σ1 s 2 < σ * 1 s 2 <σ2 s 2 < σ * 2 s 2 <σ2 p x 2 =π2 p y 2 =π2 p z 2 < π * 2 p y 2 = π * 2 p z 1
Bond Order = 6−3 2 =1.5
O 2 =σ1 s 2 < σ * 1 s 2 <σ2 s 2 < σ * 2 s 2 <σ2 p x 2 <π2 p y 2 =π2 p z 2 < π * 2 p y 1 = π * 2 p z 1
Bond order = 6−2 2 =2
N 2 =σ1 s 2 < σ * 1 s 2 <σ2 s 2 < σ * 2 s 2 <π2 p z 2 =π2 p z 2 <σ2 p x 2
Bond order = 6−0 2 =3
According to MOT ; electronic configuration of N2+ is:
N 2 + =σ1 s 2 < σ * 1 s 2 <σ2 s 2 < σ * 2 s 2 <π2 p x 2 =π2 p y 2 <σ2 p z 1
Bond order =129-4=2.5
i.e., 2Ï bonds and 0.5âÏ bonds.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Two plane mirror M1 and M2 are joined about there edges and then this structure is moving with velocity v0 as shown in the figure. A point object P is also moving with same speed v0 in the direction shown in the figure. Determine the magnitude of the velocity of the image of P in M1 wrt image of P in M2?

Options:
A v03
B v06
C v03
D v032+v0

The correct answer is:
v06

Solution:
Velocity of image of of the point in the first mirror with respect to the image of the point in the second mirror is

V→I1I2=-V03i^-V03j^
V→I1I2=V06

Topic: Physics, Ray Optics, JEE Main
Two plane mirrors are inclined at $70^{\circ}$. A ray incident on one mirror at angle, $\theta$ after reflection falls on second mirror and is reflected from there parallel to first mirror. The value of $\theta$ is:

Options:
A $45^{\circ}$
B $30^{\circ}$
C $55^{\circ}$
D $50^{\circ}$

The correct answer is:
$50^{\circ}$

Solution:
The actual situation is shown in the given degree
$$
\begin{aligned}
\theta+40^{\circ} & =90^{\circ} \\
\Rightarrow \quad \theta & =90-40=50^{\circ}
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main
Two plane mirrors are inclined to each other at an angle of \( 60^{\circ} \). If a ray of light incident on the first mirror is parallel to the second mirror, then it is reflected from the second mirror,

Options:
A perpendicular to the first mirror.
B parallel to the first mirror.
C parallel to the second mirror.
D perpendicular to the second mirror.

The correct answer is:
parallel to the first mirror.

Solution:
When light ray is incident on the first mirror parallel to the second mirror, the angle of incidence at first mirror becomes 30°.

From the laws of reflection, we know that the angle of reflection and angle of incidence are equal. So, the light ray reflects from the first mirror and is incident on the second mirror with an angle of incidence 30° and finally reflects, as shown in figure. As final reflected ray makes an angle of 60° with the second mirror then i can be concluded that reflected ray from second mirror is parallel to first mirror.

Topic: Physics, Ray Optics, JEE Main
Two plane mirrors are inclined to each other at some angle. A ray of light incident at an angle of 30Â° on one of the mirrors, after reflection from the other, retraces its path. The angle (in degree) between the mirrors

Options:

The correct answer is:
30

Solution:
â i = â r = 3 0 o
â´ â OIQ = 6 0 o
â´â IOQ = 9 0 o - 6 0 o = 3 0 o

Topic: Physics, Ray Optics, JEE Main
Two plane mirrors are inclined to each other such that a ray of light incident on the first mirror M1 and parallel to the second mirror (M2) is finally reflected from the second mirror (M2) and parallel to the first mirror M1. The angle between the two mirrors will be:

Options:
A 60o
B 45o
C 90o
D 75o

The correct answer is:
60o

Solution:
As given in question AB||CO
∴∠ABF=θ
From laws of reflection
∠OBC=θ
Similarly CD‖OB
∴∠DCE=θ
Again from laws of reflection,
∠OCB=θ
∴3θ=180o
⇒θ=60o

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (09 Jan Shift 2)
Two plane mirrors are placed parallel to each other at a distance 4d as shown in the figure. The point source of light S is having wavelength equal to Î» . The light is reaching the screen (right of the source) only after reflection. For point P to be 2nd maxima, the value of Î»2=Kd2D . Where K is an integer. Find value of K? (assume Dâ«d and dâ«Î»)

Options:
A 6
B 2
C 3
D 4

The correct answer is:
6

Solution:
At P: Path difference âx=8dÃ3dD
For 2nd maxima âx=2Î»=24d2D;Î»=12d2D
Î»2=6d2D=kd2Dâk=6

Topic: Physics, Wave Optics, JEE Main
Two planets A and B of equal mass are having their period of revolutions TA and TB such that TA=2TB. These planets are revolving in the circular orbits of radii rA and rB respectively. Which out of the following would be the correct relationship of their orbits?

Options:
A 2rA2=rB3
B rA3=2rB3
C rA3=4rB3
D TA2-TB2=π2GMrB3-4rA3

The correct answer is:
rA3=4rB3

Solution:
From Kepler's law of periods,

T2∝R3

Therefore,

⇒TATB2=rArB3

⇒22=rA3rB3, ⇒rA3=4rB3

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (28 Jun Shift 1)
Two planets A and B of radii R and 1.5 R have densities ρ and ρ2 respectively. The ratio of acceleration due to gravity at the surface of B to A is:

Options:
A 2 : 3
B 2 : 1
C 3 : 4
D 4 : 3

The correct answer is:
3 : 4

Solution:
The acceleration due to gravity is

g=GMR2=4πGRρ3⇒g∝ρ·R

Thus, the ratio is

⇒gBgA=1.5Rρ2ρR=12×1.5=0.75⇒gBgA=34

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Two planets $A$ and $B$ of radii $R$ and $1.5 R$ have densities $\rho$ and $\rho / 2$ respectively. The ratio of acceleration due to gravity at the surface of $B$ to A is :

Options:
A $2: 3$
B $2: 1$
C $3: 4$
D $4: 3$

The correct answer is:
$3: 4$

Solution:
Acceleration due to gravity,
$\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}=\frac{4}{3} \pi \mathrm{G} \rho \mathrm{R}$
$\therefore \frac{g_2}{g_1}=\frac{\rho_2}{\rho_1} \times \frac{R_2}{R_1}=\frac{1}{2} \times 1.5=\frac{3}{4}$

Topic: Physics, Gravitation, JEE Main
Two planets have the same average density but their radii are $R_1$ and $R_2$. If acceleration due to gravity on these planets be $g_1$ and $g_2$ respectively, then

Options:
A $\frac{g_1}{g_2}=\frac{R_1}{R_2}$
B $\frac{g_1}{g_2}=\frac{R_2}{R_1}$
C $\frac{g_1}{g_2}=\frac{R_1^2}{R_2^2}$
D $\frac{g_1}{g_2}=\frac{R_1^3}{R_2^3}$

The correct answer is:
$\frac{g_1}{g_2}=\frac{R_1}{R_2}$

Solution:
According to Gravitational law
\(F=\frac{G M, M_2}{r^2}\) ...(i)
Where \(m_1 \rightarrow\) mass of planet
\(m_2 \rightarrow\) mass of any body
Now, According to Newton's second law, a body of Mass \(m_2\) fuels gravitational acceleration ' \(g\) ' which is
\(F=M_2 g\) ...(ii)
So, from (i) & (ii) we get,
\(g=\frac{G M_1}{\gamma^2}\)
So, the ratio of gravitational acceleration due to two planet is
\(\begin{aligned}
& \frac{g_1}{g_2}=\frac{m_1}{r_1^2} \times \frac{r_2^2}{m_2}=\frac{(4 / 3) \pi r_1^3 \times \rho}{r_1^2} \times \frac{r_2^2}{(4 / 3) \pi r_2^3 \times \rho} \\
& \therefore \frac{g_1}{g_2}=\frac{r_1}{r_2} \\
& \text {Answer:- } \frac{g_1}{g_2}=\frac{r_1}{r_2}
\end{aligned}\)

Topic: Physics, Gravitation, JEE Main
Two plano-convex lenses of focal lengths 20 cm and 30 cm are placed together to form a double convex lens. The final focal length will be

Options:
A 12 cm
B 60 cm
C 20 cm
D 30 cm

The correct answer is:
12 cm

Solution:
Equivalent focal length
1F=1f1+1f2
=120+130
â´F=20Ã3020+30
=60050=12 cm

Topic: Physics, Ray Optics, JEE Main