datavorous/entrance-exam-dataset huggingface dataset:
Two satellites S1 and S2 are revolving in circular orbits around a planet with radius RI=3200 km and R2=800 km respectively. The ratio of speed of satellite S1 to the speed of satellite S2 in their respective orbits would be 1x where x=

Options:

The correct answer is:
2

Solution:
Orbital velocity of a satellite in circular motion around a planet,

vo=GMR∝1R

Now, v1v2=R2R1=8003200=12

Hence, the value of x=2.

Topic: Physics, Gravitation, JEE Main, JEE Main 2022 (25 Jun Shift 2)
Two second after projection a projectile is travelling in a direction inclined at 30o with horizontal, after one more second it is travelling horizontally. Angle of projection (in degree) with horizontal divided by 10 is.

Options:

The correct answer is:
6

Solution:
Let angle made by v→ initially and after time t be θ and α respectively.
tan⁡30o=usin⁡θ-g×2ucos⁡θ…….…(i)
tan⁡0o=usin⁡θ-g×3ucos⁡θ …….…(ii)
∴θ=60o

Topic: Physics, Motion In Two Dimensions, JEE Main
Two seconds after projection, a projectile is traveling in a direction inclined at 30° to the horizontal. After one more sec, it is traveling horizontally, the magnitude and direction of its velocity are

Options:
A 22 m s-1, 60°
B 203 m s-1, 60°
C 640 m s-1, 30°
D 406 m s-1, 30°

The correct answer is:
203 m s-1, 60°

Solution:
Let in 2s body reaches upto point A and after one more second upto point B.
Total time of ascent for a body is given as 3s
i.e., usin⁡θg=3
∴ usin⁡θ=10×3 =30 ......(i)
Horizontal component of velocity remains always constant
ucos⁡θ=vcos⁡30° .......(ii)
For vertical upward motion between point O and A
vsin⁡30o=usin⁡θ-g×2 [Using v = u - gt]
vsin⁡30o=30-20 As usin⁡θ=30
∴ v=20 m s-1
Substituting this value v in equation (ii), we get
ucos⁡θ=20cos⁡30°=103 .....(iii)
From equations (i) and (ii)
u=203 m s-1 and θ=60°

Topic: Physics, Motion In Two Dimensions, JEE Main
Two separate bulbs contain ideal gases $A$ and $B$. The density of gas $A$ is twice that of gas $B$. The molecular mass of $A$ is half that of gas $B$. The two gases are at the same temperature. The ratio of the pressure of $A$ to that of gas $B$ is

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

The correct answer is:
4

Solution:
$d_a=2 d_b ; 2 M_a=M_b$
$P V=n R T=\frac{m}{M} R T ; \quad P=\frac{m}{V} \cdot \frac{R T}{M}=\frac{d R T}{M}$
$\frac{P_a}{P_b}=\frac{d_a}{d_b} \frac{M_b}{M_a}=\frac{2 d_b}{d_b} \times \frac{2 M_a}{M_a}=4$

Topic: Chemistry, States of Matter, JEE Main
Two separate wires A and B are stretched by $2 \mathrm{~mm}$ and $4 \mathrm{~mm}$ respectively, when they are subjected to a force of $2 \mathrm{~N}$. Assume that both the wires are made up of same material and the radius of wire $B$ is 4 times that of the radius of wire $A$. The length of the wires $A$ and $B$ are in the ratio of $\mathrm{a}: \mathrm{b}$. Then $\mathrm{a} / \mathrm{b}$ can be expressed as $1 / x$ where $x$ is $\qquad$

Options:
A $8$
B $16$
C $32$
D $64$

The correct answer is:
$32$

Solution:
For $A \frac{E}{\pi r^2}=y \frac{2 m m}{a}$ ...(1)
For B $\frac{E}{\pi \cdot 16 r^2}=y \frac{4 m m}{b}$ ...(2)
$\therefore(1) /(2)$
$16=\frac{2 b}{4 a}$
$\frac{a}{b}=\frac{1}{32}$
$\therefore \text {Answer }=32$

Topic: Physics, Mechanical Properties of Solids, JEE Main
Two separate wires A and B are stretched by 2 mm and 4 mm respectively, when they are subjected to a force of 2 N. Assume that both the wires are made up of same material and the radius of wire B is 4 times that of the radius of wire A. The length of the wires A and B are in the ratio of a: b. Then ab can be expressed as 1x, where x is _____.

Options:

The correct answer is:
32

Solution:
For AEπr2=y2 mma ...1
For B Eπ·16r2=y4 mm b ...2
∴ (1)/(2)
16=2 b4 a

ab=132
∴ =32

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Two sets A and B are as under: A=a, b∈R×R :a-5<1 and b-5<1;
B=a, b∈R×R:4a-62+9b-52≤36. Then :

Options:
A neither A⊂B nor B⊂A
B BâA
C AâB
D Aâ©B=Ï (an empty set)

The correct answer is:
A⊂B

Solution:
Given sets, A=a, b∈R×R :a-5<1 and b-5<1

A=a, b∈R×R :-1<a-5<1 and -1<b-5<1

A=a, b∈R×R :4<a<6 and 4<b<6
B=a-62 9+b-524≤1

For set A,

For corner coordinates of the set A is 4,4,4,6,6,4,6,3.

For set B,

Now, Put 4,4 in the equation of the set B,

49+14≤1, 4,4 lies in the set B.
Similarly, 4,6,6,4,6,3 also lies in the set B.

So, A⊂B .

Topic: Mathematics, Ellipse, JEE Main
Two sets A and B are as under: A=a, b∈R×R :a-5<1 and b-5<1;
B=a, b∈R×R:4a-62+9b-52≤36. Then :

Options:
A neither A⊂B nor B⊂A
B BâA
C AâB
D Aâ©B=Ï (an empty set)

The correct answer is:
A⊂B

Solution:
Given sets, A=a, b∈R×R :a-5<1 and b-5<1

A=a, b∈R×R :-1<a-5<1 and -1<b-5<1

A=a, b∈R×R :4<a<6 and 4<b<6
B=a-62 9+b-524≤1

For set A,

For corner coordinates of the set A is 4,4,4,6,6,4,6,3.

For set B,

Now, Put 4,4 in the equation of the set B,

49+14≤1, 4,4 lies in the set B.
Similarly, 4,6,6,4,6,3 also lies in the set B.

So, A⊂B .

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2018 (08 Apr)
Two ships A and B are sailing straight away from a fixed point $\mathrm{O}$ along routes such that $\angle \mathrm{AOB}$ is always $120^{\circ}$. At a certain instance, $\mathrm{OA}=8 \mathrm{~km}, \mathrm{OB}=6 \mathrm{~km}$ and the ship $\mathrm{A}$ is sailing at the rate of $20 \mathrm{~km} / \mathrm{hr}$ while the ship B sailing at the rate of $30 \mathrm{~km} / \mathrm{hr}$. Then the distance between $\mathrm{A}$ and $\mathrm{B}$ is changing at the rate (in $\mathrm{km} / \mathrm{hr}$ ):

Options:
A
$\frac{260}{\sqrt{37}}$

B
$\frac{260}{37}$

C
$\frac{80}{\sqrt{37}}$

D
$\frac{80}{37}$

The correct answer is:
$\frac{260}{\sqrt{37}}$

Solution:
Let $\mathrm{OA}=x \mathrm{~km}, \mathrm{OB}=y \mathrm{~km}, \mathrm{AB}=\mathrm{R}$ $(\mathrm{AB})^2=(\mathrm{OA})^2+(\mathrm{OB})^2-2(\mathrm{OA})(\mathrm{OB}) \cos$ $120^{\circ}$
$$
\mathrm{R}^2=x^2+y^2-2 x y\left(-\frac{1}{2}\right)=x^2+y^2+x y
$$
$\mathrm{R}$ at $x=6 \mathrm{~km}$, and $y=8 \mathrm{~km}$
$$
\mathrm{R}=\sqrt{6^2+8^2+6 \times 8}=2 \sqrt{37}
$$
Differentiating equation (1) with respect to $t$
$$
\begin{gathered}
2 R \frac{d R}{d t}=2 \frac{d x}{d t}+2 y \frac{d y}{d t}+\left(x \frac{d y}{d t}+y \frac{d x}{d t}\right) \\
=\frac{1}{2 R}[2 \times 8 \times 20+2 \times 6 \times 30+(8 \times 30+6 \times 20)] \\
\frac{d R}{d t}=\frac{1}{2 \times 2 \sqrt{37}}[1040]=\frac{260}{\sqrt{37}}
\end{gathered}
$$

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2014 (11 Apr Online)
Two ships are sailling in the sea on the two sides of a lighthouse. If the distance between the ships is 103+1 meters and their angle of elevations of the top of the lighthouse are 60o and 45o, then the height of the lighthouse is (The two ships and the foot of lighthouse are in a straight line)

Options:
A 20 meters
B 203 meters
C 103 meters
D 53 meters

The correct answer is:
103 meters

Solution:
Let, AC is the lighthouse and the ships are at B and D.
In ÎABC,
ACBC=tanâ¡60o {where, AC=h meters = height of lighthouse}
hBC=3âBC=h3 meters
In ÎACD,
ACCD=tanâ¡45oâhCD=1âCD=h
BC+CD=h3+h=103+1
âh=103 meters

Topic: Mathematics, Heights and Distances, JEE Main
Two short bar magnets have their magnetic moments $1.2 \mathrm{Am}^2$ and $1.0 \mathrm{Am}^2$. They are placed on a horizontal table parallel to each other at a distance of $20 \mathrm{~cm}$ between their centres, such that their north poles pointing towards geographic south. They have common magnetic equatorial line. Horizontal component of earth's field is $3.6 \times 10^{-5} \mathrm{~T}$. Then, the resultant horizontal magnetic induction at mid point of the line joining their centers is $\left(\frac{\mu_0}{4 \pi}=10^{-7} \mathrm{~N} / \mathrm{m}\right)$

Options:
A $3.6 \times 10^{-5} \mathrm{~T}$
B $1.84 \times 10^{-4} \mathrm{~T}$
C $2.56 \times 10^{-4} \mathrm{~T}$
D $5.8 \times 10^{-5} \mathrm{~T}$

The correct answer is:
$2.56 \times 10^{-4} \mathrm{~T}$

Solution:
We knows, $B=\frac{\mu_0}{4 \pi} \frac{M}{r^3}$
Hence the resultant horizontal magnetic induction point of the line joining their conters is
$\begin{aligned}
B & =B_1+B_2+B_H \\
& =\frac{10^{-7} \times 1.2}{\left(10 \times 10^{-2}\right)^3}+\frac{10^{-7} \times 1}{\left(10 \times 10^{-2}\right)^3}+3.6 \times 10^{-5} \\
& =1.2 \times 10^{-4}+1 \times 10^{-4}+0.36 \times 10^{-4} \\
& =2.56 \times 10^{-4} \mathrm{~T}
\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
Two short bar magnets of dipole moments M and M3 are joined at right angles to form a cross as depicted in the figure. The value of Î¸ for which the system remains in equilibrium in a uniform external magnetic field B, is

Options:
A θ=30°
B θ=45°
C θ=60°
D θ=15°

The correct answer is:
θ=60°

Solution:
Torque τ acting on magnet (1),
τ1=MBsin⁡θ
Similarly, τ2=3 MBsin⁡θ
But for equilibrium, τ1=τ2
∴ MBsin⁡θ= 3 MBcos⁡θ
tan⁡θ=3 =tan⁡60°
∴ θ=60°

Topic: Physics, Magnetic Properties of Matter, NEET
Two short bar magnets of dipole moments M and M3 are joined at right angles to form a cross as depicted in the figure. The value of Î¸ for which the system remains in equilibrium in a uniform external magnetic field B is

Options:
A θ=30°
B θ=45°
C Î¸=60Â°
D θ=15°

The correct answer is:
θ=60°

Solution:
Torque τ acting on magnet (1),
τ1=MBsin⁡θ
Similarly, τ2=3 MBsin⁡θ
But for equilibrium, τ1=τ2
∴ MBsin⁡θ= 3 MBcos⁡θ
tan⁡θ=3 =tan⁡60°
∴ θ=60°

Topic: Physics, Magnetic Properties of Matter, NEET
Two short bar magnets of dipole moments M and M3 are joined at right angles to form a cross as depicted in the figure. The value of Î¸ for which the system remains in equilibrium in a uniform external magnetic field B, is

Options:
A θ=30°
B θ=45°
C θ=60°
D θ=15°

The correct answer is:
θ=60°

Solution:
Torque τ acting on magnet (1),
τ1=MBsin⁡θ
Similarly, τ2=3 MBsin⁡θ
But for equilibrium, τ1=τ2
∴ MBsin⁡θ= 3 MBcos⁡θ
tan⁡θ=3 =tan⁡60°
∴ θ=60°

Topic: Physics, Magnetic Properties of Matter, JEE Main
Two short bar magnets of length 1cm each have magnetic moments 1.20 A m2 and 1.00 A m2 respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the south. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centres is close to
( Horizontal component of earth's magnetic induction is 3.6×10-5 Wb m-2 )

Options:
A 3.50×10-4 Wb m-2
B 5.80×10-4 Wb m-2
C 3.6×10-5 Wb m-2
D 2.56×10-4 Wb m-2

The correct answer is:
2.56×10-4 Wb m-2

Solution:
B net = B 1 + B 2 + B H
= Î¼ 0 4 Ï M 1 x 3 + Î¼ 0 4 Ï M 2 x 3 + B H
= { 10 â7 ( 1.2+1 ) } ( 0.1 ) 3 +3.6Ã 10 â5
= 2.56 Ã 10 - 4 Wb / m 2

Topic: Physics, Magnetic Properties of Matter, JEE Main
Two short bar magnets of length 1cm each have magnetic moments 1.20 A m2 and 1.00 A m2 respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the south. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centres is close to
( Horizontal component of earth's magnetic induction is 3.6×10-5 Wb m-2 )

Options:
A 3.50×10-4 Wb m-2
B 5.80×10-4 Wb m-2
C 3.6×10-5 Wb m-2
D 2.56×10-4 Wb m-2

The correct answer is:
2.56×10-4 Wb m-2

Solution:
B net = B 1 + B 2 + B H
= Î¼ 0 4 Ï M 1 x 3 + Î¼ 0 4 Ï M 2 x 3 + B H
= { 10 â7 ( 1.2+1 ) } ( 0.1 ) 3 +3.6Ã 10 â5
= 2.56 Ã 10 - 4 Wb / m 2

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2013 (07 Apr)
Two short magnetic dipoles m1 and m2 each having magnetic moment of 1 A m2 are placed at point O and P respectively. The distance between OP is 1 m. The torque experienced by the magnetic dipole m2 due to the presence of m1 is ________×10-7N m

Options:

The correct answer is:
1

Solution:
τ→=M→2×B→1

τ=M2B1sin90∘=1×μ04πM(1)3×1=10−7N m

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (26 Aug Shift 1)
Two short magnets $A B$ and $C D$ are in the $X-Y$ plane and are parallel to $X$-axis and co-ordinates of their centres respectively are $(0,2)$ and $(2,0)$. Line joining the north-south poles of $C D$ is opposite to that of $A B$ and lies along the positive $X$-axis. The resultant field induction due to $A B$ and $C D$ at a point $P(2,2)$ is $100 \times 10^{-7} \mathrm{~T}$. When the poles of the magnet $C D$ are reversed, the resultant field induction is $50 \times 10^{-7} \mathrm{~T}$. The value of magnetic moments of $A B$ and $C D$ (in $\mathrm{Am}^2$ ) are :

Options:
A $300 ; 200$
B $600 ; 400$
C $200 ; 100$
D $300 ; 150$

The correct answer is:
$300 ; 200$

Solution:
As shown in figure,
$B=B_1+B_2=\frac{\mu_0}{4 \pi}\left(\frac{2 M_1}{r_1^3}+\frac{M_2}{r_2^3}\right)$
$100 \times 10^{-7}=10^{-7}\left(\frac{2 M_1}{8}+\frac{M_2}{8}\right)$
$\Rightarrow \quad 2 M_1+M_2=800$ $\ldots$ (i)
If the poles of the magnet $C D$ are reversed, then
$50 \times 10^{-7}=10^{-7}\left(\frac{2 M_1}{8}-\frac{M_2}{8}\right)$
$\Rightarrow \quad 2 M_1-M_2=400$ $\ldots$ (ii)
Solving Eqs. (i) and (ii), we obtain
$M_1=300 \mathrm{Am}^2, M_2=200 \mathrm{Am}^2$

Topic: Physics, Magnetic Effects of Current, JEE Main
Two sides of a parallelogram are along the lines 4x+5y=0 and 7x+2y=0 . If the equation of one of the diagonals of the parallelogram is 11x+7y=9, then other diagonal passes through the point:

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

The correct answer is:
2,2

Solution:
Both the lines 4x+5y=0,7x+2y=0 pass through origin.

D is the point of intersection of 4x+5y=0 & 11x+7y=9

So upon solving we get the coordinates of point D=53,-43

Also, point B is the point of intersection of 7x+2y=0 & 11x+7y=9

So, coordinates of point B=-23,73

The diagonals of parallelogram bisect each other, so the mid point of BD is

53-232,-43+732=12,12

The equation of diagonal AC is

y-0=12-012-0x-0 i.e., y=x

Therefore, the diagonal AC passes through 2,2.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (27 Jul Shift 2)
Two sides of a parallelogram are along the lines, x+y=3 and x-y+3=0. If its diagonals intersect at 2,4, then one of its vertex is:

Options:
A 3,6
B 2,6
C 2,1
D 3,5

The correct answer is:
3,6

Solution:
As BC∥AD⇒SlopeBC=SlopeAD

Hence, slope of BC=1⇒3-h-5h-4=1
⇒-h-2=h-4⇒ 2h=2⇒h=1
∴ B1,2 and D3,6

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Two sides of a rhombus are along the lines, x-y+1=0 and 7x-y-5=0 . If its diagonals intersect at -1, -2 , then which one of the following is a vertex of this rhombus ?

Options:
A 13, -83
B -103,Â -73
C -3,Â -9
D -3,Â -8

The correct answer is:
13, -83

Solution:
d1 :y-2=-2-2-1-1 x-1
y-2=2x-1
y-2x=0
d2⊥d1 ⇒ 2y+x=k , -1, -2 Lies on it ∴ k=-5
⇒2y+x+5=0
Now point of intersection of d2 and L1
x-y+1=0 ......1
x+2y+5=0 .......2

on solving 1 & 2
-3y-4=0
y= -43
And point of intersection of d2 and L2
x+2y+5=0 ......3
14x-2y-10=0 ........4
On solving 3 & 4
And y= -83
15x-5=0
⇒ x=13

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2016 (03 Apr)
Two sides of a triangle are $\sqrt{3}+1$ and $\sqrt{3}-1$ and the included angle is $60^{\circ}$, then the difference of the remaining angles is

Options:
A $30 ^{\circ}$
B $45 ^{\circ}$
C $60 ^{\circ}$
D $90 ^{\circ}$

The correct answer is:
$90 ^{\circ}$

Solution:
Let $\mathrm{a}=\sqrt{3}+1, \mathrm{~b}=\sqrt{3}-1, \mathrm{C}=60^{\circ}$
Using cosine Rule,
$\begin{aligned}
& \mathrm{c}^2=\mathrm{a}^2+\mathrm{b}^2-2 \mathrm{ab} \cos \mathrm{C} \\
& \mathrm{c}^2=(\sqrt{3}+1)^2+(\sqrt{3}-1)^2-2(\sqrt{3}+1)(\sqrt{3}-1) \frac{1}{2} \\
& \mathrm{c}^2=6
\end{aligned}$
$\begin{aligned}
\therefore \quad c^2 & =6 \\
\mathrm{c} & =\sqrt{6}
\end{aligned}$
Using sine Rule,
$\begin{aligned}
& \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \\
& \text { Consider } \\
& \frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \\
& \frac{\sqrt{3}-1}{\sin \mathrm{B}}=\frac{\sqrt{6}}{\frac{\sqrt{3}}{2}}
\end{aligned}$
$\begin{array}{ll}
& \frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \\
\therefore & \frac{\sqrt{3}-1}{\sin \mathrm{B}}=\frac{\sqrt{6}}{\frac{\sqrt{3}}{2}} \\
\therefore & \sin \mathrm{B}=\frac{\sqrt{3}-1}{2 \sqrt{2}} \\
\therefore & \angle \mathrm{B}=15^{\circ} \\
\therefore & \angle \mathrm{A}=105^{\circ}
\end{array}$
...[Remaining angle of a Triangle]
$\therefore \quad$ Difference $=90^{\circ}$

Topic: Mathematics, Properties of Triangles, MHT CET, MHT CET 2023 (09 May Shift 1)
Two sides of a triangle are given by the roots of the equation x 2 - 2 3 x + 2 = 0 and the angle between the sides is Ï 3 . The perimeter of the triangle is

Options:
A 2 3 units
B 6 units
C 2 3 + 6 units
D 2 3 + 6 units

The correct answer is:
2 3 + 6 units

Solution:
Let the two sides a and b are the roots of the equation x 2 - 2 3 x + 2 = 0 .
â´ a + b = 2 3 and ab = 2 . Also, â C = Ï 3
Using cosine rule,
cosC=a2+b2-c22abâ12=a2+b2-c22ab
â a 2 + b 2 - c 2 = ab â a + b 2 - c 2 = 3 ab
â 2 3 2 - c 2 = 3 Ã 2 â c = 6
â´ Perimeter =a+b+c=23+6 units

Topic: Mathematics, Properties of Triangles, JEE Main
Two similar circular loops carry equal currents in the same direction. On moving coils further apart, the electric current will

Options:
A increase in both
B decrease in both
C remain unaltered
D increases in one and decreases in the second

The correct answer is:
increase in both

Solution:
On moving the coils further apart initially the flux linked will reduced.
Then, according to Lenz's law current will increase in both the coils to increase the linked flux.

Topic: Physics, Electromagnetic Induction, KCET, KCET 2008
Two similar circular loops carry equal currents in the same direction. On moving the coils further apart, the electric current will

Options:
A increase in both
B decrease in both
C remain unaltered
D increases in one and decreases in the second

The correct answer is:
increase in both

Solution:
On moving the circular loops further apart, flux linked with loops will be reduced. Therefore, according to Lenz's law, current will increase in both circular loops to increase the linked flux.

Topic: Physics, Electromagnetic Induction, COMEDK, COMEDK 2018
Two similar plano-convex lenses are combined together in three different ways as shown in the adjoining figure. The ratio of the focal lengths in three cases will be

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

The correct answer is:
1:1:1

Solution:
In each case two plane-convex lens are placed close to each other, and $\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$.

Topic: Physics, Ray Optics, JEE Main
Two simple harmonic motions are given by $x_{1}=a \sin \omega t+a \cos \omega t$ and $x_{2}=a \sin \omega t+\frac{a}{\sqrt{3}} \cos \omega t$ The ratio of the amplitudes of first and second motion and the phase difference between them are respectively

Options:
A $\sqrt{\frac{3}{2}}$ and $\frac{\pi}{12}$
B $\frac{\sqrt{3}}{2}$ and $\frac{\pi}{12}$
C $\frac{2}{\sqrt{3}}$ and $\frac{\pi}{12}$
D $\sqrt{\frac{3}{2}}$ and $\frac{\pi}{6}$

The correct answer is:
$\sqrt{\frac{3}{2}}$ and $\frac{\pi}{12}$

Solution:
The given situation can be shown as

Fon second SHM
Ratio of amplitude $=\frac{a_{1}}{a_{2}}=\frac{\sqrt{3}}{\sqrt{2}}$
and phase difference, $\frac{\pi}{4}-\frac{\pi}{6}=\frac{\pi}{12}$

Topic: Physics, Oscillations, JEE Main
Two simple harmonic motions are represented by the equation $\mathrm{y}_1=0.1$ $\sin \left(100 \pi t+\frac{\pi}{3}\right)$ and $y_2=0.1 \cos \pi t$. The phase difference of the velocity of particle 1 w.r.t. the velocity of the particle 2 is

Options:
A
$-\pi / 6$

B
$\pi / 3$

C
$-\pi / 3$

D
$\pi / 6$

The correct answer is:
$-\pi / 6$

Solution:
Phase difference $(\phi)=99 \pi \mathrm{t}+\pi / 3-\pi / 2$ at $t=0, \phi=-\pi / 6$.

Topic: Physics, Oscillations, JEE Main, JEE Main 2005
Two simple harmonic motions are represented by the equations x1=5sin2πt+π4 and x2=52(sin2πt+cos2πt). The amplitude of the second motion is _____ times the amplitude in the first motion.

Options:

The correct answer is:
2

Solution:
A1=5

x2=52[sin2πt+cos2πt]

x2=10sin2πT+π4

A2=10

A2=2 A1

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (26 Aug Shift 2)
Two simple harmonic motions, as shown, are at right angles. They are combined to form Lissajous figures.
$x(t)=A \sin (a t+\delta)$
$y(t)=B \sin (b t)$
Identify the correct match below

Options:
A
Parameters: $\mathrm{A}=\mathrm{B}, \mathrm{a}=2 \mathrm{~b} ; \delta=\frac{\pi}{2}$; Curve: Circle

B
Parameters: $\mathrm{A}=\mathrm{B}, \mathrm{a}=\mathrm{b} ; \delta=\frac{\pi}{2} ;$ Curve: Line

C
Parameters: $\mathrm{A} \neq \mathrm{B}, \mathrm{a}=\mathrm{b} ; \delta=\frac{\pi}{2} ;$ Curve: Ellipse

D
Parameters: $\mathrm{A} \neq \mathrm{B}, \mathrm{a}=\mathrm{b} ; \delta=0$; Curve: Parabola

The correct answer is:
Parameters: $\mathrm{A} \neq \mathrm{B}, \mathrm{a}=\mathrm{b} ; \delta=\frac{\pi}{2} ;$ Curve: Ellipse

Solution:
From the two mutually perpendicular S.H.M.'s, the general equation of Lissajous figure,
$$
\frac{x^2}{A^2}+\frac{y^2}{B^2}-\frac{2 x y}{A B} \cos \delta=\sin ^2 \delta
$$
$$
\begin{aligned}
&x=A \sin (a t+\delta) \\
&y=B \sin (b t+r)
\end{aligned}
$$
Clearly $A \neq B$ hence ellipse.

Topic: Physics, Oscillations, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Two simple harmonic motions of angular frequency 100 and $1000 \mathrm{rad} \mathrm{s}^{-1}$ have the same displacement amplitude. The ratio of their maximum acceleration is

Options:
A $1: 10$
B $1: 10^{2}$
C $1: 10^{3}$
D $1: 10^{4}$

The correct answer is:
$1: 10^{2}$

Solution:
Acceleration of simple harmonic motion is
$\begin{aligned} & a_{\max }=-\omega^{2} A \\ \text { or } \frac{\left(a_{\max }\right)_{1}}{\left(a_{\max }\right)_{2}}=\frac{\omega_{1}^{2}}{\omega_{2}^{2}} \text { (as A remains same) } \\ \text { or } \frac{\left(a_{\max }\right)_{1}}{\left(a_{\max }\right)_{2}}=& \frac{(100)^{2}}{(1000)^{2}}=\left(\frac{1}{10}\right)^{2} \\=& 1: 10^{2} \end{aligned}$

Topic: Physics, Oscillations, MHT CET, MHT CET 2008
Two Simple Harmonic Motions of angular frequency 100 and $1000 \mathrm{rad} \mathrm{s}^{-1}$ have the same displacement amplitude. The ratio of their maximum accelerations is

Options:
A $1: 10^4$
B $1: 10$
C $1: 10^2$
D $1: 10^3$

The correct answer is:
$1: 10^2$

Solution:
$$
\frac{\left(a_{\max }\right)_1}{\left(a_{\max }\right)_2}=\frac{\omega_1^2 A}{\omega_2^2 A}=\left(\frac{100}{1000}\right)^2=\frac{1}{10^2}
$$

Topic: Physics, Oscillations, JEE Main
Two Simple Harmonic Motions of angular frequency 100 and $1000 \mathrm{rad} \mathrm{s}^{-1}$ have the same displacement amplitude. The ratio of their maximum accelerations is

Options:
A $1: 10^4$
B $1: 10$
C $1: 10^2$
D $1: 10^3$

The correct answer is:
$1: 10^2$

Solution:
$$
\frac{\left(a_{\max }\right)_1}{\left(a_{\max }\right)_2}=\frac{\omega_1^2 A}{\omega_2^2 A}=\left(\frac{100}{1000}\right)^2=\frac{1}{10^2}
$$

Topic: Physics, Oscillations, NEET, NEET 2008 (Mains)
Two simple harmonic motions of angular frequency 100 rad/s and 1000 rad/s have the same displacement amplitude. The ratio of their maximum acceleration is

Options:
A $1: 10$
B $1: 10^2$
C $1: 10^3$
D $1: 10^4$

The correct answer is:
$1: 10^2$

Solution:
Acceleration of simple harmonic motion is
$a_{\max }=-\omega^2 A$
or $\frac{\left(a_{\max }\right)_1}{\left(a_{\max }\right)_2}=\frac{\omega_1^2}{\omega_2^2}$ (as $A$ remains same)
or $\frac{\left(a_{\max }\right)_1}{\left(a_{\max }\right)_2}=\frac{(100)^2}{(1000)^2}=\left(\frac{1}{10}\right)^2=1: 10^2$

Topic: Physics, Oscillations, NEET, NEET 2008 (Screening)
Two simple harmonic waves having equal amplitudes of 8 cm and equal frequency of 10 Hz are moving along the same direction. The resultant amplitude is also 8 cm. The phase difference between the individual waves is _____ degree.

Options:

The correct answer is:
120

Solution:
If two waves have the same amplitude, frequency, and direction, but are out of phase by ϕ degrees, then the amplitude of the resultant waves is given by AR=2AcosΔϕ2

Here, A1=A2=AR=8 cm

So, cosΔϕ2=12

Or phase difference between the individual waves is Δϕ2=60°⇒Δϕ=120°.

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (29 Jan Shift 1)