datavorous/entrance-exam-dataset huggingface dataset:
A particle is projected at $60^{\circ}$ to the horizontal with a kinetic energy $\mathrm{K}$. The kinetic energy at the highest point is

Options:
A K
B zero
C $\frac{\mathrm{K}}{4}$
D $\frac{\mathrm{K}}{2}$

The correct answer is:
$\frac{\mathrm{K}}{4}$

Solution:
At highest point kinetic energy $=1 / 2 \mathrm{~m}\left(\mathrm{v} \cos 60^{\circ}\right)^2=1 / 4 \times 1 / 2 \mathrm{~m} \mathrm{v} v^2=\mathrm{K} / 4$

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle is projected at $60^{\circ}$ to the horizontal with a kinetic energy $\mathrm{K}$. The kinetic energy at the highest point is

Options:
A
$\mathrm{K}$

B
Zero

C
$\mathrm{K} / 2$

D
$\mathrm{K} / 4$

The correct answer is:
$\mathrm{K} / 4$

Solution:
No Solution Available

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2007
A particle is projected at an angle of 60Â° above the horizontal with a speed of 10 m s-1. After some time the direction of its velocity makes an angle of 30Â° above the horizontal. The speed of the particle at this instant is

Options:
A 53 m s-1
B 53 m s-1
C 5 m s-1
D 103 m s-1

The correct answer is:
103 m s-1

Solution:
Let v be the velocity of projectile at this instant.
Horizontal component of velocity remains unchanged.
Therefore,
vcos30°=10cos60°
v32=102
∴v=103 m s-1

Topic: Physics, Motion In Two Dimensions, NEET
A particle is projected at an angle of 60Â° above the horizontal with a speed of 10 m s-1. After some time the direction of its velocity makes an angle of 30Â° above the horizontal. The speed of the particle at this instant is

Options:
A 53 m s-1
B 53 m s-1
C 5 m s-1
D 103 m s-1

The correct answer is:
103 m s-1

Solution:
Let v be the velocity of point where it make angle 30 ο with
horizontal
thus , horizontal velocity remains uncharged, so
v cos 30 ο = 10 cos 60 ο
v 3 2 = 10 2
v=103 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle is projected at an angle of 60Â° to the horizontal with a speed of 10 m s-1. After some time the direction of its velocity makes an angle of 30Â° to the horizontal. The speed of the particle at this instant is

Options:
A 53 m s-1
B 53 m s-1
C 5 m s-1
D 103 m s-1

The correct answer is:
103 m s-1

Solution:
Let v be the velocity of the projectile at this instant.
The horizontal component of velocity remains unchanged.
Therefore,
vcos30°=10cos60°
v32=102
∴ v=103 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle is projected at time $t=o$ from a point $P$ on the ground with a speed $V_o$ at an angle of $45^{\circ}$ to the horizontal. What is the magnitude of the angular momentum of the particle about $\mathrm{P}$ at timet $\mathrm{t}=\mathrm{v}_{\mathrm{o}} / \mathrm{g}$

Options:
A $\frac{m v_o^2}{2 \sqrt{2} g}$
B $\frac{\mathrm{mv}_{\mathrm{o}}^3}{\sqrt{2} \mathrm{~g}}$
C $\frac{m v_o^2}{\sqrt{2} g}$
D $\frac{\mathrm{mv}_{\mathrm{o}}^3}{2 \sqrt{2} \mathrm{~g}}$

The correct answer is:
$\frac{\mathrm{mv}_{\mathrm{o}}^3}{2 \sqrt{2} \mathrm{~g}}$

Solution:
$\begin{aligned} & \mathrm{u}_{\mathrm{x}}=\mathrm{v}_{\mathrm{o}} \cos \theta=\frac{\mathrm{v}_{\mathrm{o}}}{\sqrt{2}} \\ & u_y=v_o \sin \theta-g t \quad=\frac{v_o}{\sqrt{2}}-\frac{v_o}{g} g \\ & =v_0\left[\frac{1-\sqrt{2}}{\sqrt{2}}\right] \\ & \bar{P}=\frac{m v_0}{\sqrt{2}} i+m v_0\left[\frac{1-\sqrt{2}}{\sqrt{2}}\right] \hat{\jmath} \\ & \overline{\mathrm{r}}=\mathrm{x} \hat{\mathbf{i}}+\mathrm{y} \hat{j} \\ & \mathrm{x}=\mathrm{uxt} \\ & =\frac{\mathrm{v}_0}{\sqrt{2}} \cdot \frac{\mathrm{v}_{\mathrm{O}}}{\mathrm{g}} \\ & =\frac{v_0^2 v_0}{\sqrt{2} g} \\ & =\frac{v_o^2}{\sqrt{2} g} \\ & \end{aligned}$
$\begin{aligned} & \mathrm{y}=\mathrm{v}_{\mathrm{o}} \sin \theta \mathrm{t}-\frac{1}{2} \mathrm{gt}^2 \\ & =\frac{v_0 v_0}{\sqrt{2} g}-\frac{1}{2} g \frac{v_o^2}{g^2} \\ & =\frac{\mathrm{v}_0^2}{\mathrm{~g}^2}\left[\frac{1}{\sqrt{2}}-\frac{1}{2}\right] \\ & =\frac{v_0^2}{g^2}\left[\frac{2-\sqrt{2}}{2 \sqrt{2}}\right] \\ & \mathrm{L}=\overline{\mathrm{r}} \times \overline{\mathrm{P}} \\ & =\left[\frac{v_0^2}{\sqrt{2} g}+\frac{v_0^2}{g^2}\left[\frac{2-\sqrt{2}}{2 \sqrt{2}}\right] \hat{\jmath}\right] \times\left[\frac{m v_0}{\sqrt{2}} \hat{i}+m v_0\left[\frac{1-\sqrt{2}}{\sqrt{2}}\right] \hat{\jmath}\right] \\ & =-\frac{m v_0^3}{2 \sqrt{2} g} \\ & |\overline{\mathrm{L}}|=\frac{\mathrm{mv}_{\mathrm{o}}^3}{2 \sqrt{2} \mathrm{~g}} \\ & \end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A particle is projected directly along a rough plane of inclination θ with velocity u. If after coming to the rest the particle returns to the starting point with velocity v, the coefficient of friction between the particle and the plane is

Options:
A u2v2tanθ
B u2âv2u2+v2tanÎ¸
C v2u2tanθ
D u2+v2u2–v2tanθ

The correct answer is:
u2–v2u2+v2tanθ

Solution:
u2 = 2as
For upward motion u2=2(μgcosθ+gsinθ)s
For downward motion v2=2(gsinθ-μgcosθ)s
u2v2=μcosθ+sinθsinθ–μcosθ
u2sinθ-u2μcosθ=μv2cosθ+v2sinθ
(u2-v2)sinθ=μ(v2+u2)cosθ
μ=u2–v2u2+v2tanθ

Topic: Physics, Laws of Motion, JEE Main
A particle is projected from a point $\mathrm{O}$ with velocity $u$ at an angle of $60^{\circ}$ with the horizontal. When it is moving in a direction at right angles to its direction at $O$, its velocity then is given by

Options:
A
$\frac{u}{3}$

B
$\frac{\mathrm{u}}{2}$

C
$\frac{2 u}{3}$

D
$\frac{\mathrm{u}}{\sqrt{3}}$

The correct answer is:
$\frac{\mathrm{u}}{\sqrt{3}}$

Solution:
$\mathrm{u} \cos 60^{\circ}=\mathrm{v} \cos 30^{\circ}$
$$
v=\frac{4}{\sqrt{3}}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2005
A particle is projected from point A, that is at a distance 4R from the centre of the earth, with speed v1 in a direction making 30Â° with the line joining the centre of the earth and point A, as shown. Find the speed v1 if particle grazes the surface of the earth as shown in the figure. Consider gravitational interaction only between these two. (use GMR=6.4Ã107 m2 s-2)

Options:
A 42 km s-1
B 32 km s-1
C 62 km s-1
D 52 km s-1

The correct answer is:
42 km s-1

Solution:
Conserving angular momentum w.r.t centre of earth,
(V1cos⁡60°) 4R=mV2R

⇒V2V1=2
Conserving mechanical energy of the system,
-GMm4R+12mV12=-GMmR+12mV22
12V22-12 V12=34GMR
V12=12GMR
V1=12 64×106=80002 m s-1=42 km s-1

Topic: Physics, Gravitation, JEE Main
A particle is projected from the earth's surface with a velocity of 50 m s-1 at an angle Î¸ with the horizontal. After 2 s it just clears a wall 5 m high. What is the value of 55 sinÎ¸? (g=10 m s-2)

Options:

The correct answer is:
13.75

Solution:
Using equation of motion in y-dir s=ut+12at2
Height after 2 s,
h=5=50 sinθt-12gt2
5 = 50 sin θ 2 - 1 2 · 10 · 2 2
25 100 = sin θ
sin θ = 1 4

55 sinθ=554=13.75

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle is projected from the ground and simultaneously a wedge starts moving towards the right, as shown in the figure. The maximum height of the wedge for which the particle will not hit the wedge is g=10 ms-2

Options:
A 289 m
B 283 m
C 73 m
D 74 m

The correct answer is:
289 m

Solution:
Initial velocities of the particle
u x =10cosâ¡53=6 m/s
u y =10sinâ¡53=8 m/s
6t=1+t+x.35 [where x is the length of the hypotenuse of the wedge]
5t=1+3x5 (in x - direction)
8t-5t2=4.x5 (in y - direction)
x=359m
Height =359Ã45=289m

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle is projected from the surface of the earth with a velocity of 5 ms-1 at an angle θ with the horizontal. Another particle is projected on the surface of some other planet with a velocity of 3 ms-1 at the same angle and it follows a trajectory which is identical to the trajectory of the projectile which is fired on earth. The value of the acceleration due to gravity on that planet is [given that the acceleration due to gravity on the surface of earth, g=9.8 ms-2]

Options:
A 16.3 ms-2
B 1.8 ms-2
C 3.5 ms-2
D 5.9 ms-2

The correct answer is:
3.5 ms-2

Solution:
y=x tan Î¸-gx22u2cos2Î¸
For equal trajectories for same angle of projection gu2= constant
â9.852=g32
g'=9.8Ã925=3.528 m/s2=3.5 m/s2

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle is projected on a frictionless inclined plane of inclination 37Â° at an angle of projection 45Â° from the inclined plane as shown in the figure. If after the first collision from the plane, the particle returns to its point of projection, then what is the value of the reciprocal of the coefficient of restitution between the particle and the plane?

Options:

The correct answer is:
3

Solution:
While going upward the time of flight is

T=2usinαgcosβ

So the range on the inclined plane is

R=ucosαT-12gsinβT2

Just after the collision of the projectile with the inclined plane, the components of its velocity down the inclined plane and perpendicular to the inclined plane are

v∥=gsinβT-ucosα, v⊥=eusinα

The time of flight while going down the inclined plane is

T'=2eusinαgcosβ=eT

The range of the projectile while going down is

R=gsinβT-ucosαeT+12gsinβe2T2

⇒gsinβT-ucosαeT+12gsinβe2T2=ucosαT-12gsinβT2

⇒ucosα1+e=gsinβT12+e+e22

Substituting the value as T=2usinαgcosβ we get

⇒cotα1+e=2tanβ12+e+e22

⇒3e2+2e-1=0

⇒e=13 or 1e=3

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle is projected towards the north with speed 20 ms at an angle 45o with horizontal. Ball gets horizontal acceleration of 7.5 ms2 towards east due to wind. Range of ball (in meter) minus 42 m will be

Options:

The correct answer is:
8

Solution:
Time of flight : T=2usinâ¡Î¸g=22 s
Range (along north) =u2sinâ¡2Î¸g=40 m
Range (along east) =12aT2=30 m
â´ Range =302+402=50 m

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle is projected up from a point at an angle $\theta$ with the horizontal direction. At any time $t$, if $p$ is the linear momentum, $y$ is the vertical displacement, $x$ is horizontal displacement, the graph among the following which does not represent the variation of kinetic energy KE of the particle is

Options:
A $\operatorname{graph}(\mathrm{A})$
B graph (B)
C $\operatorname{graph}(\mathrm{C})$
D graph (D)

The correct answer is:
$\operatorname{graph}(\mathrm{A})$

Solution:
Momentum, $p=m \cdot v \Rightarrow v=\left(\frac{p}{m}\right)$
Kinetic energy, KE
$$
\begin{array}{l}
=\frac{1}{2} m v^{2}=\frac{1}{2} m\left(\frac{p^{2}}{m^{2}}\right)=\frac{1}{2 m} p^{2} \\
\text { or, } \mathrm{KE} \propto p^{2} \quad\left(\because \frac{1}{2 m}=\text { constant }\right)
\end{array}
$$
Hence, the graph between $\mathrm{KE}$ and $p^{2}$ will be linear.
Now, kinetic energy $\mathrm{KE}=\frac{1}{2} m v^{2}$
The velocity component at point $P$,
$v_{y}=(u \sin \theta-g t)$ and $v_{x}=u \cos \theta$
Resultant velocity at point $P$
$$
\begin{aligned}
\vec{v} &=v_{y} \hat{j}+v_{x} \hat{i} \\
&=(u \sin \theta-g t) \hat{j}+u \cos \theta \hat{i} \\
&|\vec{v}|=\sqrt{(u \cos \theta)^{2}+(u \sin \theta-g t)^{2}} \\
=\sqrt{u^{2} \cos ^{2} \theta+u^{2} \sin ^{2} \theta+g^{2} t^{2}-2 u g t \sin \theta} \\
&=\sqrt{u^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)+g^{2} t^{2}-2 u g t \sin \theta}
\end{aligned}
$$
$$
\therefore \mathrm{KE}=\frac{1}{2} m\left(u^{2}+g^{2} t^{2}-2 u g t \sin \theta\right)ii/
$$
i. e., $\mathrm{KE} \propto t^{2}$
Hence, graph will be parabolic intercept on $y$-axis.
Hence, the graph between $\mathrm{KE}$ and $t$. Now, in case of height
$$
\begin{array}{l}
\mathrm{KE}=\frac{1}{2} m\left(v^{2}\right) \text { and } v^{2}=\left(u^{2}-2 g y\right) \\
\therefore \mathrm{KE}=\frac{1}{2} m\left(u^{2}-2 g y\right) \\
\mathrm{KE}=-m g y+\frac{1}{2} m u^{2}
\end{array}
$$
Intercept on $y$-axis $=\frac{1}{2} m u^{2}$
$$
\mathrm{KE}=\frac{1}{2} m\left(\frac{x}{t}\right)^{2}
$$
i.e., $\mathrm{KE} \propto x^{2}$. Thus graph between $\mathrm{KE}$ and $x$ will be parabolic.

Topic: Physics, Work Power Energy, JEE Main
A particle is projected up from a point at an angle $\theta$ with the horizontal direction. At any time $t$, if $p$ is the linear momentum, $y$ is the vertical displacement, $x$ is horizontal displacement, the graph among the following which does not represent the variation of kinetic energy KE of the particle is

Options:
A graph (A)
B graph (B)
C graph (C)
D graph (D)

The correct answer is:
graph (B)

Solution:
Momentum, $\quad p=m \cdot v$
$\begin{aligned}
& \Rightarrow \quad v=\left(\frac{p}{m}\right) \\
& \text { Kinetic energy, } \mathrm{KE}=\frac{1}{2} m v^2 \\
& =\frac{1}{2} m\left(\frac{p^2}{m^2}\right)=\frac{1}{2 m} p^2 \\
& \Rightarrow \quad \mathrm{KE} \propto p^2 \quad\left(\because \frac{1}{2 m}=\text { constant }\right) \\
&
\end{aligned}$
Hence, the graph between KE and $p^2$ will be linear as shown below
Now, kinetic energy KE $=\frac{1}{2} m v^2$
The velocity component at point $P$,
and
$\begin{aligned}
& v_y=(u \sin \theta-g t) \\
& v_x=u \cos \theta
\end{aligned}$
Resultant velocity at point $P$,
$\begin{aligned}
& \overrightarrow{\mathbf{v}}=v_y \hat{\mathbf{j}}+v_x \hat{\mathbf{i}} \\
& =(u \sin \theta-g t) \hat{\mathbf{j}}+u \cos \theta \hat{\mathbf{i}} \\
& |\overrightarrow{\mathbf{v}}|=\sqrt{(u \cos \theta)^2+(u \sin \theta-g t)^2} \\
& =\sqrt{u^2 \cos ^2 \theta+u^2 \sin ^2 \theta+g^2 t^2-2 u g t \sin \theta} \\
& \therefore=\sqrt{u^2\left(\cos ^2 \theta+\sin ^2 \theta\right)+g^2 t^2-2 u g t \sin \theta} \\
& \mathrm{KE}=\frac{1}{2} m\left(u^2+g^2 t^2-2 u g t \sin \theta\right) \\
& \Rightarrow \mathrm{KE} \propto t^2 \\
&
\end{aligned}$
Hence, graph will be parabolic with intercept on $y$-axis.
Hence, the graph between KE and $t$
Now, in case of height
$\mathrm{KE}=\frac{1}{2} m\left(v^2\right)$
and
$v^2=\left(u^2-2 g y\right)$
$\begin{aligned}
\therefore \quad \mathrm{KE} & =\frac{1}{2} m\left(u^2-2 g y\right) \\
\mathrm{KE} & =-m g \mathrm{y}+\frac{1}{2} m u^2
\end{aligned}$
Intercept on $y$-axis $=\frac{1}{2} m u^2$
Now,
$\begin{aligned}
\mathrm{KE} & =\frac{1}{2} m v^2 \\
\mathrm{KE} & =\frac{1}{2} m\left(\frac{x}{t}\right)^2
\end{aligned}$
Now,
$\begin{aligned}
\mathrm{KE} & =\frac{1}{2} m v^2 \\
\mathrm{KE} & =\frac{1}{2} m\left(\frac{x}{t}\right)^2
\end{aligned}$
$\mathrm{KE} \propto x^2$. Thus graph between KE and $x$ will be parabolic.

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle is projected vertically upwards with a speed of 16 m s-1, after some time, when it again passes through the point of projection, its speed is found to be 8 m s-1. It is known that the work done by air resistance is the same during upward and downward motion. Then the maximum height attained by the particle is (Take g=10 m s-2) :

Options:
A 8 m

B 4.8 m
C 17.6 m
D 12.8 m

The correct answer is:
8 m

Solution:
From work-energy theorem
For upward motion 1 2 m 1 6 2 = mgh + W (work by air resistance)
For downward motion 1 2 m 8 2 = mgh - W ⇒ 1 2 1 6 2 + 8 2 = 2 gh

or h=8 m

Topic: Physics, Work Power Energy, JEE Main
A particle is projected with velocity 2gh so that it just clears two walls of equal height h, which are at a distance of 2h from each other. What is the time interval of passing between the two walls?

Options:
A 2hg
B 2hg
C hg
D 2hg

The correct answer is:
2hg

Solution:
Let t be the time interval. Then,
2h=ux(t)
or ux=2ht ...(i)
Further, h=uyt-12gt2
or gt2-2uyt+2h=0
t1=2uy+4uy2-8gh2g
and t2=2uy-4uy2-8gh2g
t=t1-t2=4uy2-8ghg
or uy2=g2t24+2gh ...(ii)
Given, ux2+uy2=2gh2
4h2t2+g2t24+2gh=4gh
g24t4-2ght2+4h2=0
t2=2gh±4g2h2-4g2h2g2/2=4hg
or t=2hg

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle is projected with velocity v0 along x-axis. The deceleration on the particle is proportional to the square of the distance from the origin i.e. a=-x2. The distance at which particle stops is -

Options:
A 3v02
B v02313
C 2v02313
D 3v02213

The correct answer is:
3v02213

Solution:
Given initial velocity = v0
final velocity = 0
a = - x 2
now we know that a = dv dt and v = dx dt
thus a = dv dt = dv dt Â· dx dx = vdv dx
vdv dx = - x 2
vdv = - x 2 dx
integrating we get
â« v 0 0 vdv = â« 0 x - x 2 dx
v 2 2 v 0 0 = - x 3 3 0 x
- v 0 2 2 = - x 3 3 â x = 3 v 0 2 2 1 3

Topic: Physics, Mathematics in Physics, JEE Main
A particle is released from a height h, At a certain height; its kinetic enegy is two times its potential energy. Height and speed of the particle at that instant are

Options:
A h3,2gh3
B h3,2gh3
C 2h32gh3
D h3,2gh

The correct answer is:
h3,2gh3

Solution:
Total mechanical energy= mgh
As, KEPE=21
KE=23mgh
and PE=13mgh
Height from the ground at this instant,
h,=h3

and speed of the particle at this instant,
v=2gh-h′
=2g2h3
=2gh3

Topic: Physics, Work Power Energy, JEE Main
A particle is released from a height H . At certain height its kinetic energy is two times its potential energy. Height and speed of particle at that instant are

Options:
A H3, 2gH3
B H3 , 2gH3
C 2H3 , 2gH3
D H3 , 2gH

The correct answer is:
H3 , 2gH3

Solution:
∴K+U=mgH & K=2U
∴2U+U=mgH
⇒U=mgH3 ⇒mgh=mgH3 ⇒ h=H/3
∴12 mv2=2U=2mgH3 ⇒v=2gH3

Topic: Physics, Work Power Energy, JEE Main
A particle is released from a height S. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively

Options:
A S 4 ,  3gS 2
B S 4 ,  3gS 2
C S2,3gS2
D S4,3gS2

The correct answer is:
S4,3gS2

Solution:
We can realize the situation as shown. Let at point C distance x from highest point A , the particle's kinetic energy is three times its potential energy.
Velocity C,
v2=0+2gx
Or v2=2gx .....(i)
Potential energy at ,C=mg S-x â¦..(ii)
At Point C,
Kinetic energy =3Ã potential energy
ie, 1 2mÃ2gx=3Ãmg(S-x)
or x=3S-3x
or 4x=3S
or S=43x
or x=34S
Therefore, from Eq.(i)
v2=2gÃ34S
Or v2=32gS or v=32gS
Height of the particle from the ground
= S-x=S-34S=s4

Topic: Physics, Work Power Energy, JEE Main
A particle is released on a vertical smooth semicircular track from point X so that, OX makes angle θ from the vertical (see figure). The normal reaction of the track on the particle vanishes at the point Y where OY makes an angle ϕ with the horizontal. Then

Options:
A sinϕ=23cosθ
B sinϕ=34cosθ
C sinϕ=12cosθ
D sinϕ=cosθ

The correct answer is:
sinϕ=23cosθ

Solution:
mgRcosθ-Rsinϕ=12mV2 .......(1)

On losing contact N=0⇒mgsinϕ=mV2R ......(2)

⇒mgRcosθ-sinϕ=12mgRsinϕ

⇒2cosθ=3sinϕ

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2014 (19 Apr Online)
A particle is rotating about a vertical axis in the horizontal plane such that the angular velocity of the particle about the axis is constant and is equal to 1 rad/s. Distance of the particle from axis is given by R=R0-βt where t stands for time. The speed of the particle as a function of time is

Options:
A β2+1
B R0-βt
C β2+R0-βt2
D β

The correct answer is:
β2+R0-βt2

Solution:
Vradial=dRdt=-β
Vtangential=ωR=ωR0-βt
V=Vr2+Vt2=β2+R0-βt2

Topic: Physics, Laws of Motion, JEE Main
A particle is subjected simultaneously to two SHMs, one along the x-axis and the other along the y-axis. The two vibrations are in phase and have unequal amplitudes. The particle will execute

Options:
A Straight line motion
B Circular motion
C Elliptic motion
D Parabolic motion

The correct answer is:
Straight line motion

Solution:
Let the equations of the two mutually perpendicular SHM's of the same frequency and same phase be
x=a1 sin ⁡ωt and y=a2 sin ⁡ωt

Then,

yx=a2sinωta1sinωt=a2a1

y=a2a1x, which is the equation of a straight line

Topic: Physics, Oscillations, JEE Main
A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motion, the phase difference $(\delta)$ between the two motions is

Options:
A $\delta=\frac{\pi}{3}$
B $\delta=\frac{2 \pi}{3}$
C $\delta=\pi$
D $\delta=\frac{\pi}{2}$

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

Solution:
$A_{\text {res }}^2=A_1^2+A_2^2+2 A_1 A_2 \cos \delta \quad A^2=A^2+A^2+2 A^2 \cos \delta, \cos \delta=-\frac{1}{2} \delta=120^{\circ}$

Topic: Physics, Oscillations, JEE Main
A particle is suspended by a light vertical inelastic string of length l from a fixed support. At its equilibrium position, it is projected horizontally with a speed 6gl. What is the ratio of the tension in the string in its horizontal position to when the particle is vertically above the point of support?

Options:

The correct answer is:
4

Solution:
For a given u=6gl
At the horizontal position,
T1=mv12l=4 mg .......... (i)
At the highest point
T2+mg=mv22l=2 mgT2=mg ......... (ii)
From the above equations, we get
T1T2=4

Topic: Physics, Laws of Motion, JEE Main
A particle is thrown over a triangle from one end of a horizontal base and after grazing the vertex falls on the other end of the base. If Î± and Î² be the base angles and Î¸ be the angle of projection. Find the value of tan Î¸ .

Options:
A tan α + tan β
B sin α + sin β
C tan α + sin β
D cos α + cos α

The correct answer is:
tan α + tan β

Solution:
The situation is shown in diagram

tan α + tan β = y x + y R - x
tan α + tan β = yR x R - x ...(i)
Equation of trajectory is y = x tan θ 1 - x R
or, tan θ = yR x R - x ...(ii)
From Eqs. (i) and (ii), we have
tan θ = tan α + tan β

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle is thrown vertically upwards. If its velocity at half of the maximum height is $10 \mathrm{~m} / \mathrm{s}$, then maximum height attained by it is (Take $\left.g=10 \mathrm{~m} / \mathrm{s}^2\right)$

Options:
A 8 m
B 10 m
C 12 m
D 16 m

The correct answer is:
10 m

Solution:
Let particle thrown with velocity $u$ and its maximum height is $H$ then $H=\frac{u^2}{2 g}$ When particle is at a height $H / 2$, then its speed is $10 \mathrm{~m} / \mathrm{s}$
From equation $v^2=u^2-2 g h$
$\begin{aligned} & (10)^2=u^2-2 g\left(\frac{H}{2}\right)=u^2-2 g \frac{u^2}{4 g} \Rightarrow u^2=200 \\ & \text { Maximum height } \Rightarrow H=\frac{u^2}{2 g}=\frac{200}{2 \times 10}=10 \mathrm{~m} \\ & \end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main
A particle is travelling 4 times as fast as an electron. Assuming the ratio of de-Broglie wavelength of a particle to that of electron is 2: 1, the mass of the particle is :-

Options:
A 116 times the mass of e-
B 8 times the mass of e-
C 16 times the mass of e-
D 18 times the mass of e-

The correct answer is:
18 times the mass of e-

Solution:
λ=hp
λpλe=pepp=mevempVp
2=mempve4ve
∴ mp=me8

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (18 Mar Shift 1)
A particle is travelling along a straight line $O X$. The distance $x$ (in metre) of the particle from 0 at a time t is given by $x=37+27 t-t^{3}$, where t is time in seconds. The distance of the particle from $O$ when it comes to rest is

Options:
A $81 \mathrm{m}$
B $91 \mathrm{m}$
C $101 \mathrm{m}$
D $111 \mathrm{m}$

The correct answer is:
$91 \mathrm{m}$

Solution:
Given, $x=37+27 t-t^{3}$
$$
v=\frac{d x}{d t}=27-3 t^{2}
$$
According to problem,
$$
v=0 \Rightarrow 27-3 t^{2}=0
$$
Here, $t=3 \mathrm{s}$
$$
x=37+27 \times 3-(3)^{2}=91 \mathrm{m}
$$

Topic: Physics, Motion In One Dimension, JEE Main
A particle is undergoing SHM. Displacement vs time curve is as shown in the figure
The correct statement is

Options:
A Phase of the oscillator is same at \( \mathrm{t}=0 \mathrm{~s} \) and \( \mathrm{t}=2 \mathrm{~s} \).
B Phase of the oscillator is same at \( \mathrm{t}=2 \mathrm{~s} \) and \( \mathrm{t}=5 \mathrm{~s} \).
C Phase of the oscillator is same at \( \mathrm{t}=1 \mathrm{~s} \) and \( \mathrm{t}=7 \mathrm{~s} \).
D Phase of the oscillator is same at \( \mathrm{t}=1 \mathrm{~s} \) and \( \mathrm{t}=5 \mathrm{~s} \).

The correct answer is:
Phase of the oscillator is same at \( \mathrm{t}=1 \mathrm{~s} \) and \( \mathrm{t}=5 \mathrm{~s} \).

Solution:
The phase of a particle executing SHM is defined as the state of a particle as regard to its position and direction of motion at any instant of time. In the given curve, phase is same when t=1 s and t=5 s. Also phase is same when t=2 s and t=6 s .

Topic: Physics, Oscillations, JEE Main
A particle is vibrating simple harmonically with an amplitude a. The displacement of the particle when its energy is half
kinetic and half potential is

Options:
A \( \frac{a}{2} \)
B \( \frac{a}{\sqrt{2}} \)
C \( \frac{a}{4} \)
D Zero

The correct answer is:
\( \frac{a}{\sqrt{2}} \)

Solution:
Kinetic energy = Potential energy
⇒12mω2a2-y2=12m ω2y2
Or a2-y2=y2
Or 2y2=a2
Or y2=a22
Or y=a2

Topic: Physics, Oscillations, JEE Main