datavorous/entrance-exam-dataset huggingface dataset:
A particle of mass M originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation F=F01-t-TT2 where F0 and T are constants. The force acts only for the time interval 2 T. The velocity v of the particle after time 2 T is:

Options:
A 2 F0 TM
B F0 T2M
C 4 F0 T3M
D F0 T3M

The correct answer is:
4 F0 T3M

Solution:
t=0, u=0

a=FoM-FoMT2(t-T)2=dvdt

∫0vdv=∫t=02 TFoM-FoMT2(t-T)2dt

V=FoMto2 T-FoMT2t33-t2 T+T2t02 T

V=4 FoT3M

Topic: Physics, Motion In One Dimension, JEE Main
A particle of mass M originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation F=F01-t-TT2 where F0 and T are constants. The force acts only for the time interval 2 T. The velocity v of the particle after time 2 T is:

Options:
A 2 F0 TM
B F0 T2M
C 4 F0 T3M
D F0 T3M

The correct answer is:
4 F0 T3M

Solution:
t=0, u=0

a=FoM-FoMT2(t-T)2=dvdt

∫0vdv=∫t=02 TFoM-FoMT2(t-T)2dt

V=FoMto2 T-FoMT2t33-t2 T+T2t02 T

V=4 FoT3M

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (27 Jul Shift 2)
A particle of mass ' $m$ ' performs uniform circular motion of radius ' $r$ ' with linear speed ' $v$ ' under the application of force ' $F$ '. If $\mathrm{m}, \mathrm{v}$ and $\mathrm{r}$ are all increased by $50 \%$, the necessary change in force required to maintain the particle in uniform circular motion is

Options:
A $125 \%$
B $150 \%$
C $100 \%$
D $225 \%$

The correct answer is:
$125 \%$

Solution:
Given that, Mass $=\mathrm{m}$, Velocity $=\mathrm{v}$, Radius $=\mathrm{r}$
We know that, The force is
Now, the increment is $50 \%$ in mass, velocity and radius
Now, new force is
$F^{\prime}=\frac{\left(m+\frac{m}{2}\right)\left(v+\frac{v}{2}\right)^2}{\left(r+\frac{r}{2}\right)}=\frac{\frac{3}{2} m \times \frac{9}{4} v^2}{\frac{3}{2} r}=F \times \frac{9}{4}$
Now, the \% change in force
$\frac{F^{\prime}-F}{F} \times 100=\left(\frac{9}{4}-1\right) \times 100=125 \%$

Topic: Physics, Motion In Two Dimensions, MHT CET, MHT CET 2022 (07 Aug Shift 2)
A particle of mass m projected with a velocity u making an angle of 30° with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is :

Options:
A 316mu3g
B 32mu2g
C mu32g
D zero

The correct answer is:
316mu3g

Solution:
Angular momentum at maximum height will be, L=mucosθh, where h=maximum height

Therefore, L=mucosθ×u2sin2θ2g

=mu32g×32×122=3mu316g

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2024 (30 Jan Shift 1)
A particle of mass $M$ starting from rest undergoes uniform acceleration. If the speed acquired in time $T$ is $v$, the power delivered to the particle is

Options:
A $\frac{\mathrm{Mv}^2}{\mathrm{~T}}$
B $\frac{1}{2} \frac{\mathrm{Mv}^2}{\mathrm{~T}^2}$
C $\frac{\mathrm{Mv}^2}{\mathrm{~T}^2}$
D $\frac{1}{2} \frac{\mathrm{Mv}^2}{\mathrm{~T}}$

The correct answer is:
$\frac{1}{2} \frac{\mathrm{Mv}^2}{\mathrm{~T}}$

Solution:
The kinetic energy of particle $=\frac{1}{2} \mathrm{Mv}^2$
$$
\begin{aligned}
\text { Power } & =\frac{\text { Energy }}{\text { Time }} \\
\mathrm{P} & =\frac{1}{2} \frac{\mathrm{Mv}^2}{\mathrm{~T}}
\end{aligned}
$$

Topic: Physics, Work Power Energy, NEET, NEET 2010 (Mains)
A particle of mass m with an initial velocity ui^ collides perfectly elastically with a mass 3 m at rest. It moves with a velocity vj^ after collision, then, v is given by

Options:
A v=23u
B v=u3
C v=u2
D v=16u

The correct answer is:
v=u2

Solution:
From momentum conservation

mui^+0=mvj^+3mv→'

v→'=u3i^-v3j^

From kinetic energy conservation 12mu2=12mv2+12(3m)u32+v32

solving v=u2

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2020 (02 Sep Shift 1)
A particle of mass \( \mathrm{m} \) is attached to three identical springs \( \mathrm{A}, \mathrm{B} \) and \( \mathrm{C} \) each of force constant \( \mathrm{k} \) as shown in figure. If the particle of mass \( \mathrm{m} \) is pushed slightly against the spring \( \mathrm{A} \) and released, then the time period of oscillation is -

Options:
A \( 2 \pi \sqrt{\frac{2 \mathrm{~m}}{\mathrm{k}}} \)
B \( 2 \pi \sqrt{\frac{\mathrm{m}}{2 k}} \)
C \( 2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \)
D \( 2 \pi \sqrt{\frac{\mathrm{m}}{3 \mathrm{k}}} \)

The correct answer is:
\( 2 \pi \sqrt{\frac{\mathrm{m}}{2 k}} \)

Solution:
When particle is displaced by a distance x in the downward direction, lower spring is compressed by x distance in downward direction and both upper string is displaced by xcos45° at an angle of 45° from vertical.Due to these compressions and expansions, net force start working on the particle which tries to bring the particle back to its original position.

Writing net spring force on the particle-
Fnet=F+F1cos45°+F2cos45°
⇒Fnet=kx+kxcos45°cos45°+kxcos45°cos45°
⇒Fnet=2kx

From newton's second law-
ma=-2kx
⇒a=-2kmx

Above equation represents SHM.Comparing it with equation of SHM
a=-ω2x
⇒ω=2km
Since time period T=2πω
∴T=2πm2k

Topic: Physics, Laws of Motion, JEE Main
A particle of mass \( \mathrm{m} \) oscillates with simple harmonic motion between points \( \mathrm{x}_{1} \) and \( \mathrm{x}_{2} \), the equilibrium position being \( \mathrm{O} \). Its potential energy is plotted. It will be as given below in the graph

Options:
A
B
C
D

The correct answer is:

Solution:
Potential energy of particle performing SHM varies parabolically in such a way that at mean position it becomes zero and maximum at extreme position.

Topic: Physics, Oscillations, JEE Main
A particle of specific charge qm=Ï C kg-1 is projected from the origin towards positive x-axis with a velocity of 10 m s-1 in a uniform magnetic field Bâ=-2 k^ T. The velocity vâ of the particle after time t=112 s will be (in m s-1)

Options:
A 5 i ^ + 3 j ^
B 5 3 i ^ + j ^
C 5 3 i ^ - j ^
D 5 i ^ + j ^

The correct answer is:
5 3 i ^ + j ^

Solution:
Time period, T = 2 Ï m qB = 2 Ï Ï Ã 2 = 1 s
Thus, the particle will be at point P after t=112 s
v â = 1 0 cos 3 0 i ^ + sin 3 0 j ^
v â = 1 0 3 2 i ^ + 1 2 j ^ = 5 3 i ^ + j ^ m s - 1

Topic: Physics, Magnetic Effects of Current, JEE Main
A particle of unit mass is moving along the x-axis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in column I (a and U0 are constants). Match the potential energies in column I to the corresponding statement(s) in column II.

Column I	Column II

A.	U1x=U021-xa22	P.	The force acting on the particle is zero at x = a

B.	U2x=U02xa2	Q.	The force acting on the particle is zero at x = 0 .

C.	U3x=U02xa2exp⁡-xa2	R.	The force acting on the particle is zero at x= -a

D.	U4x=U02xa-13xa3	S.	The particle experiences an attractive force towards x = 0 in the region x<a

T.	The particle with total energy U04 can oscillate about the point x= -a

Options:
A a-s,t;b-s;c-s,t;d-t;
B a-s;b-r,s;c-r,s,t;d-s,t;
C a-r,s;b-q;c-q,r;d-t;
D a-p,q,r,t;b-q,s;c-p,q,r,s;d-p,r,t;

The correct answer is:
a-p,q,r,t;b-q,s;c-p,q,r,s;d-p,r,t;

Solution:
U1x=U021-x2a22
F=-U0221-x2a2-2xa2
=2U0a4a2-x2x
F=2U0a4xa-xa+x
F=0 , at x=0, a, -a
x=-a, U=0,x=0, U=U02â Oscillate when total ME is less then U02
AâP,Q,R,T
B- U0x=U02xa2
F=-U0xa2
F=0, x=0
BâQ,S
C- U3x=U02xa2expâ¡-x2a2
F=-U022xa2expâ¡-x2a2+x2a2expâ¡-x2a2-2xa2
=-U022xa4expâ¡-x2a2a2-x2
=U0a4ex2a2 xx+ax-a
x=0, x=-a, x=+a, F=0
x=0, U=0 even function hence minima
CâP,Q,R,S
D- U4x=U02xa-x33a3
F=-U0a1a-x2a3=U02a3x2-a2
x=+a, x=-a, F=0
x=-a, U=-U03, x=+a, U=+U03
Hence oscillate about x=-a if T.M.E<U03
DâP,R,T

Topic: Physics, Work Power Energy, JEE Main
A particle 'P' is formed due to a completely inelastic collision of particles 'x' and 'y' having de-Broglie wavelengths 'λx' and 'λy' respectively. If x and y were moving in opposite directions, then the de-Broglie wavelength of 'P' is:

Options:
A λx-λy
B λxλyλx+λy
C λxλyλx-λy
D λx+λy

The correct answer is:
λxλyλx-λy

Solution:
hλ=hλx-hλy
1λ=1λx-1λy
λ=λxλyλx-λy

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (09 Apr Shift 2)
A particle $P$ is moving in a circle of radius ' $a$ ' with a uniform speed $v . C$ is the centre of the circle and $A B$ is a diameter. When passing through $B$ the angular velocity of $P$ about $A$ and $C$ are in the ratio

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

The correct answer is:
1:2

Solution:
Angular velocity of particle $P$ about point $A$,
$\omega_A=\frac{v}{r_{A B}}=\frac{v}{2 r}$
Angular velocity of particle $P$ about point $C$,
$\begin{aligned} & \omega_C=\frac{v}{r_{B C}^r}=\frac{v}{r} \\ & \text { Ratio } \frac{\omega_A}{\omega_C}=\frac{v / 2 r}{v / r}=\frac{1}{2} .\end{aligned}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle P is projected from a point on the surface of a smooth inclined plane (see figure). Simultaneously another particle Q is released on the smooth inclined plane from the same position. P and Q collide on the inclined plane after t=4 seconds. Find the speed of projection in m/s .

Options:

The correct answer is:
10

Solution:
It can be observed from figure that P and Q shall collide if the initial component of velocity of P and Q on inclined plane i.e along incline should be equal. That is particle is projected perpendicular to incline.
â´ Time of flight T = 2 u â¥ g cos Î¸ = 2 u g cos Î¸
â´ u = g T cos Î¸ 2 = 1 0 m / s

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle performing S.H.M. starts from the equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is Ï m s-1. Amplitude of oscillation is cosâ¡45o=1â2

Options:
A 22 m
B 42 m
C 62 m
D 82 m

The correct answer is:
82 m

Solution:
Displacement of the particle =x=Asinâ¡Ït
The velocity of the particle =v=dxdt=AÏcosâ¡Ït
v=Ï m s-1, T=16 s, Ï=2ÏT=2Ï16=Ï8 rad s-1
â´ Ï=AÃÏ8Ãcosâ¡Ï8Ã2
â´ 1=A8cosâ¡Ï4=A8.1â2
â´ A=82 m

Topic: Physics, Oscillations, JEE Main
A particle performing S.H.M. when displacement is ' $x$ ', the potential energy and restoring force acting on it are denoted by ' $\mathrm{E}$ ' and ' $\mathrm{F}$ ' respectively. The relation between $\mathrm{x}, \mathrm{E}$ and $\mathrm{F}$ is

Options:
A $\frac{2 E}{F}-x^2=0$
B $\frac{2 E}{F}+x^2=0$
C $\frac{2 E}{F}+x=0$
D $\frac{2 E}{F}-x=0$

The correct answer is:
$\frac{2 E}{F}+x=0$

Solution:

Topic: Physics, Oscillations, JEE Main
A particle performing uniform circular motion has angular frequency is doubled \& its kinetic energy halved, then the new angular momentum is

Options:
A
$\frac{\mathrm{L}}{4}$

B
$2 \mathrm{~L}$

C
$4 \mathrm{~L}$

D
$\frac{\mathrm{L}}{2}$

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

Solution:
Angular momentum $\propto \frac{1}{\text { Angular frequency }} \propto$ Kinetic energy $\Rightarrow \vec{L}=\frac{\mathrm{K} . \mathrm{E} .}{\mathrm{w}}$
$\frac{\mathrm{L}_1}{\mathrm{~L}_2}=\left(\frac{\mathrm{K} . \mathrm{E}_1}{\mathrm{w}_1}\right) \times \frac{\mathrm{w}_2}{\mathrm{KE}_2}=4 \Rightarrow \mathrm{L}_2=\frac{\mathrm{L}}{4}$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2003
A particle performs linear SHM. At a particular instant, the velocity of the particle is u and acceleration is Î± (both having the same direction). At another instant velocity is v and acceleration is Î² 0<Î±<Î² (both in opposite direction to each other).The distance between the two positions is

Options:
A u2-v2α+β
B u2+v2α+β
C u2-v2α-β
D u2+v2α-β

The correct answer is:
u2-v2α+β

Solution:
Let distance be x1 when velocity is u and acceleration Î±.
Let distance be x2 when velocity is v and acceleration Î².
If Ï is the angular frequency then
Î±= Ï2 x1
And Î²=Ï2 x2
â´ Î±+Î²=Ï2(x1+x2) ....... (i)
Also u2=Ï2A2-Ï2x12
And v2=Ï2 A2-Ï2 x22
v2-u2=Ï2(x12-x22)
v2-u2=Ï2(x1-x2)(x1+x2) ........... (ii)
By Equation. (i) we get v2-u2=(x1-x2)(Î±+Î²)
â´ x1-x2=v2-u2Î±+Î²
Or x2-x1=u2-v2Î±+Î²

Topic: Physics, Oscillations, JEE Main
A particle performs SHM with an amplitude of 5 m. If at x=4 m, the magnitude of velocity and acceleration are equal, then the time period of SHM (in seconds) is

Options:
A 2π3
B 4π3
C 8π3
D 6π3

The correct answer is:
8π3

Solution:
v=ω52-42=3ω
a=ω2×4
a=v
4ω2=3ω
ω=34=2πT
T=8π3 s

Topic: Physics, Oscillations, JEE Main
A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is 1a s. The value of a to the nearest integer is

Options:

The correct answer is:
6

Solution:
t=212=16
∴ Correct answer =6.00

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (18 Mar Shift 1)
A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance 2A3 from equilibrium position. The new amplitude of the motion is:

Options:
A A3
B 7A3
C A341
D 3A

The correct answer is:
7A3

Solution:
For Original SHM,
using v=ωA2-X2
v=ωA2-4A29
=ω5A3
New SHM will be,
3v = ωAN2-XN2
3ω5A3=ωAN2-4A29
5A2=AN2-4A29
AN2=49A29
AN=7A3

Topic: Physics, Oscillations, JEE Main, JEE Main 2016 (03 Apr)
A particle shows distance-time curve as given in this figure. The maximum instantaneous velocity of the particle is around the point

Options:
A A
B B
C C
D D

The correct answer is:
C

Solution:
Maximum slope is at $\mathrm{C}$.

Topic: Physics, Motion In One Dimension, NEET, NEET 2008 (Mains)
A particle slides down on a smooth incline of inclination 30o , fixed in an elevator going up with an acceleration 2 m/s2. The box of incline has width 4m. The time taken by the particle to reach the bottom will be

Options:
A 893s
B 983s
C 4332s
D 3432s

The correct answer is:
4332s

Solution:
In the frame of elevator,
a = acceleration of the particle with respect to the elevator
â´ msinâ¡30 g+2=ma
a=g+2sinâ¡30o
=10+2 .12
=6 m/s2
â´ The distance travelled by the particle from the top to the bottom,
d=4cosâ¡30o=432=833m
Using s=ut+12 at2 ,
833=0Ãt+126t2
â t2=1633Ã6
â´ t=4332 s

Topic: Physics, Laws of Motion, JEE Main
A particle starting from the origin $(0,0)$ moves in a straight line in the $(x, y)$ plane. Its coordinates at a later time are $(\sqrt{3}, 3)$. The path of the particle makes with the $x$-axis an angle of:

Options:
A $45^{\circ}$
B $60^{\circ}$
C $0^{\circ}$
D $30^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
Particle starts moving from \((0, 0)\) and goes till \((\sqrt{3}, 3)\).
So total displacement along x axis is \(\sqrt{3}\) unites and along y axis is 3 units.
We know that:
\(\begin{aligned}
& \Rightarrow \tan (\theta)=\frac{y}{x} \\
& \Rightarrow \tan (\theta)=\frac{3}{\sqrt{3}} \\
& \Rightarrow \tan (\theta)=\sqrt{3} \\
& \Rightarrow \theta=\frac{\pi}{3}=60^{\circ}
\end{aligned}\)

Topic: Physics, Mathematics in Physics, JEE Main
A particle starts at the origin and moves along the $x$-axis in such a way that its velocity at the point $(x, 0)$ is given by the formula $\frac{d x}{d t}=\cos ^2 \pi x$.
Then the particle never reaches the point on

Options:
A $x=\frac{1}{4}$
B $x=\frac{3}{4}$
C $x=\frac{1}{2}$
D $x=1$

The correct answer is:
$x=\frac{1}{2}$

Solution:
Given $\frac{d x}{d t}=\cos ^2 \pi x$. Differentiate w.r.t. t,
$\begin{aligned} & \frac{d^2 x}{d t^2}=-2 \pi \sin 2 \pi x=-v e \\ & 8 \frac{d^2 x}{d t^2}=0 \Rightarrow-2 \pi \sin 2 \pi x=0 \Rightarrow \sin 2 \pi x=\sin \pi \\ & \Rightarrow 2 \pi x=\pi \Rightarrow x=1 / 2\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
A particle starts executing simple harmonic motion (SHM) of amplitude a and total energy E. At any instant, its kinetic energy is 3E4, then its displacement y is given by:

Options:
A y=a
B y=a2
C y=a32
D y=a2

The correct answer is:
y=a2

Solution:
E=12Ka2
3E4=12 Ka2-y2v
34×12Ka2=12 Ka2-y2
y2=a2-3a24
y=a2

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (27 Jul Shift 1)
A particle starts from mean position and performs S.H.M. with period 4 second. At what time its kinetic energy is $50 \%$ of total energy?

Options:
A $0.1 \mathrm{~s}$
B $0.2 \mathrm{~s}$
C $0.4 \mathrm{~s}$
D $0.5 \mathrm{~s}$

The correct answer is:
$0.5 \mathrm{~s}$

Solution:
The total energy is given by T.E $=\frac{1}{2} \mathrm{kA}^2$ Kinetic energy is given by K.E $=\frac{1}{2} k\left(A^2-x^2\right)$
$$
\begin{array}{ll}
& \text { K.E }=\frac{1}{2} P . E \\
\therefore \quad & \frac{1}{2} k\left(A^2-x^2\right)=\frac{1}{2}\left(\frac{1}{2} \mathrm{kA}^2\right) \\
\therefore \quad & \left(\mathrm{A}^2-\mathrm{x}^2\right)=\frac{1}{2} \mathrm{~A}^2 \\
\therefore \quad & \left(\mathrm{A}^2-\mathrm{x}^2\right)=\frac{1}{2} \mathrm{~A}^2 \\
\therefore \quad & \mathrm{x}=\frac{\mathrm{A}}{\sqrt{2}}
\end{array}
$$
The equation of displacement in SHM is
$$
\begin{array}{ll}
& \mathrm{x}=\mathrm{A} \sin \frac{2 \pi \mathrm{t}}{\mathrm{T}} \\
\therefore \quad & \frac{\mathrm{A}}{\sqrt{2}}=\mathrm{A} \sin \frac{2 \pi \mathrm{t}}{4} \\
\therefore \quad \frac{1}{\sqrt{2}}=\sin \frac{\pi \mathrm{t}}{2} & \quad \ldots .(\because \mathrm{T}=4 \mathrm{sec}) \\
\therefore \quad \frac{\pi \mathrm{t}}{2} & =\sin ^{-1} \frac{1}{\sqrt{2}} \\
& \frac{\pi \mathrm{t}}{2}=\frac{\pi}{4} \\
\therefore \quad & \mathrm{t}=0.5 \mathrm{~s}
\end{array}
$$

Topic: Physics, Oscillations, JEE Main
A particle starts from origin O from rest and moves with a uniform acceleration along the positive x-axis . Identify all figures that correctly represent the motion qualitatively.
( a= acceleration, v= velocity, x= dispalcement, t= time)
A
B
(C)
(D)

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

The correct answer is:
A,B,(D)

Solution:
A. Acceleration = constant; a v→t graph will be constant line parallel to time axis. Option A is correct.
B. v=u+at ; velocity vs time graph will be straight line passing through origin. Option B is correct.
u=0; initially at rest
v= at
C. X=12at2 (particle was initially at origin) v→t graph will be parabola passing through origin. Option D is correct.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2019 (08 Apr Shift 2)
A particle starts from rest, accelerates at $2 \mathrm{~m} / \mathrm{s}^2$ for $10 \mathrm{~s}$ and then goes for constant speed for $30 \mathrm{~s}$ and then decelerates at $4 \mathrm{~m} / \mathrm{s}^2$ till it stops. What is the distance travelled by it

Options:
A 750 m
B 800 m
C 700 m
D 850 m

The correct answer is:
750 m

Solution:
Velocity acquired by body in $10 \mathrm{sec}$
$v=0+2 \times 10=20 \mathrm{~m} / \mathrm{s}$
and distance travelled by it in $10 \mathrm{sec}$
$S_1=\frac{1}{2} \times 2 \times(10)^2=100 \mathrm{~m}$
then it moves with constant velocity $(20 \mathrm{~m} / \mathrm{s}) \quad$ for $30 \mathrm{sec}$
$S_2=20 \times 30=600 \mathrm{~m}$
After that due to retardation $\left(4 \mathrm{~m} / \mathrm{s}^2\right)$ it stops
$S_3=\frac{v^2}{2 a}=\frac{(20)^2}{2 \times 4}=50 \mathrm{~m}$
\(\Rightarrow\) Total distance travelled by it is \(100+600+50=750\)

Topic: Physics, Motion In One Dimension, JEE Main
A particle starts from rest and traverses a distance 2x with uniform acceleration, then moves uniformly over a further distance 4x and finally comes to rest after moving a further distance 6x under uniform retardation. Assuming entire motion to be rectilinear motion, the ratio of average speed over the journey to the maximum speed on its way is

Options:
A 45
B 35
C 25
D 15

The correct answer is:
35

Solution:
Let t1,t2 and t3 be the time taken by the particles to cover the distance 2x, 4x and 6x respectively. Let v be the velocity of the particle at B ie, maximum velocity. The particle moves with uniform acceleration from A to B.
For motion from A to B.
Average velocity =0+v2=v2
Time taken, t1=2xv2
Particle moves with uniform retardation from C TO D, Vavg=0+v2=v2
Time taken ,t3=6x(v2)
Total time =t1+t2+t3
t=2xv2+4xv+6xv2=20xv
Average Velocity over = 6x+2x+4x20xv=35v
Required Ratio =AverageSpeedv=35

Topic: Physics, Motion In One Dimension, JEE Main
A particle starts from rest and traverses a distance 2x with uniform acceleration, then moves uniformly over a further distance 4x and finally comes to rest after moving a further distance 6x under uniform retardation. Assuming entire motion to be rectilinear motion, the ratio of average speed over the journey to the maximum speed on its way is

Options:
A 45
B 35
C 25
D 15

The correct answer is:
35

Solution:
Let t1,t2 and t3 be the time taken by the particles to cover the distance 2x, 4x and 6x respectively. Let v be the velocity of the particle at B i.e, maximum velocity. The particle moves with uniform acceleration from A to B.
For motion from A to B.
Average velocity =0+v2=v2
Time taken, t1=2xv2
Particle moves with uniform retardation from C TO D, Vavg=0+v2=v2
Time taken ,t3=6x(v2)
Total time = t1+t2+t3
t=2xv2+4xv+6xv2=20xv
Average Velocity over= 6x+2x+4x20xv=35v
Required Ratio = AverageSpeedv=35

Topic: Physics, Motion In One Dimension, JEE Main
A particle starts from the origin at t=0 with an initial velocity of 3.0i^m/s and moves in the x-y plane with a constant acceleration 6.0i^+4.0j^m/s2. The x- coordinate of the particle at the instant when its y- coordinate is 32m is D meters. The value of D is:

Options:
A 32
B 50
C 60
D 40

The correct answer is:
60

Solution:
Sy=uyt+12ayt2
32=0+12×4t2⇒t=4 s
Sx=uxt+12axt2
=3×4+12×6×16
=60 m

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2020 (09 Jan Shift 2)
A particle starts from the origin at time t=0 and moves along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t=5s?

Options:
A 10 m
B 9 m
C 6m
D 3m

The correct answer is:
9 m

Solution:
Area enclosed by V-T graph=displacement (As particle moving in same direction)
12×2×2+2×2+1×3= Displacement
∴ So, displacement =9m

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2019 (10 Jan Shift 2)
A particle starts from the origin of coordinates at time t=0 and moves in the x-y plane with a constant acceleration Î± in the y-diretion. Its equation of motion is y= Î²x2. Its velocity component in the x-direction is

Options:
A Variable
B 2 α β
C α 2 β
D α 2 β

The correct answer is:
α 2 β

Solution:
Equation of motion is y = β x 2
d 2 y dt 2 = α
d 2 x dt 2 = 0
from y = x
dy dt = 2 x dx dt β
d 2 y dt 2 = 2 dx dt 2 + 2x d 2 x dt 2 β
a y = 2 dx dt 2 + 0 β
a y = 2 v x 2 β
v x = a y 2 β = α 2 β

Topic: Physics, Motion In One Dimension, JEE Main
A particle starts its motion from rest under the action of a constant force. If the distance covered in first $10 \mathrm{~s}$ is $\mathrm{s}_1$ and that covered in the first $20 \mathrm{~s}$ is $\mathrm{s}_2$, then

Options:
A $\mathrm{s}_2=2 \mathrm{~s}_1$
B $\mathrm{s}_2=3 \mathrm{~s}_1$
C $s_2=4 s_1$
D $s_2=s_1$

The correct answer is:
$s_2=4 s_1$

Solution:
Key Idea If the particle is moving in a straight line under the action of a constant force then distance covered $\mathrm{s}=\mathrm{ut}+\frac{1}{2}$ at $^2$.
Since the body start from rest $\mathrm{u}=0$
$\therefore \quad \mathrm{s}=\frac{1}{2} \mathrm{at}^2$
Now, $\quad \mathrm{s}_1=\frac{1}{2} \mathrm{a}(10)^2$
and
$$
\mathrm{s}_2=\frac{1}{2} \mathrm{a}(20)^2
$$
Dividing Eq. (i) and Eq. (ii), we get
$$
\begin{aligned}
\frac{\mathrm{s}_1}{\mathrm{~s}_2} & =\frac{(10)^2}{(20)^2} \\
\Rightarrow \quad s_2 & =4 \mathrm{~s}_1
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, JEE Main
A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is $\mathrm{S}_1$ and that covered in the first 20 seconds is $\mathrm{S}_2$ then :

Options:
A $\mathrm{S}_2=\mathrm{S}_1$
B $\mathrm{S}_2=2 \mathrm{~S}_1$
C $\mathrm{S}_2=3 \mathrm{~S}_1$
D $\mathrm{S}_2=4 \mathrm{~S}_1$

The correct answer is:
$\mathrm{S}_2=4 \mathrm{~S}_1$

Solution:
$\mathrm{s}=\frac{1}{2} \mathrm{at}^2$
$$
\begin{aligned}
& \frac{\mathrm{s}_2}{\mathrm{~s}_1}=\left(\frac{20}{10}\right)^2=4 \\
& \mathrm{~s}_2=4 \mathrm{~s}_1
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, NEET, NEET 2009 (Mains)