datavorous/entrance-exam-dataset huggingface dataset:
A particle is whirled in a vertical circle of radius $1.0 \mathrm{~m}$ using a string with one end fixed. If the ratio of maximum to minimum tensions in the string is $5 / 3$, then the minimum velocity (in $\mathrm{m} / \mathrm{s}$ ) of the particle during circular motion is

Options:
A $10 \mathrm{~m} / \mathrm{s}$
B $12 \mathrm{~m} / \mathrm{s}$
C $14 \mathrm{~m} / \mathrm{s}$
D $20 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$10 \mathrm{~m} / \mathrm{s}$

Solution:
Step1: Draw the rough diagram for given situation
Step2: Find ratio of maximum and minimum tension.
Given, $T_{\max }: T_{\min }=5:3$
At the lower most point,
$T_{\max }-\mathrm{mg}=\mathrm{mv2maxR} \ldots$..(i)
At the upper most point,
$\mathrm{mg}+T_{\min }=\mathrm{mv} 2 \min R \ldots$...(ii)
Using equation (i),(ii)
$\frac{T_{\max }}{T_{\min }}=\frac{\left(\frac{m v_{\max }^2+m g}{R}\right)}{\left(\frac{m v_{\min }^2}{R}-m g\right)} \ldots(iii)$
step3: Find minimum velocity of the particle.
Apply conservation of energy at lower most point and upper most point,
$\begin{aligned}& \frac{1}{2} m v_{\text {max }}^2=m g(2 R)+\frac{1}{2} m v_{\text {min }}^2 \\& \Rightarrow v_{\text {max }}^2=2 g(2 R)+v_{\text {min }}^2 \ldots \text { (iv) }\end{aligned}$
Using equation (iii) and (iv)
$\begin{aligned}& \frac{5}{3}=\frac{\frac{v_{\text {min }}^2+2 \cdot g \cdot 2 R}{R}+g}{\frac{v_{\text {min }}^2}{R}-g} \\& \Rightarrow v_{\min }=10 \mathrm{~m} / \mathrm{s}\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main
A particle just clears a wall of height $b$ at distance $a$ and strikes the ground at a distance $c$ from the point of projection. The angle of projection is

Options:
A
$\tan ^{-1} \frac{b}{a c}$

B
$45^{\circ}$

C
$\tan ^{-1} \frac{b c}{a(c-a)}$

D
$\tan ^{-1} \frac{b c}{a}$

The correct answer is:
$\tan ^{-1} \frac{b c}{a(c-a)}$

Solution:
$a=(u \cos \alpha) t$ and $b=(u \sin \alpha) t-\frac{1}{2} g t^2$
$b=a \tan \alpha-\frac{1}{2} g \frac{a^2}{u^2 \cos ^2 \alpha}$
also, $c=\frac{u^2 \sin 2 \alpha}{g}$
$b=a \tan \alpha-\frac{a^2 g}{2}\left(\frac{\sin 2 \alpha}{c g}\right) \sec ^2 \alpha$
$b=a \tan \alpha-\frac{a^2}{2 c} 2 \tan \alpha$
$\Rightarrow\left(a-\frac{a^2}{c}\right) \tan \alpha=b$
$\tan \alpha=\frac{b c}{a(c-a)}$.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2007
A particle leaves the origin with an initial velocity v â = 3.00 i ^ ms - 1 and a constant acceleration a â = - 1.00 i ^ - 0.5 j ^ ms - 2 . When the particle reaches it maximum x-coordinate, what is the magnitude of its velocity (in m/s) in y-direction? If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
3

Solution:
Velocity along x direction is
v x = v 0 2 + at
v x = 3 + - 1 t
so when particle reach maximum x coordinate v x = 0 â t = 3 sec
similarly v y = v 0 y + a y t
v y = 0 + - 1 2 t
vyatt=3=-32=-1.5 m s-1

Topic: Physics, Motion In One Dimension, JEE Main
A particle leaves the origin with an initial velocity v â = 3.00 i ^ ms - 1 and a constant acceleration a â = - 1.00 i ^ - 0.5 j ^ ms - 2 . When the particle reaches it maximum x-coordinate, what is the magnitude of its velocity (in m/s) in y-direction?

Options:

The correct answer is:
1.5

Solution:
Velocity along x direction is
v x = v 0 2 + at
v x = 3 + - 1 t
so when particle reach maximum x coordinate v x = 0 â t = 3 sec
similarly v y = v 0 y + a y t
v y = 0 + - 1 2 t
vyatt=3=-32=-1.5 m s-1

Topic: Physics, Motion In One Dimension, JEE Main
A particle located at $x=0$ at time $t=0$, starts moving along the positive $x$-direction with a velocity ' $v$ ' that varies as $v=\alpha \sqrt{x}$. The displacement of the particle varies with time as

Options:
A
$t^3$

B
$t^2$

C
$t$

D
$t^{1 / 2}$

The correct answer is:
$t^2$

Solution:
$\frac{\mathrm{d} x}{\mathrm{dt}}=\alpha \int \frac{\mathrm{dx}}{\sqrt{\mathrm{x}}}=\int \alpha \mathrm{dt} \Rightarrow \mathrm{x} \alpha \mathrm{t}^2$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2006
A particle m=1kg slides down a frictionless track AOC starting from rest at a point A (height 2m ). After reaching C, the particle continues to move freely in air as a projectile. When it reaching its highest point P (height 1m ), the kinetic energy of the particle (in J ) is: (Figure drawn is schematic and not to scale; take g=10ms-2 ) ______________.

Options:

The correct answer is:
10

Solution:
KEP-KEA=PEP-PEA=mghP-mghA
=1×10×2-1×10×1=10J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2020 (07 Jan Shift 1)
A particle (mass = m and charge = q) is moving in a region in which there exists a uniform electric field Ei^ and a uniform magnetic field Bk^. At t=0, the particle is at 0, a and is moving with vi^. After some time, the particle is located at 2a, 0 and has a velocity â2vj^, then which of the following is true?

Options:
A E=32mv2qa
B at the initial moment, the rate of work done by electric field is 34mv3a
C E=38mv2qa
D at the initial moment, the rate of work done by electric field is 316mv3a

The correct answer is:
at the initial moment, the rate of work done by electric field is 34mv3a

Solution:
By work energy theorem

Wmag+Wele=12m(2v)2-12m(v)2

0+qE02a=32mv2
E0=34mv2qa

Rate of work done at initial instant = power of electric force

=qE0V
=34mv3a

Topic: Physics, Magnetic Effects of Current, JEE Main
A particle mass $m$, charge $Q$ and kinetic energy $T$ enters a transverse uniform magnetic field of induction $\overrightarrow{\mathbf{B}}$. After 3 s the kinetic energy of the particle will be

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

The correct answer is:
$T$

Solution:
After passing through a magnetic field, the magnitude of its mass and velocity of the particle remain same, so its energy does not change, ie., kinetic energy will remain $T$.

Topic: Physics, Magnetic Properties of Matter, JEE Main
A particle mass $m$, charge $Q$ and kinetic energy $T$ enters a transverse uniform magnetic field of induction $\overrightarrow{\mathbf{B}}$. After 3 s the kinetic energy of the particle will be

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

The correct answer is:
$T$

Solution:
After passing through a magnetic field, the magnitude of its mass and velocity of the particle remain same, so its energy does not change, ie., kinetic energy will remain $T$.

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2008 (Screening)
A particle moves according to the law x=rcosâ¡Ït2. The distance covered by it in the time interval between t=0 and t=3 s is nr. What is the value of n?

Options:

The correct answer is:
3

Solution:
Given, x=rcos⁡πt2
⇒v=dxdt
=-π2rsin⁡πt2
Graphically it can be represented as shown in figure.
Distance covered by the particle
= Magnitude of area of x-t curve between
0 to 2 s and 2 to 3 s
=∫02vdt+∫23vdt
=-∫02π2rsin⁡π2t dt+∫23-π2r sin⁡π2t dt
=-π2r∫02sin⁡πt2dt+-π2r∫23sin⁡π2t dt
=πr2cos⁡πt2π202+πr2cos⁡π2tπ223
=rcos⁡π-cos⁡0+rcos⁡3π2-cos⁡π
=r-1-1+r0--1=2r+r=3r

Topic: Physics, Motion In One Dimension, JEE Main
A particle moves along a circle $\left(\frac{20}{\pi}\right) \mathrm{m}$ with constant tangential acceleration. If the velocity of the particle is $80 \mathrm{~m} / \mathrm{s}$ at the end of the second revolution after after motion has begun, the tangential acceleration is:

Options:
A $40 \mathrm{~m} / \mathrm{s}^2$
B $640 \pi \mathrm{m} / \mathrm{s}^2$
C $160 \pi \mathrm{m} / \mathrm{s}^2$
D $40 \pi \mathrm{m} / \mathrm{s}^2$

The correct answer is:
$40 \mathrm{~m} / \mathrm{s}^2$

Solution:
From question
$$
\begin{aligned}
& r=\frac{20}{\pi} m, v=80 \mathrm{~m} / \mathrm{s} \\
& \therefore \theta=2 \text { revolutions }=2 \times 2 \pi=4 \pi \mathrm{rad} \\
& \therefore \omega^2=\omega_0{ }^2+2 \alpha \theta \\
& \therefore \omega^2=2 \alpha \theta \ldots(\mathrm{i}) \text { since } \omega_0=0 \\
& \text { Since } \alpha=\frac{a}{r} \text { and } \omega=\frac{\mathrm{V}}{r}
\end{aligned}
$$
Putting these values in eq. (i)
$$
\begin{aligned}
& \frac{V^2}{r^2}= \frac{2 a \theta}{r} \\
& \Rightarrow a= \frac{V^2}{2 r \theta}=\frac{80 \times 80}{2 \times\left(\frac{20}{\pi}\right) \times 4 \pi} \\
&=40 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle moves along a circular path with decreasing speed. Hence

Options:
A its resultant acceleration is towards the centre.
B it moves in a spiral path with decreasing radius.
C the direction of angular momentum remains constant.
D its angular momentum remains constant

The correct answer is:
the direction of angular momentum remains constant.

Solution:
The direction of angular moment remains constant as the angular momentum is a vector quantity, and its direction is perpendicular to the plane of motion. As the speed decreases, the linear momentum decreases, but the angular momentum remains constant due to the conservation of angular momentum.

Topic: Physics, Motion In Two Dimensions, MHT CET, MHT CET 2023 (11 May Shift 1)
A particle moves along a circular path with decreasing speed. Hence

Options:
A its resultant acceleration is towards the centre.
B its angular momentum remains constant.
C the direction of angular momentum remains constant.
D it moves in a spiral path with decreasing radius.

The correct answer is:
the direction of angular momentum remains constant.

Solution:
A particle moves along a circular path with speed, then the angular momentum is given by
$\overrightarrow{\mathrm{L}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{r}} \times \mathrm{m} \overrightarrow{\mathrm{v}}$
When the speed v decrease, the magnitude of of angular momentum decreases. But the direction of angular momentum remains constant.

Topic: Physics, Motion In Two Dimensions, MHT CET, MHT CET 2020 (14 Oct Shift 2)
A particle moves along a straight line such that its displacement at any time t is given by
$S=t^3-6 t^2+3 t+4$ meters
The velocity when the acceleration is zero is

Options:
A $3 \mathrm{~ms}^{-1}$
B $-12 \mathrm{~ms}^{-1}$
C $42 \mathrm{~ms}^{-1}$
D $-9 \mathrm{~ms}^{-1}$

The correct answer is:
$-9 \mathrm{~ms}^{-1}$

Solution:
$v=\frac{d s}{d t}=3 t^2-12 t+3$ and $a=\frac{d v}{d t}=6 t-12$
For $a=0$, we have $t=2$ and at $t=2, v=-9 \mathrm{~ms}^{-1}$

Topic: Physics, Motion In One Dimension, JEE Main
A particle moves along a straight line with a variable acceleration given in the acceleration-displacement a-S curve as shown in the figure. Determine the velocity (in m s-1) of the particle after it has travelled a distance of 30 m. The initial velocity is 10 m s-1.

Options:

The correct answer is:
20

Solution:
Area under curve is =12Ã10Ã30
=150 â¦â¦..(i)
Area under the curve is also equal to =v2-u22 â¦â¦.(ii)
From (i) and (ii)
12v2-u2=150
v2-u2=300
v2=u2+300
v2=102+300
v=400=20 m s-1

Topic: Physics, Motion In One Dimension, JEE Main
A particle moves along $x$ -axis and its displacement at any time is given by $x(t)=2 t^{3}-3 t^{2}+4 t$ in SI units. The velocity of the particle when its acceleration is zero, is

Options:
A $2.5 \mathrm{ms}^{-1}$
B $3.5 \mathrm{ms}^{-1}$
C $4.54 \mathrm{ms}^{-1}$
D $8.5 \mathrm{ms}^{-1}$

The correct answer is:
$2.5 \mathrm{ms}^{-1}$

Solution:
Given, $x(t)=2 t^{3}-3 t^{2}+4 t$
So, velocity, $v=\frac{d x}{d t}=\left(6 t^{2}-6 t+4\right)$
and acceleration $a=\frac{d v}{d t}=(12 t-6)$
when acceleration is zero, $i . e . .(12 t-6)=0$
or
$$
t=\frac{6}{12}=\frac{1}{2} \mathrm{s}
$$
$\therefore$ Velocity of the particle at zero acceleration is
$$
v=6\left(\frac{1}{2}\right)^{2}-6\left(\frac{1}{2}\right) 4=2.5 \mathrm{m} / \mathrm{s}
$$

Topic: Physics, Motion In One Dimension, JEE Main
A particle moves along $x$-axis as
$x=4(t-2)+a(t-2)^2$
Which of the following is true?

Options:
A The initial velocity of particle is 4
B The acceleration of particle is $2 a$
C The particle is at origin at $t=0$
D None of these

The correct answer is:
The acceleration of particle is $2 a$

Solution:
$\begin{aligned} & x=4(t-2)+a(t-2)^2 \\ & \text { At } t=0, x=-8+4 a=4 a-8\end{aligned}$
$\begin{aligned} & v=\frac{d x}{d t}=4+2 a(t-2) \\ & \text { At } t=0, v=4-4 a=4(1-a)\end{aligned}$
At $t=0, v=4-4 a=4(1-a)$
But acceleration, $a=\frac{d^2 x}{d t^2}=2 a$

Topic: Physics, Motion In One Dimension, JEE Main
A particle moves along \( X \)-axis from \( x=0 \) to \( x=5 \mathrm{~m} \) under the influence of force given by \( F=\left(7-2 x+3 x^{2}\right) \mathrm{N} \). The
work done in the process is

Options:
A \( 70 \mathrm{~J} \)
B \( 270 \mathrm{~J} \)
C \( 35 \mathrm{~J} \)
D \( 135 \mathrm{~J} \)

The correct answer is:
\( 135 \mathrm{~J} \)

Solution:
The given force varies with distance
Therefore, work done,
W=â«0x=5F.dx
=â«057-2x+3x2 dx
=7x-2x22+3x3305
=7Ã5-52+53-0
=135 J

Topic: Physics, Work Power Energy, JEE Main
A particle moves for 20 seconds with velocity $3 \mathrm{~m} / \mathrm{s}$ and then velocity $4 \mathrm{~m} / \mathrm{s}$ for another 20 seconds and finally moves with velocity $5 \mathrm{~m} / \mathrm{s}$ for next 20 seconds. What is the average velocity of the particle

Options:
A 3 m/s
B 4 m/s
C 5 m/s
D Zero

The correct answer is:
4 m/s

Solution:
Time average velocity $=\frac{v_1+v_2+v_3}{3}=\frac{3+4+5}{3}=4 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Motion In One Dimension, JEE Main
A particle moves from the point $(2.0 \hat{i}+4.0 \hat{j}) \mathrm{m},$ at $\mathrm{t}=0$,

with an initial velocity $(5.0 \hat{i}+4.0 \hat{j}) \mathrm{ms}^{-1} .$ It is acted upon by a constant force which produces a constant acceleration

$(4.0 \hat{i}+4.0 \hat{j}) \mathrm{ms}^{-2} .$ What is the distance of the particle

from the origin at time $2 \mathrm{~s} ?$

Options:
A $15 \mathrm{~m}$
B $20 \sqrt{2} \mathrm{~m}$
C $5 \mathrm{~m}$
D $10 \sqrt{2} \mathrm{~m}$

The correct answer is:
$20 \sqrt{2} \mathrm{~m}$

Solution:
As $\overrightarrow{\mathrm{S}}=\overrightarrow{\mathrm{ut}}+\frac{1}{2} \overrightarrow{\mathrm{a}} \mathrm{t}^{2}$

$\overrightarrow{\mathrm{S}}=(5 \hat{i}+4 \hat{j}) 2+\frac{1}{2}(4 \hat{i}+4 \hat{j}) 4$

$=10 \hat{i}+8 \hat{j}+8 \hat{i}+8 \hat{j}$

$\overrightarrow{\mathrm{r}}_{\mathrm{f}}-\overrightarrow{\mathrm{r}}_{\mathrm{i}}=18 \hat{\mathrm{i}}+16 \hat{\mathrm{j}}$

$\left[\mathrm{as} \overrightarrow{\mathrm{s}}=\right.$ change in position $\left.=\overrightarrow{\mathrm{r}}_{\mathrm{f}}-\overrightarrow{\mathrm{r}}_{\mathrm{i}}\right]$

$\overrightarrow{\mathrm{r}}_{\mathrm{r}}=20 \hat{\mathrm{i}}+20 \hat{\mathrm{j}}$

$\left|\overrightarrow{\mathrm{r}}_{\mathrm{r}}\right|=20 \sqrt{2}$

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A particle moves in a circle of radius $5 \mathrm{~cm}$ with constant speed and time period $0.2 \pi \mathrm{s}$. The acceleration of the particle is

Options:
A $25 \mathrm{~m} / \mathrm{s}^2$
B $36 \mathrm{~m} / \mathrm{s}^2$
C $5 \mathrm{~m} / \mathrm{s}^2$
D $15 \mathrm{~m} / \mathrm{s}^2$

The correct answer is:
$5 \mathrm{~m} / \mathrm{s}^2$

Solution:
Given, $r=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$ and $T=0.2 \pi \mathrm{s}$
We know that acceleration
$$
\begin{aligned}
a & =r \omega^2 \\
& =\frac{4 \pi^2}{T^2} r \\
& =\frac{4 \times \pi^2 \times 5 \times 10^{-2}}{(0.2 \pi)^2}=5 \mathrm{~ms}^{-2}
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle moves in a circle with a uniform speed. When it goes from a point B to a diametrically opposite point A, the momentum of the particle changes by PâA-PâB=2j^ kg m s-1 and the centripetal force acting on it changes by F â A - F â B = 8 i ^ N . Where i^ and j^ are unit vectors along x and y axes respectively. The angular velocity of the particle is

Options:
A 8 rad s-1
B 4 rad s-1
C 2 rad s-1
D 16 rad s-1

The correct answer is:
4 rad s-1

Solution:
PâAââPâB=m(VâAâVâB)=mv(j^â(âj^))
2mvj^â=2 kg m sâ1âj^â.....(i)
FâAâFâB=mv2R(âi^)âmv2R(+i^)
=2mv2R-i^ ...(ii)

Equating magnitudes (Also using the force relation given in question)

2mv2R=8 N

From the above two equations, we get
vR=Ï=4 rad s-1

Topic: Physics, Laws of Motion, JEE Main
A particle moves in a circular path with decreasing speed. Choose the correct statement.

Options:
A Angular momentum remains constant
B Acceleration $\vec{a}$ is towards the centre
C Particle moves in a spiral path with decreasing radius
D The direction of angular momentum remains constant

The correct answer is:
The direction of angular momentum remains constant

Solution:
Angular momentum is an axial vector, so its direction is along the axis, perpendicular to the plane of motion which is not changing because of change of speed. Therefore, the direction of angular momentum remains the same and its magnitude may vary.

Topic: Physics, Rotational Motion, AIIMS, AIIMS 2014
A particle moves in a plane along an elliptic path given by $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 .$ At point $(0, b)$, the $x$-component of velocity is $\mathrm{u}$. The y-component of acceleration at this point is-

Options:
A $-\mathrm{bu}^{2} / \mathrm{a}^{2}$
B $-u^{2} / b$
C $-\mathrm{au}^{2} / \mathrm{b}^{2}$
D $-\mathrm{u}^{2} / \mathrm{a}$

The correct answer is:
$-\mathrm{bu}^{2} / \mathrm{a}^{2}$

Solution:
$\begin{array}{l}
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \\
u_{x}=u \text { at }(0, b) \\
u_{y}=0 \\
\frac{2 x}{a^{2}} \frac{d x}{d t}+\frac{2 y}{b^{2}} \frac{d y}{d t}=0
\end{array}$
Again diff. w.r.t. to time
$\frac{2 x}{a^{2}} \frac{d^{2} x}{d t^{2}}+\frac{2}{a^{2}}\left(\frac{d x}{d t}\right)^{2}+\begin{array}{ll}
2 y & d^{2} y \\
b^{2} & {d t^{2}}+\frac{2}{b^{2}}\left(\begin{array}{l}
d y \\
d t
\end{array}\right)=0
\end{array}$
acceleration at $(0, b)$ is
$a_{y}=\frac{-b}{a^{2}} u^{2}$

Topic: Physics, Motion In One Dimension, JEE Main
A particle moves in a straight line so that its displacement $x$ at any time $t$ is given by $x^2=1+t^2$. Its acceleration at any time $\mathrm{t}$ is $x^{-\mathrm{n}}$ where $\mathrm{n}=$ ___________

Options:

The correct answer is:
3

Solution:
$\begin{aligned} & \mathrm{x}^2=1+\mathrm{t}^2 \\ & 2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{t} \\ & \mathrm{xv}=\mathrm{t} \\ & \mathrm{x} \frac{\mathrm{dv}}{\mathrm{dt}}+\mathrm{v} \frac{\mathrm{dx}}{\mathrm{dt}}=1 \\ & \mathrm{x} \cdot \mathrm{a}+\mathrm{v}^2=1 \\ & \mathrm{a}=\frac{1-\mathrm{v}^2}{\mathrm{x}}=\frac{1-\mathrm{t}^2 / \mathrm{x}^2}{\mathrm{x}} \\ & \mathrm{a}=\frac{1}{\mathrm{x}^3}=\mathrm{x}^{-3}\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (06 Apr Shift 2)
A particle moves in a straight line with a constant acceleration. It changes its velocity from $10 \mathrm{~ms}^{-1}$ to $20 \mathrm{~ms}^{-1}$ while passing through a distance $135 \mathrm{~m}$ in t second. The value of $t$ is

Options:
A 9
B 10
C 1.8
D 12

The correct answer is:
9

Solution:
$a=\frac{V_f^2-V_i^2}{2 S}, t=\frac{V_f-V_i}{a}$
or $S=\frac{1}{2}(u+v) t$

Topic: Physics, Motion In One Dimension, JEE Main
A particle moves in a straight line with a constant acceleration. It changes its velocity from $10 \mathrm{~ms}^{-1}$ to $20 \mathrm{~ms}^{-1}$ while passing through a distance $135 \mathrm{~m}$ in t second. The value of $t$ is

Options:
A 9
B 10
C 1.8
D 12

The correct answer is:
9

Solution:
$a=\frac{V_f^2-V_i^2}{2 S}, t=\frac{V_f-V_i}{a}$
or $S=\frac{1}{2}(u+v) t$

Topic: Physics, Motion In One Dimension, NEET, NEET 2008 (Mains)
A particle moves in a straight line with a velocity given by $\frac{d x}{d t}=x+1$ ( $x$ is the distance described). The time taken by a particle to traverse a distance of 99 meter is

Options:
A $\log _{10} e$
B $2 \log _e 10$
C $2 \log _{10} e$
D $\frac{1}{2} \log _{10} e$

The correct answer is:
$2 \log _e 10$

Solution:
$\frac{d x}{d t}=x+1 \Rightarrow \log (x+1)=t+c$
Putting $t=0, x=0$, we get ${ }^{\log 1-c} \Rightarrow c=0$
$\therefore t=\log (x+1) . \quad$ For $x=99, t=\log _e 100=2 \log _e 10$

Topic: Mathematics, Differential Equations, JEE Main
A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $\mathrm{x}$ is proportional to

Options:
A
$x^3$

B
$e^x$

C
$x$

D
$\log _e x$

The correct answer is:
$x^3$

Solution:
$\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}=\frac{\mathrm{mk}}{2} \mathrm{x}^2$
$\mathrm{~K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}} \propto \mathrm{x}^2$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2004
A particle moves in space along the path z=ax3+by2 in such a way that dâ¡ x dâ¡ t = c = dâ¡ y dâ¡ t where a, b and c are constants. The acceleration of the particle is

Options:
A 6 a c 2 x + 2 b c 2 k ^
B 2 a x 2 + 6 b y 2 k &Hat;
C 4 b c 2 x + 3 a c 2 k ^
D b c 2 x + 2 b y k &Hat;

The correct answer is:
6 a c 2 x + 2 b c 2 k ^

Solution:
Given that
dâ¡ x dâ¡ t = dâ¡ y dâ¡ t = c
∴ dâ¡ 2 x dâ¡ t 2 = dâ¡ 2 y dâ¡ t 2 = 0
Further z = ax3 + by2
∴ dâ¡ z dâ¡ t = 3 a x 2 dâ¡ x dâ¡ t + 2 b y dâ¡ y dâ¡ t
= 3 a c x 2 + 2 b c y dâ¡ x dâ¡ t = c = dâ¡ y dâ¡ t
∴ dâ¡ 2 z dâ¡ t 2 = 6 a c x dâ¡ x dâ¡ t + 2 b c dâ¡ y dâ¡ t
= 6ac2x + 2bc2
Now acceleration of particle is
a → = dâ¡ 2 x dâ¡ t 2 i ^ + dâ¡ 2 y dâ¡ t 2 j ^ + dâ¡ 2 z dâ¡ t 2 k ^
= 6 a c 2 x + 2 b c 2 k ^

Topic: Physics, Motion In Two Dimensions, JEE Main
A particle moves in the x-y plane under the action of a force F such that the coordinates of its linear momentum Pâ at any time t is px=2 cos t,py= 2sinâ¡t
The angel Î¸ between Fâ and Pâ at a given time t will be

Options:
A 90°
B 0°
C 180°
D 30°

The correct answer is:
90°

Solution:
P=px2+py2
=2cos⁡t2+2sin⁡t2=2
If m be the mass of the body, then kinetic energy
=p22m=222m=2m
Since kinetic energy does not change with time, both work done and power are zero
Now Power = Fvcos⁡θ=0
As F≠0,v≠0
∴ cos⁡θ=0
Or θ=90°
As direction of p→ is same that v→(∵p→=mv→) hence angle between F→ and p→ is equal to 90°

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A particle moves in the $x-y$ plane under the action of a force $\vec{F}$ such that the value of its linear momentum ( $\vec{P}$ at anytime $t$ is $P_x=2 \cos t, p_y=2 \sin t$. The angle $\theta$ between $\vec{F}$ and $\vec{P}$ at a given time $t$. will be

Options:
A $\theta=0^{\circ}$
B $\theta=30^{\circ}$
C $\theta=90^{\circ}$
D $\theta=180^{\circ}$

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

Solution:
$P_x=2 \cos t$, $P_y=2 \sin t$ $\therefore$ $\vec{P}=2 \cos t \hat{i}+2 \sin t \hat{j}$
$\vec{F}=\frac{\vec{dP}}{\mathrm{dt}}$=-$-2 \sin t \hat{i}+2 \cos t \hat{j}$
$\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{P}}$=0. $\theta=90^{\circ}$

Topic: Physics, Mathematics in Physics, JEE Main
A particle moves in $x-y$ plane according to rule $\mathrm{x}=\mathrm{a} \sin \omega \mathrm{t}$ and $\mathrm{y}=\mathrm{a} \cos \omega \mathrm{t}$. The particle follows

Options:
A an elliptical path
B a circular path
C a parabolic path
D a straight line path inclined equally to $x$ and $y$-axes

The correct answer is:
a circular path

Solution:
$$
\begin{aligned}
& \mathrm{x}=\mathrm{a} \sin \omega \mathrm{t} \Rightarrow \frac{\mathrm{x}}{\mathrm{a}}=\sin \omega \mathrm{t} \\
& \mathrm{x}=\mathrm{a} \cos \omega \mathrm{t} \Rightarrow \frac{\mathrm{x}}{\mathrm{a}^2}=\cos \omega \mathrm{t} \\
& \therefore \quad \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \\
& \text { or } \quad \mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2
\end{aligned}
$$
This is equation of circle, so the particle follows a circular path.

Topic: Physics, Oscillations, JEE Main
A particle moves in $x-y$ plane according to rule $\mathrm{x}=\mathrm{a} \sin \omega \mathrm{t}$ and $\mathrm{y}=\mathrm{a} \cos \omega \mathrm{t}$. The particle follows

Options:
A an elliptical path
B a circular path
C a parabolic path
D a straight line path inclined equally to $x$ and $y$-axes

The correct answer is:
a circular path

Solution:
$$
\begin{aligned}
& \mathrm{x}=\mathrm{a} \sin \omega \mathrm{t} \Rightarrow \frac{\mathrm{x}}{\mathrm{a}}=\sin \omega \mathrm{t} \\
& \mathrm{x}=\mathrm{a} \cos \omega \mathrm{t} \Rightarrow \frac{\mathrm{x}}{\mathrm{a}^2}=\cos \omega \mathrm{t} \\
& \therefore \quad \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \\
& \text { or } \quad \mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2
\end{aligned}
$$
This is equation of circle, so the particle follows a circular path.

Topic: Physics, Oscillations, NEET, NEET 2010 (Mains)