datavorous/entrance-exam-dataset huggingface dataset:
A particle at rest suddenly disintegrates into two particles of equal masses which start moving. The two fragments will

Options:
A Move in the same direction with equal speeds
B Move in any directions with any speed
C Move in opposite directions with equal speeds
D Move in opposite directions with unequal speeds

The correct answer is:
Move in opposite directions with equal speeds

Solution:
A particle at rest suddenly disintegrates into two particles of equal masses which start moving. The two fragments will move in opposite directions with equal speeds (From momentum conservation).

Topic: Physics, Center of Mass Momentum and Collision, NEET
A particle describes a horizontal circle in a conical funnel whose inner surface is smooth with the speed of 0.5 m s-1 . What is the height of the plane of the circle from the vertex of the funnel?

Options:
A 0.25Â cm
B 2Â cm
C 4Â cm
D 2.5 cm

The correct answer is:
2.5 cm

Solution:
Nsinâ¡Î¸=mg
Ncosâ¡Î¸=mv2/r
Dividing above equations
â tanâ¡Î¸=rg/v2
â rh=rgv2
â h=v2g=2.5 cm

Topic: Physics, Laws of Motion, NEET
A particle enters uniform constant magnetic field region with its initial velocity parallel to the field direction. Which of the following statements about its velocity is correct? (neglect the effects of other fields)

Options:
A There is change in both magnitude and direction
B There is no change
C There is change only in magnitude
D There is change only in direction

The correct answer is:
There is no change

Solution:
Magnetic force when the particle, initial velocity parallel to the field direction
Fâ=qvâÃBâ=qvâBâsinâ¡Î¸=qvâBâsinâ¡Î¸=0
(âµ Direction of the velocity of the electron is parallel to direction of the magnetic field)
So, velocity is constant. Hence, there is no change in direction and magnitude of the velocity.

Topic: Physics, Magnetic Effects of Current, NEET
A particle executes a simple harmonic motion of time period T. Find the time taken by the particle to go directly from its mean position to half the amplitude.

Options:
A T2
B T4
C T8
D T12

The correct answer is:
T12

Solution:
y=Asin⁡ωt=Asin2πTt
⇒A2=Asin⁡2πtT
⇒t=T12

Topic: Physics, Oscillations, NEET
A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Options:
A 5π
B 52π
C 4π5
D 2π3

The correct answer is:
4π5

Solution:
At position x from the mean position in S.H.M.
velocity, v=ωA2-x2
acceleration, a=xω2

here ω is the angular frequency and A is amplitude.
Given in the problem, v=a
⇒ ωA2-x2=xω2
⇒ 32-22=22πT
5=4πT
Time period, T=4π5

Topic: Physics, Oscillations, NEET, NEET 2017
A particle executes SHM along a straight line so that its period is 12 s. The time it takes in traversing a distance equal to half its amplitude from its equilibrium position is-

Options:
A 6Â s
B 4Â s
C 2Â s
D 1 s

The correct answer is:
1 s

Solution:
for T = 12 s, ω= 2π T = 2π 12 = π 6
Using x=Asin( ωt ) (from equilibrium position)
A 2 =Asin( π 6 t )
⇒ π 6 t=30°= π 6
⇒ t=1 s

Topic: Physics, Oscillations, NEET
A particle executes simple harmonic motion (amplitude = A) between x = - A and x = + A. The time taken for it to go from 0 to A2 is T1 and to go from A2 to A is T2 . Then

Options:
A T1<T2
B T1>T2
C T1=T2
D T1=2T2

The correct answer is:
T1<T2

Solution:
x=Asinωt
For x=A2 , sinωT1=12 ⇒ T1=π6ω
For x = A, sinωT1+T2=1 ⇒T1+T2=π2ω
⇒T2=π2ω-T1=π2ω-π6ω=π3ω
i.e. T1<T2

Topic: Physics, Oscillations, NEET
A particle executes simple harmonic motion with a frequency f. The frequency with which its kinetic energy oscillates is

Options:
A f2
B f
C 2f
D 4f

The correct answer is:
2f

Solution:
In SHM, the frequency with which kinetic energy oscillates is two times the frequency of oscillation of displacement. Correct answer is 2f.

Topic: Physics, Oscillations, NEET
A particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:

Options:
A $T / 8$
B $T / 12$
C $T / 2$
D $T / 4$.

The correct answer is:
$T / 12$

Solution:
For equilibrium position
$\begin{aligned}
x(t) & =a \sin \omega t \\
\text {At } x(t) & =\frac{a}{2} \\
\therefore \sin \left(\frac{\pi}{6}\right) & =\sin \omega t\left\{\because \omega=\frac{2 \pi}{\mathrm{T}}\right\} \\
\text {or } \frac{\pi}{6} & =\frac{2 \pi t}{T} \\
\Rightarrow t & =\frac{T}{12}
\end{aligned}$

Topic: Physics, Oscillations, NEET, NEET 2007
A particle executes simple harmonic oscillation with an amplitude A. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

Options:
A T / 8
B T / 12
C T / 2
D T / 4

The correct answer is:
T / 12

Solution:
Using the equation of SHM,
x t = A sin Ï t (from the equilibruim position)
At x t = A / 2
â´ A 2 = A sin Ï t
sin Ï 6 = sin Ï t or, Ï 6 = 2 Ï t Tâ¡ âµ Ï = 2 Ï Tâ¡
or, t = T / 12 .

Topic: Physics, Oscillations, NEET
A particle executes the motion described by x(t)= xo(1-e-γt). Where does the particle start

Options:
A xo+e
B xo-e
C xo
D 0

The correct answer is:
0

Solution:
Correct Option is : (D)
0

Topic: Chemistry, Motion In One Dimension, NEET
A particle executing a simple harmonic motion has a period of 6 s. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is

Options:
A 14 s
B 34 s
C 12 s
D 32 s

The correct answer is:
12 s

Solution:
y=Asin⁡2πTt
A2=Asin⁡2πTt=2πTt=π/6
Time period t=T12=612=12 s

Topic: Physics, Oscillations, NEET
A particle executing SHM has a maximum speed of 0.5 m s-1 and a maximum acceleration of 1.0 m s-2. The angular frequency of oscillation is

Options:
A 2 rad s-1
B 0.5 rad s-1
C 2π rad s-1
D 0.5π rad s-1

The correct answer is:
2 rad s-1

Solution:
In SHM,
Maximum velocity vmax=ωA
Maximum acceleration amax=Aω2
⇒ ω=Aω2Aω=amaxvmax
vmax=0.5 m s-1
amax=1.0 m s-1
ω=10.5=105 =2 rad s-1

Topic: Physics, Oscillations, NEET
A particle executing S.H.M has amplitude 0.01 and frequency 60 Hz. The maximum acceleration of the particle is

Options:
A 144 Ï 2 m/s 2
B 120 π 2 m/s 2
C 80 π 2 m/s 2
D 60 π 2 m/s 2

The correct answer is:
144 Ï 2 m/s 2

Solution:
Given : Amplitude, a = 0.01 m and frequency,
f = 60 Hz.
Maximum acceleration of the particle
= Ï 2 a = 2 Ï f 2 . a
= 2 Ï Ã 6 0 2 Ã 0.01
= 144 Ï 2 m/s 2

Topic: Physics, Oscillations, NEET
A particle executing SHM of amplitude 4 cm and T=4 s. The time taken by it to move from positive extreme position to half the amplitude is

Options:
A 1Â s
B 13 s
C 23 s
D 32 s

The correct answer is:
23 s

Solution:
Equation of motion y=acosâ¡Ït
â a2=acosâ¡Ït
âcosâ¡Ït=13
â Ït=Ï3
â2ÏtT=Ï3
â t=Ï3Ãt2Ï=43Ã2=23 s

Topic: Physics, Oscillations, NEET
A particle executing simple harmonic motion along y-axis has its motion described by the equation y=Asinωt+B. The amplitude of the simple harmonic motion is

Options:
A A
B B
C A+B
D A+B

The correct answer is:
A

Solution:
y=Asinωt+B
The amplitude of the simple harmonic motion is A

Topic: Physics, Oscillations, NEET
A particle executing simple harmonic motion of amplitude $5 \mathrm{~cm}$ has maximum speed of $31.4 \mathrm{~cm} / \mathrm{s}$. The frequency of its oscillation is:

Options:
A $4 \mathrm{~Hz}$
B $3 \mathrm{~Hz}$
C $2 \mathrm{~Hz}$
D $1 \mathrm{~Hz}$

The correct answer is:
$1 \mathrm{~Hz}$

Solution:
Here $a=5 \mathrm{~cm}, V_{\max }=\frac{31.4 \mathrm{~cm}}{\mathrm{~s}}$
$\begin{aligned}
& V_{\max }=\omega a=31.4=2 \pi \mathrm{v} \times 5 \\
& \Rightarrow 31.4=10 \times 3.14 \times V \\
& \Rightarrow \quad V=1 Hz
\end{aligned}$

Topic: Physics, Oscillations, NEET, NEET 2005
A particle executing simple harmonic motion with amplitude $A$ has the same potential and kinetic energies at the displacement

Options:
A $2 \sqrt{A}$
B $\frac{A}{2}$
C $\frac{A}{\sqrt{2}}$
D $A \sqrt{2}$

The correct answer is:
$\frac{A}{\sqrt{2}}$

Solution:
Potential energy $=\frac{1}{2} k x^2$
Kinetic energy $=\frac{1}{2} k A^2-\frac{1}{2} k x^2$
According to given condition
$\frac{1}{2} k x^2=\frac{1}{2} k A^2-\frac{1}{2} k x^2$
$k 2 x^2=k A^2$
$x^2=\frac{A^2}{2}$
$x=\frac{A}{\sqrt{2}}$

Topic: Physics, Oscillations, NEET, NEET 2024 (Re-NEET)
A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle θ with speed vo and rebounds elastically (Fig). Find the distance along the plane where if will hit second time.

Options:
A $\frac{A{{v}_{o}}^{2}}{2g}sin\theta $
B $\frac{A{{v}_{o}}^{2}}{2g}cos\theta $
C $\frac{A{{v}_{o}}^{2}}{g}sin\theta $
D $\frac{A{{v}_{o}}^{2}}{g}cos\theta $

The correct answer is:
$\frac{A{{v}_{o}}^{2}}{g}sin\theta $

Solution:
Correct Option is : (C)
$\frac{A{{v}_{o}}^{2}}{g}sin\theta $

Topic: Chemistry, Motion In Two Dimensions, NEET
A particle has initial velocity $(2 \mathbf{i}+3 \mathbf{j})$ and acceleration $(0.3 \mathbf{i}+0.2 \mathbf{j})$. The magnitude of velocity after $10 \mathrm{~s}$ will be

Options:
A $9 \sqrt{2}$ units
B $5 \sqrt{2}$ units
C 5 units
D 9 units

The correct answer is:
$5 \sqrt{2}$ units

Solution:
\(\begin{aligned}
& \text {Here, } \vec{u}=2 \hat{i}+3 \hat{j} \text {, } \\
& \vec{a}=0.3 \hat{i}+0.2 \hat{j}, t=10 \mathrm{~s} \\
& \text { As } \bar{v}=\bar{u}+\vec{a} t \\
& \bar{v}=(2 \hat{i}+3 \hat{j})+(0.3 \hat{i}+0.2 \hat{j})(10) \\
& =2 \hat{i}+3 \hat{j}+3 \hat{i}+2 \hat{j}=5 \hat{i}+5 \hat{j} \\
& |\vec{v}|=\sqrt{(5)^2+(5)^2} \\
& =5 \sqrt{2} \text { units }
\end{aligned}\)

Topic: Physics, Motion In One Dimension, NEET, NEET 2012 (Screening)
A particle has initial velocity $(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}})$ and has acceleration $(0.1 \hat{\mathrm{i}}+0.3 \hat{\mathrm{j}})$. Its speed after $10 \mathrm{~s}$ is

Options:
A 7 unit
B $7 \sqrt{2}$ unit
C 8.5 unit
D 10 unit

The correct answer is:
$7 \sqrt{2}$ unit

Solution:
From first equation of motion
$$
\begin{aligned}
\mathrm{v} & =\mathrm{u}+\mathrm{at} \\
\mathrm{v} & =3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+10(0.4 \hat{\mathrm{i}}+0.3 \hat{\mathrm{j}}) \\
\mathrm{v} & =7 \hat{\mathrm{i}}+7 \hat{\mathrm{j}} \\
|\mathrm{v}| & =7 \sqrt{2}
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2010 (Screening)
A particle has velocity 3rg at the highest point in vertical circle. Find the ratio of tensions at the highest and lowest point

Options:
A 1 : 6
B 1 : 4
C 1 : 3
D 1 : 2

The correct answer is:
1 : 4

Solution:
In vertical circular motin tension at top is mg - mv2/r and tension at bottom is mg + mv2/r
Ttop=mv2r-mg=2mg
TtopTbottom=2mg2mg+6mg=14

Topic: Physics, Laws of Motion, NEET
A particle is at a distance r from the axis of rotation. A given torque τ produces some angular acceleration in it. If the mass of the particle is doubled and its distance from the axis is halved, the value of torque to produce the same angular acceleration is -

Options:
A τ / 2
B τ
C 2 τ
D 4 τ

The correct answer is:
τ / 2

Solution:
I ′ = 2 1 2 2 I = I 2
τ = I α
⇒ τ ′ τ = I ′ I = τ 2

Topic: Physics, Rotational Motion, NEET
A particle is dropped from a height H. The de-Broglie wavelength of the particle depends on height as

Options:
A H-1∕2
B H0
C H1∕2
D H

The correct answer is:
H-1∕2

Solution:
de-Broglie wavelength λ=hmv
λ=hm2gH
Hence, λ∝H-1∕2

Topic: Physics, Dual Nature of Matter, NEET
A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β. Then, its time period of vibration will be:

Options:
A αβ
B β2α
C 2πβα
D β2α2

The correct answer is:
2πβα

Solution:
ω2A=α
ωA=β
⇒ω=αβ
⇒T=2πω=2πβα

Topic: Physics, Oscillations, NEET, NEET 2015 (Phase 2)
A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean position are V1 and V2 respectively. Its time period is:

Options:
A 2πx12+x22V12+V22
B 2πx22-x12V12-V22
C 2πV12+V22x12+x22
D 2πV12-V22x12-x22

The correct answer is:
2πx22-x12V12-V22

Solution:
For particle undergoing SHM,
V=ÏA2-x2âV2=Ï2 A2-x2
V1=ÏA2-x12âV12=Ï2A2-x12 ...(i)
V2=ÏA2-x22 âV22=Ï2A2-x22 ...(ii)
V12-V22=Ï2x22-x12
Ï=V12-V22x22-x12
T=2Ïx22-x12V12-V22

Topic: Physics, Oscillations, NEET, NEET 2015 (Phase 1)
A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean position are V1 and V2 respectively. Its time period is:

Options:
A 2πx12+x22V12+V22
B 2πx22-x12V12-V22
C 2πV12+V22x12+x22
D 2πV12-V22x12-x22

The correct answer is:
2πx22-x12V12-V22

Solution:
For particle undergoing SHM,
V=ÏA2-x2âV2=Ï2 A2-x2
V1=ÏA2-x12âV12=Ï2A2-x12 ...(i)
V2=ÏA2-x22 âV22=Ï2A2-x22 ...(ii)
V12-V22=Ï2x22-x12
Ï=V12-V22x22-x12
T=2Ïx22-x12V12-V22

Topic: Physics, Oscillations, NEET
A particle is executing simple harmonic motion with an amplitude A and time period T. The displacement of the particles after 2T period from its initial position is

Options:
A A
B 4A
C 8A
D zero

The correct answer is:
zero

Solution:
Particle doing SHM repeats its motion after each Time period.

Since 2T is a multiple of time period of the SHM hence particle will be at the position where it stared it's motion time 2T ago.

∴ displacement of the particle will be zero

Topic: Physics, Oscillations, NEET
A particle is given an initial speed u inside a smooth spherical shell of radius R=1 m such that it is just able to complete the circle. Acceleration of the particle, when its velocity is vertical, is

Options:
A g10
B g
C g2
D 3g

The correct answer is:
g10

Solution:
Speed at lowest point, to just complete the circle,
u2=5gR
â´ Speed, when velocity becomes vertical, is
v2=u2-2gR
v2=5gR-2gR=3gR
Tangential acceleration at B is
at=g (downwards)
Centripetal acceleration at B is
ac=v2R=3g
â´ Total acceleration will be,
a=aC2+at2=g10

Topic: Physics, Laws of Motion, NEET
A particle is given an initial speed u inside a smooth spherical shell of the radius R=1 m that it is just able to complete the circle. Acceleration of the particle when its velocity is vertical is

Options:
A g10
B g
C g 2
D 3g

The correct answer is:
g10

Solution:
u2=5gR
∴ v2=u2-2gR
=5gR-2gR=3gR
Tangential acceleration at B is
at=g (downwards)
Centripetal acceleration at B is
ac=v2R=3g
∴ Total acceleration will be
a=aC2+at2=g 10

Topic: Physics, Laws of Motion, NEET
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at 2 cm away from B going towards A,

Options:
A +,+,+
B -,-,-
C +,-,+
D 0,+,+

The correct answer is:
-,-,-

Solution:
Correct Option is : (B)
-,-,-

Topic: Chemistry, Oscillations, NEET
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at the mid-point of AB going towards A

Options:
A +,+,+
B -,0,0
C +,-,+
D 0,+,+

The correct answer is:
-,0,0

Solution:
Correct Option is : (B)
-,0,0

Topic: Chemistry, Oscillations, NEET
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at 3 cm away from A going towards B,

Options:
A +,+,+
B -,-,-
C +,-,+
D 0,+,+

The correct answer is:
+,+,+

Solution:
Correct Option is : (A)
+,+,+

Topic: Chemistry, Oscillations, NEET
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at the end A

Options:
A +,+,+
B -,-,-
C +,-,+
D 0,+,+

The correct answer is:
0,+,+

Solution:
Correct Option is : (D)
0,+,+

Topic: Chemistry, Oscillations, NEET
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at 4 cm away from B going towards A.

Options:
A +,+,+
B -,-,-
C +,-,+
D 0,+,+

The correct answer is:
-,-,-

Solution:
Correct Option is : (B)
-,-,-

Topic: Chemistry, Oscillations, NEET
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at the end B

Options:
A +,+,+
B -,-,-
C 0,-,-
D 0,+,+

The correct answer is:
0,-,-

Solution:
Correct Option is : (C)
0,-,-

Topic: Chemistry, Oscillations, NEET
A particle is kept at rest at a distance R (Earth's radius) above the earth's surface. The minimum speed with which it should be projected so that it does not return is

Options:
A GM 4R
B GM 2R
C GM R
D 2GM R

The correct answer is:
GM R

Solution:
According to the conservation of energy,
K.E.= U G
1 2 m v 2 = GMm R+h
1 2 m v 2 = GMm R+R = GMm 2R
Therefore, the escape velocity is,
v= GM R

Topic: Physics, Gravitation, NEET
A particle is moving along a circular path with a constant speed. The acceleration of the particle is constant in

Options:
A magnitude
B direction
C both magnitude and direction
D neither magnitude nor direction

The correct answer is:
magnitude

Solution:
For constant speed along circular path acceleration is centripetal which is constant in magnitude ac=v2r and always directed towards the center so the direction changes continuously.

Topic: Physics, Laws of Motion, NEET
A particle is moving along $x$-axis with its position (x) varying with time $(t)$ as $x=\alpha t^4+\beta t^2+\gamma t+\delta$. The ratio of its initial velocity to its initial acceleration, respectively, is:

Options:
A $2 \alpha: \delta$
B $\gamma: 2 \delta$
C $4 \alpha: \beta$
D $\gamma: 2 \beta$

The correct answer is:
$\gamma: 2 \beta$

Solution:
Position of particle, $x=\alpha t^4+\beta t^2+\gamma t+\delta$
Velocity $v=\frac{d x}{d t}=4 \alpha t^3+2 \beta t+\gamma$
Initial velocity $=v(t=0)=\gamma$
Acceleration $a=\frac{d v}{d t}=12 \alpha t^2+2 \beta$
Initial acceleration $=a(t=0)=2 \beta$
$\therefore \frac{v(t=0)}{a(t=0)}=\frac{\gamma}{2 \beta}$

Topic: Physics, Motion In One Dimension, NEET, NEET 2024 (Re-NEET)
A particle is moving in a circle of radius 1 cm and with a constant speed of 8 cm s-1. Centre of the circle lies on principal axis of a converging lens of focal length 50 cm and at a distance of 75 cm from the lens. Plane of the circle is perpendicular to principal axis,. The correct statement is

Options:
A Speed of the image is 4 cm s-1
B Speed of the image is 8 cm s-1
C Speed of the image is 16 cm s-1
D Speed of the image is 3.2 cm s-1

The correct answer is:
Speed of the image is 16 cm s-1

Solution:
m=ff+u=-2
⇒ radius of image = 2 cm
∵ ω is same
⇒81=v2⇒v=16 cm s-1 .

Topic: Physics, Ray Optics, NEET
A particle is moving in a circle with uniform speed. Its motion is

Options:
A periodic and simple harmonic
B periodic but not simple harmonic
C not periodic
D none of the above

The correct answer is:
periodic but not simple harmonic

Solution:
Simple harmonic motion is a special type of periodic motion where the restoring force on the moving object is directly proportional to, and opposite of, the object's displacement vector. It results in an oscillation which, if uninhibited by friction or any other dissipation of energy, continues indefinitely.
Hence we can say if a particle is moving in a circle with uniform speed then motion is periodic but not simple harmonic.

Topic: Physics, Oscillations, NEET
A particle is moving in a circular path and its acceleration vector is making an angle of 30° with the velocity vector, then the ratio of centripetal acceleration to its tangential acceleration is

Options:
A 12
B 32
C 13
D 3

The correct answer is:
13

Solution:
For a body undergoing circular motion, let net acceleration makes an angle θ with the tangent, then
tan⁡θ=acat
∴ acat=tan⁡30°=13

Topic: Physics, Laws of Motion, NEET
A particle is moving in a force field given by potential energy U=-Î»x+y+z from point 1, 1, 1 to 2, 3, 4. The work done in the process is

Options:
A 3λ
B 1.5λ
C 6λ
D 12λ

The correct answer is:
6λ

Solution:
As U=-Î»x+y+z
At point 1,1,1 potential energy is,
U1=-Î»1+1+1
U1=-3Î»
At point 2,3,4 potential energy is,
U2=-Î»2+3+4
U2=-9 Î»
Charge in potential energy,
ÎU=U2-U1
ÎU=-9Î»--3Î»
ÎU=-6Î»
â´ Work done =-ÎU
W=6Î»

Topic: Physics, Work Power Energy, NEET
A particle is moving in x - y plane as shown in the figure. Angular velocity of the particle with respect to the origin is

Options:
A 825k^ rad s-1
B -825k^ rad s-1
C 1925k^ rad s-1
D -1925k^ rad s-1

The correct answer is:
825k^ rad s-1

Solution:
ω=2sin⁡θ5
=2×425=825
ω→=825k^ rad s-1

Topic: Physics, Laws of Motion, NEET
A particle is moving on rough horizontal ground with an initial velocity v0. If 34th of its kinetic energy is lost due to friction in time t0, then the coefficient of friction between the particle and the ground is

Options:
A v02gt0
B v04gt0
C 3v04gt0
D v0gt0

The correct answer is:
v02gt0

Solution:
34th energy is lost i.e., 14th kinetic energy is left.
Hence, its velocity becomes v02 under a retardation of Î¼g in time t0 .
â´ From the first equation of motion,
v02=v0-Î¼g t0
or Î¼gt0=v02
or Î¼=v02gt0

Topic: Physics, Laws of Motion, NEET
A particle is moving such that its position coordinates x, y are 2 m, 3 m at time t=0, 6 m, 7 m at time t=2 s and 13 m, 14 m at time t=5 s. The average velocity vector V→av from t=0 to t=5 s is:

Options:
A 1513i^+14j^
B 73i^+j^
C 2i^+j^
D 115i^+j^

The correct answer is:
115i^+j^

Solution:
The position vector of the particle at time t=0 is r→i=2i^+3j^
The position vector of the particle at time t=5 s is r→f=13i^+14j^
The displacement of the particle is s→=11i^+11j^
The average velocity vector V→=11i^+11j^5

Topic: Physics, Motion In One Dimension, NEET, NEET 2014
A particle is moving with a constant speed along a straight line path. A force is not required to

Options:
A increase its speed
B decrease the momentum
C change the direction
D keep it moving with uniform velocity

The correct answer is:
keep it moving with uniform velocity

Solution:
To increase speed force should be forward
To decrease momentum force should be opposite
To change direction force should be sidewise
But in order to keep the body with uniform speed, net force should be zero.

Topic: Physics, Laws of Motion, NEET
A particle is moving with velocity v=kyi^+xj^ where k is a constant. The general equation for its path is

Options:
A y=x2+constant
B y2=x+constant
C xy=constant
D y2=x2 +constant

The correct answer is:
y2=x2 +constant

Solution:
v=kyi^+xj^
dxdt=ky
dydt=kx
dydx=dydt·dtdx=kxky
dyky=dx.kx
ydy=xdx
y22=x22+constant
⇒ y2=x2+constant

Topic: Physics, Motion In Two Dimensions, NEET
A particle is placed at the origin and a force F=kx is acting on it (where k is a positive constant). If U 0=0 , the graph of U x versus x will be: (where U is the potential energy function)

Options:
A
B
C
D

The correct answer is:

Solution:
From the equation, F=-dUdx
â« 0 UxdU=-â« 0 xFdx=-â« 0 xKxdx
â´U x=-kx22
as U 0= 0

Topic: Physics, Work Power Energy, NEET
A particle is projected along a line of greatest slope on a rough plane inclined at an angle of 45° with the horizontal. If the coefficient of friction is 12, then the retardation is

Options:
A g2
B g22
C g21+12
D g21-12

The correct answer is:
g21+12

Solution:
The acceleration of the body on the incline is
a=gsin⁡θ+μcos⁡θ
a=gsin⁡45°+12×cos⁡45°
a=g12+12×12
a=g21+12
a=3g22

Topic: Physics, Laws of Motion, NEET