datavorous/entrance-exam-dataset huggingface dataset:
A particle moves in a circle of the radius 0.5 m at a speed that uniformly increases. Find the angular acceleration of particle if its speed changes from 2.0 m s-1 to 4.0 m s-1 in 4.0 s.

Options:
A 1 rad s-2
B 2 rad s-2
C 4 rad s-2
D 0 rad s-2

The correct answer is:
1 rad s-2

Solution:
The tangential acceleration of the particle is
at=dvdt=4.0-2.04.0
=0.5 m s-2
The angular acceleration is
α=atr=0.50.5=1 rad s-2

Topic: Physics, Laws of Motion, NEET
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 10
B 1.8
C 12
D 9

The correct answer is:
9

Solution:
Key Idea : The problem requires kinematics equations of motion.
Let $u$ and $v$ be the first and final velocities of particle and $a \& s$ be the constant acceleration and distance covered by it.
From third equation of motion
$\begin{array}{l}
y^2 =u^2+2 a s \\
\Rightarrow (20)^2 =(10)^2+2 a \times 135 \\
\text {or } a =\frac{300}{2 \times 135}=\frac{10}{9} \mathrm{~ms}^{-2}
\end{array}$
Now using first equation of motion,
$\begin{aligned}
& v=u+a t \\
& \text {or } t=\frac{v-u}{a}=\frac{20-10}{(10 / 9)}=\frac{10 \times 9}{10}=9 \mathrm{~s}
\end{aligned}$

Topic: Physics, Motion In One Dimension, NEET, NEET 2008 (Screening)
A particle moves in a straight line with retardation proportional to its displacement. The loss of kinetic energy during a displacement x is proportional to

Options:
A x2
B ex
C x
D log ex

The correct answer is:
x2

Solution:
dv dt ∝ - x
dv dx · dx dt = - cx
Vdv = - cx dx
v 2 2 = - cx 2 2
Kinetic energy = mv 2 2
Kinetic energy = mcx 2 2
k ∝ x 2

Topic: Physics, Work Power Energy, NEET
A particle moves in the plane xy with constant acceleration a directed along the negative y-axis. The equation of motion of the particle has the form y=px-qx2 where p and q are positive constants. Find the velocity of the particle at the origin of coordinates.

Options:
A a 1+ p 2 2q
B a 1+ q 2 2p
C a 1+ q 2 p
D a 1+ p 2 q

The correct answer is:
a 1+ p 2 2q

Solution:
Comparing the given equation with the equation of a projectile motion,
y=px−q x 2
and y = x tan Î¸ - gx 2 2 u 2 1 + tan 2 Î¸
we find that g=a, tanÎ¸=p
and g2u21+tan2â¡Î¸=q
âu2=g2q1+tan2â¡Î¸
⇒u= a 2q ( 1+ p 2 )

Topic: Physics, Motion In Two Dimensions, NEET
A particle moves in the X- Y plane under the influence of a force such that its linear momentum is P â t = A i ^ cos k t - j ^ sin k t , where A and k are constants. The angle between the force and the momentum is

Options:
A 00
B 300
C 450
D 900

The correct answer is:
900

Solution:
F → = d P → dt = - k A sin k t i ^ - k A cos k t j ^
P → = A cos k t i ^ - A sin k t j ^
Since, F → · P → = 0
∴ Angle between F → and P → should be 9 0 0

Topic: Physics, Center of Mass Momentum and Collision, NEET
A particle moves on x-axis according to the equation x=x0sin2⁡ωt , the motion is simple harmonic

Options:
A with amplitude x0
B with amplitude 2x0
C with time period 2πω
D with time period πω

The correct answer is:
with time period πω

Solution:
x=x0sin2â¡Ït=x02Â 1-cosâ¡2Ït=x02-x02cosâ¡2Ït
Frequency Ï′=2Ï
â2ÏT′=2Â Ï
âT′=ÏÏ
Amplitude = x02

Topic: Physics, Oscillations, NEET
A particle moves so that its acceleration a is given by a=-bx, where x is displacement from equilibrium position and b is a non-negative real constant. The time period of oscillation of the particle is

Options:
A 2πb
B 2πb
C 2πb
D 2πb

The correct answer is:
2πb

Solution:
The relation between acceleration (a) and displacement (x) for a body in SHM is
a=-ω2x
Given, a=-bx
On comparing the two equations, we get
ω2=b
∴ ω=b
Since, ω=2πT
∴ 2πT=b
⇒T=2πb

Topic: Physics, Oscillations, NEET
A particle moves so that its acceleration is given by a=-bx where x is the displacement from the equilibrium position and b is a positive constant. The period of oscillation is

Options:
A 2πb
B 2πb
C 2πb
D 2πb

The correct answer is:
2πb

Solution:
Comparing with a=-ω2x
ω2=b
ω=b
T=2πω
T=2πb

Topic: Physics, Oscillations, NEET
A particle moves so that its position vector is given by r→=cosωt x^ +sinωt y^, where ω is a constant. Which of the following is true?

Options:
A Velocity and acceleration both are perpendicular to r→.
B Velocity and acceleration both are parallel to r→.
C Velocity is perpendicular to r→ and acceleration is directed towards the origin.
D Velocity is perpendicular to r→ and acceleration is directed away from the origin.

The correct answer is:
Velocity is perpendicular to r→ and acceleration is directed towards the origin.

Solution:
Position vector is, r→=cos⁡ωtx^+sin⁡ωty^
Velocity of particle is v→=dr→dt
v→=-sinωt×ωx^+cosωt×ωy^
⇒v→=ω-sinωtx^+cosωty^
Acceleration of the particle is
a→=dv→dt
⇒a→=-ω2cosωtx^+sinωty^
⇒a→=-ω2r→ ,
So direction of r→ and a→ are opposite.
As v→·a→=0⇒v→⊥a→ and
v→·r→=0⇒v→⊥r→,
So, velocity is perpendicular to r→ and acceleration is directed towards the origin.

Topic: Physics, Motion In One Dimension, NEET, NEET 2016 (Phase 1)
A particle moves such that its acceleration is given by

a=-βx-2

Here : β is a positive constant and x is the position from origin. Time period of oscillations is

Options:
A 2πβ
B 2π1β
C 2πβ+2
D 2π1β+2

The correct answer is:
2π1β

Solution:
a=-βx-2
as a=-ω2x-x0
by comparing
ω2=β
⇒T=2π1β

Topic: Physics, Oscillations, NEET
A particle moves under the effect of a force F=cx from x=0 to x=x1. The work done in the process is

Options:
A cx12
B 12cx12
C cx13
D zero

The correct answer is:
12cx12

Solution:
F=cx
W=∫F.dx
W=∫0x1cx.dx
F=cx220x1
F=cx122

Topic: Physics, Work Power Energy, NEET
A particle moves with a speed v in a circle of radius R. When the particle reaches from A to B as shown in the figure, then Y component of average velocity is

Options:
A 2vπ
B -vπ
C -2vπ
D zero

The correct answer is:
-vπ

Solution:
Time taken=πRv
Displacement in Y direction=-R
Component of Average velocity
=-RπRv=-vπ

Topic: Physics, Laws of Motion, NEET
A particle moves with a velocity 5iâ¡^-3jâ¡^+6k^m s-1 under the influence of a constant force Fâ¡â=10iâ¡^-10jâ¡^+20k^N. The instantaneous power applied to the particle is

Options:
A 140 W

B 40 W

C 200 W

D 170 W

The correct answer is:
200 W

Solution:
v→=5i⁡^-3j⁡^+6k^m s-1
F⁡→=10i⁡^-10j⁡^+20k^N
∴Power=F⁡→·v→=10i⁡^-10j⁡^+20k^·5i⁡^-3j⁡^+6k^
= 50 + 30 + 120 = 200 W

Topic: Physics, Work Power Energy, NEET
A particle moves with a velocity $(5 \hat{i}-3 \hat{j}+6 \hat{k}) \mathrm{ms}^{-1}$ horizontally under the action of constant force $(10 \hat{i}+10 \hat{j}+20 \hat{k}) \mathrm{N}$. The instantaneous power supplied to the particle is:

Options:
A $100 \mathrm{~W}$
B $140 \mathrm{~W}$
C $200 \mathrm{~W}$
D Zero

The correct answer is:
$140 \mathrm{~W}$

Solution:
Instantaneous power is given by
$P=\vec{F} \cdot \vec{V}$
$\begin{aligned} P & =[10 \hat{i}+10 \hat{j}+20 \hat{k}] \cdot[5 \hat{i}-3 \hat{j}+6 \hat{k}] \mathrm{W} \\ & =[50-30+120] \mathrm{W}=140 \mathrm{~W}\end{aligned}$

Topic: Physics, Work Power Energy, NEET, NEET 2023 (Manipur)
A particle moving along $x$-axis has acceleration $f$, at time $t$, given by $f=$ $f_0\left(1-\frac{t}{T}\right)$, where $f_0$ and $T$ are constants.
The particle at $t=0$ has zero velocity. In the time interval between $t=0$ and the instant when $f=0$, the particle's velocity $\left(v_x\right)$ is:

Options:
A $\frac{1}{2} f_0 T^2$
B $f_0 T^2$
C $\frac{1}{2} f_0 T$
D $f_0 T$

The correct answer is:
$\frac{1}{2} f_0 T$

Solution:
Acceleration $\frac{d v}{d t}=f=f_0\left(1-\frac{t}{T}\right)$
$\begin{aligned}
\Rightarrow{ }_0^v d v & =f_0^T 1-\frac{t}{T} d t \\
\Rightarrow \quad v & =f_0\left(t-\frac{t^2}{2 T}\right)_0^T \\
& v =f_0\left(T-\frac{T^2}{2 T}\right)=\frac{1}{2} f_0 T
\end{aligned}$

Topic: Physics, Motion In One Dimension, NEET, NEET 2007
A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle θ to the horizontal, the maximum height attained by it equals 4R. The angle of projection θ is then given by___

Options:
A θ=cos-1gT2π2R12
B θ=cos-1π2RgT212
C θ=sin-1π2RgT212
D θ=sin-12g T2π2R12

The correct answer is:
θ=sin-12g T2π2R12

Solution:
T=2πω=2πRv⇒v=2πRT

H=v2sin2θ2g⇒4R=4π2R2sin2θ2gT2

⇒sin2θ=2gT2π2R

⇒sinθ=2gT2π2R1/2

⇒θ=sin-12gT2π2R12

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2021
A particle moving with a velocity 110th of that of light will cross a nucleus in about

Options:
A 10-47s
B 10-21s
C 10-12 s
D 10-8 s

The correct answer is:
10-21s

Solution:
t=2rv=2Ã10â15 m3Ã107 m s-1=23Ã10â22 s

Topic: Physics, Nuclear Physics, NEET
A particle of charge -16Ã10-18 C moving with velocity 10 m s-1 along the x-axis enters a region where a magnetic field of induction B is along the y-axis and an electric field of magnitude 104 V m-1 is along the negative z-axis. If the charged particle continues moving along the x-axis, the magnitude of B is

Options:
A 103 Wb m-2
B 105 Wb m-2
C 1016 Wb m-2
D 10-3 Wb m-2

The correct answer is:
103 Wb m-2

Solution:
The force on a particle is
F=qE+v×B
or F=Fe+Fm
Fe=qE
=-16×10-18×104-k^
=16×10-14k^
and Fm=-16×10-18(10i^×Bj^)
=-16×10-17×B(+k^)
=-16×10-17B×k^
Since particle will continue to move along +x-axis, so resultant force is equal to 0.
Fe+Fm=0
∴ 16×10-14=16×10-17B
⇒ B=16×10-14 16×10-17=103
B=103 Wb m-2

Topic: Physics, Magnetic Effects of Current, NEET
A particle of charge âqâ and mass âmâ is projected from the origin with velocity u0i^+v0j^ in a gravity free region where uniform electric field -E0i^ and uniform magnetic field -B0i^ exist. Find the condition so that the particle would return to origin at least for once.

Options:
A μ0B02πE0 is an integer
B u02+v02 B0πE0 is an integer
C v0B0πB0 is an integer
D u0B0πE0 is an integer

The correct answer is:
u0B0πE0 is an integer

Solution:
Time for one complete rotation T0=2πmqB0
Time for retracing motion in x direction T1=2u0mqE0
∴ T1T0 must be an integer. So, 2u0mqE0×qB02πm=u0B0πE0

Topic: Physics, Magnetic Effects of Current, NEET
A particle of charge q and mass m is projected with a velocity V0 towards a circular region having uniform magnetic field B perpendicular and into the plane of paper from point P as shown in the figure. R is the radius and O is the centre of the circular region. If the line OP makes angle θ with the direction of V0 , then the value of V0 so that particle passes through O is

Options:
A qBRmsinθ
B qBR2msinÎ¸
C 2qBRmsinθ
D 3qBR2msinθ

The correct answer is:
qBR2msinθ

Solution:
r=mV0qB
R=2r sinÎ¸
âR=2mV0qBsinÎ¸
âV0=qBR2msinÎ¸

Topic: Physics, Magnetic Effects of Current, NEET
A particle of charge q and mass m is projected with a velocity V0 towards a circular region having uniform magnetic field B perpendicular and into the plane of paper from point P as shown in the figure. R is the radius and O is the centre of the circular region. If the line OP makes angle θ with the direction of V0, then the value of V0 so that particle passes through O is -

Options:
A qBRmsin⁡θ
B qBR2msinâ¡Î¸
C 2qBRmsin⁡θ
D 3qBR2msin⁡θ

The correct answer is:
qBR2msin⁡θ

Solution:
r=mV0qB
R=2 r sinâ¡Î¸
â R=2mV0qBsinâ¡Î¸
â V0=qBR2msinâ¡Î¸

Topic: Physics, Magnetic Effects of Current, NEET
A particle of mass 0.1 kg is subjected to a force which varies with distance as shown in the figure. If it starts its journey from rest at x=0, its velocity at x=12 m is

Options:
A 0 m s-1

B 40 m s-1

C 202m s-1
D 20 m s-1

The correct answer is:
40 m s-1

Solution:
Net work done by the force = area under the curve
=4×102+4×10+4×102=80 J
According to work-energy theorem
ΔK.E. = Wnet
1 2 m v 2 - 0 = 8 0 or 1 2 × 0 · 1 v 2 = 8 0
v=1600=40m s-1

Topic: Physics, Work Power Energy, NEET
A particle of mass 1 g moving with a velocity uâ1=3i^-2j^ m s-1 experiences a perfectly inelastic collision with another particle of mass 2 g moving with a velocity uâ2=4j^-6k^ m s-1 . The velocity (in m s-1 ) of the combined particle is

Options:
A i^+2j^-4k^
B i^-2j^+4k^
C i^-2j^-4k^
D i^+3.33j^+4k^

The correct answer is:
i^+2j^-4k^

Solution:
From conservation of momentum
13i^-2j^+24j^-6k^=1+2vâ
â 3i^+6j^-12k^=3vâ
â vâ=i^+2j^=4k^ m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
A particle of mass 1 kg is moving in SHM with a path length of 0.01 m and frequency of 50 Hz. The maximum force in newton, acting on the particle is

Options:
A 150Â Ï2
B 200Â Ï2
C 100 Ï2
D 50Â Ï2

The correct answer is:
100 Ï2

Solution:
F=mr 4π2v2
⇒F=1×0.01×4×π2×502
Or F=1×0.01×4×π2×2500
Or F=100 π2 N

Topic: Physics, Oscillations, NEET
A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of $3 \times 10^6 \mathrm{~ms}^{-1}$. The velocity of the particle is
(Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ )

Options:
A $2.7 \times 10^{-18} \mathrm{~ms}^{-1}$
B $9 \times 10^{-2} \mathrm{~ms}^{-1}$
C $3 \times 10^{-31} \mathrm{~ms}^{-1}$
D $2.7 \times 10^{-21} \mathrm{~ms}^{-1}$

The correct answer is:
$2.7 \times 10^{-18} \mathrm{~ms}^{-1}$

Solution:
Wavelength of a particle is given by
$\lambda=\frac{h}{p}$
where $h$ is Planck's constant.
and wavelength of an electron is given by
$\begin{aligned}
\lambda_e & =\frac{h}{p_e} \\
\lambda & =\lambda_e \\
p & =p_e \\
m v & =m_e y_e \\
y & =\frac{m_e v_e}{m}
\end{aligned}$
So, Putting the undergiven data, we get
$\begin{aligned}
& m_e=9.1 \times 10^{-31} \mathrm{~kg}, v_e=3 \times 10^6 \mathrm{~m} / \mathrm{s} \\
& m=1 \mathrm{mg}= 1 \times 10^{-6} \mathrm{~kg} \\
& y=\frac{9.1 \times 10^{-31} \times 3 \times 10^6}{1 \times 10^{-6}} \\
&=2.7 \times 10^{-18} \mathrm{~ms}^{-1}
\end{aligned}$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2008 (Screening)
A particle of mass 2Ã10-5 kg moves horizontally between the plates of a parallel plate capacitor which produce an electric field of 200 N C-1 in the vertically upward direction. A magnetic induction of 2.0 T is applied at right angles to the electric field in a direction normal to both Eâ and vâ. If g is 9.8 m s-2 and the charge on the particle is 10-6 C, then the velocity of the charged particle so that it continues to move horizontally is

Options:
A 2 m s-1
B 20 m s-1
C 0.2 m s-1
D 100 m s-1

The correct answer is:
2 m s-1

Solution:
Net force on the particle should be zero.

qE=10-6×200=2×10-4 N
mg=2×10-5×9.8=1.96×10-4 N
Since qE>mg, so magnetic force qvB should act downwards to balance the forces.
qE=mg+qvB⇒2×10-4=1.96×10-4+10-6v×2
⇒ v=2 m s-1

Topic: Physics, Magnetic Effects of Current, NEET
A particle of mass 2 g and charge 1 Î¼C is held at rest on a frictionless horizontal surface at a distance of 1 m from a fixed charge 1 Î¼C. If the particle is released, it will be repelled. The speed of the particle when it is at a distance of 10 m from the fixed charge is

Options:
A 100 m s-1
B 90 m s-1
C 60 m s-1
D 45 m s-1

The correct answer is:
90 m s-1

Solution:
From conservation of mechanical energy,
Ui+Ki=Uf+Kf
Kq1q2r1+0=Kq1q2r2+12mv2
9×109 ×1×10–31×10–611–110=122×10–3v2
9×910×22×10–3=v
v=90 m s-1

Topic: Physics, Electrostatics, NEET
A particle of mass 2 kg is initially at rest. A force starts acting on it in one direction whose magnitude changes with time. The force-time graph is shown in the figure. Find the velocity of the particle at the end of 10 s.

Options:
A 45  m s −1
B 50  m s −1
C 40  m s −1
D 60  m s −1

The correct answer is:
50  m s −1

Solution:
Using impulse = change in linear momentum (or area under F - t graph)
We have, m(vf â vi) = Area
or 2 v f - 0 = 1 2 Ã 2 Ã 1 0 + 2 Ã 1 0 + 1 2 Ã 2 Ã 1 0 + 2 0 + 1 2 Ã 4 Ã 2 0
= 1 0 + 2 0 + 3 0 + 4 0
or 2 v f = 1 0 0
or vf=50 m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
A particle of mass 3 kg, attached to a spring with force constant 48 N m-1 execute simple harmonic motion on a frictionless horizontal surface. The time period of oscillation of the particle, in seconds, is

Options:
A π4
B π2
C 2 π
D 8 π

The correct answer is:
π2

Solution:
The time period of mass,
T=2πmk
Given, m=3 kg
k=48 N m-1
T=2π348=2π116
=2π×14=π2

Topic: Physics, Oscillations, NEET
A particle of mass 5 g is moving with a speed of 32 cm s-1 in X-Y plane along the line y=x+4 . The magnitude of its angular momentum about the origin in g cm2 s-1 is

Options:
A Zero
B 60
C 30
D 302

The correct answer is:
60

Solution:
Given y=x+4
When x=0, y=4 cm
And y=0, x=-4cm
Perpendicular distance
OC=OAsinâ¡45o=4Ã12=22cm
Angular momentum
=mvOC=53222
=60 gm cm2 s-1

Topic: Physics, Rotational Motion, NEET
A particle of mass 5 units is moving with a uniform speed 3 2 units in the XOY plane along the line y = x+ 4. The magnitude of the angular momentum about the origin is

Options:
A Zero
B 60 units
C 7.5 units
D 402 units

The correct answer is:
60 units

Solution:
Angular momentum about the origin
L=|r×mv|
=(4cos⁡45°)×(5)×32
=42×5×32
L=60 units

Topic: Physics, Rotational Motion, NEET
A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8×10-4 J by the end of the second revolution after the beginning of the motion?

Options:
A 0.18 m s-2
B 0.2 m s-2
C 0.1 m s-2
D 0.15 m s-2

The correct answer is:
0.1 m s-2

Solution:
Tangential acceleration at=rα= constant =K
α=Kr
At the end of second revoluation angular velocity is ω then
ω2-ω02=2αθ
ω 2 =2( K r )( 4π )
ω2=8πKr
K.E. of the particle is =K.E.=12mv2
K.E.=12mr2ω2
K.E.=12m(r2)(8πKr)=12mr(8πK)
8× 10 −4 = 1 2 ×10× 10 −3 ×6.4× 10 −2 ×8×3.14×K
K=26.4×3.14=0.1 m s-2

Topic: Physics, Laws of Motion, NEET, NEET 2016 (Phase 1)
A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8Ã10-4 J by the end of the second revolution after the beginning of the motion?

Options:
A 0.18âmâ s -2
B 0.2 m  s -2
C 0.1âmâ s -2
D 0.15âmâ s -2

The correct answer is:
0.1 m  s -2

Solution:
Tangential acceleration at=rα= constant =K
α=Kr
At the end of second revoluation angular velocity is ω then
ω2-ω02=2αθ
ω 2 =2( K r )( 4π )
ω2=8πKr
K.E. of the particle is =K.E.=12mv2
K.E.=12mr2ω2
K.E.=12m(r2)(8πKr)=12mr(8πK)
8× 10 −4 = 1 2 ×10× 10 −3 ×6.4× 10 −2 ×8×3.14×K
K=26.4×3.14=0.1 m s-2

Topic: Physics, Rotational Motion, NEET
A particle of mass 100 gm moves in a potential well given by U=8x2-4x+400 joule. Find its acceleration at a distance of 25 cm from equilibrium in the positive direction

Options:
A 4 m s-1
B 40 m s-1
C -40 m s-1
D -4 m s-1

The correct answer is:
-40 m s-1

Solution:
We know that F=-dUdx
F=-ddx8x2-4x+400
F=-16x-4
F=4-16x
At equilibrium 4-16x0=0 and
x0=14 m=25 cm
At the given position
x=25+25=50 cm
⇒ F=4-16×0.5=-4 N
or a=-40.1=-40 m s-2

Topic: Physics, Work Power Energy, NEET
A particle of mass 100 g tied to a string is rotated along the circle of radius 0.5 m. The breaking tension of the string is 10 N. The maximum speed with which particle can be rotated without breaking the string is

Options:
A 10 m s-1
B 9.8 m s-1
C 7.7 m s-1
D 7.07 m s-1

The correct answer is:
7.07 m s-1

Solution:
F=mÏ2r=mv2r
v= f.r m = 10Ã0.5 100Ã1 0 -3 m s -1
=7.07âââm s -1

Topic: Physics, Laws of Motion, NEET
A particle of mass $m_1$ is moving with a velocity $v_1$ and another particle of mass $m_2$ is moving with a velocity $v_2$. Both of them have the same momentum but their different kinetic energies are $E_1$ and $E_2$ respectively. If $m_1>m_2$ then:

Options:
A $E_1 < E_2$
B $\frac{E_1}{E_2}=\frac{m_1}{m_1}$
C $E_1>E_2$
D $E_1=E_2$

The correct answer is:
$E_1 < E_2$

Solution:
We know that K.E. $=\frac{P^2}{2 m}$
$$
\begin{array}{ll}
\therefore & \frac{E_1}{E_2}=\frac{P_1^2 / 2 m_1}{P_2^2 / 2 m_2} \\
\Rightarrow \quad \frac{E_1}{E_2}=\frac{m_2}{m_1}
\end{array}
$$
Here $\quad m_1=m_2$
$$
\therefore \quad E_1 < E_2
$$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2004
A particle of mass m=1 kg is projected on a horizontal surface with speed u=202 m s-1 at an angle 45o with horizontal. Find the torque of the weight of the particle about the point of projection when the particle is the highest point. Take g=10 m s-2

Options:
A 400 Nâm
B 200Â Nâm
C 150Â Nâm
D 450Â Nâm

The correct answer is:
400 Nâm

Solution:
Ï=mgR2=mgu2sinâ¡2Î¸2g
=mu22 (as Î¸=45o)
=120222
=400 N m

Topic: Physics, Rotational Motion, NEET
A particle of mass m1 makes a head-on elastic collision with another particle of mass m2 at rest. m1 rebounds straight back with 49 of its initial kinetic energy. Then m1m2 is:

Options:
A 23
B 15
C 32
D 25

The correct answer is:
15

Solution:
Let the initial speed of m1 is v0, then from the question we know that

4912m1v02=12m1v12

v1=23v0

Since the collision is elastic, e=1.

⇒v0=v2+v1

where v1 is the speed of m1 after the collision and v2 is the speed of m2 after collision.

⇒v2=v03

m1v0=-m123v0+m2v03

m1m2=15

Topic: Physics, Center of Mass Momentum and Collision, NEET
A particle of mass m=5 units is moving with uniform speed V=32 units in the XY plane along the line Y=X+4. The magnitude of the angular momentum about origin is

Options:
A Zero
B 60 units
C 7.5 units
D 40 units

The correct answer is:
60 units

Solution:
Angular momentum = momentum Ã perpendicular distance.
L=mvbcosâ¡Î¸
=5Ã32Ã4Ãcosâ¡45o
L=60 kg m2s-1

Topic: Physics, Rotational Motion, NEET
A particle of mass M and charge Q is attached to a string of length L. It is whirled into a vertical circle in field E. The speed of the particle at point A, so that the tension in the string at A is ten times the weight of particle, is - (g =acceleration due to gravity)

Options:
A 5gL
B L9g+QEm
C 11gL
D v =L11g+QEm

The correct answer is:
L9g+QEm

Solution:
(T + QE) – mg = mv2L
10 mg + QE – mg = mv2L
9 mg +QEm=mv2L, v = L9g+QEm

Topic: Physics, Electrostatics, NEET
A particle of mass m and charge q is thrown from a point in space where uniform gravitational field & electric field are present. The particle

(1) may follow a straight line

(2) may follow a circular path

(3) may follow a parabolic path

Options:
A 1 & 2 are correct
B 1 & 3 are correct
C 2 & 3 are correct
D all these are correct

The correct answer is:
1 & 3 are correct

Solution:
When force due to the gravitation field is equal to force due to the electric field, the particle must follow a straight-line path with acceleration.
when force due to gravitation field is not equal to force due to the electric field, the particle will follow the path of a parabola,

Hence 1 and 3 both are correct

Topic: Physics, Electrostatics, NEET
A particle of mass m and charge q moving with constant velocity v along positive x direction enters a region containing a uniform magnetic fileld B directed along the negative z direction, extending from x=a to x=b. For the particle to enter the region x>b, the minimum value of v is

Options:
A qbBm
B qb-aBm
C qaBm
D qb+aB2m

The correct answer is:
qb-aBm

Solution:
R=mvqB>b-a

v=qB(b-a)m

Topic: Physics, Current Electricity, NEET
A particle of mass M and moving with velocity v describes a circular path of radius R when subjected to a uniform transverse magnetic field of induction B. the work done by the field when the particle completes one full circle is

Options:
A Mv 2 R
B Zero
C 2Mv 2 R
D Mv2R

The correct answer is:
Zero

Solution:
When particle describes circular path in a magnetic field, its velocity is always perpendicular to the magnetic force.
Power, P = F.V = FV cos θ
Here θ = 9 0
P = 0
But P = W t
⇒ W = P.t
Hence, work done
W = 0 (Everywhere)

Topic: Physics, Magnetic Effects of Current, NEET
A particle of mass M, at rest decays into two masses m 1 and m 2 with non-zero velocities. The ratio of de Broglie wavelengths of the particles Î» 1 Î» 2 is m n, then the minimum value of m+n is

Options:
A 3
B 4
C 2
D 5

The correct answer is:
2

Solution:
As there is no external force
Linear momentum will be conserved
P1=P2=P
Now, de-Broglie wavelength
Î»1=hP1 or Î»2=hP2
âÎ»1Î»2=P2P1=PP=1 = m n
âm+n=1+1=2

Topic: Physics, Dual Nature of Matter, NEET