datavorous/entrance-exam-dataset huggingface dataset:
A particle of mass 200 MeV c-2 collides with a hydrogen atom at rest. Soon after the collision, the particle comes to rest, and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in eV) is N4. The value of N is: (Given the mass of the hydrogen atom to be 1 GeV c-2).........

Options:

The correct answer is:
51

Solution:
Given, M=200 MeV c-2 and m=1 GeV c-2, now use the concept of conservation of linear momentum, Mv0=mv⇒v=Mv0m ...1.

Now use the conservation of energy,

12Mv02=12mv2+∆E ..2, here ∆E=13.6 eV, for the hydrogen atom in the ground state, now from equation, 1 & 2, 12Mv02=143∆E×108 eV=N4,

⇒N4=514 eV⇒N=51.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2020 (05 Sep Shift 1)
A particle of mass 250 g executes a simple harmonic motion under a periodic force F=(–25x) N. The particle attains a maximum speed of 4 m s-1 during its oscillation. The amplitude of the motion is ______cm.

Options:

The correct answer is:
40

Solution:
Given, force F=(–25x) N

⇒0.25a=-25x

⇒a=-100x

Now comparing it with standard equation, a=-ω2x we get

ω=10.

Now, vmax=ωA

⇒ωA=4

⇒A=410=0.4 m

⇒A=40 cm.

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (29 Jan Shift 2)
A particle of mass 500 g is moving in a straight line with velocity v=bx52. The work done by the net force during its displacement from x=0 to x=4 m is (Take b=0.25 m-32s-1).

Options:
A 2 J
B 4 J
C 8 J
D 16 J

The correct answer is:
16 J

Solution:
According to the work-energy theorem, work done by all forces will be is equal to the change in kinetic energy.

Wall=ΔK

=12mb2x25-x15 =12×0.5×116×45=16 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2022 (29 Jun Shift 1)
A particle of mass m1 is fastened to one end of a string and another particle of mass m2 is attached to the middle point, the other end of the string being fastened to a fixed point on a smooth horizontal table. If the particles are made to revolve in a horizontal circular path as shown in the figure, then the ratio of tension in the part of the string between the centre and m2 to the tension in the part of the string between m2 and m1 is

Options:
A m1m1+m2
B m1+m2m1
C 2m1+m22m1
D 2m1m1+m2

The correct answer is:
2m1+m22m1

Solution:
for m1,

T1=m1ω22r ......(i)
for m2,

T2-T1=m2ω2r ....(ii)

Dividing both the equations:

T1T2-T1=2m1m2

⇒T2T1=2m1+m22m1

Topic: Physics, Laws of Motion, JEE Main
A particle of mass m = 1 kg is dropped from a height h = 40 cm on a light horizontal platform fixed to one end of an elastic spring, the other being fixed to a base, as shown in the diagram. The particle collides with the platform and sticks to it. As a result, the spring is compressed by an amount x = 10 cm. What is the force constant of the spring?
(Take g = 10 m s-2)

Options:
A 600 N m-1

B 800 N m-1

C 1000 N m-1

D 1200 N m-1

The correct answer is:
1000 N m-1

Solution:
Since the platform is depressed by an amount x, the total work done on the spring is mg(h + x). This work is stored in the spring in the form of potential energy 12kx2.
Equating the two, we have
12kx2=mgh+x or k=2 mgh+xx2
Given, h = 0.4 m, x = 0.1 m, m = 1 kg and g = 10 m s-2. Substituting these values, we get
k = 1000 N m-1.

Topic: Physics, Work Power Energy, JEE Main
A particle of mass $m$ and charge $(-q)$ enters the region between the two charged plates initially moving along $x-$ axis with speed $v_x$. The length of plate is $L$ and an uniform electric field $E$ is maintained between the plates. Show that the vertical deflction of the particle at the far edge of the plate is $q E L^2 /\left(2 m v_x^2\right)$.

Options:

None

Solution:
Let the point at which the charged particle enters the electric field, be origin $O(0,0)$, then after travelling a horizontal displacement $L$, it gets deflected by displacement $y$ in vertical direction as it comes out of electric field. So, co-ordinates of its initial position are $x_1=0$ and $y_1=0$ and of final position on coming out of electric field are
Components of its acceleration are $a_x=0$ and $a_y=\frac{F}{m}=\frac{q E}{m}$ and of initial velocity are
$$
u_{\mathrm{x}}=v_{\mathrm{x}} \text { and } u_{\mathrm{y}}=0
$$
So, by $2^{\text {nd }}$ equation of motion in horizontal direction, $y_2-y_1=u_x t+\frac{1}{2} a_x t^2$ or $L-0=u_x t+0$, $\Rightarrow t=\frac{L}{v_x}$ and by $2^{\text {nd }}$ equation of motion in vertical direction, $y_2-y_1=u_y t+\frac{1}{2} a_y t^2$
or $y-0=0+\frac{1}{2}, \frac{q E}{m} \cdot\left(\frac{L}{v_x}\right)^2$
or $y=\frac{q E L^2}{2 m v_x^2}$
This gives the vertical deflection of the particle at the far edge of the plate.

Topic: Physics, Electrostatics
A particle of mass m and charge q has an initial velocity v→=v0j^ . If an electric field E→=E0i^ and magnetic field B→=B0i^ act on the particle, its speed will double after a time

Options:
A 2mv0qE0
B 3mv0qE0
C 3mv0qE0
D 2mv0qE0

The correct answer is:
3mv0qE0

Solution:
Acceleration produced by electric field,
a=qE0mi^

After time t, velocity of the particle,
v=u+at

⇒v=v0j^+qE0tmi^ ∵u=v=v0j^

So, speed of the particle,
v=v02+qE0tm2

Now, given v=2v0after time t, so
2v0=v02+qE0tm2

⇒ 4v02=v02+qE0tm2

⇒ 3v02=qE0tm2

So, time interval, t=3mv0qE0

Here, we must note that no change in magnitude of velocity is caused by a perpendicular magnetic field. So, we are not taking effect of magnetic field while calculating change in speed.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (07 Jan Shift 2)
A particle of mass $M$ and charge $q$ is at rest at the mid point between two other fixed similar charges each of magnitude $Q$ placed a distance $2 d$ apart. The system is collinear as shown in the figure. The particle is now displaced by a small amount $x(x< < d)$ along the joining the two charges and is left to itself. It will now oscillate about the mean position with a time period $\left(\varepsilon_{0}=\right.$ permittivity of free space)

Options:
A $2 \sqrt{\frac{\pi^{3} M \varepsilon_{0} d}{Q q}}$
B $2 \sqrt{\frac{\pi^{2} M \varepsilon_{0} d^{3}}{Q q}}$
C $2 \sqrt{\frac{\pi^{3} M \varepsilon_{0} d^{3}}{Q q}}$
D $2 \sqrt{\frac{\pi^{3} M \varepsilon_{0}}{Q q d^{3}}}$

The correct answer is:
$2 \sqrt{\frac{\pi^{3} M \varepsilon_{0} d^{3}}{Q q}}$

Solution:
Restoring force on displacement of $x$
$\begin{aligned} F &=K\left[\frac{q}{(d-x)^{2}}-\frac{Q q}{(d+x)^{2}}\right] \\ &=K Q q\left[\frac{1}{(d-x)^{2}}-\frac{1}{(d+x)^{2}}\right] \\ &=K Q q\left[\frac{4 d x}{\left(d^{2}-x^{2}\right)^{2}}\right] \\ &=K Q q\left[\frac{4 d x}{d^{4}}\right] \text { If }(d>>x) \\ F &=K Q q\left[\frac{4 x}{d^{3}}\right] \end{aligned}$
Acceleration, $a=\frac{F}{m}=\frac{4 K Q q x}{M d^{3}}$
or $\omega^{2}=\frac{4 K Q q}{M d^{3}}$
$\therefore \quad T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{M d^{3}}{4 K O q}}$
$=2 \sqrt{\frac{\pi^{3} M d^{3} \varepsilon_{0}}{Q q}}$

Topic: Physics, Oscillations, JEE Main
A particle of mass $m$ and charge $q$ is in an electric and magnetic field given by

$$

\overrightarrow{\mathrm{E}}=2 \hat{i}+3 \hat{j} ; \overrightarrow{\mathrm{B}}=4 \hat{j}+6 \hat{k}

$$

The charged particle is shifted from the origin to the point $\mathrm{P}(x=1 ; y=1)$ along a straight path. The magnitude of the total work done is :

Options:
A $(0.35) \mathrm{q}$
B $5 \mathrm{q}$
C (2.5) $\mathrm{q}$
D $(0.15) \mathrm{q}$

The correct answer is:
$5 \mathrm{q}$

Solution:
Resultant force on the charged particle

$$

\begin{array}{l}

=\overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}}) \\

=(2 \mathrm{q} \hat{\mathrm{i}}+3 \mathrm{q} \hat{\mathrm{j}})+\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})

\end{array}

$$

Also $\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})=0$

So, required work done

$$

\begin{aligned}

\mathrm{W} &=\overrightarrow{\mathrm{F}}_{\mathrm{net}} \cdot \overrightarrow{\mathrm{S}}=(2 \mathrm{q} \hat{\mathrm{i}}+3 \mathrm{q} \hat{\mathrm{j}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}) \\

&=2 \mathrm{q}+3 \mathrm{q} \\

&=5 \mathrm{q}

\end{aligned}

$$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A particle of mass $M$ and charge $q$ is released from rest in a region of uniform electric field magnitude $E$. After a time $t$, the distance travelled by the charge is $S$ and the kinetic energy attained by the particle is $T$. Then, the ratio $T / S$

Options:
A remains constant with time t
B varies linearly with the mass $M$ of the particle
C is independent of the charge $q$
D is independent of the magnitude of the electric field E

The correct answer is:
remains constant with time t

Solution:
Given, mass of the particle $=M$.
Charge on the particle $=q$
Electric field $=E$
Initial velocity, $u=0$
$\therefore$ Acceleration, $a=\frac{F}{M}=\frac{q E}{M}$
:.Distance travelled in electric field, $S=u t+\frac{1}{2} a t^{2}$
$$
S=\frac{1}{2}\left(\frac{q E}{M}\right) t^{2}
$$
Also, kinetic energy $T=\frac{1}{2} M\left(\frac{q E t}{M}\right)^{2}$
So, $\quad \frac{T}{S}=\frac{\frac{1}{2} M\left(\frac{q E t}{M}\right)^{2}}{\frac{1}{2}\left(\frac{q E}{M}\right) t^{2}}=q E$
$\Rightarrow$ Ratio of $\frac{T}{S}$ remains constant with time t.

Topic: Physics, Work Power Energy, JEE Main
A particle of mass $M$ and charge $q$ is released from rest in a region of uniform electric field magnitude $E$. After a time $t$, the distance travelled by the charge is $S$ and the kinetic energy attained by the particle is $T$. Then, the ratio $T / S$

Options:
A remains constant with time t
B varies linearly with the mass $M$ of the particle
C is independent of the charge $q$
D is independent of the magnitude of the electric field E

The correct answer is:
remains constant with time t

Solution:
Given, mass of the particle $=M$.
Charge on the particle $=q$
Electric field $=E$
Initial velocity, $u=0$
$\therefore$ Acceleration, $a=\frac{F}{M}=\frac{q E}{M}$
:.Distance travelled in electric field, $S=u t+\frac{1}{2} a t^{2}$
$$
S=\frac{1}{2}\left(\frac{q E}{M}\right) t^{2}
$$
Also, kinetic energy $T=\frac{1}{2} M\left(\frac{q E t}{M}\right)^{2}$
So, $\quad \frac{T}{S}=\frac{\frac{1}{2} M\left(\frac{q E t}{M}\right)^{2}}{\frac{1}{2}\left(\frac{q E}{M}\right) t^{2}}=q E$
$\Rightarrow$ Ratio of $\frac{T}{S}$ remains constant with time t.

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2013
A particle of mass m and charge q is released from rest in a uniform electric field. If there is no other force on the particle, the dependence of its speed v on the distance x travelled by it is correctly given by (graphs are schematic and not drawn to scale)

Options:
A
B
C
D

The correct answer is:

Solution:
The particle is at rest initially.

So, the initial velocity u=0 m s-1.

The force on the charged particle in the uniform electric field,

F=qE

⇒ ma=qE

⇒ a=qEm

Applying the velocity-displacement equation under constant acceleration,

v2=u2+2ax

⇒ v2=02+2qEmx

⇒ v2=2qEmx

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2020 (08 Jan Shift 2)
A particle of mass $m$ and charge $Q$ moving with a speed $v$ in a circular path of radius $R$ has magnetic moment $\mu$. If the mass of the particle is doubled and maintains the same speed while revolving in the same circular path, then the magnetic moment will be

Options:
A doubled
B halved
C tripled
D unchanged

The correct answer is:
unchanged

Solution:
Magnetic moment associated with moving charge on circular path is given as,
$\mu=\frac{Q v R}{2} \Rightarrow \mu \propto Q v R$
Clearly $\mu$ doesn't depend on mass of charged particle and also if $v=$ unchanged then $\mu=$ uncharged.

Topic: Physics, Magnetic Effects of Current, TS EAMCET, TS EAMCET 2020 (14 Sep Shift 1)
A particle of mass $M$ and charge Q moving with velocity $\overrightarrow{\mathrm{v}}$ describe a circular path of radius $\mathrm{R}$ when subjected to a uniform transverse magnetic field of induction $\mathrm{B}$. The work done by the field when the particle completes one full circle is

Options:
A
$\left(\frac{\mathrm{Mv}^2}{\mathrm{R}}\right) 2 \pi \mathrm{R}$

B
Zero

C
$B Q 2 \pi R$

D
$B Q$ v $2 \pi R$

The correct answer is:
Zero

Solution:
Force is $\perp^{\mathrm{r}}$ to displacement $\Rightarrow$ the work done is zero

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2003
A particle of mass $\mathrm{m}$, charge $\mathrm{Q}$ and kinetic energy $\mathrm{T}$ enters a transverse uniform magnetic field of induction $\vec{B}$. After 3 seconds the kinetic energy of the particle will be

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

The correct answer is:
$\mathrm{T}$

Solution:
Work done by magnetic field on charge particle is zero. Therefore its kinetic energy will be same i.e. $\mathrm{T}$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2008 (Mains)
A particle of mass $m$ executes simple harmonic motion with amplitude ' $a$ ' and frequency ' $v$ '. The average kinetic energy during its motion from the position of equilibrium to the end is

Options:
A
$\pi^2 \mathrm{ma}^2 v^2$

B
$\frac{1}{4} \pi^2 m a^2 v^2$

C
$4 \pi^2 \mathrm{ma}^2 \mathrm{v}^2$

D
$2 \pi^2 m a^2 v^2$

The correct answer is:
$\pi^2 \mathrm{ma}^2 v^2$

Solution:
$\frac{1}{4} m a^2 \omega^2=\pi^2 f^2 m a^2$

Topic: Physics, Oscillations, JEE Main, JEE Main 2007
A particle of mass ' $m$ ' having speed $v$ goes in a vertical circular motion such that its centre is at its origin, as shown in the figure. If at any instant the angle made by the string with a negative $y$-axis is $\theta$ then the tension in the string is:
[Take radius $=\mathrm{R}$ ]

Options:
A $m g \sin \theta+\frac{m v^2}{R}$
B $m g \cos \theta-\frac{m v^2}{R}$
C $m g \cos \theta+\frac{m v^2}{R}$
D $m g \sin \theta-\frac{m v^2}{R}$

The correct answer is:
$m g \cos \theta+\frac{m v^2}{R}$

Solution:
To find the tension in the string, we need to analyze the forces acting on the particle in the vertical circular motion.
Given:
- Mass of the particle $=m$
- Speed of the particle $=v$
- Radius of the circular path $=R$
- Angle made by the string with the negative $y$-axis $=\theta$
We need to consider two forces acting on the particle:
1. The gravitational force $(m g)$ acting downwards.
2. The tension in the string $(T)$.
At any angle $\theta$, the forces in the radial direction need to provide the centripetal force for the circular motion. The components of the forces in the radial direction are:
- The component of the gravitational force along the radius: $m g \cos \theta$
- The tension in the string: $T$
The net force providing the centripetal acceleration is:
$T-m g \cos \theta=\frac{m v^2}{R}$
Solving for $T$ :
$T=\frac{m v^2}{R}+m g \cos \theta$
Thus, the correct answer is:
$m g \cos \theta+\frac{m v^2}{R}$
Which corresponds to option 3 in the given choices.

Topic: Physics, Laws of Motion, JEE Main
A particle of mass m, initial speed u and angle of projection Î¸, is projected as shown in the figure. Average torque on the projectile between initial and final positions P and Q about the point of projection is

Options:
A mu2sin2θ2
B mu2cosθ
C mu2sinθ
D mu2cosθ2

The correct answer is:
mu2sin2θ2

Solution:
τ→αv×Δt=ΔL→ ...(1)
Here, Δt = time of flight = 2usinθg
Δ→L=L→f-L→i about the point of projection.
=musinθrange-0
=musinθu2sin2θg
∴ τ→av=ΔLΔt=mu3sinθsin2θgg2usinθ
=mu2sin2θ2

Topic: Physics, Rotational Motion, JEE Main
A particle of mass m is acted upon by a force F given by the empirical law F=Rt2 vt. If this law is to be tested experimentally by observing the motion starting from rest, the best way is to plot

Options:
A logâ¡vt against1t
B vt against t2
C log⁡vt against1t2
D log⁡vt against t

The correct answer is:
logâ¡vt against1t

Solution:
F=Rt2vt⇒mdvdt=Rt2vt
∫dvv=∫Rdtmt2
ln v=-Rmt
∴ ln v∝1t

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2016 (10 Apr Online)
A particle of mass m is allowed to oscillate near the point of minima of a vertical parabolic path having the equation x2=4ay, where x-axis is along the horizontal direction and y-axis is along the vertical direction. The angular frequency of small oscillations of the particle is

Options:
A 8ga
B 2ga
C ga
D g2a

The correct answer is:
g2a

Solution:
ma = - mg sin Î¸
a = - g sin Î¸ or a = - g tan Î¸ ...(i)
(as Î¸ is small)
Now, x 2 = 4 ay
â´ dy dx = x 2 a
â´ a = - g x 2 a
- Ï 2 x = - gx 2 a
Ï = g 2 a

Topic: Physics, Oscillations, JEE Main
A particle of mass $\mathrm{m}$ is at rest at the origin at time $\mathrm{t}=0$. It is subjected to a force $\mathrm{F}(\mathrm{t})=\mathrm{F}_0 \mathrm{e}^{-\mathrm{bt}}$ in the $x$ direction. Its speed $v(t)$ is depicted by which of the following curves?

Options:
A

B

C

D

The correct answer is:

Solution:
$\mathrm{F}=\mathrm{F}_0 \mathrm{e}^{-\mathrm{bt}}$
$\Rightarrow \quad \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{F}_0}{\mathrm{~m}} \mathrm{e}^{-\mathrm{bt}}$
$\Rightarrow \quad \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{F}_0}{\mathrm{~m}} \mathrm{e}^{-\mathrm{bt}}$
$\quad\int \mathrm{dv}=\int_0^{\mathrm{t}} \frac{\mathrm{F}}{\mathrm{m}} \mathrm{e}^{-\mathrm{bt}} \mathrm{dt}$
$\Rightarrow v=\frac{F}{m}\left[\frac{-1}{b}\right]\left[e^{-b t}\right]_0^t$
$\Rightarrow \quad v=\frac{F}{m b}\left[e^{-b t}\right]$
$v=0 \text { at } t=0$
and $\quad v \rightarrow \frac{F}{m b}$ as $t \rightarrow \infty$
So, velocity increases continuously and attains a maximum value of $v=\frac{F}{m b}$ as $t \rightarrow \infty$.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2012 (Offline)
A particle of mass $m$ is attached to a spring (of spring constant $k$ ) and has a natural angular frequency $\omega_0$. An external force $F(t)$ proportional to $\cos \omega t\left(\omega \neq \omega_0\right)$ is applied to the oscillator. The time displacement of the oscillator will be proportional to

Options:
A
$\frac{\mathrm{m}}{\omega_0^2-\omega^2}$

B
$\frac{1}{m\left(\omega_0^2-\omega^2\right)}$

C
$\frac{1}{m\left(\omega_0^2+\omega^2\right)}$

D
$\frac{m}{\omega_0^2+\omega^2}$

The correct answer is:
$\frac{1}{m\left(\omega_0^2-\omega^2\right)}$

Solution:
For forced oscillations, the displacement is given by
$$
\mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}+\phi) \text { with } \mathrm{A}=\frac{\mathrm{F}_0 / \mathrm{m}}{\omega_0^2-\omega^2}
$$

Topic: Physics, Oscillations, JEE Main, JEE Main 2004
A particle of mass m is attached to four springs with spring constant k,k,2k and 2k as shown in the figure. Four springs are attached to the four corners of a square and a particle is placed at the center. If the particle is pushed slightly towards any side of the square and released, the period of oscillation will be

Options:
A 2πm3k
B 2πm32k
C 2πm6k
D 2πm2k

The correct answer is:
2πm32k

Solution:
when the mass is displaced in the upward direction then the forces acting ar as shown in the diagram,

F1=-kx=F2 & F3=-2kx=F4

Referring to second diagram,

FR=F1+F32+F2+F42=32kx

But since the particle performing SHM,

∴ FR=mω2x

∴ mω2x=32kx

∴ ω=32km

2πT=32km⇒T=2πm32k

Topic: Physics, Oscillations, JEE Main
A particle of mass m is attached to three identical springs A, B and C each of force constant k a shown in figure. If the particle of mass m is pushed slightly against the spring A and released then the time period of oscillations is

Options:
A 2π2mk
B 2πm2k
C 2πmk
D 2πm3k

The correct answer is:
2πm2k

Solution:
When the particle of mass m at O is pushed by y in the direction of A, the spring A will be compressed by y while spring B and C will be stretched by yâ²=ycos45Â°. So that the total restoring force on the mass m along OA.
Fnet=FA+FBcos45Â°+FCcos45Â°=ky+2kyâ²cos45Â°=ky+2kycos45Â°cos45Â°=2ky
Also Fnet=kâ²y

âkâ²y=2ky

âkâ²=2k

T=2Ïmkâ²=2Ïm2k

Topic: Physics, Oscillations, JEE Main
A particle of mass m is driven by a machine that delivers a constant power k watts. If the particle starts from rest the force on the particle at time t is:

Options:
A mk 2   t − 1 2
B mk t-12
C 2mk t-12
D 12mk t-12

The correct answer is:
mk 2   t − 1 2

Solution:
Fv=constant=k
mdvdtv=k
â«vdv=â«kmdt
v22=kmt
v=2kmt
F=mdvdt
=m 2km12t-12
=mk2 t-12

Topic: Physics, Work Power Energy, JEE Main
A particle of mass m is dropped from a height h above the ground. At the same time another particle of the same mass is thrown vertically upwards from the ground with a speed of 2gh. If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of hg is:

Options:
A 12
B 34
C 12
D 32

The correct answer is:
32

Solution:
time for collision t1=h2gh
After t1
VA=0-gt1=-gh2
and VB=2gh-gt1=gh2-12
at the time of collision
p→i=pf→
⇒mV→A+mV→B=2mV→f
⇒-gh2+gh2-12=2V→f
Vf=0

and height from ground

=h-12gt12=h-h4=3h4

so time ,

t=2×3h4g=32 hg

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2020 (08 Jan Shift 2)
A particle of mass m is dropped from a height h above the ground. Simultaneously another particle of the same mass is thrown vertically upwards from the ground with a speed of 2gh. If they collide head-on completely inelastically, then the time taken for the combined mass to reach the ground is

Options:
A 3h4g
B 3h2g
C h2g
D h4g

The correct answer is:
3h2g

Solution:
The time for collision

t1=h2gh

After t1,

vA=0-gt1=-gh/2

VB=2gh-gt1=gh2-12

At the time of collision

P→i=P→f⇒mV→A+mV→B=2mV→f⇒-gh2+gh2-12=2V→fVf=0

and height from ground

=h-12gt12=h-h4=3h4

so time =2×3h4g=3h2g

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length l. The other end is fixed. The system is given an angular speed ω about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is:

Options:
A mlω2k-ωm
B mlÏ2k-mÏ2
C mlÏ2k+mÏ2
D mlÏ2k+mÏ

The correct answer is:
mlω2k-mω2

Solution:
As we know, the spring force will give the necessary centripetal force for rotation.

So,

mω2l+x=kx
lx+1=kmω2

Thus, the stretch in the spring
x=lmω2k-mω2

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2020 (08 Jan Shift 1)
A particle of mass ' $\mathrm{m}$ ' is kept at rest a height $3 \mathrm{R}$ from the surface of earth, where ' $R$ ' is radius of earth and ' $M$ ' is the mass of earth. The minimum speed with which it should be projected upward, so that it does not return back is
( $\mathrm{g}=$ acceleration due to gravity on the earth's surface)

Options:
A $\left[\frac{\mathrm{GM}}{2 \mathrm{R}}\right]^{1 / 2}$
B $\left[\frac{\mathrm{gR}}{4}\right]^{1 / 2}$
C $\left[\frac{2 g}{R}\right]^{1 / 2}$
D $\left[\frac{\mathrm{GM}}{\mathrm{R}}\right]^{1 / 2}$

The correct answer is:
$\left[\frac{\mathrm{GM}}{2 \mathrm{R}}\right]^{1 / 2}$

Solution:
$$
\mathrm{K}_{\mathrm{i}}+\mathrm{U}_{\mathrm{i}}=\mathrm{K}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}}
$$
$$
\frac{1}{2} \mathrm{mv}^2-\frac{\mathrm{GMm}}{4 \mathrm{R}}=0
$$
$$
\therefore \mathrm{v}=\left[\frac{\mathrm{GM}}{2 \mathrm{R}}\right]^{1 / 2}
$$

Topic: Physics, Gravitation, JEE Main
A particle of mass m is moving along a trajectory given by
x=x0+a cosω1t
y=y0+b sinω2t
The torque, acting on the particle about the origin, at t=0 is:

Options:
A +my0aω12k^
B -mx0bω22-y0aω12k^
C Zero
D m-x0b+y0aω12k^

The correct answer is:
+my0aω12k^

Solution:
x=x0+a cosω1t
y=y0+b sin ω2t
ax=-a ω12cos ω1t
ay=-b ω22sin ω2t
At t=0 ,
F→=ma→=-a ω12 i^
And
r→=x0+ a i^+y0 j^
Torque, τ=r→×F→
τ=x0+ a i^+y0 j^×-a ω12 i^
=my0 aω12 k^

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2019 (10 Apr Shift 1)
A particle of mass m is moving along the side of a square of side 'a', with a uniform speed υ in the x-y plane as shown in the figure:
Which of the following statements is false for the angular momentum L→ about the origin?

Options:
A Lâ=mÏR2+ak^
when the particle is moving from B to C.
B Lâ=mÏ2Rk^
when the particle is moving from D to A.
C Lâ=-mÏ2Rk^
when the particle is moving from A to B.
D L→=mυR2+ak^
when the particle is moving from C to D.

The correct answer is:
L→=mυ2Rk^
when the particle is moving from D to A.

Solution:
L→=r→×P→ or L→=rpsin⁡θn^
For D to A, L→=R2mv-k^
For A to B, L→=R2mv-k^
For C to D, L→=R2+amvk^
For B to C, L→=R2+amvk^

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2016 (03 Apr)