datavorous/entrance-exam-dataset huggingface dataset:
A projectile of mass $M$ is fired so that the horizontal range is $4 \mathrm{~km}$. At the highest point the projectile explodes in two parts of masses $M / 4$ and $3 M / 4$ respectively and the heavier part starts falling down vertically with zero initial speed. The horizontal range (distance from point of firing) of the lighter part is :

Options:
A
$16 \mathrm{~km}$

B
$1 \mathrm{~km}$

C
$10 \mathrm{~km}$

D
$2 \mathrm{~km}$

The correct answer is:
$10 \mathrm{~km}$

Solution:
No Solution Available

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2013 (23 Apr Online)
A projectile of mass m is fired with velocity a v from a point P making an angle of 45Â° with the horizontal. Neglecting air resistance, the magnitude of the change in momentum between the points P and arriving at Q is

Options:
A zero
B 12mv
C mv2
D 2 mv

The correct answer is:
mv2

Solution:
Since there is no resistance, the trajectory of motion will be parabolic, and as a result, the angle of projection will be equal to the angle of striking the ground.

Also, we know that the horizontal component of velocity remains unaffected throughout the motion. So it will not contribute to the change in momentum. The components contributing to the change in momentum will be the vertical components.
Initial vertical momentum component is mvsin45Â° and the final vertical momentum component is -mvsin45Â°.
Change in momentum,âp=mvsinâ¡45Â°--mvsinâ¡45Â°=2mvsinâ¡45Â°= 2 mv12=2mv.

Topic: Physics, Motion In Two Dimensions, NEET
A projectile thrown with a speed $v$ at an angle $\theta$ has a range $R$ on the surface of earth. For same $v$ and $\theta$, its range on the surface of moon will be

Options:
A R/6
B 6R
C R/36
D 36R

The correct answer is:
6R

Solution:
Range is given by $R=\frac{u^2 \sin 2 \theta}{g}$
On moon. $g_m=\frac{g}{6}$. Hence $R_m=6 R$

Topic: Physics, Motion In Two Dimensions, JEE Main
A protein 'A' contains 0.30% of glycine (molecular weight 75). The minimum molar mass of the protein 'A' is____×103 g mol-1 [nearest integer]

Options:

The correct answer is:
25

Solution:
% of glycine =Number of molecules of glycine×molecular weight of glycinMolecular mass of protein×100=75×1GMM×100=0.3

For minimum molecular mass of protein we can take number of molecules of glycine as one. since it can not have less than one molecule.

GMM=75×1000.3=25×103

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (25 Jun Shift 2)
A protein ‘X’ with molecular weight of 70,000 u, on hydrolysis gives amino acids. One of these amino acid is

Options:
A
B
C
D

The correct answer is:

Solution:
Amino acids contain amino (-NH2) and carboxyl (-COOH) functional groups. Depending upon the relative position of the amino group with respect to the carboxyl group, the amino acids can be classified as α, β, γ, δ and so on. Only α-amino acids are obtained on the hydrolysis of proteins.

So, only in option (B) α-amino acid given all the other options are not α-amino acids.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2023 (31 Jan Shift 1)
A proton, a deuteron and an α particle are moving with same momentum in a uniform magnetic field. The ratio of magnetic forces acting on them ______ is and their speed is ______ in the ratio.

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

The correct answer is:
2:1:1 and 4:2:1

Solution:
F=qv→×B→=qmP→×B→

⇒F∝qm

thus,
F1:F2:F3=q1 m1:q2 m2:q3 m3=emp:e2mp:2e4mp=11:12:24=2:1:1

Now for speed calculation, P=constant⇒v∝1 m

thus v1:v2:v3=1 mp:12mp:14mp=11:12:14=4:2:1.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (25 Feb Shift 1)
A proton, a deuteron and an α-particle with same kinetic energy enter into a uniform magnetic field at right angle to magnetic field. The ratio of the radii of their respective circular paths is :

Options:
A 1 : 2 : 2
B 1:2:1
C 2 :1 :1
D None of these

The correct answer is:
1:2:1

Solution:
Radius of circular path in magnetic field is given by

R=mvqB=m2KEqBm∝mq

RP:RD:Rα=11:21:42=1:2:1

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (24 Jun Shift 2)
A proton, a deuteron (nucleus of ${ }_1 \mathrm{H}^2$ ) and an $\alpha$-particle with same kinetic energy enter a region of uniform magnetic field moving at right angles to the field. The ratio of the radii of their circular paths is :

Options:
A $1: 2: 4$
B $1: \sqrt{2}: 1$
C $2: \sqrt{2}: 1$
D $1: 1: 2$

The correct answer is:
$1: \sqrt{2}: 1$

Solution:
$r=\frac{m v}{q B}=\frac{\sqrt{2 m K}}{q B}$
$\Rightarrow \quad r \propto \frac{(m)^{1 / 2}}{q}$
$\therefore \quad r_p=\frac{\left(m_p\right)^{1 / 2}}{q_p}, r_d=\frac{\left(m_d\right)^{1 / 2}}{q_d}$
and $\quad r_\alpha=\frac{\left(m_\alpha\right)^{1 / 2}}{q_\alpha}$
$\therefore \quad r_p: r_d: r_\alpha=\frac{\sqrt{m}}{q}: \frac{\sqrt{2 m}}{q}: \frac{\sqrt{4 m}}{2 q}$
$=1: \sqrt{2}: 1$

Topic: Physics, Magnetic Effects of Current, JEE Main
A proton, a deutron and an $\alpha$-particle having the same momentum, enters a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is

Options:
A $1: 2: 8$
B $1: 2: 4$
C $1: 1: 2$
D None of these

The correct answer is:
$1: 2: 8$

Solution:
The deflection suffered by charged particle in an electric field is
$y=\frac{q E L D}{m \mu^2}=\frac{q E L D}{p^2 / m} \quad(p=m u)$
$\Rightarrow \quad y \propto \frac{q m}{p^2}$
$\Rightarrow y_p: y_d: y_\alpha=\frac{q_p m_p}{p_p^2}: \frac{q_d m_d}{p_d^2}: \frac{q_\alpha m_\alpha}{p_a^2}$
Since, $\quad p_\alpha=p_d=p_p \quad$ (given)
$m_p: m_d: m_\alpha=1: 2: 4$
and $\quad q_1: q_d: q_\alpha=1: 1: 2$
$\Rightarrow \quad y_b: y_d: y_\alpha=1 \times 1: 1 \times 2: 2 \times 4$
$=1: 2: 8$

Topic: Physics, Electrostatics, JEE Main
A proton accelerated by a potential difference V, moves through a transverse magnetic field B as shown in the figure. Then the angle Î¸ through which the proton deviates from the initial direction of its motion is

Options:
A θ=sin-1dBqV2m
B θ=sin-1dBq2mV
C θ=tan-1dBqV2m
D θ=tan-1dBq2mV

The correct answer is:
θ=sin-1dBq2mV

Solution:
Bqv=mv2r
12mv2=Vq
And sin⁡θ=dr
r=mvqB=mqB 2Vqm
sin⁡θ=dBq2mV

θ=sin-1dBq2mV

Topic: Physics, Magnetic Effects of Current, JEE Main
A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicular to the plane. Let rp,re and rHe be their respective radii, then,

Options:
A re>rp=rHe
B re<rp=rHe
C re<rp<rHe
D re>rp>rHe

The correct answer is:
re<rp=rHe

Solution:
We know, r=mvqB∵ K.E.=12mv2
=2mK.E.qB
∴ K.E. is same for all particle and magnetic field B is also some.
r∝mq
For proton: rp∝mpe
For electron: re∝mee
For Helium nuclei: rHe∝4mp2e=mpe

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (10 Apr Shift 1)
A proton, an electron and a Helium nucleus have the same kinetic energy. They are in circular orbits in a plane due to a magnetic field which is perpendicular to the plane. Let rp, re and rHe be their respective radii, then

Options:
A r e > r p = r He
B r e > r p > r He
C r e < r p < r He
D r e < r p = r He

The correct answer is:
r e < r p = r He

Solution:
Radius of circle is given by r=mvqB=pqB=2mKqB=2mqBK (where K is kinetic energy)

For proton, rp=2mpeBK

For electron, re=2meeBK

For He2+, rHe=2×4mp2eBK=2mpeBK

Clearly, rHe=rp>re

Topic: Physics, Magnetic Effects of Current, JEE Main
A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as :

Options:
A $\lambda_\alpha < \lambda_{\mathrm{p}} < \lambda_{\mathrm{e}}$
B $\lambda_{\mathrm{e}}>\lambda_\alpha>\lambda_{\mathrm{p}}$
C $\lambda_{\mathrm{p}}>\lambda_{\mathrm{e}}>\lambda_\alpha$
D $\lambda_{\mathrm{p}} < \lambda_{\mathrm{e}} < \lambda_\alpha$

The correct answer is:
$\lambda_\alpha < \lambda_{\mathrm{p}} < \lambda_{\mathrm{e}}$

Solution:
$\begin{aligned} & \lambda_{\mathrm{DB}}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}} \\ & \Rightarrow \lambda_{\mathrm{DB}} \alpha \frac{1}{\sqrt{\mathrm{m}}} \\ & \Rightarrow \lambda_{\mathrm{a}} < \lambda_{\mathrm{p}} < \lambda_{\mathrm{e}}\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2024 (09 Apr Shift 1)
A proton and a deuteron are both accelerated through the same potential difference and enter in a magnetic field perpendicular to the direction of the field. If the deuteron follows a path of radius $R$, assuming the neutron and proton masses are nearly equal, the radius of the proton's path will be

Options:
A
$\sqrt{2} R$

B
$\frac{R}{\sqrt{2}}$

C
$\frac{R}{2}$

D
$R$

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

Solution:
As charge on both proton and deuteron is same i.e. ' $e$ '
Energy acquired by both, $E=\mathrm{eV}$ For Deuteron.
Kinetic energy, $\frac{1}{2} m V^2=e V$
[ $V$ is the potential difference]
$$
v=\sqrt{\frac{2 e V}{m_d}}
$$
But $m_d=2 m$
Therefore, $v=\sqrt{\frac{2 e V}{2 m}}=\sqrt{\frac{e V}{m}}$
Radius of path, $R=\frac{m v}{e B}$
Substituting value of ' $v$ ' we get
$$
\begin{array}{r}
R=\frac{2 m \sqrt{\frac{e v}{m}}}{e B} \\
\frac{R}{2}=\frac{m \sqrt{\frac{e v}{m}}}{e B}
\end{array}
$$
For proton :
$$
\begin{aligned}
& \frac{1}{2} m V^2=e V \\
& V=\sqrt{\frac{2 e V}{m}} \\
& \text { Radius of path, } R^{\prime}=\frac{m V}{e B}=\frac{m \sqrt{\frac{2 e V}{m}}}{e B} \\
& R^{\prime}=\sqrt{2} \times \frac{R}{2} \quad \text { [From eq. (i)] } \\
& R^{\prime}=\frac{R}{\sqrt{2}}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2012 (19 May Online)
A proton and a deutron ( $q=+\mathrm{e}, m=2.0 \mathrm{u})$ having same kinetic energies enter a region of uniform magnetic field $\vec{B}$, moving perpendicular to $\vec{B}$. The ratio of the radius $r_d$ of deutron path to the radius $r_p$ of the proton path is:

Options:
A $\sqrt{2}: 1$
B $1: 1$
C $1: \sqrt{2}$
D $1: 2$

The correct answer is:
$\sqrt{2}: 1$

Solution:
In uniform magnetic field,
$\mathrm{R}=\frac{\mathrm{m} \nu}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{~K} . \mathrm{E})}}{\mathrm{qB}}$
Since same K.E
$\begin{aligned}& \mathrm{R} \propto \frac{\sqrt{\mathrm{m}}}{\mathrm{q}} \\& \therefore \frac{\mathrm{R}_{\text {deutron }}}{\mathrm{R}_{\text {proton }}}=\sqrt{\frac{\mathrm{m}_{\mathrm{d}}}{\mathrm{m}_{\mathrm{p}}}} \times \frac{\mathrm{q}_{\mathrm{p}}}{\mathrm{q}_{\mathrm{d}}} \\& =\sqrt{2} \times 1 \\& \therefore \gamma_{\mathrm{d}}: \gamma_{\mathrm{p}}=\sqrt{2}: 1\end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (09 Apr Shift 2)
A proton and a Li3+ nucleus are accelerated by the same potential. If λLi and λp denote the de Broglie wavelengths of Li3+ and proton respectively, then the value of λLiλp is x×10-1. The value of x is _____ (Rounded off to the nearest integer) [Mass of Li3+=8.3 mass of proton]

Options:

The correct answer is:
2

Solution:
λ=h2mqV

λLiλp=mpeVmLi3eV mLi=8.3 mp

λLiλp=18.3×3=15=0.2=2×10-1

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2021 (24 Feb Shift 1)
A proton and an alpha particle are subjected to same potential difference $V$. Their de-Broglie wavelengths $\lambda_{p}, \lambda_{\alpha}$ will be in the ratio

Options:
A $2: 1$
B $2 \sqrt{2}: 1$
C $1: 1$
D $1: 2$

The correct answer is:
$2 \sqrt{2}: 1$

Solution:
T
he de-Broglie wavelength of a particle is given by
$\lambda=\frac{h}{m v}=\frac{h}{p}$
The momentum of proton,
$p_{p}=\sqrt{2 m E}=\sqrt{2 m e V}$
Similarly, for $\alpha$ - particle,
$\begin{aligned}
p_{\alpha} &-\sqrt{2 \times 4 m \times 2 e \times V} \\
&=\sqrt{16 m e V} \\
\therefore \quad \frac{\lambda_{p}}{\lambda_{\alpha}} &=\frac{p_{\alpha}}{p_{p}}=\frac{\sqrt{16 m e V}}{\sqrt{2 m e V}}=\sqrt{8} \\
&=2 \sqrt{2}: 1
\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main
A proton and an \( \alpha \)-particle, having kinetic energies \( K_{\mathrm{p}} \) and \( K_{\alpha} \), respectively, enter into a magnetic field at right angles. The ratio of the radii of the trajectory of proton to that of \( \alpha \)-particle is \( 2: 1 \). The ratio of \( K_{\mathrm{p}}: K_{\alpha} \) is:

Options:
A \( 1: 8 \)
B \( 8: 1 \)
C \( 1: 4 \)
D \( 4: 1 \)

The correct answer is:
\( 4: 1 \)

Solution:
r=mvqB=pqB mαmp=4

rprα=ppqpqαpα=21

pppa=2qpqa=212

pppα=1

KpKα=pp2pαpmαmp=4:1

Topic: Physics, Magnetic Effects of Current, JEE Main
A proton and an antiproton come close to each other in vacuum such that the distance between them is 10 cm. Consider the potential energy to be zero at infinity. The velocity at this distance will be:

Options:
A 1.17 m/s
B 2.3 m/s
C 3.0 m/s
D 23 m/s

The correct answer is:
1.17 m/s

Solution:
Applying mechanical energy conservation
Ki+Ui=Kf+Uf
0+0=12mv2+12mv2+kq1q2r
q1=q2=e, q1=+e
q2=-e
and r=0.1 m
∴mv2=ke2r

∴v2=ke2mr=9×109×1.6×10-1921.67×10-27(0.1)
v≃1.17 m/s

Topic: Physics, Work Power Energy, JEE Main
A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is:
(Assume h=6.63 $\times 10^{-34} \mathrm{~J} \mathrm{~s}, \mathrm{~m}_{\mathrm{e}}=9.0 \times 10^{-31} \mathrm{~kg}$ and $\mathrm{m}_{\mathrm{p}}=1836$ times $\mathrm{m}_{\mathrm{e}}$ )

Options:
A $1: \sqrt{1836}$
B $1: \frac{1}{1836}$
C $1: \frac{1}{\sqrt{1836}}$
D 1: 1836

The correct answer is:
1: 1836

Solution:
$\lambda$ is same for both
$P=\frac{h}{\lambda}$ same for both
$\mathrm{P}=\sqrt{2 \mathrm{mK}}$
Hence,
$\begin{aligned} & \mathrm{K} \propto \frac{1}{\mathrm{~m}} \\ & \Rightarrow \frac{\mathrm{KE}_{\mathrm{p}}}{\mathrm{KE}_{\mathrm{e}}}=\frac{\mathrm{m}_{\mathrm{e}}}{\mathrm{m}_{\mathrm{p}}}=\frac{1}{1836}\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2024 (08 Apr Shift 1)
A proton and an electron have the same de Broglie wavelength. If $\mathrm{K}_{\mathrm{p}}$ and $\mathrm{K}_{\mathrm{e}}$ be the kinetic energies of proton and electron respectively, then choose the correct relation :

Options:
A $K_p>K_e$
B $\mathrm{K}_{\mathrm{p}} < \mathrm{K}_{\mathrm{e}}$
C $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{e}}$
D $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{e}}{ }^2$

The correct answer is:
$\mathrm{K}_{\mathrm{p}} < \mathrm{K}_{\mathrm{e}}$

Solution:
De Broglie wavelength of proton \& electron $=\lambda$
$\begin{aligned} & \because \lambda=\frac{\mathrm{h}}{\mathrm{p}} \\ & \therefore \mathrm{p}_{\text {proton }}=\mathrm{p}_{\text {electron }} \\ & \because \mathrm{KE}=\frac{\mathrm{p}^2}{2 \mathrm{~m}} \\ & \therefore \mathrm{KE}_{\text {proton }} < \mathrm{KE}_{\text {electron }} \\ & {\left[\mathrm{K}_{\mathrm{p}} < \mathrm{K}_{\mathrm{c}}\right]}\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2024 (08 Apr Shift 2)
A proton and an α-particle are accelerated from rest by 2 V and 4 V potentials, respectively. The ratio of their de-Broglie wavelength is :

Options:
A 8 : 1
B 2 : 1
C 4 : 1
D 16 : 1

The correct answer is:
4 : 1

Solution:
The formula to calculate the de Broglie wavelength of the proton is given by

λp=h2mpqpV ...1

The formula to calculate the de Broglie wavelength of the alpha particle is given by

λα=h2mαqαV' ...2

Divide equation (1) by equation (2) and simplify to obtain the required ratio.

λpλα=mαmp×qαV'qpV ...3

Substitute the values of the known parameters into equation (3) to calculate the required ratio.

λpλα=41×2×41×2=4:1

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (12 Apr Shift 1)
A proton and an α-particle, having kinetic energies Kp and Kα, respectively, enter into a magnetic field at right angles. The ratio of the radii of the trajectory of proton to that of α -particle is 2:1. The ratio of Kp:Kα is:

Options:
A 1:8
B 8:1
C 1:4
D 4:1

The correct answer is:
4:1

Solution:
r=mvqB=pqB mαmp=4

rprα=ppqpqαpα=21

pppa=2qpqa=212

pppα=1

KpKα=pp2pαpmαmp=14

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (18 Mar Shift 2)
A proton and an α- particle (with their masses in the ratio of 1:4 and charges in the ratio of 1:2 ) are accelerated from rest through a potential difference V. If a uniform magnetic field B is set up perpendicular to their velocities, the ratio of the radii rp:rα of the circular paths described by them will be:

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

The correct answer is:
1:2

Solution:
The kinetic energy of a charge q of mass m in potential is given by,

12mv2=qV

mv=2mqV

Then, the radius of the charge will be,

r=2mqVqB

⇒r∝mq

Hence, the ratio of radius of proton to α-particle,
rprα=14×21=12.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (12 Jan Shift 1)
A proton carrying $1 \mathrm{MeV}$ kinetic energy is moving in a circular path of radius $R$ in uniform magnetic field. What should be the energy of an $\alpha$-particle to describe a circle of same radius in the same field?

Options:
A $2 \mathrm{MeV}$
B $1 \mathrm{MeV}$
C $0.5 \mathrm{MeV}$
D $4 \mathrm{MeV}$

The correct answer is:
$1 \mathrm{MeV}$

Solution:
For proton
$r=\frac{\sqrt{2 m(\mathrm{KE})}}{q B}$
So $\quad q \propto \sqrt{m(\mathrm{KE})}$
Hence
$$
\begin{aligned}
\frac{e}{2 e} & =\sqrt{\frac{\left(m_p\right)(1 \mathrm{MeV})}{\left(4 m_p\right)(\mathrm{KE})}} \\
\frac{1}{4} & =\frac{1 \mathrm{MeV}}{4 \mathrm{KE}} \\
\mathrm{KE} & =1 \mathrm{MeV}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2012 (Mains)
A proton collides with a stationary deuteron to form a He3 nucleus. For this reaction to take place, the proton must have a minimum kinetic energy K0. If instead, a deuteron collides with a stationary proton to make a He3 nucleus, then it must have minimum kinetic energy equal to

Options:
A 2K0
B 1.5 K0
C K0
D K02

The correct answer is:
2K0

Solution:
In case 1:
Conservation of linear momentum: mv0=3mv
â âv=v03
Conservation of mechanical energy: 12mv02=123mv032+U0

Here U0 is the minimum energy required for the reaction to happen
ââU0=K0-K03=2K03
In case 2:
Conservation of linear momentum: 2mv0'=3mv' ââv'=23v0'
Conservation of mechanical energy: 122mv0'2=123m23v0'2+2K03
â mv0'2-23mv0'2=2K03
ââ13mv0'2=2K03
ââ122mv0'2=2K0

Topic: Physics, Nuclear Physics, JEE Main
A proton is bombarded on a stationary lithium nucleus. As a result of the collision two Î±-particles are produced. If the direction of motion of the Î±-particles with the initial direction of motion makes an angle cos-114, then the kinetic energy of the striking proton is [Given, binding energies per nucleon of Li7=5.60 MeV and He4=7.60 MeV, mprotonâmneutron]

Options:
A 17.28 MeV
B 17.36 MeV
C 17.58 MeV
D 17.44 MeV

The correct answer is:
17.28 MeV

Solution:
Q value of the reaction is,
Q = (2 Ã 4 Ã 7.06 â 7 Ã 5.6) MeV = 17.28 MeV
Applying conservation of energy for collision,
Kp + Q = 2 KÎ± ....(i)
(Here, Kp and KÎ± are the kinetic energies of proton and Î± - particle respectively)
From the conservation of linear momentum (As there is no external force) ....(ii)
[Here P=2mk]
â Kâ¡ pâ¡ = 1 6 Kâ¡ Î± cos 2 Î¸ = 1 6 Kâ¡ Î± 1 4 2 as mâ¡ Î± = 4 mâ¡ pâ¡
â´ KÎ± =Kp ....(iii)
Solving eqs. (i) and (iii) with Q = 17.28 MeV
we get Kp = 17.28 MeV

Topic: Physics, Nuclear Physics, JEE Main
A proton is fired from very far away towards a nucleus with charge $Q=120 e$, where $e$ is the electronic charge. It makes a closest approach of $10 \mathrm{fm}$ to the nucleus. The de Broglie wavelength (in units of $\mathrm{fm}$ ) of the proton at its start is: (take the proton mass, $m_{p}=(5 / 3) \times 10^{-27} \mathrm{~kg} ; h / e=4.2 \times$ $10^{-15} \mathrm{~J} . \mathrm{S} / \mathrm{C}$;

$\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{~m} / \mathrm{F} ; 1 \mathrm{fm}=10^{-15} \mathrm{~m}$

Options:

The correct answer is:
7

Solution:
From energy conservation Loss in K.E. of proton = gain in potential energy of the proton - nucleus system

$\begin{array}{c}

\frac{1}{2} m v^{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r} \quad \therefore \frac{p^{2}}{2 m}=\frac{1}{4 \pi \in_{0}} \frac{q_{1} q_{2}}{r} \\

\therefore \frac{1}{2 m}\left(\frac{h^{2}}{\lambda^{2}}\right)=\frac{1}{4 \pi \in_{0}} \frac{q_{1} q_{2}}{r} \quad \therefore \lambda=\sqrt{\frac{4 \pi \epsilon_{0} r \cdot h^{2}}{q_{1} q_{2}(2 m)}}

\end{array}$

Putting the values of $4 \pi \varepsilon_{0}, r, h, q_{1}, q_{2}$ and $m$ we get, deBroglie wavelength of proton, $\lambda=7 \mathrm{fm}$

Topic: Physics, Dual Nature of Matter, JEE Main
A proton is fired from very far away towards a nucleus with charge Q = 120e, where e is the electronic charge. It makes the closest approach of 10 fm to the nucleus. The de Broglie wavelength (in fm) of the proton at its start is: (mp=5/3Ã10-27 kg;he=4Â·2Ã10-15 J s C-1; 14ÏÎµ0=9Ã109N m2 C-2; 1fm=10-15 m)

Options:
A 7
B 5
C 9
D 3

The correct answer is:
7

Solution:
0 + 1 2 mv 2 = K Q e 1 0 Ã 1 0 - 1 5 = K 1 2 0 e e 1 0 Ã 1 0 - 1 5
1 2 Ã 5 3 Ã 1 0 - 2 7 v 2 = 9 Ã 1 0 9 Ã 1 2 0 Ã 1 Â· 6 Ã 1 0 - 1 9 2 1 0 Ã 1 0 - 1 5
v = 9 Ã 6 Ã 1 0 9 Ã 1 2 0 Ã 2 Â· 5 6 Ã 1 0 - 3 8 5 0 Ã 1 0 - 4 2
v = 3 3 1 Â· 7 7 6 Ã 1 0 1 3
Î» = h mv
Î» = 4 Â· 2 Ã 1 0 - 1 5 Ã 1 Â· 6 Ã 1 0 - 1 9 5 3 Ã 1 0 - 2 7 Ã 3 3 1 Â· 7 7 6 Ã 1 0 1 3 = 4 Â· 2 Ã 4 Â· 8 Ã 1 0 - 3 4 5 7 Â· 6 Ã 5 Ã 1 0 - 2 1 = 0 Â· 0 7 Ã 1 0 - 1 3
Î» = 7 Ã 1 0 - 1 5 = 7 fm

Topic: Physics, Atomic Physics, JEE Main
A proton is moving along $Z$-axis in a magnetic field. The magnetic field is along $X$-axis. The proton will experience a force along

Options:
A $X$-axis
B $Y$-axis
C Z-axis
D Negative Z-axis

The correct answer is:
$Y$-axis

Solution:
Lorent's Force is given by
\(\overrightarrow{\mathrm{F}}=\mathrm{q} \vec{v} \times \vec{B}\)
Therefore for a proton
\(\hat{\mathrm{F}}=+\hat{\mathrm{v}} \times \hat{\mathrm{B}}=\hat{\mathrm{k}} \times \hat{\mathrm{i}}=\hat{\mathrm{j}} \rightarrow \text { Y-Axis }\)

Topic: Physics, Magnetic Effects of Current, JEE Main
A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B. The field occupies a region of space by width d. If α be the angle of deviation of proton from the initial direction of motion (see figure), the value of sin⁡α will be:

Options:
A B2qdmV
B Bdq2mV
C Bdq2mV
D qV Bd2m

The correct answer is:
Bdq2mV

Solution:
Due to potential difference V speed acquired by proton is v0,
⇒qV=12mv02=K
∵Radius, R=mv0qB=2mKqB
∵K=qV
R=2meVqB
sinα=dR=dqB2meV=Bdq2mV

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2015 (10 Apr Online)
A proton moves with a speed of 5.0Ã106 m s-1 along the x- axis. It enters a region where there is a magnetic field of magnitude 2.0 T directed at an angle of 30Â° to the x- axis and lying in the xy- plane. The magnitude of the magnetic force on the proton is

Options:
A 0.8×10-13 N
B 1.6Ã10-13 N
C 8.0Ã10-13 N
D 16.0Ã10-13 N

The correct answer is:
8.0Ã10-13 N

Solution:
Given, The speed of proton,
v=5.0×106 m s-1
The magnetic field, B=2.0 T
Angle θ=30o
Charge on the proton, q=1.6×10-19 C
When the proton enters in magnitude field, it experiences a magnetic force
F=qv×B
F=qvBsin⁡θ
F=1.6×10-19×5×106×2×sin⁡30°
=8.0×10-13 N

Topic: Physics, Magnetic Effects of Current, JEE Main
A proton moving in perpendicular magnetic field possess energy $E$. The strength of magnetic field is increases four times. But the proton is constrained to move in the path of same radius. The kinetic energy of proton will increase

Options:
A 4 times
B 12 times
C 8 times
D 16 times

The correct answer is:
4 times

Solution:
In orthogonal magnetic field, the changed particle undergoes uniform circular motion.
$\begin{aligned} & \therefore \frac{m v^2}{R}=q v B \\ & \Rightarrow v^2\left(\frac{q B}{m}\right) r\end{aligned}$
If B increases by four times $\& r$ remains same
$\begin{aligned} & v_{\text {new }}^2=\left[\left(\frac{q B}{m}\right) r\right] 4=(2 v)^2 \\ & \therefore K E_{\text {new }}=\frac{1}{2} v_{\text {new }}^2=\left(\frac{1}{2} m v^2\right) 4\end{aligned}$

Topic: Physics, Magnetic Effects of Current, MHT CET, MHT CET 2022 (06 Aug Shift 2)