The radius of the proton is about ${10^{ - 15}}m$. The radius of the observable universe is ${10^{26}}$ m. identify the distance which is half-way between these two extremes on a logarithmic scale.

${r_1} = {10^{ - 15}}{\rm{m}},\;{r_2} = {10^{26}}m$ , Log $r = \frac{1}{2}\left[ {{\rm{log}}{{10}^{ - 15}} + {\rm{log}}{{10}^{26}}} \right]$ , $ = \frac{1}{2}\left[ { - 15 + 26} \right] = 5.5 \approx 6 \Rightarrow r = {10^6}m$

The correct Option is "${10^6}m$"

Proton is referred as ${{\rm{H}}^ + }$.

The correct Option is "Ionised hydrogen atom"

Penetration power of proton is:

A proton requires more energy for penetration due to its relatively higher mass and positive charge than electron.

The correct Option is "Less than electron"

The volume of a proton is approximately;

$1.5 \times {10^{ - 30}}{\rm{c}}{{\rm{m}}^3}$

The correct Option is "$1.5 \times {10^{ - 38}}{\rm{c}}{{\rm{m}}^3}$"

$1.5 \times {10^{ - 34}}{\rm{c}}{{\rm{m}}^3}$

A proton is revolving in a circular path with constant speed. Angle between angular momentum and magnetic dipole moment associated with it is

$\vec M = \frac{q}{{2m}}\,\vec L \Rightarrow \theta = 0^\circ $

The correct Option is "$0^\circ $"

A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of $1\;kV,$ its kinetic energy will be

$KE = QV = e \times {10^3}V = 1KeV$

The correct Option is "$1\;keV$"

When a proton is accelerated through $1V,$ then its kinetic energy will be

$\Delta KE = qV = eV = e \times 1V = 1eV$

The correct Option is "$1\;eV$"

In a hydrogen atom, the distance between the electron and proton is $2.5\,\, \times \,\,{10^{ - 11}}\,\,m$. The electrical force of attraction between them will be

$\begin{array}{l} F = k{q_1}{q_2}/{r^2},i.e.,\\ F = \frac{{9\,\, \times \,\,{{10}^9}\,\, \times \,\,1.6\,\, \times \,\,{{10}^{ - 19}} \times \,\,1.6\,\, \times \,\,{{10}^{ - 19}}}}{{{{\left( {2.5\,\, \times \,\,{{10}^{ - 11}}} \right)}^2}}} = 3.7\,\, \times \,\,{10^{ - 7}}\,\,N \end{array}$

$2.8\,\, \times \,\,{10^{ - 7}}\,\,N$

The correct Option is "$3.7\,\, \times \,\,{10^{ - 7}}\,\,N$"

$6.2\,\, \times \,\,{10^{ - 7}}\,\,N$

$9.1\,\, \times \,\,{10^{ - 7}}\,\,N$

One electron and one proton is accelerated by equal potential. Ratio in their de-Broglie wavelength is

If a charge particle of mass $m$ and charge $q$ is accelerated through a potential difference $V$ and $E$ is the energy acquired by the particle , then $E = qV$ If $v$ is velocity of particle, then $E = \frac{1}{2}m{v^2}$ Or $v = \sqrt {\left( {\frac{{2E}}{m}} \right)} $ Now, de-Broglie wavelength of particle is $\lambda = \frac{h}{{mv}} = \frac{h}{{m\sqrt {\left( {2E\;/m} \right)} }}$ Substituting the value of E, we get $\lambda = \frac{h}{{\sqrt {2mqV} }}$ For electron, ${\lambda _e} = \frac{h}{{\sqrt {2{m_e}eV} }}$ For proton, ${\lambda _p} = \frac{h}{{\sqrt {2{m_p}eV} }}$ $\therefore \;\;\;\;\;\;\frac{{{\lambda _e}}}{{{\lambda _p}}} = \sqrt {\left( {\frac{{{m_p}}}{{{m_e}}}} \right)} $

The correct Option is "$\sqrt {\frac{{{m_p}}}{{{m_e}}}} $"

$\sqrt {\frac{{{m_e}}}{{{m_p}}}} $

The kinetic energy of electron and proton is ${10^{ - 32}}J.$ Then the relation between their de-Broglie wavelength is

By using $\lambda = \frac{h}{{\sqrt {2mE} }}\;E = {10^{ - 32}}J = $ Constant for both particles. Hence $\lambda \propto \frac{1}{{\sqrt m }}$ Since ${m_p} > {m_e}$ so ${\lambda _p} < {\lambda _e}$

The correct Option is "${\lambda _p} < {\lambda _e}$"

If the uncertainty in the position of proton is $6 \times {10^8}$m, then the minimum uncertainty in its speed will be

${\rm{\Delta }}p = m{\rm{\Delta }}v = \frac{\hbar }{{{\rm{\Delta }}x}}$ or ${\rm{\Delta }}v = \frac{\hbar }{{m{\rm{\Delta }}x}} = \frac{{1.034 \times {{10}^{ - 34}}}}{{1.67 \times {{10}^{ - 27}} \times 6 \times {{10}^{ - 8}}}}$ $ \approx 1{\rm{m}}{{\rm{s}}^{ - 1}}$

$1{\rm{\;cm}}{{\rm{s}}^{ - 1}}$

The correct Option is "$1{\rm{\;m}}{{\rm{s}}^{ - 1}}$"

$1{\rm{\;mm}}{{\rm{s}}^{ - 1}}$

$100{\rm{\;m}}{{\rm{s}}^{ - 1}}$

When a proton is accelerated with $1\;volt$ potential difference, then its kinetic energy is

$K = Q.V = 1e \times 1\;Volt = 1eV$

The correct Option is "$1\;eV$"

The specific charge of a proton is $9.6 \times {10^7}\;{\rm{C\;K}}{{\rm{g}}^{ - 1}}$. The specific charge of an alpha particle will be

Specific charge on proton =${\left( {\frac{e}{m}} \right)_p}$ $ = 9.6 \times {10^7}{\rm{Ck}}{{\rm{g}}^{ - 1}}$ specific charge on $ \propto $-particle, ${\left( {\frac{q}{m}} \right)_ \propto } = \frac{{2e}}{{4m}} = \frac{1}{2}{\left( {\frac{e}{m}} \right)_p} = \frac{1}{2} \times 9.6 \times {10^7}$ $ = {\rm{\;}}4.8 \times {10^7}\;{\rm{Ck}}{{\rm{g}}^{ - 1}}$

$9.6 \times {10^7}{\rm{\;C\;K}}{{\rm{g}}^{ - 1}}$

$19.2 \times {10^7}{\rm{\;C\;K}}{{\rm{g}}^{ - 1}}$

The correct Option is "$4.8 \times {10^7}{\rm{\;C\;K}}{{\rm{g}}^{ - 1}}$"

$2.4 \times {10^7}\;{\rm{C\;K}}{{\rm{g}}^{ - 1}}$

When the momentum of a proton is changed by an amount ${P_0},$ the corresponding change in the de-Broglie wavelength is found to be 0.25%. Then, the original momentum of the proton was

$\lambda \propto \frac{1}{p} \Rightarrow \frac{{\Delta p}}{p} = - \frac{{\Delta \lambda }}{\lambda } \Rightarrow \left| {\frac{{{\rm{\Delta }}p}}{p}} \right| = \left| {\frac{{{\rm{\Delta }}\lambda }}{\lambda }} \right|$ $ \Rightarrow \frac{{{p_0}}}{p} = \frac{{0.25}}{{100}} = \frac{1}{{400}}\; \Rightarrow p = 400\;{p_0}$

The correct Option is "$400\;{p_0}$"

Isomerism among compounds due to the migration of a proton is known as :

The correct Option is "Tautomerism"

In a cyclotron, if the frequency of proton is $5MHz$, the magnetic field necessary for resonance is

$F\, = \,qVB$ $\frac{{m{v^2}}}{r} = \,qVB$ $v\, = \,\frac{{qBr}}{m}$ …. (1) $\begin{array}{l} T\, = \,\frac{{2\pi r}}{v}\, = \,\frac{{2\pi m}}{{Bq}}\\ f\, = \,\frac{1}{T}\, = \,\frac{{qB}}{{2\pi m}}\\ B\, = \,\frac{{f2\pi m}}{q}\\ B\, = \,\frac{{50 \times {{10}^5} \times 2 \times 3.14 \times 1.67 \times {{10}^{ - 27}}}}{{1.6 \times {{10}^{ - 19}}}}\\ B\, = \,0.327\,T \end{array}$

The correct Option is "$0.327T$"

A 2 MeV proton is moving perpendicular to a uniform magnetic field of 2.5 tesla. The force on the proton (in N)

$F = qV\,B = 1.6 \times {10^{ - 19}} \times \sqrt {\frac{{2E}}{m}} \times 2.5$ $F = 4 \times {10^{ - 19}}\sqrt {\frac{{2 \times 2 \times 1.6 \times {{10}^{ - 19}} \times {{10}^6}}}{{1.66 \times {{10}^ - }27}}} = 7.6 \times {10^{ - 12}}N$

The correct Option is "$8 \times {10^{ - 12}}$"

A 2 MeV proton is moving perpendicular to a uniform magnetic field of 2.5 tesla. The force on the proton is.

$F = qvB = 1.6 \times {10^{ - 19}} \times \left[ {\sqrt {\frac{{2E}}{m}} } \right]2.5 = 4 \times {10^{ - 19}}\sqrt {\frac{{2 \times 2 \times 1.6 \times {{10}^{ - 19}} \times {{10}^6}}}{{1.66 \times {{10}^{ - 27}}}}} = 7.6 \times {10^{ - 12}}\,N$

The correct Option is "$7.6 \times {10^{ - 12}}N$"

A 2MeV proton is moving perpendicular to uniform magnetic field of 2.5T. The magnetic force on the proton:-

$F = qvB,\,\,\,E = \frac{1}{2}m{v^2}$ $F = qB\sqrt {\frac{{2E}}{m}} $ $ = 1.6 \times {10^{ - 19}} \times 2.5 \times \sqrt {\frac{{2 \times 2 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}{{1.67 \times {{10}^{ - 31}}}}} $ $F = 8 \times {10^{ - 12}}N$

The correct Option is "$8 \times {10^{ - 12}}N$"

The specific charge of proton is $9.6 \times {10^7}\,C\,.\,k{g^{ - 1}}$, then for an $\alpha - $ particle is

$\frac{e}{m}$ of proton $ = 2 \times \frac{e}{m}$ of $\alpha - $ particle

The correct Option is "$4.8 \times {10^7}\,C/kg$"

A 2 MeV proton is moving perpendicular to a uniform magnetic field. The force on the proton is $1.6 \times {10^{ - 12}}\,\,N.$ The intensity of the magnetic field will be

$E = \frac{1}{2}m{v^2} \Rightarrow {\text{velocity}}\,\,\,v = \sqrt {\frac{{2E}}{m}} $ $v = \sqrt {\frac{{2 \times 2 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 27}}}}} = 20 \times {10^6}\,\,\,\,m/s$ $B = \frac{F}{{qv}} = \frac{{1.6 \times {{10}^{ - 12}}}}{{1.6 \times {{10}^{ - 19}} \times 20 \times {{10}^6}}} = 0.5\,\,T$

The correct Option is "$0.5\,\,\,T$"

(A): e/m of proton is constant. (R): e/m of anode ray depend on the nature of the gas used in the discharge tube.

R can’t explain (e/m) ratio of proton.

Both A & R are true and R is the correct explanation of A

The correct Option is "Both A & R are true and R is not the correct explanation of A"

The mass of a proton is 1847 times that of an electron. An electron and a proton are injected into a uniform electric field at right angle to the direction of the field with the same initial K.E.

The kinetic energy of a particle is given by, ${K}=\frac{1}{2} {mV}^2$ The trajectory of curvature is given by, ${y}=\frac{{qEx}}{2 {mV}^2}$ Substitute $2 {K}$ for ${mV}^2$. $y=\frac{q E x^2}{2(2 K)}=\frac{q E}{4 K} x^2$ The trajectory of curvature is directly proportional to $\frac{{qE}}{4 {K}}$. So, the trajectories of both the particle will be same.

The electron trajectory will be less curved than the proton trajectory

Both the trajectories will be straight

The proton trajectory will be less curved than the electron trajectory

The correct Option is "Both the trajectories will be equally curved"

The potential of the electric field produced by point charge at any point $\left( {x,y,z} \right)$ is given by $V = 3{x^2} + 5,$ where $x,y$ are in metres and V is in volts. The intensity of the electric field at (-2,1,0) is

Potential $\left( V \right) = 3{x^2} + 5$ Intensity of the electric field= $\frac{{dV}}{{dx}} = 6x$ $\therefore E\;{\rm{at}}\;x = - 2$ $ = 6\left( { - 2} \right) = - 12{\rm{V}}{{\rm{m}}^{ - 1}}$

$ + 17{\rm{\;V}}{{\rm{m}}^{ - 1}}$

$ - 17{\rm{\;V}}{{\rm{m}}^{ - 1}}$

The correct Option is "$ + 12{\rm{\;V}}{{\rm{m}}^{ - 1}}$"

$ - 12{\rm{\;V}}{{\rm{m}}^{ - 1}}$

A hollow charged metal sphere has a radius r. If the potential difference between its surface and a point at a distance 3r from the centre is V, then electrical intensity at distance 3r from the centre is

The potential at a distance $r$, due to charge $q$ is $V = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{q}{r}$ Potential at a distance (3r) is $V = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{q}{{3r}}$ Difference in potential $ = \frac{q}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{r} - \frac{1}{{3r}}} \right]$ $ \Rightarrow V = \frac{{2q}}{{4\pi {\varepsilon _0} \times 3r}}$ Intensity of electric field $E = \frac{q}{{4\pi {\varepsilon _0}}}\frac{1}{{{{\left( {3r} \right)}^2}}}$ $\therefore \frac{E}{V} = \frac{q}{{4\pi {\varepsilon _0}q{r^2}}} \times \frac{{4\pi {\varepsilon _0}3r}}{{2q}}$ $ \Rightarrow \frac{E}{V} = \frac{1}{{6r}}\;$ $ \Rightarrow E = \frac{V}{{6r}}$

The correct Option is "$\frac{V}{{6r}}$"

A uniform slender rod of length $L$, cross-sectional area $A$ and Young’s modulus $Y$ is acted upon by the forces shown in the figure. The elongation of the rod is .

Net elongation of the rod is $\begin{array}{l} l = \frac{{3F\left( {\frac{{2L}}{3}} \right)}}{{AY}} + \frac{{2F\left( {\frac{L}{3}} \right)}}{{AY}}\\ l = \frac{{8\,FL}}{{3\,AY}} \end{array}$ .

The correct Option is "$\frac{{8\,\,FL}}{{3\,\,AY}}$"

The energy required to separate a hydrogen atom into a proton and an electron 13.6 eV. Then the velocity of electron in a hydrogen atom is.

Given $E = 13.6\,eV$ ${m_e} = 9.1 \times {10^{ - 31}}\,kg$ $KE = 13.6eV$ $\frac{1}{2}{m_e}{V^2} = 13.6\,eV$ ${V^2} = \frac{{13.6\,eV \times 2}}{{{m_e}}}$ ${V^2} = \frac{{13.6 \times 1.6 \times {{10}^{ - 19}} \times 2}}{{9.1 \times {{10}^{ - 31}}}}$ $V = \sqrt {4.78 \times {{10}^{12}}} $ $V = 2.2 \times {10^6}m/s$

The correct Option is "$2.2 \times {10^6}m/s$"

The de-Broglie wavelength is proportional to

$\lambda = \frac{h}{p} \Rightarrow \lambda \propto \frac{1}{p}$

The correct Option is "$\lambda \propto \frac{1}{p}$"

If the electric potential at a certain distance from a point charge is 900 V and electric intensity is 225 V/m. The charge is

$V = \frac{{Kq}}{r}\,;\,E = \frac{{Kq}}{{{r^2}}}$ $900 = \frac{{Kq}}{r}\,\,\, - (1)\,$ $225 = \frac{{Kq}}{{{r^2}}}\,\,\, - (2)$ Solving eqs (1) and (2 $225 = \frac{{Kq\left( {900} \right) \times \left( {900} \right)}}{{\left( {Kq} \right)\left( {Kq} \right)}}$ $ \Rightarrow q = 4 \times {10^{ - 7}}C$

The correct Option is "$4 \times {10^{ - 7}}C$"

A water molecule has an electric dipole moment $6.4 \times {10^{ - 30}}\;cm$ cm when it is in vapour state. The distance in metre between the centre of positive and negative charge of the molecule is

Electric dipole moment of a water molecule $p = 6.4 \times {10^{ - 30}}Cm$ $p = qd$ Where $d$ is the distance between the centre of positive and negative charge of the molecule $ \Rightarrow d = \frac{p}{q} = \frac{{6.4 \times {{10}^{ - 30}}Cm}}{{1.6 \times {{10}^{ - 19}}C}} = 4 \times {10^{ - 11}}m$

The correct Option is "$4 \times {10^{ - 11}}$"

A block weighing W is held against a vertical wall by applying a horizontal force F. The minimum value of F needed to hold the block is

Here applied horizontal force F acts as normal reaction. For holding the block. Force of friction = Weight of block. $f = W \Rightarrow \mu \;R = W \Rightarrow \mu F = W$ , $ \Rightarrow F = \frac{W}{\mu }\,As\,\mu W$