Physics M.C.Q (1 Marks)

$\text { So } \phi_{\text {in }}=\phi_{\text {out }}$

Number of field lines entering is equal number of field lines leaving.

$\tau= pEsin \theta$

$4= q \times \ell \times E \times \sin 30^{\circ}$

$4= q \times 2 \times 10^{-2} \times 2 \times 10^5 \times \frac{1}{2}$
$q =2 \times 10^{-3}$

$q =2\,mC$

$2.4 \times 10^{39}=\frac{ K }{ G } \times \frac{\left(1.6 \times 10^{-19}\right)^2}{\left(9.11 \times 10^{-31} \times 1.67 \times 10^{-27}\right)}$

$\frac{ K }{ G }=\frac{2.4 \times 10^{39} \times 15.2137 \times 10^{-58}}{2.56 \times 10^{-38}}$

$=14.26 \times 10^{19}$

$=1.426 \times 10^{20}$

$\approx 10^{20}$

$E=\frac{K P}{R^{3}} \sqrt{1+3 \cos ^{2} \theta}$

$\therefore E \propto \frac{1}{R^{3}}$

As field lines are closer at charge $+q$.

So, net force on the dipole acts towards right side. A system always moves to decrease it's potential energy.

$a_{e}=\frac{F}{m_{e}}$

$a_{e}=\frac{e^{2}}{4 \pi \varepsilon_{0} m_{e} r^{2}}$

$a_{e}=\frac{9 \times 10^{9} \times(1.6)^{2} \times 10^{-38}}{9 \times 10^{-31} \times(1.6)^{2} \times 10^{-20}}$

$a_{n}=10^{22} m / s ^{2}$

$E =1.28 \times 10^{5} N / C$

$E=-\frac{1}{4 \pi \in_{0}} \frac{\bar{P}}{r^{3}}$

$25 \%$ charge from $A$ is transferred to $B$

New force $(\mathrm{F})=\frac{\mathrm{K}\left(\frac{3 \mathrm{q}}{4}\right)\left(\frac{-3 \mathrm{q}}{4}\right)}{\mathrm{r}^{2}}=\frac{-9 \mathrm{kq}^{2}}{16 \mathrm{r}^{2}}=\frac{9 \mathrm{F}}{16}$

$\mathrm{E}=\mathrm{E}_{1}+\mathrm{E}_{2}$

$E=\frac{\lambda}{2 \pi \in_{0} R}+\frac{\lambda}{2 \pi \in_{0} R}$

$\mathrm{E}=\frac{\lambda}{\pi \in_{0} \mathrm{R}} \mathrm{N} / \mathrm{C}$

$\therefore \,\,t = \sqrt {\frac{{2hm}}{{eE}}} $

$\therefore \,\,t \propto \sqrt m $ as $'e'$ is same for electron and proton.

$\because $ Electron has smaller mass so it will take smaller time.

For $t=0$ to $t=1 \,\mathrm{s}$

$S_{1}=\frac{1}{2} \times 6(1)^{2}=3 \,{m}.........(i)$

For $t=1$ $s$ to $t=2 \,s$

$S_{2}=6.1-\frac{1}{2} \times 6(1)^{2}=3\,{m}.........(ii)$

For $t=2$ $s$ to $t=3\, s$

$S_{3}=0-\frac{1}{2} \times 6(1)^{2}=-3 \,{m}$

Total displacement $\mathrm{S}=\mathrm{S}_{1}+\mathrm{S}_{2}+\mathrm{S}_{3}=3\, \mathrm{m}$

Average velocity $=\frac{3}{3}=1\,{ms}^{-1}$

Total distance travelled $=9 \,{m}$

Average speed $=\frac{9}{3}=3 \,{ms}^{-1}$

$\therefore$ Charge on one hydrogen atom

$=q_{e}+q_{p}=-e+(e+\Delta e)=\Delta e$

Since a hydrogen atom carry a net charge $\Delta e$

$\therefore \quad$ Electrostatic force,

$F_{e} \frac{1}{4 \pi \varepsilon_{o}} \frac{(\Delta e)^{2}}{d^{2}}.........(i)$

will act between two hydrogen atoms.

The gravitational force between two hydrogen atoms is given as

$F_{g}=\frac{G m_{h} m_{h}}{d^{2}}.........(ii)$

since, the net force on the system is zero, $F_{e}=F_{g}$ Using eqns. $(i)$ and $(ii)$, we get

${\frac{(\Delta e)^{2}}{4 \pi \varepsilon_{o} d^{2}}=\frac{G m_{h}^{2}}{d^{2}}}$

${(\Delta e)^{2}=4 \pi \varepsilon_{o} G m_{h}^{2}}$

${=6.67 \times 10^{-11} \times\left(1.67 \times 10^{-27}\right)^{2} /\left(9 \times 10^{9}\right)}$

${\Delta e=10^{-37} \,\mathrm{C}}$

$r f ={ PE } \sin \theta$

$\tau= PE \sin \theta$

$\text { But } f = mg \sin \theta$

$mgr \sin \theta= PE \sin \theta$

$E =\frac{ mgr }{ P }=\frac{ mgr }{ q \times 2 r }$

$P = q \times 2 r (\text { dipole moment })$

$E =\frac{ mg }{2 q }$

$\tau  = 4\,{\text{Nm}},l = 2\,{\text{cm}} = 0.02\,{\text{m}},q = ?$

$\tau  = pE\sin \theta  = (ql)E\sin \theta $

$\therefore q =\frac{\tau}{E l \sin \theta}=\frac{4}{2 \times 10^{5} \times 0.02 \times \frac{1}{2}}$

$ = \frac{4}{{2 \times {{10}^3}}} = 2 \times {10^{ - 3}}{\text{C}} = 2\,{\text{mC}}$

$T \sin \theta=\frac{k q^{2}}{x^{2}}.........(ii)$

From eqns. $(i)$ and $(ii)$, $tan \theta=\frac{k q^{2}}{x^{2} m g}$

since $\theta$ is small, $\therefore \tan \theta  \approx \sin \theta  = \frac{x}{{2l}}$

$\therefore \quad \frac{x}{2 l}=\frac{k q^{2}}{x^{2} m g} \Rightarrow q^{2}=x^{3} \frac{m g}{2 l k}$ or $q \propto x^{3 / 2}$

$\Rightarrow \frac{d q}{d t} \propto \frac{3}{2} \sqrt{x} \frac{d x}{d t}=\frac{3}{2} \sqrt{x} v$

since, $\frac{d q}{d t}=$ constant

$\therefore v \propto \frac{1}{\sqrt{x}}$

$2=n \times 1.6 \times 10^{-19}$

Removed ${e^ - }s.$ $\mathrm{n}=\frac{2}{1.6 \times 10^{-19}}$

So mass transfer $=\mathrm{n} \times \mathrm{Me}$

$=\frac{2}{1.6 \times 10^{-19}} \times 9.1 \times 10^{-31} \mathrm{\,kg}$

so $11.37 \times 10^{-12} \mathrm{\,kg}$ will be the answer.

Thus, one mole $\left( =6.02 \times 10^{23} \text { molecules }\right)$ Therefore the number of molecules in one cup of water is $(250 / 18) \times 6.02 \times 10^{23}$

Each molecule of water contains two hydrogen atoms and one oxygen atom. i.e... $10$ electrons and $10$ protons. Hence the total positive and total negative charge has the same magnitude. It is equal to $(250 / 18) \times 6.02 \times 10^{23} \times 10 \times 1.6 \times 10^{-19}\, C$$ =1.34 \times 10^{7} \;C$

A charged gold leaf electroscope has its leaves apart by certain amount having enclosed air. When an electroscope is subjected to $x$-rays, then the $x$-ray will ionize the air around the leaves into positive and negative charges. Hence, the charge created in the air and opposite to the gold leaf will move towards it. Thus, the opposite charges neutralize the charge on the leaf, causing the leaf to collapse.

New charge on sphere A is $Q'_A $$= Q\,[ \frac {r_A}{r_A + r_B} ]$ $ = 120\,\left[ {\frac{4}{{4 + 6}}} \right]\, = 48\,\mu \,C$.

Initially it was $80\,\mu C$ i.e., $32\,\mu \,C$ charge flows from $A$ to $B$.

If a positive charge is placed at $P$ and distributed, the positive charge either goes towards, $-Q$ or moves away from $-Q$ but won't oscillate

$\left(\because \frac{d^{2} v_{p}}{d x^{2}}>0\right)$

(unstable equilibrium) while negative charge oscillate

$\left(\because \frac{d^{2} U_{n}}{d^{2} x}\right.$$\left.=-\frac{\mathrm{d}^{2} \mathrm{U}_{\mathrm{P}}}{\mathrm{d}^{2} \mathrm{x}}<0\right)$ (Stable equilibrium)

$\mathrm{q}$ is $-ve$ then there will be net attractice force which will send the charge particle back to its original position so it will be an $S.H.M.$

$F_{1}=\frac{K q_{1} q_{2}}{r^{2}}$

$F_{2}=\frac{K q^{2}}{r^{2}}=\frac{K}{r^{2}}\left(\frac{q_{1}+q_{2}}{2}\right)^{2}$

$F_{2}-F_{1}=\frac{K}{r^{2}}\left\{\left(\frac{q_{1}+q_{2}}{2}\right)^{2}-q_{1} q_{2}\right\}$

$=\frac{K}{4 r^{2}}\left(q_{1}-q_{2}\right)^{2} \Rightarrow F_{2}-F_{1}>0$ or $F_{2}>F_{1}$

$=\frac{\sigma^{2}}{2 \varepsilon_{0}}\left(\pi r^{2}\right)=\left(\frac{Q}{4 \pi R^{2}}\right)^{2}\left(\frac{1}{2 \varepsilon_{0}}\right) \pi R^{2}$

$\left[ {\frac{{{{\rm{Q}}^2}}}{{32\pi {\varepsilon _0}{{\rm{R}}^2}}}} \right]$

$(2)$ Charge is invariant with velocity

$(3)$ Charge requires mass for existance

$(4)$ Repultion shows charge of both bodies because attraction can be there between charged and uncharged body.

$\therefore \frac{2 q^{2}}{4 \pi \varepsilon_{0} x^{2}}=\frac{6 q^{2}}{4 \pi \varepsilon_{0}(d+x)^{2}}$

$\therefore(d+x)^{2}=3 x^{2}$

$\therefore 2 x^{2}-2 d x-d^{2}=0$

$x=\frac{d}{2} \pm \frac{\sqrt{3} d}{2}$

($-ve$ sign would be between $\mathrm{q}$ and $-3 \mathrm{q}$ and hence is unacceptable.)

$x=\frac{d}{2}+\frac{\sqrt{3} d}{2}=\frac{d}{2}(1+\sqrt{3})$ to the left of $q$

$\mathrm{F}_{2} \propto\left(\frac{\mathrm{q}_{1}+\mathrm{q}_{2}}{2}\right)^{2}$

So $\mathrm{F}_{2}>\mathrm{F}_{1}$

$Q=2 L \sqrt{4 \pi \varepsilon_{0} K\left(L-L_{0}\right)}$