Physics 3 Marks Question

1. $\mu = \frac{\sin(\frac{\text{A} + \text{D}}{2})}{\sin\frac{\text{A}}{2}}$

$ = \frac{\sin(\frac{2\text{A}}{2})}{\sin\frac{\text{A}}{2}} =2 \cos\text{A}\sqrt{2} = \sqrt{3}$

$\therefore\text{A} = 60 ^{o}$

2. $\mu = \sqrt{3} = \frac{1}{\sin\text{i}_{c}}$

$\therefore\sin\text{i}_{c} = \frac{1}{\sqrt{3}}\cong0.58$

Lies between 30^o and 45^o

Hence, TIR takes place.

Alternate Answer

$\sin\text{C} = \frac{1}{\sqrt{3}}$ which is less than $\frac{1}{\sqrt{2}}$

$\therefore $ Angle of incidence > i_c

_{$\therefore\text{TIR}$}

1. Electric fields decreases Justification.

Because initially $\text{E}_{1} = \frac{\sigma}{\varepsilon}_{\circ}$ and finally $\text{E}_{2} =\frac{1}{\text{k}}.\frac{\sigma}{\varepsilon}_{\circ}$

$\text{E} = \frac{\text{E}_{1}}{\text{K}}$

2. Energy stored increases:

New capacitance$\text{C} = \bigg(\frac{\varepsilon_\circ\text{A}}{2\text{d}}\bigg)\text{k}$

$ = \frac{\text{K}}{2}\text{C}_{\circ} , $

$\therefore\text{C} < \text{C}_{\circ}$

Initially Energy $ = \frac{\text{Q}^{2}}{2\text{c}}$ and Energy $ = \frac{\text{Q}^{2}}{\text{c}}.\frac{2}{\text{k}}\text{as }1 < \text{K} < 2.$

1. Maintain Earth’s warmth through green house effect.
2. Produce heat.

Let Q be the total charge on the shell.

1. For the point M outside the shell, we have

$\text{E}.4\pi\text{r}^{2} = \frac{\text{Q}}{\in_{0}}$

$\therefore\text{E} = \frac{1}{4\pi\in_{0}}\frac{\text{Q}}{\text{r}^{2}}$

2. For the point N inside the shell, as charge enclosed inside the shell is zero.

$\text{E}.4\pi\text{r}^{2}_{1} = 0 $

$\therefore\text{E} = 0 $

The graph is as shown

1. Let the applied voltage be

$\text{V}= \text{V}_{o}\sin\omega\text{t}$

The current through an ideal capacitor, would then be

$\text{I} = \text{I}_{o}\sin(\omega\text{t} + \frac{\pi}{2}) = \text{I}_{0}\cos\omega\text{t}$

$\therefore\text{P}_{inst} = \text{VI}$

$\therefore\text{P}_{AV} = \frac{1}{\text{T}}\int^{T}_{0}\text{VIdt}$

$\therefore\text{P}_{AV} = \frac{\text{V}_{0}\text{I}_{0}}{2}\langle\sin2\omega\text{t}\rangle$

$=0.$

2. $\text{X}_{c} = \frac{1}{\omega\text{C}}$

$\therefore\text{X}_{c}$ increases as C decreases. Hence, with decreasing C, the brightness of the bulb would decrease.

1. Equivalent capacitance $(\text{C}_{n}) = \frac{\text{C}}{3} + \text{C}$

$ = \frac{4\text{C}}{3} = \frac{40}{3}\mu\text{F}$

2. $\text{Charge on } \text{C}_{4}, \text{q}_{4} = \text{C}_{4}\times\text{V}=10\times500 \mu\text{C}$

$ = 5\times10^{-3}\text{C} = 5 \text{mC}$

$\text{Charge on } \text{C}_ {1},\text{C}_{2},\text{C}_{3}\text{ is same and is equal to} \frac{\text{C}}{3}\times\text{V}$

$ = \frac{5}{3}\times10^{-3}\text{C}$

$ = 1.67\text{mC}$.

1. Emf induced $\int^{1}_{0}\text{Bwrdr}$

$ = \frac{1}{2}\text{Bwl}^{2}$

$\because \omega = 2\pi\text{v}$

$\therefore\varepsilon = \pi\text{Bvl}^{2}$

2. $\text{P} = \frac{\varepsilon^2}{\text{R}} = \frac{(\pi\text{Bvl}^{2})^{2}}{\text{R}}$

$ = \frac{\pi^{2}\text{B}^{2}\text{V}^{2}\text{l}^{4}}{\text{R}}.$

1. When a ray of light travels from a denser medium into a rarer medium at an angle greater than the critical angle, it reflects back into the denser medium. This phenomenon is called total internal reflection. Conditions for total internal reflection.
   1. Light should travel from denser medium to rarer medium.
   2. Angle of incidence should be greater than critical angle.
2. $\frac{1}{\mu} = \frac{\sin\text{ i}}{\sin\text{ r}} , $ for total internal reflection to occur $i \geq i _{c}$ at critical angle, angle of refraction$\text{𝑟}=90^{o}$, hence, $\frac{1}{\mu} = \frac{\sin\text{i}_{c}}{\sin90^{o}}$

$\Rightarrow\mu = \frac{1}{\sin\text{i}_{c}}$

3. Mirage/sparkling of diamond/ optical fiber/totally reflecting Prism/shinning of air bubbles in water.

Resolving power of compound microscope $ = \frac{2\mu\sin\theta}{1.22\lambda}.$

Resolving power can be increased by decreasing wavelength and by increasing refracting index of medium.

1. In LCR circuit: $\tan\varphi= \frac{\text{X}_{L} - \text{X}_{c}}{\text{R}} = \frac{\text{wL} - \frac{1}{\text{wC}}}{\text{R}}$

Now $\text{X}_{L} = \text{wL} = ( 1000\times100\times10^{-3})\Omega$

$= 100\Omega$

and $\text{X}_{c} = \frac{1}{\text{wc}} = \bigg(\frac{1}{100\times2\times10^{-6}}\bigg)\Omega$

$\therefore\text{X}_{c} = 500 \Omega$

$\therefore\tan\varphi = \frac{500 - 100}{400} = 1 $

$\tan\varphi = 1 $

$\varphi = 45^{o}$

2. Power Factor:

When power factor = 1, we have X_L=X_C

$\therefore\text{X'}_{c} = \frac{1}{\omega\text{C}'} = 100 \Omega$

This gives $\text{C'} = \frac{1}{100\omega} = 10 \mu\text{F}$

We, therefore, need to add a capacitor of capacitance (10-2) μF = 8μF in parallel with the given capacitor.

Alternate Answer

Let addition capacitance C₁ be connected:

$\text{X}'_{c} = \frac{1}{1000(2 + \text{C}_{1})\times10^{-6}}$

$\therefore100 = \frac{1}{1000(2 + \text{C}_{1})\times10^{-6}}$

$\therefore2 + \text{C}_{1} = 10 $

$\text{C}_{1} = 8 \mu\text{F}$.

1. The energy of a photon is hv.
2. Each photon is completely absorbed by a single electron.

2. $\text{E}_{k} = \text{hv} - \text{W}$

Alternate Answer

$\text{hv} =\text{hv}_{o} + \frac{1}{2}\text{mv}_{max^{2}} \text{ or }\text{hv} = \text{hv}_{o} + \text{eV}_{o}\text{ or }\text{E}_{k} = \text{h}(\text{v - v}_{o})$

1. When Incident frequency < Threshold frequency, there will be no emission of electrons. Hence , frequency of incident radiation should be greater than threshold frequency. $\bigg(\text{v}_{o} = \frac{\text{w}}{\text{h}}\bigg)$

$\text{E}_{k} = \text{eV}_{o} = \text{hv} - \text{W}$

$\therefore\text{V}_{o} = \frac{\text{h}}{\text{e}}\text{v} - \frac{\text{W}}{\text{e}}$

2. At $\text{v} =\text{v}_{0} , \text{E}_{k} = \text{eV}_{o} = 0 $

V_o is called stopping potential.

Alternate Answer

Alternate Answer

1. $\text{R.P}\propto\frac{1}{\lambda}$.
2. $\text{R.P}\propto\mu$.

2. $\text{A}_{v} = \beta_{ac}.\frac{\text{R}_{L}}{\text{r}}$

$\therefore\beta_{ac} = \text{A}_{c}\frac{\text{r}}{\text{R}_{L}}$

Alternate Answer

$\beta_{ac} = \frac{\delta\text{l}_{c}}{\Delta\text{I}_{B}}$

1. The magnitude of the total field is obtained due to small elements

$\text{dB} = \frac{\mu_{o}\text{ndxla}^{2}}{2[(\text{r} - \text{x})^{2} + \text{a}^{2}]\frac{3}{2}}$

x varies from $\text{x} = - l \text{ to } \text{x} = + l $

$\text{B} = \frac{\mu_{o}\text{nla}^{2}}{2} \int\limits^{l}_{-l}\frac{\text{dx}}{[(\text{r} - \text{x})^{2} +\text{a}^{2}] \frac{3}{2}}$

For $\text{r} >> \text{a} \text{ and } , \text{we have } \text{r} > > \text{x}$

$\text{B}\simeq\frac{\mu_{o}n\text{la}^{2}}{2\text{r}^3}\int\limits^{l}_{-l}\text{dx} =\text{B} = \frac{\mu_{o}\text{nla}^{2}(2\ell)}{2\text{r}^{3}}$

Here magnetic moment m = n2l $(\pi\text{a}^{2}) $

$\text{Thus }\text{B} = \frac{\mu_{o}2\text{m}}{4\pi\text{r}^{3}}$

This is also the far axial magnetic field of a bar magnet. Hence, the magnetic field, due to current carrying solenoid along its axial line is similar to that of a bar magnet for far off axial points.

$\therefore$ Intensity $ = 4 \text{I}_{o}\cos^{2}(\frac{\phi}{2}) , \text{where } \text{I}_{o} =\text{a}^{2}$ is the intensity of each harmonic wave

1. Sharper for R = R₂

Sharpness of resonance $ = \frac{\omega_{\circ}\text{L}}{\text{R}}\propto\frac{1}{\text{R}}.$

2. More power dissipation for $\text{R} = \text{R}_{2}$

At Resonance, power dissipation  $ = \frac{\text{V}^{2}}{\text{R}}\propto\frac{1}{\text{R}}$ (for same V).

1. Intensity of radiation is determined by the number of photons incident per unit area per unit time.
2.  

1. Red Light.

Reason: Energy of photon of red light is less than that of a photon of blue light.

Alternate Answer

$\text{hv}_{red} < \text{hv}_{blue}$

2. Blue Light.

Reason: Energy of photon of Blue light is more than that of a photon of red light.

Alternate Answer

$\text{hv}_{red} > \text{hv}_{blue}$

Note: [The Einstein's photoelectric equation:

$\text{hv} = \text{hv}_{\circ} + \frac{1}{2}\text{mv}^{2}_{max}$

Instead of the reason in part.]

Alternate Answer

1. Transverse nature.
2. Have a definite speed (for all frequencies) in vacuum.
3. Can be polarised.
4. Can show the phenomenon of interference and diffraction.
5. Can transport energy from one point to another.
6. Have oscillating electric and magnetic fields along mutually perpendicular directions.
7. Have a momentum associated with them.
8. Their speed, in a medium, depends upon the values of $\mu$ and $\varepsilon$ for that medium.

1. For a radiation of frequency$\upsilon$, each photon has an energy, $\text{E} = \text{h}v,$ associated with it.
2. The energy of a photon is independent of the intensity of incident radiation.
3. During the collision of a photon, with an electron, the total energy of the photon gets absorbed by the electron.

Einstein’s photoelectric equation is,

$\text{K}_{max} = \text{hv} - \phi_{0}$

or $\text{eV}_{0} = \text{hv} -\phi_{0}$

1. Stopping potential, $\text{V}_{0}$ equals that value of the negative potential for which

$|\text{eV}_{0}| =\text{K}_{max}$

Alternate Answer

(The stopping potential ($\text{V}_{0}$) equals that (least) value of the (negative) plate potential that just stops the most energetic emitted photoelectrons from reaching the plate.)

2. Threshold frequency $(\vartheta_{o})$ equals that value of the frequency of incident radiation for which $\text{K}_{max} = 0 . $

Alternate Answer

(For a given photosensitive surface, its threshold frequency is the minimum value of the frequency of incident radiation for which photoelectrons can be just emitted from that surface or that maximum frequency of incident radiation below which no photo emission takes place.)

The plot, between $\text{V}_{0}$ and$\vartheta , $ , has the form shown:

Alternate Answer

1. A magnetic field to make the charged particles move in a circular path.
2. An alternating electric field which accelerates the charged particles as they repeatedly cross it in a way that makes them gain energy continuously.

Alternate Answer

We have BC $ = \vartheta_{1}\tau\text{ and }\text{AE} = \vartheta_{2}\tau$

Also $\sin\text{i} = \frac{\text{BC}}{\text{AC}}\text{ and } \sin\text{r} = \frac{\text{AE}}{\text{AC}}$

$\therefore\frac{\sin\text{i}}{\sin\text{r}} = \frac{\text{BC}}{\text{AE}} = \frac{\vartheta_{1}}{\vartheta_{2}} = \frac{\text{n}_{2}}{\text{n}_{1}}$

= a constant

This is Snell’s law.

2. Plane wavefront:

Spherical wavefront: